crux mathematicorum 1998
TRANSCRIPT
1
THE ACADEMY CORNERNo. 16
Bruce Shawyer
All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7
This month, we present some more of the solutions to a university en-trance scholarship examination paper from the 1940's, which appeared in theApril 1997 issue of CRUX with MAYHEM [1997: 129].
4. (a) Suppose that a 6= 0 and c 6= 0, and that ax3 + bx+ c has a factorof the form x2 + px+ 1. Show that a2 � c2 = ab.
(b) In this case, prove that ax3 + bx + c and cx3 + bx2 + a have acommon quadratic factor.
Solution.
(a) The given conditions on the coe�cients show that x = 0 is not asolution. Also, the other factor must be linear, and, on examiningcoe�cients, must be of the form ax+ c. Therefore
ax3+bx+c = (x
2+px+1)(ax+c) = ax
3+(c+ap)x
2+(a+cp)x+c:
Comparing coe�cients gives
c+ ap = 0;
a+ cp = b:
Thus, p = � ca, so that a� c2
a= b, giving the required result.
(b) Since x 6= 0, we write y = 1
x, and examine the corresponding
equations in y to get this part.
5. Prove that all the circles in the family de�ned by the equation
x2 + y2 � a(t2 + 2)x� 2aty � 3a2 = 0
(a �xed, t variable) touch a �xed straight line.
2
Solution.
We re-write the equation as
�x� a
�1 +
t2
2
��2+ (y � at)
2=
�a
�2 +
t2
2
��2:
This shows that the line x = �a is a tangent line. (The y{coordinate is2at.)
6. Find the equation of the locus of a point P which moves so that thetangents from P to the circle x2 + y2 = r2 cut o� a line segment oflength 2r on the line x = r.
Solution.
From the circle, x2 + y2 = a2, we �nd the points of intersection withthe line (y� q) = m(x� p).
Substituting y = mx �mp+ q, we get x2 + (mx�mp+ q)2 = a2,leading to the two solutions:
x =m(mp� q)�
pa2(m2 + 1)�m2p2 + q(2mp� q)
m2 + 1:
For the line to be a tangent, the discriminant, � = a2(m2 + 1) �m2p2 + q(2mp� q), must be zero. So we solve � = 0, to get
m =pq
p2 � a2� a
pp2 + q2� a2
p2 � a2:
So, we have that the equations of the tangent lines from (p; q) to thecircle x2 + y2 = a2 are:
y =
� pq
a2 � p2� a
pa2 � p2 � q2
p2 � a2
!x
� �pq=(a2 � p2) +�
pa2 � p2 � q2
p2 � a2
!p+ q:
Setting x = a in these, and subtracting the two values leads to
(a+ p)2 = p2 + q2 � a2:
Thus the locus of P is given by
y2 = 2a(a+ x):
This is a parabola opening to the right with central axis, the x{axis andnose at x = �a. There are, of course, three points that should beexcluded: (�a; 0), (a; 2a), and (a;�2a).
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7. If the tangents at A, B, C, to the circumcircle of triangle4ABC meetthe opposite sides at D, E, F , respectively, prove that D, E, F , arecollinear.
Solution.
D Q
E
F
CB
AR
P
For notational ease, we write PB = PC = �, QC = QA = � andRB = RA = .
Apply Menelaus' Theorem for triangle4PQR with transversals BDC,EAC and BAF
�1 =PB
BR
RD
DQ
QC
CP=
QC
BR
RD
DQ;
�1 =PE
ER
RA
AQ
QC
CP=
PE
ER
RA
CP;
�1 =PB
BR
RA
AQ
QF
FP=
PB
AQ
QF
FP:
These lead to
+ � + QD
QD= �BR
QC= �
�; (1)
�+ +ER
ER= �CP
RA= ��
; (2)
QF
FQ+ �+ �= �AQ
PB= ��
�: (3)
Now consider the product of ratios to prove D, E and F collinear bythe converse of Menelaus:
PE
ER
RD
DQ
QF
FP:
This is equal to
�+ � + ER
ER
+ � +QD
QD
QF
FQ+ �+ �:
Using (1), (2) and (3) above, we get a value of �1, and the result isproved.
4
THE OLYMPIAD CORNERNo. 187
R.E. Woodrow
All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.
Another year has passed|How the time seems to y. I would particu-larly like to thank Joanne Longworth for her excellent help in pulling togetherthe information for the column and her work in producing a good LATEX copyfor the Editor-in-Chief { usually under enormous time pressure because I ambehind schedule.
The Corner could not exist without its readers who supplyme with goodOlympiad materials and send in their interesting solutions and comments toproblems in the Corner. I hope we have missed no one in the following listof contributors.
Miguel Amengual Covas
S�efket Arslanagi �c
Mansur Boase
Christopher Bradley
Sabin Cautis
Adrian Chan
Byung-Kuy Chun
Mihaela Enachescu
George Evagelopoulos
Shawn Godin
Joanne Juszu �nska
Joel Kamnitzer
Deepee Khosla
Derek Kisman
Murray Klamkin
Marcin Kuczma
Andy Liu
Beatriz Margolis
Vedula Murty
Richard Nowakowski
Colin Percival
Bob Prielipp
Toshio Seimiya
Michael Selby
Zun Shan
D.J. Smeenk
Daryl Tingley
Panos Tsaoussoglou
Ravi Vakil
Dan Velleman
Stan Wagon
Edward Wang
Thank you all, and all the best for 1998!
The �rst Olympiad we give for the new year is the 17th Austrian-PolishMathematics Competition, written in Poland, June 29{July 1, 1994. Mythanks go to Richard Nowakowski, Canadian Team Leader to the 35th IMO inHong Kong and to Walther Janous, Ursulinengymnasium, Innsbruck, Austriafor sending a copy to me.
17th AUSTRIAN{POLISH MATHEMATICS
COMPETITION
Poland, June 29{July 1, 1994
1. The function f : R! R satis�es for all x 2 R the conditions
f(x+ 19) � f(x) + 19 and f(x+ 94) � f(x) + 94:
5
Show that f(x+ 1) = f(x) + 1 for all x 2 R.2. The sequence (an) is de�ned by the formulas
a0 =1
2and an+1 =
2an
1 + a2nfor n � 0;
and the sequence (cn) is de�ned by the formulas
c0 = 4 and cn+1 = c2n � 2cn + 2 for n � 0:
Prove that
an =2c0c1 : : : cn�1
cnfor all n � 1:
3. A rectangular building consists of two rows of 15 square rooms (situ-ated like the cells in two neighbouring rows of a chessboard). Each room hasthree doors which lead to one, two or all the three adjacent rooms. (Doorsleading outside the building are ignored.) The doors are distributed in sucha way that one can pass from any room to any other one without leaving thebuilding. How many distributions of the doors (in the walls between the 30rooms) can be found so as to satisfy the given conditions?
4. Let n � 2 be a �xed natural number and let P0 be a �xed vertex ofthe regular (n+1)-gon. The remaining vertices are labelled P1; P2; : : : ; Pn,in any order. To each side of the (n + 1)-gon assign a natural number asfollows: if the endpoints of the side are labelled Pi and Pj , then i� j is thenumber assigned. Let S be the sum of all the n+ 1 numbers thus assigned.(Obviously, S depends on the order in which the vertices have been labelled.)
(a) What is the least value of S available (for �xed n)?
(b) How many di�erent labellings yield this minimum value of S?
5. Solve the equation
1
2(x+ y)(y+ z)(z + x) + (x+ y+ z)3 = 1� xyz
in integers.
6. Let n > 1 be an odd positive integer. Assume that the integersx1; x2; : : : ; xn � 0 satisfy the system of equations
(x2 � x1)2 + 2(x2 + x1) + 1 = n2
(x3 � x2)2 + 2(x3 + x2) + 1 = n2
......................................(x1 � xn)
2 + 2(x1 + xn) + 1 = n2:
Show that either x1 = xn or there exists j with 1 � j � n � 1 such thatxj = xj+1.
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7. Determine all two-digit (in decimal notation) natural numbersn = (ab)10 = 10a+ b (a � 1) with the property that for every integer x thedi�erence xa � xb is divisible by n.
8. Consider the functional equation f(x; y) = a f(x; z) + b f(y;z)
with real constants a, b. For every pair of real numbers a, b give the generalform of functions f : R2! Rsatisfying the given equation for all x; y; z 2 R.
9. On the plane there are given four distinct points A, B, C, D lying(in this order) on a line g, at distances AB = a, BC = b, CD = c.
(a) Construct, whenever possible, a point P , not on g, such that theangles \APB, \BPC, \CPD are equal.
(b) Prove that a point P with the property as above exists if and only ifthe following inequality holds: (a+ b)(b+ c) < 4ac.
As a second source of problem pleasure for winter evenings (that is ifyou are havingwinter this January!), we give the Second Round of the IranianNational Mathematical Olympiad. My thanks go to Richard Nowakowski,Canadian Team Leader to the 35th IMO in Hong Kong for collecting the prob-lems and forwarding them to me.
IRANIAN NATIONALMATHEMATICAL OLYMPIAD
February 6, 1994
Second Round
1. Suppose that p is a prime number and is greater than 3. Prove that7p � 6p � 1 is divisible by 43.
2. Let ABC be an acute angled triangle with sides and area equal toa, b, c and S respectively. Show that the necessary and su�cient conditionfor existence of a point P inside the triangle ABC such that the distancebetween P and the vertices of ABC be equal to x, y and z respectivelyis that there be a triangle with sides a, y, z and area S1, a triangle withsides b, z, x and area S2 and a triangle with sides c, x, y and area S3 whereS1 + S2 + S3 = S.
3. Let n and r be natural numbers. Find the smallest natural num-ber m satisfying this condition: For each partition of the set f1; 2; : : : ;mginto r subsets A1; A2; : : : ; Ar there exist two numbers a and b in some Ai
(1 � i � r) such that 1 < ab� 1 + 1
n.
4. G is a graph with n vertices A1; A2; : : : ; An such that for eachpair of nonadjacent vertices Ai and Aj there exists another vertex Ak that isadjacent to both Ai and Aj.
(a) Find the minimum number of edges of such a graph.
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(b) If n = 6 and A1; A2; A3; A4; A5; A6 form a cycle of length 6, �ndthe number of edges that must be added to this cycle such that the abovecondition holds.
5. Show that if D1 and D2 are two skew lines, then there are in�-nitely many straight lines such that their points have equal distance from D1
and D2.
6. f(x) and g(x) are polynomials with real coe�cients such that for
in�nitely many rational values x, f(x)
g(x)is rational. Prove that f(x)
g(x)can be
written as the ratio of two polynomials with rational coe�cients.
Now we turn to readers' solutions and comments for problems posed inthe Corner. First solutions to a problem of the 25th United States of AmericaMathematical Olympiad [1996: 203{204].
1. Prove that the average of the numbers
n sinn�; n = 2; 4; 6; : : : ; 180
is cot 1�.
Solution by D.J. Smeenk, Zaltbommel, the Netherlands.Let
x = 2 sin2� + 4 sin4� + � � �+ 90 sin90� + � � �+ 178 sin178�
= (2 + 178) sin 2� + (4 + 176) sin 4� + � � �= 180(sin2� + sin4� + � � �+ sin88�) + 90 sin 90�:
Then�x =
x
90= 2 sin2� + 2 sin4� + � � �+ 2 sin88� + 1
�x sin 1� = 2sin2� sin 1� + 2 sin4� sin 1� + � � �+ 2 sin88� sin 1� + sin1�:
Now2 sin 2� sin1� = cos 1� � cos 3�
2 sin 4� sin1� = cos 3� � cos 5�
� � �2 sin88� sin 1� = cos 87� � cos 89�:
Hence
�x sin 1� = cos 1� � cos 89� + sin1�
= cos 1�:
Thus �x = cot 1�, as required.
8
Next we give a comment on a problem, and one solution to anotherfrom the 1994 Italian Mathematical Olympiad.
2. [1996:204] Italian Mathematical Olympiad 1994.Find all integer solutions of the equation
y2 = x3 + 16:
Comment by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain.The equation has no solutions in integers di�erent from x = 0, y = �4.
For a proof see Theorem 20, page 102 of W. Sierpinski's Elementary Theoryof Numbers.
4. [1996: 204] Italian Mathematical Olympiad 1994
Let r be a line in the plane and let ABC be a triangle contained in oneof the half-planes determined by r. Let A0, B0, C0 be the points symmetricto A, B, C with respect to r; draw the line through A0 parallel to BC, theline throughB0 parallel to AC and the line through C0 parallel to AB. Showthat these three lines have a common point.
Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain.
q
q
q
q
q
q
r
C0
B0
A0
A
B
C There is no need that 4ABC becontained in one of the half planesdetermined by r.If the coordinates of A are (x1; y1),the coordinates ofB are (x2; y2) andthose ofC are (x3; y3) in a Cartesiancoordinate system with r as x-axis,then A0 has coordinates (x1;�y1),B0 has coordinates (x2;�y2) and C0has coordinates (x3;�y3).
We have:The equation of the line through A0 parallel to BC is
(y2� y3)x� (x2 � x3)y� x1(y2 � y3)� y1(x2 � x3) = 0:
The equation of the line through B0 parallel to CA is
(y3� y1)x� (x3 � x1)y� x2(y3 � y1)� y2(x2 � x1) = 0:
The equation of the line through C0 parallel to AB is
(y1� y2)x� (x1 � x2)y� x3(y1 � y2)� y3(x1 � x2) = 0:
Since the three equations when added together vanish identically, thelines represented by them meet in a point.
9
Its coordinates are found, by solving between any two, to be�(x2
1+ x2x3)(y2� y3) + (x2
2+ x3x1)(y3� y1) + (x2
3+ x1x2)(y1� y2)
x1(y2� y3) + x2(y3� y1) + x3(y1� y2);
(y21+ y2y3)(x2 � x3) + (y2
2+ y3y1)(x3 � x1) + (y2
3+ y1y2)(x1 � x2)
x1(y2 � y3) + x2(y3 � y1) + x3(y1 � y2)
�:
Now we jump to the October 1996 number of the Corner, and twosolutions to problems of the 10th IBEROAMERICANMathematical Olympiad[1996: 251{252].
1. (Brazil). Determine all the possible values of the sum of the digitsof the perfect squares.
Solution by Mansur Boase, student, St. Paul's School, London, Eng-land.
The squares can only be 0, 1, 4 or 7 mod 9.Thus the sum of the digits of a perfect square cannot be 2, 3, 5, 6 or
8 mod 9, since the number itself would then be 2, 3, 5, 6 or 8 mod 9.We shall show that the sum of the digits of a perfect square can take
every value of the form 0, 1, 4 or 7 mod 9.
(10m � 1)2 = 102m � 2 � 10m + 1 = 9 9 � � � 9| {z }m�1
8 0 � � � 0| {z }m�1
1; m � 1:
The sum of the digits is 9m, giving all the values greater than or equal to 9
congruent to 0 mod 9
(10m � 2)2 = 102m � 4 � 10m + 4 = 9 9 � � � 9| {z }m�1
6 0 � � � 0| {z }m�1
4; m � 1:
The sum of the digits is 9m+ 1, which gives all values greater than or equalto 10 congruent to 1 mod 9.
(10m � 3)2 = 102m � 6 � 10m + 9 = 9 9 � � � 9| {z }m�1
4 0 � � � 0| {z }m�1
9; m � 1:
The sum of the digits is 9m+4, which takes every value greater than or equalto 13 which is congruent to 4 mod 9
(10m � 5)2 = 102m � 10m+1 + 25 = 9 � � � 9| {z }m�1
0 0 � � � 0| {z }m�1
2 5:
The sum of the digits is 9(m� 1) + 7 = 9m� 2, from which we get everyvalue greater than or equal to 7 congruent to 7 mod 9.
We have taken care of all the integers apart from 0, 1, 4, which are thesums of the digits of 02, 12, and 22 respectively.
10
5. (Spain). The inscribed circumference in the triangle ABC is tan-gent to BC, CA and AB at D, E and F , respectively. Suppose that thiscircumference meets AD again at its mid-pointX; that is, AX = XD. Thelines XB and XC meet the inscribed circumference again at Y and Z, re-spectively. Show that EY = FZ.
Solution by Toshio Seimiya, Kawasaki, Japan.
A
B C
FE
D
X
ZY
Since \BFY = \BXF and\FBY = \XBF we have 4BFYand4BXF are similar, so that
FY : FX = BF : BX: (1)
Similarly we get
DY : DX = BD : BX: (2)
As BF = BD, we have from (1) and(2) that
FY : FX = DY : DX:
Since AX = DX we get
FY : FX = DY : AX: (3)
Since X, F , Y , D are concyclic we have
\FYD = \AXF: (4)
Thus we get from (3) and (4) that 4FYD is similar to 4FXA.Hence \Y FD = \XFA = \XDF so that FY kXD. Similarly we
have EZkXD. Thus FY kEZ.Therefore FYZE is an isosceles trapezoid and then EY = FZ.
We now turn to two solutions to the Maxi Finale 1994 of the OlympiadeMath �ematique Belge [1996: 253{254].
1. Un pentagone plan convexe a deux angles droits non adjacents. Lesdeux cot �es adjacents au premier angle droit out des longueurs �egales. Lesdeux cot �es adjacents au second angle droit ont des longueurs �egales. En rem-pla�cant par leur point milieu les deux sommets du pentagone situ �es sur unseul cot �e de ces angles droits, nous formons un quadrilat �ere. Ce quadrilat �ereadmet-il n �ecessairement un angle droit?
11
Solution by Mansur Boase, student, St. Paul's School, London, Eng-land.
Let the pentagon be ABCDE and let \ABC = \AED = 90�. Letthe midpoint of CD be M .
Then the problem is equivalent to proving that given any triangleACDwith isosceles right triangles constructed onAC andAD, giving pointsB andE, \BM = 90� and so quadrilateral BMEA must have a right angle.
This result is well known. One very nice proof is that if L and N
are the midpoints of AC and AD respectively then 4BLM � 4MNE
(BL = LC = MN , LM = ND = EN and
\BLM = 90� + \CLM = 90� + \CAD
= 90� + \MND = \MNE):
And sinceMLkND andND ? EN ,ML ? EN and similarlyBL ? NM .Therefore BM ? EM since the other two pairs of sides are perpendicular.
4. Le plan contient-il 1994 points (distincts) non tous align �es tels quela distance entre deux quelconques d'entre eux soit un nombre entier?
Solution by Mansur Boase, student, St. Paul's School, London, Eng-land.
There are 1994 pairs of solutions (m;n) withm > n satisfying
2mn = 23988
[(20; 23987); : : : ; (21993; 21994)]:
So there are 1994 Pythagorean triples of the form (m2�n2; 2mn;m2+n2)
with one side of length 23988.Hence we can form a line of 1993 collinear points and place a 1994th
point a distance 23998 above this line so that, with the correct arrangementof the points, if it is joined to any other point, the resulting line is the hy-potenuse corresponding to a Pythagorean triple and is therefore integral.Also the distance between any two points on the line will be integral.
a1993 : : : a3 a2 a1
� � �
a1994
23988
That completes the Corner for this issue. Olympiad season is fast uponus. Send me your contest materials and your nice solutions for use in theCorner.
12
BOOK REVIEWS
Edited by ANDY LIU
Mini-Reviews Update
Andy Liu
This is an update of earlier Mini-Reviews (see [1] to [9]), and is an ab-breviated form of what appeared in the journals Delta K and AGATE.Written permission for reprint has been granted by the Alberta Teachers'Association which publishes the above two journals.
A. Mir Publishers' Little Mathematics Library Series (see [1])
This excellent series has become an unfortunate casualty of the demiseof the former Soviet Union. Lost also are Mathematics Can Be Fun andFun with Maths and Physics featured in Section I.
B. New Mathematical Library of the Mathematical Associ-ation of America (see [2])
In additional to the new titles listed below, there is also a revised edi-tion of an earlier volume, Graphs and Their Uses. It was #10 in the series,but is now #34. Two further volumes have also been published recently.
USA Mathematical Olympiads: 1972-1986, by Murray Klamkin, 1988.ISBN# 0-88385-634-4.
This book collects the problems of the �rst �fteen USA Mathemati-cal Olympiads. While they are presented chronologically, the solutions aregrouped according to subject matters, which facilitates using this book fortraining sessions. There is a very useful 10-page glossary of mathematicalterms and results, and a most extensive bibliography.
Exploring Mathematics with your Computer, by Arthur Engel, 1993.(see [16])
Game Theory and Strategy, by Philip Stra�n, 1993. (see [17])
C. Martin Gardner's Scienti�c American Series (see [3])
Two more volumes have appeared, and there will be a �fteenth and�nal volume, about to be released by Springer-Verlag. Several earlier vol-umes have also changed publishers. The Mathematical Association of Amer-ica has acquired Martin Gardner's New Mathematical Diversions from Sci-enti�c American, Martin Gardner's 6th Book of Mathematical Diversionsfrom Scienti�cAmerican, Mathematical Carnival, Mathematical Magic Show
13
and Mathematical Circus. The University of Chicago Press has acquiredThe Scienti�c American Book of Mathematical Puzzles and Diversions,The 2nd Scienti�c American Book of Mathematical Puzzles and Diversionsand The Unexpected Hanging and Other Mathematical Diversions.
Penrose Tiles to Trapdoor Ciphers, 1989, W. H. Freeman, 1997, Math-ematical Association of America. ISBN# 0-88385-521-6.
Topics covered are Penrose tilings,Mandebrot's fractals, Conway's sur-real numbers, mathematical wordplay, Wytho�'s version of the game \Nim",mathematical induction, negative numbers, dissection puzzles, trapdoor ci-phers, hyperbolas, the new version of the game \Eleusis", Ramsey theory,the mathematics of Berrocal's sculptures, curiosities in probability, RaymondSmullyan's logic puzzles, as well as two collections of short problems. Thebook contains a surprise ending, the resurrection of Dr. Matrix! There is alsoan update chapter. Unfortunately, Figures 3 to 6 are inadvertently left out.They are reproduced below.
Figure 3. Figure 4.
Figure 5. Figure 6.
Fractal Music, Hypercards and More, 1992, W.H. Freeman.ISBN# 0-7167-2189-9.
Topics covered are fractal music, the Bell numbers, mathematical zoo,Charles Sanders Peirce, twisted prismatic rings, coloured cubes, Egyptianfractions, minimal sculptures, tangent circles, time, generalized tick-tack-toe, psychic wonders and probability, mathematical chess problems, Hofs-tadter's G�odel, Escher, Bach, imaginary numbers, some accidental patterns,packing squares, Chaitin's irrational number , as well as one collection ofshort problems.
14
D. Books from W.H. Freeman & Company, Publishers (see [4])
Note that the most recent book inMartin Gardner's Scienti�c AmericanSeries is a Freeman publication. The �rst of the books listed below has ac-tually gone out of print, but fortunately Dover Publications Inc. has decidedto pick it up.
The Puzzling Adventures of Dr. Ecco, by Dennis Shasha, 1988.
The title character calls himself an omniheurist, solver of all prob-lems (mathematical). The narrative is by a Watsonesque companion, Prof.Scarlet. Ecco's clients range from government o�cials, industrialists, eccen-tric millionaires to no less than the President of a Latin American country.They brought him important, instructive and interesting problems in dis-crete mathematics, all of which Ecco solves to their satisfaction. The bookconcludes with the mysterious disappearance of Dr. Ecco.
Codes, Puzzles, and Conspiracy, by Dennis Shasha, 1992. (see [11])
New Book of Puzzles, by Jerry Slocum and Jack Botermans, 1992.(see [12])
Another Fine Math You've Got Me Into : : : , by Ian Stewart, 1992.(see [13])
E. Oxford University Press Series on Recreations inMathemat-ics (see [5])
The Mathematics of Games, by John D. Beasley, 1989.ISBN# 0-19-853206-7.
This book analyses mathematically some card and dice games, nim-type games, a version of John Conway's \Hackenbush", as well as providinga mathematical model for the study of some sports games. The principaltechniques are counting, probability and game theory. Some mathematicalpuzzles are also considered.
The PuzzlingWorld of PolyhedralDissections, by Stewart T. Co�n, 1990.ISBN# 0-19-853207-5.
This is a labour of love from an expert craftsman. Starting with twochapters of two-dimensional geometric puzzles, the author eases the read-ers gently into the third dimension and soon launches into his specialty,the burrs, which are assemblies of interlocking notched sticks. The bookis profusely illustrated with black-and-white line drawings and photographs.It concludes with a chapter on woodworking techniques.
More Mathematical Byways, by Hugh ApSimon, 1990.ISBN# 0-19-217777-X.
This book contains fourteen chapters. The �rst three form a sequencebut the others are independent of each other. Unlike the earlier volumeby the same author, the problems are of uneven level of di�culty, rangingfrom the relatively simple Alphametics to others which require a considerable
15
amount of what the author calls \slog". One of the chapters, titled PotentialPay, is not really a problem but a commentary on a classic paradox.
F. Raymond Smullyan's Logic Series (see [6])
Satan, Cantor, and In�nity, 1992, Alfred A. Knopf.ISBN# 0-679-40688-3.
In this book, a remarkable character known as the Sorcerer makes hisdebut. He escorts the readers on a wonderful guided tour, visiting familiargrounds such as the domains of the knights and the knaves, and those bor-dering on the land of G �odel. There are also ventures into new territories,including an island where intelligent robots create others which can continuethis process ad in�nitum. This eventually leads to the pioneering discoverieson in�nity of the great mathematician, Georg Cantor. The readers may beamused to discover how Satan got into the picture.
G. Dolciani Mathematical Expositions Series of theMathemat-ical Association of America (see [7])
In addition to the new titles listed below, three further volumes havebeen published recently.
More Mathematical Morsels, by Ross Honsberger, 1991. (see [10])ISBN# 0-88385-313-2.
This is a collection of 57 problems, almost all of which are taken fromthe Canadian Mathematical Society's journal Crux Mathematicorum, plusfurther \gleanings" from its famed Olympiad Corner.
Old and New Unsolved Problems in Plane Geometry and Number Theory,by Victor Klee and Stan Wagon, 1991. ISBN# 0-88385-315-9.
This book is divided into two halves, as suggested by the title, thoughthe second half also covers problems about some interesting real numbers.Each half consists of two parts. In the �rst, twelve problems are presented,giving the statement, known results and background information. In the sec-ond, the same twelve problems are reexamined for further results and exten-sions. Each half concludes with a comprehensive bibliography. Although theproblems are unsolved, and therefore hard, it is not impossible for them toyield to an inspired attack. Even if this does not happen, gifted students whoare willing to attempt them will �nd their mathematical talent enhanced.
Problems for Mathematicians Young and Old, by Paul Halmos, 1992.ISBN# 0-88385-320-5.
The fourteen chapters of this book are titled Combinatorics, Calculus,Puzzles, Numbers, Geometry, Tilings, Probability, Analysis, Matrices, Alge-bra, Sets, Spaces, Mappings and Measures. The author, a ranking mathe-matician and master expositor, wrote this book for fun, and hoped that itwill be read the same way.
16
Excursions in Calculus, by Robert Young, 1992. ISBN# 0-88385-317-5.
The subtitle of this book is An Interplay of the Continuous and the Dis-crete. Using calculus as a unifying theme, the author branches into numbertheory, algebra, combinatorics and probability. The book contains a largecollection of exercises and problems.
The Wohascum County Problem Book, by George Gilbert, MarkKrusemeyer and Loren Larson, 1993. (see [14])
Lion Hunting & Other Mathematical Pursuits, edited by GeraldAlexanderson and Dale Muggler, 1995. (see [18])
The Linear Algebra Problem Book, by Paul Halmos, 1995.ISBN# 0-88385-322-1.
The whole book is a sequence of structured problems. Like the preced-ing volume, most of this book is beyond high school level. However, the in-troductory problems are certainly not intimidating, and inquisitive studentsmay be lured into a most rewarding exploration, laying a good foundationfor their undergraduate studies.
H. Books from Dover Publications, Inc. (see [8])
Excursions in Number Theory, by Stanley Ogilvy and John Anderson,1988. ISBN# 0-486-25778-9.
This book covers the basics of classical number theory. Topics includeprime numbers, congruences, Diophantine equations and Fibonacci num-bers. The narrative style is very soothing. It concludes with 20 pages ofelaborations and commentary on some �ner points raised in the text.
Excursions in Geometry, by Stanley Ogilvy, 1990.ISBN# 0-486-26530-7.
The �rst half of this book is on inversive geometry, and the secondhalf on projective geometry. These two topics are linked by the concept ofcross-ratio and the study of the conic sections. It is in the same style as thepreceding volume.
Excursions in Mathematics, by Stanley Ogilvy, 1994.ISBN# 0-486-28283-X.
The original title of this volume was Through the Mathescope. Theopening chapter is titled What Do Mathematicians Do? It is followed bylively tours of number theory, algebra, geometry and analysis. The last chap-ter is titled Topology and Apology.
I. Books from Various Publishers (see [9])
Selected Problems and Theorems in Elementary Mathematics has beenacquired by Dover and renamed The USSR Olympiad Problem Book. Doverhas also picked up The Moscow Puzzles. The Mathematical Association ofAmerica has published Five Hundred Mathematical Challenges, comprising
17
the �rst �ve booklets of the project, 1001 Problems in High School Math-ematics. (see [19]) The Canadian Mathematical Society has published TheCanadian Mathematical Olympiad, 1969{1993, edited by Michael Doob andClaude La amme. This book combines two earlier volumes, The First TenCanadian Mathematics Olympiads, 1969{1978 and The CanadianMathemat-ics Olympiads, 1979-1985, and adds the contests from 1986 to 1993. (see[15])
Cross References to other entries in CruxMathematicorum:
1. A. Liu, Mini-reviews, MIR Publishers' Little Mathematics Library Se-ries, 1989, 142-147.
2. A. Liu, Mini-reviews, MAA's New Mathematical Library Series, 1989,171-176.
3. A. Liu, Mini-reviews, Martin Gardner's Mathematical Games Series,1989, 202-206
4. A. Liu, Mini-reviews, Books on popular mathematics from W. H. Free-man, 295-297.
5. A. Liu, Mini-reviews, Oxford University Press' Recreation in Mathe-matics Series, 1990, 42-43.
6. A. Liu,Mini-reviews, Raymond Smullyan's books on logic puzzles, 1990,106-108.
7. A. Liu, Mini-reviews, MAA's Dolciani Mathematical Exposition Series,1990, 237-238.
8. A. Liu, Mini-reviews, Books on popular mathematics from Dover, 1991,11-13.
9. A. Liu, Mini-reviews, Books on popular mathematics from various pub-lishers, 1991, 74-77.
10. A. Liu, Review of More Mathematical Morsels, 1991, 235-236.
11. A. Liu, Review of Codes, Puzzles and Conspiracy, 1992, 204-205.
12. A. Liu, Review of New Books of Puzzles, 1993, 13-14.
13. R.K. Guy, Review of Another Fine Math You've Got Me Into, 1993,46-47.
14. A. Liu, Review of The Wohascum County Problem Book, 1993, 202.
15. R. Geretschl�ager and G. Perz, Review of The Canadian MathematicalOlympiad Book, 1969-1993, 1994, 15-16.
16. G.C. Denham, Review of Exploring Mathematics with Your Computer,1994, 134-135.
17. M. Larsen, Review of Game Theory and Strategy, 1995, 88.
18. A. Liu, Review of LionHunting and OtherMathematical Pursuits, 1995,304-305.
19. M. Kuczma, Review of Five Hundred Mathematical Challenges, 1996,313-316.
18
CANADIAN MATHEMATICAL SOCIETY
AWARD FOR CONTRIBUTIONS TO
MATHEMATICAL EDUCATION
The Adrien Pouliot Award
Eric Muller, Chair, CMS Education Committee
In 1995, in celebration of its 50th Anniversary, the Canadian MathematicalSociety (CMS) instituted an Award for Contributions to Mathematics Edu-cation. This award is to recognize individuals or teams of individuals whohave made signi�cant and sustained contributions to mathematics educa-tion within Canada. Such contributions are to be interpreted in the broadestpossible sense and might include: community outreach programmes, the de-velopment of a new programme in either an academic or industrial setting,publicizing mathematics so as to make it accessible to the general public,developing mathematics displays, establishing and supporting mathematicsconferences and competitions for students, etc..
The prize is named after Adrien Pouliot. But who was Adrien Pouliot?Pouliot is certainly one of the main scienti�c �gures of the twentieth centuryin the Province of Quebec. Trained as a civil engineer at the Ecole Polytech-nique de Montreal he started teaching mathematics at the Universite Lavalin the early 1920s. He completed a `licence' in sciences (mathematics) atthe Sorbonne in Paris (1928). In the words of Professor Hodgson of Laval[1] \Pouliot was the main person behind the growth of science and mathe-matics in the Province of Quebec. To understand this one has to recall thatthe �rst part of the century, the education system in Quebec was prepar-ing more for traditional jobs (priests, physicians) than for science. Pouliotled (with success) a public movement to have science and math become amajor part of the education system. (Because of that he was accused to beanti-humanities. His response was to learn Greek; when he died in 1980 hewas working on a Greek-French dictionary)". Pouliot's in uence extended toother parts of Canada and beyond through his work as a governor and o�cerof Radio-Canada and through his participation on many missions to othercountries. Pouliot was awarded a number of honorary doctorates in Canada,France and Italy.
Recipients of the �rst three awards are known to many Crux readers.
� The �rst award wasmade in 1995, andwas awarded to Professor EdwardBarbeau of the University of Toronto, Ed as he is known to students,mathematics teachers, mathematicians and university colleagues, thepress, local school board trustees, and all who come into contact with
19
him. Ed was recognized for his numerous contributions to mathemat-ics education, sustained over so many years and in so many di�erentforums. He has always been willing to spend the time to communicatewith interested mathematics teachers, students and the general pub-lic. He continues to be involved with the International MathematicalOlympiad, an involvement that started some 16 years ago. Many havebene�ted from his innovative and challengingmathematics publicationsaccessible to secondary and post secondary students.
� The 1996 recipient was Professor Bruce Shawyer of MemorialUniversity of Newfoundland. Bruce is of course the present Editor-in-Chief of Crux. He was recognized for his substantial contributionsto many di�erent mathematical competitions and challenges. He pro-posed andwas the foundingmember of the Newfoundland and LabradorMathematical Association and was involved in the development of theNewfoundland Teachers' Association Senior Math League and theJunior High Math Challenge. Bruce was most visible as the Chief Op-erating O�cer of the 36th International Mathematical Olympiad thatwas held in Canada in 1995.
� The third award, in 1997, went to a team from the University ofWaterloo, a team of university and school mathematics educators, EdAnderson, Don Attridge, Ron Dunkley and Ron Scoins. They were in-strumental in starting the CanadianMathematics Competition that grewout of their local competition activities in 1962. They have continued tonurture and sustain these competitions that have grown and, in the pastyear, involved over 200,000 students. These contests not only involvestudents but also support a network of mathematics teachers through-out Canada. They provide an environment that stimulates professionaldevelopment and provide opportunities for mathematics teachers fromschools and universities to meet and exchange ideas.
Reference
[1] Hodgson, Bernard, CMS Notes, Jan-Feb 1996
20
Cyclic ratio sums and products
Branko Gr �unbaum
The well known classical theorems of Menelaus and Ceva deal withcertain properties of triangles by relating them to the products of three ratiosof directed lengths of collinear segments. Less well known is a theorem ofEuler [2] which states, in the notation of Figure 1, that
3Xj=1
jjQBj=AjBj jj = 1
for every triangle T = [A1A2A3].
Q
t t
t
t
t
t
t
A2 A3
A1
B2B3
B1
Figure 1.A theorem of Euler states that if Bj is the intersection of the line AjQ with the side
of the triangle A1A2A3 opposite to Aj, thenP
3
j=1jjQBj=AjBjjj = 1. Here, and
throughout this note, jjMN=RSjj means the ratio of signed lengths of the collinear
segmentsMN and RS.
However, while the theorems of Menelaus and Ceva have been gener-alized to arbitrary polygons, and in many other ways | see, for example, [4][5] [6] | until very recently there have been no analogous generalizationsof Euler's result. One explanation for this situation may be that attemptsat straightforward generalizations lead to invalid statements. An example ofsuch a failed \theorem" is given by the question whether, in the notation ofFigure 2,
nXj=1
jjQBj=AjBj jj
equals 1 or some other constant independent of Q and the polygon.
Research supported in part by NFS grant DMS-9300657.
Copyright c 1998 Canadian Mathematical Society
21
t
t
t
t
t
t
t
t
tt
t
Q
A3 B1 A4
B2
A5
B3
A1
B4
A2
B5
Figure 2.Attempts to generalize Euler's theorem in the form
Pn
j=1jjQBj=AjBj jj = const
necessarily fail for n > 3 (here n = 5). However, as shown by Shephard [9], it is
possible to �nd weights wj which depend on the polygon but not on the position
of Q, such thatPn
j=1wjjjQBj=AjBjjj = 1.
Recently, Shephard [9] had the idea, apparently not considered previously,of attaching to the ratios rj = jj(QBj)=(AjBj)jj certain weights wj, whichdepend on the polygon P = [A1; A2; : : : ; An] but not on the point Q, suchthat
Pnj=1wjrj = 1. (In fact, Shephard established a much more general
result in this spirit; its complete formulation would lead us too far from thepresent aims.)
By sheer chance, the same day I received from Shephard a preprintof [9], I happened to read [7], in which two di�erent sums of ratios appear,one in Bradley's solution, the other in Kone �cn �y's comments. This coinci-dence led me to consider whether these results could be generalized alongShephard's idea. As it turns out, the answer is a�rmative, and leads to a
number of other results.
Let P = [A1; A2; : : : ; An] be an arbitrary n{gon, and Q an arbitrarypoint, subject only to the condition that all the points Bj mentioned be-low are well determined. On each side AjAj+1 of P (understood as theunbounded line) the point Bj is the intersection with the line through Q
parallel to Aj+1Aj+2. (Here, and throughout the present note, subscriptsare understood mod n). This is illustrated in Figure 3 by an example withn = 5. We are interested in the ratios
rj = jjBjAj+1=AjAj+1jj:
We denote by 4(UVW ) the signed area of the triangle4UVW withrespect to an arbitrary orientation of the plane and, more generally, by4(P)
the signed area of any polygon P, calculated with appropriate multiplicitiesfor the di�erent parts if P has self-intersections.
22
s
s
s
s
s
s
s
s
s
s
s
Q
A1
A2
A3
A4
A5
B1
B2
B3
B4
B5
Figure 3.The point Bj is the intersection of the line AjAj+1 with the parallel through Q to
the line Aj+1Aj+2. The ratios rj = jjBjAj+1=AjAj+1jj of directed segments are
considered in Theorem 1.
Theorem 1. For each polygonP we havePn
j=1wjrj = 4(P) for allQ, wherewj = 4(AjAj+1Aj+2) are weights that depend on the polygon P but areindependent of the point Q.
For a proof it is su�cient to note that
(i) by straightforward calculations or by easy geometric arguments it can beshown that rj = 4(QAj+1Aj+2)=4(AjAj+1Aj+2); and
(ii) therefore the sumPn
j=1wjrj is equal toPn
j=14(QAj+1Aj+2) = 4(P),since the triangles with vertex Q triangulate the polygon P.
As a corollary we deduce at once thatPn
j=1wjsj =�Pn
j=1wj
��4(P),
where sj = jjAjBj=AjAj+1jj = 1� rj.
In the special case that P is a regular (n=d){gon, all the weights wj
are equal to the value w = 4 sin3(d�=n) cos(d�=n). (The regular(n=d){gon has n vertices and surrounds its centre d times. Successive ver-tices are obtained by rotation through 2�d=n, see [1]. It is usually assumed
23
that n and d are coprime, but this is a restriction that is unnecessary hereand in most other contexts, and downright harmful in some cases | see, forexample, [3]). Hence, in this case one can divide throughout by w, and theresult becomes
nXj=1
rj =4(P)
w=
n
4 sin2(d�=n): (1)
Since the ratios rj involve only collinear lengths, the sum is invari-ant under a�nities, and so the result (1) remains valid for all a�ne-regular(n=d){gons P. (An (n=d){gon is a�ne-regular if it is the image of a reg-ular (n=d){gon under a nonsingular a�nity.) Thus in this special case weactually achieve the analogue of the generally invalid statement mentionedabove. Since all triangles are a�ne-regular, this establishes the condition forconcurrency found by Kone �cn �y, mentioned in [7]. (We note that Shephardobtains in [9] the analogous generalization of Euler's result to a�ne-regularn{gons.) In the a�ne case, the above corollary can be simpli�ed in the sameway. For n = 3 this yields the condition for concurrency obtained by Bradleyin [7].
From the above it follows that in the case of a�ne-regular polygons(but not for general polygons) we have
nXj=1
jjBjCj=AjAj+1jj = � n cos(2d�=n)
2 sin2(d�=n); (2)
where theCj is the intersection of the lineAjAj+1 with the parallel throughQ to the line Aj�1Aj (see Figure 4). For n = 3, the right-hand side of (2)equals 1, and the result coincides with Problem 16 in [8].
It may be observed that for n = 3 and d = 1, the right-hand side of
condition (1) equals 1, and the equality to 1 of the ratio sum is necessary andsu�cient for the three parallels to the sides of the triangle to be concurrent,just as the equality to 1 of the product in Ceva's theorem for triangles isnecessary and su�cient for the concurrence of the Cevians. However, forn > 3 it is not obvious that the weights given above are the only ones whichyield the right-hand constants for all Q, although one may conjecture thatthis is the case. Naturally, for particular choices of P and Q other weightsmay be used.
The expression for rj obtained in (i), together with the analogous for-mula for the ratio tj = jjAjCj=AjAj+1jj (in the notation of Figure 4) leadsat once to the following:
24
s
s
s
s
s
s
s
s
s
s
s
Q
A1
A2
A3
A4
A5
B1
B2
B3
B4
B5s
s
s
s
sC1
C2 C3
C4
C5
Figure 4.The point Bj is obtained as in Figure 3, while the point Cj is the intersection of the
line AjAj+1 with the parallel through Q to the line Aj�1Aj . The ratios
rj = jjBjAj+1=AjCj jj of directed segments are considered in Theorem 2.
Theorem 2. For each polygon P with we have
nYj=1
rj
tj=
nYj=1
jjBjAj+1=AjCj jj = 1
for all Q.
Finally, since QBjAj+1Cj+1 is a parallelogram for every j, we alsohave
Qnj=1 jjBjQ=QCj+2jj = 1.
This last is a Ceva-type result which seems not to have been noticedpreviously.
A referee's suggestions for improved presentation are acknowledgedwith thanks.
25
References
1 H.S.M. Coxeter, Introduction to Geometry, Wiley, New York 1969.
2 L. Euler, Geometrica et sphaerica quedam, M �emoires de l'Acad �emiedes Sciences de St. Petersbourg 5 (1812), pp. 96{114. (This was sub-mitted to the Academy on 1 May 1789; the actual publication date is1815.) Reprinted in: L. Euleri Opera Omnia, Ser. 1, vol. 26, pp. 344{358; F �ussli, Basel 1953.
3 B. Gr �unbaum,Metamorphoses of polygons. The Lighter Side ofMathe-matics, Proc. Eug �ene Strens Memorial Conference, R.K. Guy andR.E. Woodrow, eds., Math. Assoc. of America, Washington, D.C. 1994,pp. 35{48.
4 B. Gr �unbaum and G.C. Shephard, Ceva, Menelaus, and the area prin-ciple, Math. Magazine 68 (1995), 254{268.
5 B. Gr �unbaum and G.C. Shephard, A new Ceva-type theorem, Math.Gazette 80 (1996), 492{500.
6 B. Gr �unbaum and G.C. Shephard, Ceva, Menelaus and selftransversal-ity, Geometriae Dedicata 65 (1997), 179{192.
7 H. G �ulicher, Problem 1987. Crux Math. 20 (1994), p. 250. Solution,ibid. 21 (1995), pp. 283{285.
8 J.D.E. Konhauser, D. Velleman and S. Wagon,WhichWay Did The Bicy-cle Go? Dolciani Math. Expositions No. 18. Math. Assoc. of America,Washington, DC, 1996.
9 G.C. Shephard, Cyclic sums for polygons, Preprint, August 1997.
University of Washington, Box 354350,Seattle, WA 98195-4350e-mail: [email protected]
26
THE SKOLIAD CORNERNo. 27
R.E. Woodrow
To start the new year we give the problems of the Preliminary round ofthe British Columbia Colleges Junior High School Mathematics Contest for1997. This round of the contest is written in the schools. The top studentsare invited to a mathematics day at which the �nal round is written, and theyget to participate in talks, tours, and a presentation ceremony. My thanksgo to John Grant McLoughlin, now of the Faculty of Education, MemorialUniversity of Newfoundland, who participated in the writing process whilehe was at the Okanagan University College.
BRITISH COLUMBIA COLLEGES
Junior High School Mathematics Contest
Preliminary Round 1997
1. A number is prime if it is greater than one and is divisible only byitself and one. The number 1997 is prime but the sum of the digits of 1997is not. The sum of the prime divisors of the sum of the digits of 1997 is:
(a) 6 (b) 8 (c) 10 (d) 13 (e) 15
2. If x = 3, which expression has a value di�erent from the other four?
(a) 2x2 (b) x2 + 9x (c) 12x (d) x2(x� 1)2 (e) 2x2(x� 1)
3. Suppose that you lace your shoes using the pattern shown belowand that the horizontal spacing between the eyelets is 4 centimetres and thevertical spacing is 3 centimetres. If there is a total of ten eyelets and thereare 10 centimetres of lace left at each of the upper eyelets, then the totallength, in centimetres, of the lace used for one shoe is:
(a) 44 (b) 54 (c) 63 (d) 64 (e) 73
27
4. Consider two unequal numbers. If we subtract half the smallernumber from both numbers, the result with the larger number is three timesas large as the result with the smaller number. How many times is the largernumber greater than the smaller number?
(a) 2 (b) 3 (c) 6 (d) 8 (e) 9
5. Suppose your hockey card collection grew by 20% last year and lastyear it was 30% larger than the year before. By what percentage has yourcollection grown during the last two years?
(a) 25 (b) 50 (c) 56 (d) 60 (e) 66
6. Antonino, the chocolate freak, eats x bars of chocolate every y days.In a week he eats how many bars of chocolate?
(a) 7xy
(b) 7y
x(c) 7xy (d) 1
7xy(e) x
7y
7. Six mattresses, each of which was originally 20 centimetres thick,are piled in a stack in a warehouse. Each mattress is compressed by one tentheach time an additional mattress is added to the stack. The height h of thestack, in centimetres, satis�es:
(a) h < 70 (b) 70 � h < 86 (c) 86 � h < 92
(d) 92 � h < 110 (e) h � 110
8. If a man walks to work and rides back home it takes him an hourand a half. When he rides both ways, it takes 30 minutes. How long wouldit take him to make the round trip by walking?
(a) 212hrs (b) 11
4hrs (c) 11
2hrs (d) 31
2hrs (e) 23
4hrs
9. Tanya was asked to add 14 to a certain number and then divide theresult by 4. Instead she �rst added 4 and then divided the answer by 14.Her result was 5. If Tanya had followed the instructions correctly, her resultwould have been:
(a) 5 (b) 20 (c) 25 (d) 66 (e) 70
10. In the diagram, a circle with centre at point C and radius 15 over-laps a circle with centre at point D and radius 20. How many square unitslarger is the shaded region on the right than that on the left?
DC
2015
(a) 400� (b) 400� � 400 (c) 175� � 400 (d) 175� (d) 225� � 400
28
11. ABCD is a rectangle in which AB is twice as long as BC. E isa point such that ABE is an equilateral triangle. M is the midpoint of BE.The measure, in degrees, of \CMB is:
A B
D C
E
M
(a) 30 (b) 60 (c) 75 (d) 90 (e) 150
12. The numbers 2800, 3600, 5400 and 6200 listed in increasing orderare:
(a) 2800; 3600; 5400; 6200 (b) 6200; 2800; 3600; 5400 (c) 6200; 2800; 5400; 3600
(d) 2800; 5400; 3600; 6200 (e) 3600; 6200; 2800; 5400
13. Think about a stopped clock. It shows the correct time twice aday. Now consider a working clock that gains �ve seconds per day. If it isnever adjusted, it will �rst show the correct time after:
(a) 48 days (b) 360 days (c) 720 days (d) 8640 days (e) 17280 days
14. The diagrams below show �ve pieces of rope. Imagine graspingthe two loose ends of each piece �rmly, then imagine pulling them until youhave a straight piece of rope | either with a knot or without one. Which ofthe �ve will give you a knot?
(a) (b) (c)
(d) (e)
29
15. In how many ways can 75 be expressed as the sum of at least twopositive integers, all of which are consecutive?
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5
Last issue we gave the thirty problems of the 1996 Kangourou desMath �ematiques [1997: 473{478]. Here are solutions.
1. B 2. B 3. E 4. D 5. E6. C 7. B 8. E 9. B 10. E11. B 12. C 13. B 14. B 15. C16. E 17. A 18. B 19. D 20. C21. A 22. D 23. B 24. D 25. D26. A 27. D 28. B 29. B 30. C
Note: A number of typos were missed in proof-reading the problems. In theFigure for Number 2, apparently four, not three, of the triangles should havebeen shaded | but didn't look to be the case on the photocopy I had!
In the Figure for Number 16 the six angles not sharing a common vertexshould be marked.
Number 18 should start \Les cot �es".
More seriously in Number 26: AD = DC = CB.
That completes the Skoliad Corner for this number. I need good contestmaterials at this level from around the world. Please send me materials foruse in the Corner as well as suggestions for future directions for the column.
30
MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by
High School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.
All material intended for inclusion in this section should be sent to the
Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,
PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronicaddress is still
The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).The rest of the sta� consists of Adrian Chan (Upper Canada College), JimmyChui (Earl Haig Secondary School), Richard Hoshino (University ofWaterloo),David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).
Shreds and Slices
The K Method
In this brief article, we will present a useful geometrical tool, which wecoin \The K Method". As is convention, let [P ] denote the area of polygonP . We will further stipulate that if P is labelled counter-clockwise, then [P ]
is positive, and negative otherwise. This sign convention will matter.
LetABC be a triangle, labelled counter-clockwise, and let P be a pointin the plane. Let K = [ABC], KA = [PBC], KB = [PCA], and KC =
[PAB] (see Figure 1). Because of the signed areas, we have that for all points
P , inside or outside the triangle ABC,
K = KA +KB +KC :
B C
A
P
B C
A
PC0
B0
A0
KC
KB
KA
Figure 1 Figure 2
31
Now, extend AP to A0 onBC, and de�neB0 andC0 similarly (see Figure 2).Recall that triangles with the same height have areas in proportion to theirbases. Then we have
AP
PA0=
[PAB]
[PBA0]=
[PCA]
[PA0C]=
[PAB] + [PCA]
[PBA0] + [PA0C]=KB +KC
KA
; and
BA0
A0C=
[ABA0]
[AA0C]=
[PBA0]
[PA0C]=
[ABA0]� [PBA0]
[AA0C]� [PA0C]=KC
KB
:
We can similarly derive that
BP
PB0
=KA +KC
KB
;CP
PC0=KA +KB
KC
;CB0
B0A=KA
KC
; andAC0
C0B=KB
KA
:
Again, these hold regardless of whether P is inside or outside triangleABC,but only with directed line segments (that is, if PQ and PR are in di�erentdirections, then their ratio will be negative).
These expressions can be very useful in problems which involve theseratios. In fact, one half of Ceva's theorem is now trivial (and with someconsideration, so is the other half).
Problem 1. Consider triangle P1P2P3 and a point P within the trian-gle. Lines P1P , P2P , P3P intersect the opposite sides in points Q1, Q2, Q3
respectively. Prove that, of the numbers
P1P
PQ1
;P2P
PQ2
;P3P
PQ3
;
at least one is less than or equal to 2 and at least one is greater than or equalto 2. (1961 IMO, Problem #4)
Solution. Let us use the same understood notation as above. Withoutloss of generality, we can assume thatKA � KB � KC (re-label the triangleif necessary). Then
P3P
PQ3
=KA +KB
KC
� KC +KC
KC
= 2; and
P1P
PQ1
=KB +KC
KA
� KA +KA
KA
= 2:
Problem 2. In triangle ABC, A0, B0, and C0 are on sides BC, AC,and AB, respectively. Given that AA0, BB0, and CC0 are concurrent at the
point O, and thatAO
OA0+
BO
OB0
+CO
OC0= 92, �nd the value of
AO
OA0� BOOB0
� COOC0
. (1992 AIME)
32
Solution. The given implies
KA +KB
KC
+KA +KC
KB
+KB +KC
KA
= 92;
and further that
K2
AKB +KAK2
B +K2
AKC +KAK2
C +K2
BKC +KBK2
C = 92KAKBKC :
Then
AO
OA0� BOOB0
� COOC0
=
�KB +KC
KA
��KA +KC
KB
��KA +KB
KC
�
=K2
AKB +KAK2
B +K2
AKC +KAK2
C +K2
BKC +KBK2
C + 2KAKBKC
KAKBKC
=92KAKBKC + 2KAKBKC
KAKBKC
= 94:
Problems
1. Prove Ceva's Theorem, which states that AA0, BB0, and CC0 (as inFigure 2) are collinear if and only if
AC0
C0B� BA
0
A0C� CB
0
B0A= 1:
2. Let P be a point inside the triangle ABC. Let AP meet BC at D, BPmeet CA at E, and CP meet AB at F . Prove that
PA
PD� PBPE
+PB
PE� PCPF
+PC
PF� PAPD
� 12:
(The Red Book of Mathematical Problems, Williams and Hardy)
33
Powers of Two
Wai Ling Yeestudent, University of Waterloo
We will solve several problems involving the number 2 and, in theprocess, survey various interesting mathematical results.
Problem 1. Prove that for all positive integers n, pn(x) = xn � 2 isirreducible in Q[x].
Remark. Q[x] is the set (or more precisely, ring) of polynomials in xwith rational coe�cients. Hence, a polynomial is irreducible in Q[x] if andonly if it cannot be factored non-trivially as a product of polynomials, alsowith rational coe�cients.
Solution. This follows directly from Eisenstein's Criterion, but we willtake a more basic approach. By DeMoivre's Theorem, the n roots of pn(x)
(in C ) are 21n cis(2k�
n), k = 0; 1; 2; : : : ; n� 1, so that
pn(x) =
x� 2
1n cis 0
! x� 2
1n cis
�2�
n
�!� � � x� 2
1n cis
�2(n� 1)�
n
�!:
Suppose pn(x) = f(x)g(x) for some f(x), g(x) 2 Q[x], with deg f(x) = m
and n > m > 0. (Note: we only need to consider n � 2.) Let the rootsof f(x), respectively g(x), be !1, !2, : : : , !m, respectively !m+1, !m+2,: : : , !n, so !1, !2, : : : , !n is a permutation of the roots of pn(x). Let!i = 21=n cis �i, and let � =
Pm
k=1 �i. Note (�1)mQm
k=1 !k is the constantterm of f(x), so
(�1)mmYk=1
!k = (�1)m2mnmYk=1
cis �k = (�1)m2mn cis � 2 Q:
But cis � = cos � + i sin � 2 Q implies that sin � = 0 and further, thatcos � = �1. Therefore, the constant term of f(x) is�2mn . It is easy to showthat 2
m
n 62 Q, a contradiction. Hence, xn � 2 is irreducible in Q[x].
Problem 2. If the sum of the proper divisors of n (that is, the divisors ofn that are less than n) is n, then n is called a perfect number. Equivalently,the sum of the divisors of n, �(n), is 2n. For example, 2 � 6 = �(6) =
1 + 2 + 3 + 6. Classify all even perfect numbers.
Solution. By the Fundamental Theorem of Arithmetic, we can expressn uniquely, up to permutations, in the form pa11 p
a22 � � � pakk , where the pi are
Copyright c 1998 Canadian Mathematical Society
34
distinct primes and the ai are positive integers. Then, all divisors of n are ofthe form p�1
1p�22� � � p�kk , where 0 � �i � ai. Hence,
�(n) =
kYj=1
�1 + pj + p2j + � � �+ p
ajj
�
=
pa1+11
� 1
p1 � 1
! pa2+12
� 1
p2 � 1
!� � � pak+1k � 1
pk � 1
!:
It follows that if gcd(m;n) = 1, then �(mn) = �(m)�(n).
We claim that if 2n�1 is prime, then 2n�1(2n�1) is a perfect number.In such a case, we have
�(2n�1(2n � 1)) =2n � 1
2� 1� (2
n � 1)2� 1
(2n � 1)� 1
= (2n � 1)(2n� 1 + 1) = 2 � 2n�1(2n � 1):
Now consider an arbitrary even perfect number n. Then n = 2km,where k � 1 and gcd(2k;m) = 1. Also, 2k+1m = �(2km) = �(2k)�(m) =
(2k+1 � 1)�(m). Now, 2k+1 � 1 is odd, so (2k+1 � 1) j m.Let m = d(2k+1 � 1), so d is a divisor ofm. By the above,
�(m) =2k+1m
2k+1 � 1= m+
m
2k+1 � 1= m+ d;
but �(m) is de�ned as the sum of all of the divisors ofm. Therefore, m onlyhas the two divisorsm and d, som is prime, d is 1, and so 2k+1�1 is prime.
We have shown that all even perfect numbers are of the form2n�1(2n�1), where 2n�1 is prime. Primes of this form are calledMersenneprimes.
Problem 3. Perform the following transformation T on the vector(x1; x2; x3; : : : ; x2n), where the xi are non-negative integers:
T (x1; x2; x3; : : : ; x2n) = (jx1 � x2j; jx2 � x3j; : : : ; jx2n � x1j):
Prove that a �nite number of applications of T will transform any such vectorto the zero vector.
Solution 1. Notice that mod 2 has the following special property:1 � �1 (mod 2)) a � �a (mod 2) for all integers a. This implies jaj � a
(mod 2) for all integers a, and that subtraction is the same as addition inmod 2. Therefore, the values jx1 � x2j, jx2 � x3j, : : : , jx2n � x1j mod 2,are congruent to x1+x2, x2+x3, : : : , x2n +x1 mod 2, which will be easierto work with.
35
De�ne S(x1; x2; : : : ; x2n) = (x1+x2; x2+x3; : : : ; x2n +x1). By theprevious remarks, S and T give the same results mod 2. Notice that
S2(x1; x2; x3; : : : ; x2n)
= (x1 + 2x2 + x3; x2 + 2x3 + x4; : : : ; x2n + 2x1 + x2);
and
S3(x1; x2; : : : ; x2n)
= (x1 + 3x2 + 3x3 + x4; x2 + 3x3 + 3x4 + x5;
: : : ; x2n + 3x1 + 3x2 + x3):
The coe�cients seem to be entries in Pascal's triangle. Indeed, wecan prove by induction that the ith entry of Sk(x1; x2; x3; : : : ; x2n) iskX
j=0
�k
j
�xj+i, where xr = xs if r � s (mod 2n).
What does S2n
(x1; x2; x3; : : : ; x2n) look like? First, we claim that if
1 � j � 2n � 1, then�2n
j
�is even. We prove the more general result which
states that if p is prime and 1 � j � pk � 1, then p divides�pk
j
�.
Step 1: For p prime, we determine the highest power of p dividingm!.
There are bmpc natural numbers less than or equal to m that are divis-
ible by p, bmp2c natural numbers less than or equal to m that are divisible by
p2, and so on. For each number divisible by at most pa, we have counted itexactly a times in the sums.
Therefore, the highest power of p dividingm! is�m
p
�+
�m
p2
�+
�m
p3
�+ � � � :
Step 2: We calculate the highest power of p dividing�pk
j
�, where
1 � j � pk � 1 and p is prime, which is simply the highest power of pdividing the numerator subtracted by the highest power of p dividing thedenominator.
Applying the result from the previous step, the highest power of p di-
viding�pk
j
�=
pk!
j!(pk�j)!is
1Xi=1
$pk
pi
%�
1Xi=1
�j
pi
��
1Xi=1
$pk � j
pi
%=
kXi=1
$pk
pi
%��j
pi
��$pk � j
pi
%!:
Each term in the sum is non-negative, since bx + yc � bxc + byc forall reals x, y, and when i = k,$
pk
pk
%��j
pk
��$pk � j
pk
%= 1;
36
since 1 � j � pk � 1. Hence, there is at least one factor of p in�pk
j
�.
Now we relate this result to the problem. The parity of the ith en-try of T 2
n
(x1; x2; x3; : : : ; x2n) is the same as the parity of the ith entry ofS2
n
(x1; x2; x3; : : : ; x2n), which is
2nX
j=0
�2n
j
�xj+i �
�2n
0
�xi +
�2n
2n
�xi+2n � 2xi � 0 (mod 2):
Therefore, the ith entry of T 2n
(x1; x2; x3; : : : ; x2n) is even. We canthen pull a factor of 2 out of each of the resulting elements and apply T 2n
times to that vector (since T is linear), and again, obtain a vector with allentries divisible by 2, and so on. Hence, after t � 2n applications of T , eachentry in the vector must be divisible by 2t.
If yk is used to denote the largest of the 2n entries of
T k(x1; x2; x3; : : : ; x2n);
then fykg is a non-increasing sequence. Furthermore, there exists t such thaty1 < 2t. Then after t � 2n applications of T , the resulting vector has entriesall divisible by 2t. Since the entries are non-negative and less than 2t, theymust all be zero. Therefore, we have transformed the original vector to thezero vector with a �nite number of applications of T .
Solution 2. We shall prove by induction that for all natural n, thereexists k such that every entry of T k(x1; x2; x3; : : : ; x2n) and, equivalently,every entry of Sk(x1; x2; x3; : : : ; x2n) is even for all non-negative integersx1; x2; : : : ; x2n. When n = 0, T (x1) = x1 � x1 = 0 and our hypothesisholds. Assume the result holds for some n = r, with kr iterations of Talways producing a result whose entries are all divisible by 2. Let n = r+1.Then
S(x1; x2; x3; : : : ; x2r+1)
= (x1 + x2; x2 + x3; x3 + x4; : : : ; x2r+1 + x1);
S2(x1; x2; x3; : : : ; x2r+1)
= (x1 + 2x2 + x3; x2 + 2x3 + x4; x3 + 2x4 + x5; : : : )
� (x1 + x3; x2 + x4; x3 + x5; : : : ; x2r+1 + x2) (mod 2) :
If we take every other element of the last vector, starting with the �rst,we obtain (x1 + x3; x3 + x5; : : : ; x2r+1�1 + x1), which is also what we ob-tain when we apply T to (x1; x3; x5; : : : ; x2r+1�1), and similarly with theeven indexed elements. Therefore, by the induction hypothesis, after 2krapplications of T , we have a vector whose entries are all divisible by 2. Bymathematical induction, for all natural n, a �nite number of applications ofT will transform any vector into a vector divisible by 2. As we concluded inthe previous solution, a �nite number of applications of T will transform anyvector into the zero vector.
37
The Cantor Set and Cantor Function
Naoki Sato
We begin by considering a problem that was given out at IMO training thispast summer.
Problem. Let f : [0; 1] ! [0; 1] be a function satisfying the followingthree properties:
(i) f is non-decreasing (that is; x < y) f(x) � f(y)),
(ii) f(x) = 1� f(1� x) for all x 2 [0; 1], and
(iii) f(3x) = 2f(x) for all x 2 [0; 13].
Evaluate f(1=7) and f(1=13).
The reader at this point should stop and try to work out the values theproblem asks for; this is a good way to get a feel for the dynamics of thisfunction, which turns out to be a very special function, and well-known inanalysis and chaos theory.
What is the �rst of many remarkable facts is that the three proper-ties given above are enough to determine the value of f(x) for any pointx 2 [0; 1], even though they do not seem to be { what if x is irrational?
We will come back to these questions later. We will �rst describe aseemingly unrelated concept, the Cantor set. There are several possiblede�nitions, and this is perhaps the most straightforward.
Let A0 be the interval [0; 1]. From this, we remove the \middle-third"�1
3; 23
�; and we obtain A1 = [0; 1
3] [ [2
3; 1], which is the union of two in-
tervals. We remove the middle-third from these two intervals, and obtainA3 = [0; 1
9] [ [2
9; 13] [ [2
3; 79] [ [8
9; 1], and so forth (see Figure 1). We call C,
the set obtained in this limiting process, the Cantor set. Rigorously speaking,
An = [0; 1]nn[
k=1
3k�1[j=1
�3j � 2
3k;3j � 1
3k
�;
C = [0; 1]n1[k=1
3k�1[j=1
�3j � 2
3k;3j � 1
3k
�:
Note that An is the union of closed intervals, and that the sum of thelengths of these intervals is
�2
3
�n(since we remove one third at each step),
which goes to 0 as n approaches in�nity. In this sense, we have removed\most" of the interval [0,1] in obtaining C.
Copyright c 1998 Canadian Mathematical Society
38
We mentioned that C has alternate de�nitions. Each x 2 [0; 1] has abase 3 expansion of the form
x =
1Xi=1
ai
3i;
where ai 2 f0; 1; 2g. Then C is the set of all x 2 [0; 1] such that ai 6= 1 forall i. In other words, each digit is a 0 or 2.
Caution. There are some cases where this is not exactly true |some numbers have two possible base 3 expansions, for example1
3= 0:13 = 0:0222 : : :3. Which one should we use?
The set of numbers which have a1 = 1 is precisely the interval�1
3; 23
�,
which is what we threw away from A0 to get A1. Then, the set of numberswhich have a1 6= 1 and a2 = 1 is
�1
9; 29
� [ �79; 89
�, which is what we threw
away from A1 to get A2, and so on. Hence, the two constructions are thesame.
The Cantor set has many uses. For example, althoughmost of [0; 1] hasbeen removed to obtain C, it can be shown that there is an onto map fromC to [0; 1]; that is, every element in [0; 1] is equal to the image of x for somex 2 C. Not even Q, the rationals, can make this claim. In the language ofset theory, Q is a countable set, and C and [0; 1] are uncountable sets, whichare \bigger" (at this level, the size, or cardinality of sets are measured by theexistence of 1{1, onto, or bijective maps between them).
Also, as the reader might have suspected, the Cantor set is also oneof the most basic examples of a fractal, an object, roughly speaking, withself-similarity. In fact, the Cantor set is one of the most important fractals,despite its almost bland simplicity and lack of interesting detail. For example,certain Julia sets are modi�ed Cantor sets. But how does all this relate toour original problem?
We give a formulation for f(x). If x 2 C, then ai 2 f0; 2g for all i inthe base 3 expansion of x, and
f(0:a1a2a3 : : :3) = 0:a1
2
a2
2
a3
2: : :2 :
Note we have base 3 on the left and base 2 on the right. If x 62 C, thenai = 1 for some i. Choose the smallest such i. Then
f (0:a1a2a3 : : : ai : : :3) = 0:a1
2
a2
2
a3
2: : :
ai�1
2ai 2:
For example,
f
�4
13
�= f (0:022022 : : :3) = 0:011011 : : :2 =
3
7;
f
�38
243
�= f (0:021023) = 0:0112 =
3
8:
39
We leave it to the reader to verify that these are the correct expressions, butwe will show that the properties above are su�cient to determine f .
First, assume that x 2 C, so ai 2 f0; 2g for all i. We proceed byinduction on the number of digits of x. Let 0:b1b2b3 : : :2 be the binary ex-pansion of f(x). Using the properties, we see that f(0) = 0, f(1) = 1, andf�1
3
�= f
�2
3
�= 1
2, or f (0:0222 : : :3) = 0:0111 : : :2, and f (0:23) = 0:12.
Hence, by expressing 1
2in these two forms, and since f is non-decreasing,
we see that if a1 = 0, then b1 = 0, and if a1 = 2, then b1 = 1, so b1 = a1=2.
Now assume bi = ai=2 for all i from 1 up to some n. We must showthat bn+1 = an+1=2. First, consider the case a1 = 0. Then b1 = 0, as shownabove, and f(3x) = 2f(x), or
f(0:a2a3 : : : an+1 : : :3) = 0:b2b3 : : : bn+1 : : :2 :
By the induction hypothesis, bn+1 = an+1=2. The case where a1 = 2 isleft to the reader [Hint: This is where we must use f(x) = 1� f(1� x)].Therefore, by induction, bi = ai=2 for all i.
Also, it can be seen that if x and y are the end-points of one of theintervals removed to obtain C, then f(x) = f(y). Thus, again since f isnon-decreasing, f on each missing interval is the common value at the end-points.
We also get another remarkable property as a bonus. Note that f isonto. Pick y 2 [0; 1], express y in base 2, and it should be obvious whichx 2 [0; 1] satis�es f(x) = y. We can even choose x 2 C, so that f , whenrestricted to C, is an example (in fact, the standard example) of a map fromC to [0; 1] that is onto. Moreover, f is non-decreasing, so the only possiblediscontinuities of f are jump discontinuities. But f is onto, so it cannot haveany jump discontinuities, so f is continuous.
Figure 1. Figure 2.
When f is graphed, the features alluded to above become immediately ap-parent (see Figure 2). On the interval
�1
3; 23
�, f is indeed the constant 1
2,
40
on�1
9; 29
�, the constant 1
4, and on
�7
9; 89
�, the constant 3
4, and so on. As
mentioned, this function f , sometimes called the Devil's Staircase, has someexceptional properties. By the observation just made, f 0(x) = 0 almosteverywhere (\almost everywhere" does have a technical meaning, and it isrelated to the fact that most of [0; 1] is missing); in fact, the derivative is notde�ned precisely at points in the Cantor set. But f(0) = 0 and f(1) = 1,so f must somehow climb up abruptly at points in the Cantor set, since it is at almost everywhere.
So, a fairly innocuous problem on a functional equation turns out tohave some large rami�cations. We end with a few more miscellaneous facts.
Let T1(x) = x=3 and T2(x) = (x + 2)=3. Pick a random pointx 2 [0; 1], and recursively apply Ti to x (where i is randomly chosen ateach step). Plot each x; you can do this on your computer. What do you get?
De�ne T (x) on [0; 1] as follows:
T (x) =
(3x if 0 � x � 1=2
3� 3x if 1=2 � x � 1:
This function is sometimes called the tent function, for obvious reasons.Some points in the range of T are outside of [0; 1]; which ones? These arebad points we wish to remove from the domain of T , because we wish toapply T an arbitrarily number of times to values in [0; 1]. For which x is ittrue that T 2(x) 2 [0; 1]? What about T 3?
In fact, what is the set of all x such that T k(x) 2 [0; 1] for all k?
You guessed it, it is the Cantor set. In terms of the base 3 expansion,how does T (x) relate to x?
Problems
1. Show that for all z 2 [0; 2], there exist x, y 2 C such that x+ y = z.Symbolically, C + C = [0; 2].
2. Show that C is totally disconnected; that is, show that for all x, z 2 C,there is a y 62 C such that x < y < z.
41
J.I.R. McKnight Problems Contest 1981
1. If x, y, and z are all between 0 and �
2inclusive, solve for x, y and z if
log2(sinx siny sinz) = �3
2
log2
sin2 x sin3 y
sin z
!= �5
2
log2
sinx sin4 y
sin2 z
!= �3
2. A variable chord of an ellipse subtends a right angle at the centre. Showthat the chord always touches a �xed circle.
3. (a) Obtain the equation of the tangent with slope m to the parabolawhose equation is y2 = 4px. Assume p > 0.
(b) Obtain the equation of the tangent perpendicular to the tangent in(a).
(c) Find the equation of the locus of the points of an intersection ofpairs of perpendicular tangents to the parabola in (a).
4. Find the �rst four terms in ascending powers of x in the expansion asan in�nite series of
1� 2x
1 + x� 2x2
and state the restrictions on x.
5. A tent has the shape of a cone and has a capacity of 1000 m3. Find theradius of the base if the amount of canvas used is to be a minimum.
(No canvas is used on the oor.)
6. Find the length of the longest ladder that can be carried horizontallyaround the corner of the corridor shown in the diagram.
� -4 m
6
?
3 m
42
Mayhem ProblemsThe Mayhem Problems editors are:
Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.
Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only problems | thenext issue will feature only solutions.
We warmly welcome proposals for problems and solutions. With thenew schedule of eight issues per year, we request that solutions from thisissue be submitted by 1 April 1998, for publication in the issue 5 monthsahead; that is, issue 6 of 1998. We also request that only students submitsolutions (see editorial [1997: 30]), but we will consider particularly elegantor insightful solutions from others.
High School Problems
Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,Canada. L3S 3E8 <[email protected]>
H223. In each of the following alphametics, each letter in the additionrepresents a unique digit:
1997 and 1998
+ OLD + OLD
YEAR YEAR
For each alphametic, �nd a solution, or prove that a solution does notexist.
H224. Let ABCD be a square. Construct equilateral triangles APB,BQC, CRD, and DSA, where P , Q, R, and S are points outside of thesquare.
(a) Prove that PQRS is a square.
(b) Determine the ratio PQ
AB. (See how many ways you can solve this!)
H225. Consider a row of �ve chairs, numbered 1, 2, 3, 4, and 5. Youare originally sitting on 1. On each move, you must stand up and sit downon an adjacent chair. Make 19 moves, then take away chairs 1 and 5. Thenmake another 97 moves, with the three remaining chairs. No matter howthe moves are made, you will always end up on chair 3. Why is this the case?
43
H226. The smallest multiple of 1998 that only consists of the digits 0and 9 is 9990.
(a) What is the smallest multiple of 1998 that only consists of the digits 0and 3?
(b) What is the smallest multiple of 1998 that only consists of the digits 0and 1?
Advanced Problems
Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.M1G 1C3 <[email protected]>
A209. Are there an in�nite number of squares among the triangularnumbers? Triangular numbers are numbers of the form Tn = n(n+ 1)=2.
A210. Let P be a point inside circle C. Find the locus of the centresof all circles ! which pass through P and are tangent to C.
A211. Does there exist a convex polyhedron and a plane, not passingthrough any of its vertices, and intersecting more than 2
3of all of the edges
of the polyhedron?
(Polish Mathematical Olympiad, �rst round)
A212. Let A and B be real n�n matrices such that A2 +B2 = AB.Prove that if AB � BA is an invertible matrix, then n is divisible by 3.
(International Competition for University Students in Mathematics)
Challenge Board Problems
Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <[email protected]>
In this issue, we introduce David Savitt as the new Challenge Boardeditor. David is currently a graduate student at Harvard, and comes to uswith much problem solving experience under his belt, including high rankingon the Putnam competition. It looks like he may be asking some very toughproblems!
Ravi Vakil has recently graduated from Harvard University, and will begoing on to Princeton and the IAS for post-doctoral work, and we wish himthe best of luck.
44
C75. (a) Let n be an integer, and suppose a1, a2, a3, and a4 are inte-gers such that a1a4 � a2a3 � 1 (mod n). Show that there exist integers Ai,1 � i � 4, such that each Ai � ai (mod n) and A1A4 �A2A3 = 1.
(b) Let SL(2;Z)denote the group of 2�2matrices with integer entriesand determinant 1, and let �(n) denote the subgroup of SL(2;Z)of matriceswhich are congruent to the identity matrix modulo n. (By this we mean thatall pairs of corresponding entries are congruent modulo n.)
What is the index of �(n) in SL(2;Z)?
C76. Let X be any topological space. The nth symmetric power of X,denoted X(n), is de�ned to be the quotient of the ordinary n-fold productXn by the action of the symmetric group on n letters { that is, it is the spaceof unordered n-tuples of points of X. Show that the symmetric power C (n)
is actually homeomorphic to the ordinary product Cn .
From the �les
For the bene�t of those readers who are new toMATHEMATICALMAY-
HEM, we present a few problems from back issues:
J14. [MAYHEM 1988: Vol 1, #1, 19]
In trapezoid ABCD, we have ABkCD and jABj = 2jCDj. Supposethat AC meets BD at X.
Find the ratio BX : XD.
S12. [MAYHEM 1988: Vol 1, #1, 20]
Find all functions with domain [0;1) such that
f(x) =
Z x
0
f(t)dt :
U10. [MAYHEM 1988: Vol 1, #1, 21]
Prove that
nXr=0
�2n
r
��2n� 2r
n� r
�= 4n :
45
PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, De-partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.
In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.
To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81
2"�11" or A4 sheets of paper.
These may be typewritten or neatly hand-written, and should be mailed tothe Editor-in-Chief, to arrive no later than 1 September 1998. They may alsobe sent by email to [email protected]. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication. Please note that we do not accept submissions sent by FAX.
2301. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.
Suppose that ABC is a triangle with sides a, b, c, that P is a pointin the interior of 4ABC, and that AP meets the circle BPC again at A0.
De�ne B0 and C0 similarly.
Prove that the perimeter P of the hexagon AB0CA0BC0 satis�es
P � 2�p
ab+pbc+
pca�:
2302. Proposed by Toshio Seimiya, Kawasaki, Japan.Suppose that the bisector of angle A of triangle ABC intersects BC
at D. Suppose that AB + AD = CD and AC + AD = BC.
Determine the angles B and C.
2303. Proposed by Toshio Seimiya, Kawasaki, Japan.Suppose that ABC is a triangle with angles B and C satisfying
C = 90�+ 1
2B, that the exterior bisector of angleA intersects BC atD, and
that the side AB touches the incircle of4ABC at E.
Prove that CD = 2AE.
46
2304. Proposed by Toshio Seimiya, Kawasaki, Japan.An acute angled triangle ABC is given, and equilateral trianglesABD
and ACE are drawn outwardly on the sides AB and AC. Suppose that CDand BE meet AB and AC at F and G respectively, and that CD and BEintersect at P .
Suppose that the area of the quadrilateral AFPG is equal to the areaof the triangle PBC. Determine angle BAC.
2305. Proposed by Richard I. Hess, Rancho Palos Verdes, California,USA.
An integer sided triangle has angles p� and q�, where p and q are rel-atively prime integers. Prove that cos � is rational.
2306. Proposed by Vedula N. Murty, Visakhapatnam, India.(a) Give an elementary proof of the inequality:
�sin�x
2
�2>
2x2
1 + x2; (0 < x < 1): (1)
(b) Hence (or otherwise) show that
tan�x
8<:<
�x(1�x)
1�2x;
�0 < x < 1
2
�;
>�x(1�x)
1�2x;
�1
2< x < 1
�:
(2)
(c) Find the maximum value of f(x) =sin�x
x(1� x)on the interval (0;1).
2307. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Madrid, Spain.
It is known that every regular 2n{gon can be dissected into�n
2
�rhom-
buses with the same side length.
(a) How many di�erent classes of rhombuses are there?
(b) How many rhombuses are there in each class?
2308. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Madrid, Spain.
A sequence fvng has initial value v0 = 1 and, for n � 0, satis�es therecurrence relation
vn+1 = 2n+1 �nX
k=0
vk vn�k; :
Find a formula for vn in terms of n.
47
2309. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.
Suppose thatABC is a triangle and that P is a point of the circumcircle,distinct from A, B and C. Denote by SA the circle with centre A and radiusAP . De�ne SB and SC similarly. Suppose that SA and SB intersect at Pand PC. De�ne PB and PA similarly.
Prove that PA, PB and PC are collinear.
2310. Proposed by K.R.S. Sastry, Dodballapur, India.Let n 2 N. I call a positive integral divisor of n, say d, a unitary divisor
if gcd(d; n=d) = 1.
Let �(n) denote the sum of the unitary divisors of n.
Find a characterization of n so that �(n) � 2(mod 4).
2311. Proposed by K.R.S. Sastry, Dodballapur, India.Let �e(n) denote the sum of the even unitary divisors, and�o(n), the
sum of the odd unitary divisors, of n. Assume that �e(n)��o(n) = n.
(a) If n is composed of powers of exactly two distinct primes, show that nmust be the product of two consecutive integers, one of which is a Mersenneprime.
(b) Give an example of a natural number n that is composed of powers ofmore than two distinct primes.
2312. Proposed by K.R.S. Sastry, Dodballapur, India.
The rth n{gonal number is given by P (n; r) = (n� 2)r2
2� (n� 4)r
2,
where n � 3, r = 1, 2, : : : .
Prove that, in the interval [P (n; r); P (n; r+ 1)], there is an (n� 1){gonal number.
2313. Proposed by Heinz-J �urgen Sei�ert, Berlin, Germany.Let N be a non-negative integer and let a and b be complex numbers
with a; b 62 f0;�1;�2; : : : ;�(n� 1)g. Find a closed form expression for
nXk=0
(�1)k(a)k (b)n�k
;
where (a)k denotes the Pochhammer symbol, de�ned by (a)0 = 1,(a)k = a(a+ 1) : : : (a+ k � 1), k 2 N.
48
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.
Erratum
On page 510 of issue 8, 1997, for \Cautis", read \Howard". However,the solution printed for 2181 was incorrect | see below.
The name of RICHARD I. HESS, Rancho Palos Verdes, California, USAwas inadvertantly omitted from the list of solvers of problem 2153. He alsosumbitted a late solution to problem 2136.
2145. [1996: 170, 1997: 302] Proposed by Robert Geretschl�ager,Bundesrealgymnasium, Graz, Austria.
Prove that
nYk=1
�ak + bk�1
� � nYk=1
�ak + bn�k
�for all a, b > 1.
Comment by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta. The given inequality is a special case of the following more generalre-arrangement inequality due to Oppenheim [1]:
If 0 < a1 � a2 � : : : � an, 0 < b1 � b2 � : : : � bn, and thesequence fcng is any permutation of the sequence fbng, then
nYk=1
(ak + bk) �nY
k=1
(ak + ck) �nY
k=1
(ak + bn�k+1) :
Reference
1. A. Oppenheim, Inequalities connected with de�nite Hermition formsII, Amer. Math. Monthly 61 (1954), 263{266.
2181. [1996: 318, 1997: 509] Proposed by �Sefket Arslanagi �c, Berlin,Germany.
Prove that the product of eight consecutive positive integers cannot bethe fourth power of any positive integer.
Corrected solution| several readers of the on-line version pointed outthat the printed solution in the previous issue was incorrect | thank you!
It has been pointed out by many readers that this problem has ap-peared before. Most readers referred to theAmerican Mathematical Monthly,1936, p.310 for the solution to #3703 (posed by Victor Th �ebault in 1934,
49
p.522). Another reference was made to Honsberger's monograph Mathe-matical Morsels, where it appears on p.156 as \A Perfect 4th Power". Severalreaders also made reference to the general problem of proving that the prod-uct of (two or more) consecutive integers is never a square, which was es-tablished in 1975 by Erd }os and Selfridges (Illinois Journal ofMath 19 (1975),292-301). Because the solution appears elsewhere, we will simply refer theinterested reader to these other sources for a solution.
Comments and/or references and/or solutions were submitted byFRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADAM BROWN,Scarborough, Ontario; GORAN CONAR, student, Gymnasium Vara�zdin,Vara�zdin, Croatia; THEODORE CHRONIS, student, Aristotle University ofThessaloniki, Greece; F.J. FLANIGAN, San Jose State University, San Jose,California, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON,Beaverton, Oregon, USA; MURRAY S. KLAMKIN, University of Alberta, Ed-monton, Alberta; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece;MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's,Newfoundland; HARRY SEDINGER, St. Bonaventure University, St. Bon-aventure, NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; D.J.SMEENK, Zaltbommel, the Netherlands; DIGBY SMITH, Mount Royal Col-lege, Calgary, Alberta; DAVID R. STONE, Georgia Southern University,Statesboro, Georgia, USA; EDWARD T.H. WANG, Wilfrid Laurier Univer-sity, Waterloo, Ontario; KENNETH M. WILKE, Topeka, Kansas, USA; PAULYIU, Florida Atlantic University, Boca Raton, Florida, USA; and the pro-poser. There were two incorrect solutions submitted.
Sei�ert remarks that A. Guibert proved the result in 1862 according toL.E. Dickson in History of the Theory of Numbers, Vol. II, 1952, pp. 679-680.
2198. Proposed by Vedula N. Murty, Visakhapatnam, India.
Prove that, if a, b, c are the lengths of the sides of a triangle,
(b� c)2�2
bc� 1
a2
�+ (c� a)2
�2
ca� 1
b2
�+ (a� b)2
�2
ab� 1
c2
�� 0;
with equality if and only if a = b = c.
Editor's comment: Several solvers (indicated by a y beside their namein the list of solvers below) pointed out that a proof of this inequality hadalready appeared in the comment (given by the proposer) on problem 1940[1994: 321].
50
Solved by RICHARD I. HESS, Rancho Palos Verdes, California, USA;yWALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAELLAMBROU, University of Crete, Crete, Greece(two proofs); yHEINZ-J�URGEN SEIFFERT, Berlin, Germany; and yPANOS E. TSAOUSSOGLOU,Athens, Greece.
2199. Proposed byDavidDoster, Choate Rosemary Hall, Wallingford,Connecticut, USA.
Find the maximum value of c for which (x + y + z)2 > cxz for all0 � x < y < z.
Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.With x = 1, y = 1 + u; and z = 2, where 0 < u < 1, the in-
equality gives (4 + u)2 > 2c. Letting u approach zero, we �nd c � 8.If 0 � x < y < z, then
(x+ y+ z)2 > (2x+ z)2 = (2x� z)2 + 8xz � 8xz:
Hence, 8 is the maximum value of c.
Also solved by MANSUR BOASE, student, St. Paul's School, London,England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; PAUL-OLIVIER DEHAYE, Bruxelles, Belgium; F.J. FLANIGAN, San Jose State Uni-versity, San Jose, California, USA; ROBERT GERETSCHL �AGER, Bundesre-algymnasium, Graz, Austria; FLORIAN HERZIG, student, Perchtoldsdorf,Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUSHSIA, student, University of Toronto, Toronto, Ontario; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, Univer-sity of Alberta, Edmonton, Alberta; V �ACLAV KONE �CN �Y, Ferris State Uni-versity, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University ofCrete, Crete, Greece; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Aus-tria; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia,USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. Therewere two incomplete solutions.
Klamkin used a similar proof to verify the more general result:
The maximum value of c, for which
(x1 + x2 + : : :+ xn)2 > cx1 : : : xn
for all 0 � x1 < x2 < : : : < xn, is c = 4(n� 1).
Janous provided a more general result still (note the inequalities hereare not strict):
Let n � 2; 1 � k < l � n be natural numbers. Then the maximumconstant ck;l such that
(x1 + x2 + : : :+ xn)2 � ck;lxkxl
51
is valid for all 0 � x1 � x2 � : : : � xn is given by:
ck;l =
�4(l� k)(n� l + 1) if 2l� k � n+ 1
(n� k + 1)2 if 2l� k � n:
[In our case, this allows us to conclude c � c1;3 = 8:]
2200. Proposed by Jeremy T. Bradley, Bristol, UK and ChristopherJ. Bradley, Clifton College, Bristol, UK.
Find distinct positive integers a, b, c, d, w, x, y, z, such that
z2 � y2 = x2 � c2 = w2 � b2 = d2 � a2
andc2 � a2 = y2� w2:
I. Solution by Richard I. Hess, Rancho Palos Verdes, California, USA.Let d1; d2; d3; d4 and n=d1; n=d2; n=d3; n=d4 all be distinct divisors of
some positive integer n. Then
z =1
2
�d1 +
n
d1
�; y =
1
2
�d1 �
n
d1
�; etc.
satisfy
n = z2 � y2 = x2 � c2 = w2 � b2 = d2 � a2: (1)
The other condition in the problem becomes�d2
2� n
2d2
�2��d4
2� n
2d4
�2=
�d1
2� n
2d1
�2��d3
2+
n
2d3
�2;
or
n2�1
d22
+1
d23
� 1
d21
� 1
d24
�+ 4n+ (d2
2+ d2
3� d2
1� d2
4) = 0; (2)
or An2 + 4n+ C = 0 where
A =
�1
d22
+1
d23
� 1
d21
� 1
d24
�; C = d2
2+ d2
3� d2
1� d2
4:
A computer search exposed the following solutions.
[Editorial note. Hess searched for integer solutions of (2), which yieldrational values for z; y etc. He then multiplied all of z; y; x; c; w; b; d; a byan appropriate positive integer �, and n by �2, to clear all denominators.He gave several solutions, some very large. We give just two of the smallerones.]
52
(i) If A = 0, then a solution to (2) is
d1 = 5; d2 = 6; d3 = 9; d4 = 90; n = 2002;
which with � = 90 gives the solution
z = 18243; y = 17793; x = 15285; c = 14745;
w = 10415; b = 9605; d = 5051; a = 3049:
(ii) A solution to (2) with A 6= 0 is
d1 = 24; d2 = 20; d3 = 15; d4 = 12; n = 24
which leads (� = 10) to the solution
z = 125; y = 115; x = 106; c = 94; w = 83; b = 67; d = 70; a = 50:
II.More solutions, and editorial comments. There were only two othercontributions to this problem. The proposers give two solutions:
z = 109; y = 89; x = 81; c = 51; w = 73; b = 37; d = 63; a = 3;
and
z = 97; y = 83; x = 79; c = 61; w = 57; b = 27; d = 51; a = 9;
found by computer. They wonder if there are in�nitely many solutions. (Ofcourse, any solution can be multiplied through by a positive integer to createanother solution, but we are only interested in primitive solutions, that is,ones in which the eight numbers z; y; : : : have no common factor.) Hessclaims to generate in�nitely many primitive solutions, but this editor wasunable to untangle his argument. Does anyone have a simple proof?
V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA,uses an idea similar to Hess's, namely he �nds parametric expressions forthe eight variables of the problem which automatically satisfy the �rst givencondition, and then searches for values of the parameters so that the secondcondition is satis�ed too. In particular he notes that
z = fgh+ e; y = fgh� e; x = egh+ f; c = egh� f;
w = efh+ g; b = efh� g; d = efg+ h; a = efg� h;
satisfy (1) whenever e; f; g; h are positive integers yielding positive distinctvalues for these expressions. The second condition then becomes
(egh� f)2� (efg� h)2 = (fgh� e)2 � (efh+ g)2;
and Kone�cn �y gives the two solutions
e = 4; f = 5; g = 6; h = 11 and e = 4; f = 6; g = 5; h = 11;
53
which result in the respective solutions
z = 334; y = 326; x = 269; c = 259; w = 226; b = 214; d = 131; a = 109;
and
z = 334; y = 326; x = 226; c = 214; w = 269; b = 259; d = 131; a = 109:
Note that these solutions di�er only in that b; c have been switched andw; xhave been switched. In fact this pair of switches is always possible. Condition(1) obviously remains true whenever these switches are made; moreover, inHess's solution, the other condition (2) is symmetric in d2 and d3, whichshows that these switches do not a�ect the truth of (2) either. And Hess'ssolution is general: given any solution of (1), put z + y = d1 (and z � y =
n=d1), and similarly for the other variables, and Hess's expressions for z; yetc. follow.
As a consequence, we need only look for solutions to the problem whichsatisfy b < c and w < x, which hold for all solutions above except forKone�cn �y's second. Similarly (but more easily), given any solution to theproblem, switching z with d and y with a will always result in another solu-tion, which means we can also assume that d < z and a < y, which hold forall the above solutions.
At the moment this problem is still in an unsatisfactory state of dis-array, and needs someone to bring some order to the chaos! Readers areencouraged to try.
2201. [1997: 45] Proposed by Toshio Seimiya, Kawasaki, Japan.ABCD is a convex quadrilateral, and O is the intersection of its diag-
onals. Let L;M;N be the midpoints ofDB;BC;CA respectively. Supposethat AL;OM;DN are concurrent. Show that
either AD k BC or [ABCD] = 2[OBC];
where [F ] denotes the area of �gure F.
I. Solution by C. Festraets-Hamoir, Brussels, Belgium.Let O be the origin of a coordinate system where A;B;C;D are repre-
sented by (a; 0); (0; b); (c; 0); (0; d)with a; b positive and c; d negative. Thus
L is the point�0;
(b+d)
2
�;M is
�c
2; b2
�; N is
�(a+c)
2; 0�and
AL : (b+ d)x+ 2ay� a(b+ d) = 0
OM : bx� cy = 0
DN : 2dx+ (a+ c)y� d(a+ c) = 0.
These lines are concurrent if and only if������b �c 0
b+ d 2a �a(b+ d)
2d a+ c �d(a+ c)
������ = 0:
54
This equation reduces (after some manipulation) to
(ab� cd)[(a� c)(b� d) + 2bc] = 0:
Consequently, either
(a) ab = cd, in which case ADjjBC, or
(b) 1
2(a � c)(b� d) sin� = 2
��1
2bc sin�
�(where � = \AOB), in which
case [ABCD] = 2[OBC].
II. Comment based on a solutionbyMichael Lambrou, University of Crete,Crete, Greece.
Using barycentric coordinates one can obtain a geometric characteriza-tion of those quadrilaterals ABCD for which AL;OM , and DN are con-current at J : Either ABjjBC (in which case J always exists), or JCOB is aparallelogram whose area equalsABCD. Thus to draw an accurate �gure ofthe latter case, begin with a parallelogram JCOB and let L be any point onBO. (For ABCD to be convex, L should lie between O and the midpoint ofBO.)
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; FLORIANHERZIG, student, Perchtoldsdorf,Austria; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, Universityof Crete, Crete, Greece, (3 solutions); GERRY LEVERSHA, St Paul's School,London, England; ISTV �AN REIMAN, Budapest, Hungary; D.J. SMEENK,Zaltbommel, the Netherlands; and the proposer.
2202. [1997: 45] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.
Suppose than n � 3. Let A1; : : : ; An be a convex n-gon (as usual withinterior angles A1; : : : ; An).
Determine the greatest constant Cn such that
nXk=1
1
Ak
� Cn
nXk=1
1
� � Ak
:
Determine when equality occurs.
55
Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-sity, Waterloo, Ontario.
We assume that the intention of the problem is to determine the largestconstant Cn so that the given inequality holds for all convex n-gons.
If n � 4 then clearly Cn = 0, since we could let one of the interiorangles tend to � without forcing any other interior angle to tend to zero.Then
nXk=1
1
� � Ak
! +1 while
nXk=1
1
Ak
is bounded:
Thus the inequality cannot hold unless Cn � 0. Hence Cn = 0 and equalitycan never occur.
If n = 3, then by the arithmetic{harmonic{mean inequality,
(A1 +A2)
�1
A1
+1
A2
�� 4
and thus1
A1
+1
A2
� 4
A1 +A2
=4
� � A3
:
Similarly,
1
A2
+1
A3
� 4
� � A1
and1
A3
+1
A1
� 4
� � A2
:
Adding, we get3X
k=1
1
Ak
� 2
3Xk=1
1
� � Ak
:
It is easily seen that equality holds if and only if all theAk's are equal. HenceC3 = 2 and equality holds if and only if �A1A2A3 is equilateral.
Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria;MICHAEL LAMBROU, University of Crete, Crete, Greece (three solutions);GERRY LEVERSHA, St Paul's School, London, England; HEINZ-J�URGENSEIFFERT, Berlin, Germany; and the proposer. The case n = 3 only wassolved by GORAN CONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croa-tia. Two incorrect solutions were also sent in.
Sei�ert uses the known inequality
nXk=1
1
1� ak
nXk=1
(1� ak) �nX
k=1
1
ak
nXk=1
ak ;
where 0 < ak � 1=2 (see pages 25{27 of Mitrinovi�c, Pe�cari�c and Fink, Clas-sical and New Inequalities in Analysis, Kluwer, 1993), to prove the following
56
related inequality: if n � 4 and A1A2 : : : An is a convex n-gon such thatAk � �=2 for all k, then
nXk=1
1
Ak
� 2
n� 2
nXk=1
1
� �Ak
(put ak = 1� Ak=�). Furthermore, equality holds if and only if the n-gonis regular, since equality holds in the earlier inequality if and only if the ak'sare equal.
One reader pointed out that the proposer had also published this prob-lem, in German, in the journal Wissenschaftliche Nachrichten in January1996, and that a solution was published on page 36 of the January 1997issue.
Readers are reminded that a problem submitted to CRUX with MAY-HEM should not be submitted for publication elsewhere, unless and untilthe problem has either been rejected by CRUX with MAYHEM or withdrawnby the proposer | Editor-in-Chief.
2203. [1997: 46] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.
Let ABCD be a quadrilateral with incircle I. Denote by P , Q, Rand S, the points of tangency of sides AB, BC, CD and DA, respectivelywith I.
Determine all possible values of \(PR;QS) such that ABCD is cyclic.
Comments. A quadrilateral that is simultaneously inscribed in one cir-cle and circumscribed about another is call bicentric. Kone �cn �y and Reimanboth report that in [2, section 39] you can �nd the solution to our problemand lots more, including its converse and the fact (Brianchon's Theorem) thatAC;BD;PR, and ST all pass through the same point (see also [1, p. 79]).Bellot Rosado reminds us that our problem was part of a problem proposedby India (but not used) at the 1989 IMO.We present two of the many possiblesolutions.
I. Solution by C. Festraets-Hamoir, Brussels, Belgium.
We denote by_
XY the angle subtended at the centre of I by the arcXY . [Note that since I is an incircle, ABCD is convex and the diagonalsPR and QS intersect in the interior.]
\A = 1
2(_
PQ +_
QR +_
RS �_
SP ), and
\C = 1
2(_
RS +_
SP +_
PQ �_
QR),
so that
\A+ \C =_
PQ +_
RS.
57
ABCD is cyclic if and only if \A + \C = �, and so, if and only if\(PR;QS) = �
2.
II. Solution by Francisco Bellot Rosado, I.B. Emilio Ferrari, Val-ladolid, Spain.
Consider inversion in I. The images A0; B0; C0; D0 of A;B;C; D arethe respective midpoints of SP; PQ;QR, andRS (as in [1, Figure 5.3A]), sothat A0B0C0D0 is a parallelogram. Since circles are preserved by inversion,A0B0C0D0 is cyclic if and only if ABCD is, in which case the parallelogramA0B0C0D0 would be a rectangle. Because each side of a midpoint quadrangleis parallel to a diagonal (PR or QS) we conclude that PR ? QS if and onlyif ABCD is cyclic.
References.
[1] H.S.M. Coxeter and S.L. Greitzer, Geometry Revisited. MAA, 1967.
[2] Heinrich D�orrie, Triumph der Mathematik, W �urzburg, 1958. Englishtranslation: 100 Great Problems of Elementary Mathematics, Dover,1965.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; JORDI DOU, Barcelona, Spain; YEO KENG HEE, Hua Chong Junior Col-lege, Singapore; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRYLEVERSHA, StPaul's School, London, England; ISTV �AN REIMAN, Budapest,Hungary; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel,the Netherlandsand the proposer.
2204. Proposed by �Sefket Arslanagi�c, University of Sarajevo, Sara-jevo, Bosnia and Herzegovina.
For triangle ABC such that R(a+ b) = cpab, prove that
r <3
10a:
Here, a, b, c, R, and r are the three sides, the circumradius and theinradius of4ABC.
Solution by Kee-Wai Lau, Hong Kong.Since c = 2R sinC,
R(a+ b) = 2R sinCpab:
Using the AM{GM inequality,
sinC =a+ b
2pab
� 1:
58
Hence, sinC = 1, C = 900, a = b and c =p2a. Thus
r =Area of4ABC
Semiperimeter of 4ABC =a2
2a+p2a
=a
2 +p2<
3
10a;
as required.
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, andMAR�IA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid,Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUELANGEL CABEZ �ON OCHOA, Logro ~no, Spain; ADRIAN CHAN, student, UpperCanada College, Toronto, Ontario; GORAN CONAR, student, GymnasiumVara�zdin, Vara�zdin, Croatia; C. FESTRAETS-HAMOIR, Brussels, Belgium;FLORIAN HERZIG, student, Perchtoldsdorf, Austria; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris StateUniversity, Big Rapids, Michigan, USA; MICHAEL LAMBROU, Universityof Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA,St Paul's School, London, England; ISTV �AN REIMAN, Budapest, Hungary;HEINZ-J�URGEN SEIFFERT, Berlin, Germany; ZUN SHAN and EDWARDT.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; PANOS E.TSAOUSSOGLOU, Athens, Greece(two solutions); and the proposer.
Most of the submitted solutions are similar to the one given above.
2205. [1997: 46] Proposed by V�aclav Kone �cn �y, Ferris State Univer-sity, Big Rapids, Michigan, USA.
Find the least positive integer n such that the expression
sinn+2A sinn+1B sinnC
has a maximum which is a rational number (where A, B, C are the angles ofa variable triangle).
Many of the solvers cited the relationship between this problem and[1984: 19] proposed by M.S. Klamkin which asks for the maximum value of
P = sin�A sin� B sin C, where A;B;C are the angles of a triangle and�; �; are given. The published solution by Walther Janous [1985: 908]gives this maximum as:
Pmax =
��(�+ � + )
(� + �)(� + )
��=2 ��(� + � + )
(� + )(� + �)
��=2 � (�+ � + )
( + �)( + �)
� =2:
Substitution and simpli�cation yields
Pmax =(n+ 1)n+1(n+ 2)3n+1
2n+1(2n+ 3)n+1(2n+ 1)n�nn=2(n+ 2)n=2 � 3(n+1)=2
(2n+ 1)1=2(2n+ 3)1=2:
The �rst term is rational and the smallest integer for which the second partis rational is n = 12. (See also problem 2183 [1996: 319; 1997: 514], 2116[1996: 75; 1997: 116].
59
Solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo,Bosnia and Herzegovina; CHRISTOPHER J. BRADLEY, Clifton College, Bris-tol, UK; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; FLORIANHERZIG, student, Perchtoldsdorf, Austria; WALTHER JANOUS, Ursulinen-gymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete,Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St Paul'sSchool, London, England; HEINZ-J�URGEN SEIFFERT, Berlin, Germany;PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.
However, some of our solvers were not content to leave it here. The�rst word goes to Walther Janous who provided the solution in [1985: 908].
Generalization, Part I, by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.
We now show that there are in�nitelymany integersn such thatPmax(n)
is rational.
Let n be even. Then n=2 is an integer and
Pmax(n) 2 Q is equivalent to
s3
(2n+ 1)(2n+ 3)2 Q:
Because gcd (2n+1;2n+3) = 1, one of the following two possibilities mustoccur:
(i) 2n+ 1 = x2 and 2n+ 3 = 3y2, or
(ii) 2n+ 1 = 3x2 and 2n+ 3 = y2,
where x; y are integers. We show that (i) leads to the claimed in�nitely manyn's. Indeed, (i) implies x2 + 2 = 3y2; that is,
x2 � 3y2 = �2; (�)
which has (x; y) = (1; 1) as one solution. The associated \pure" Pellianequation
X2 � 3Y 2 = 1
has (X;Y ) = (2;1) as its fundamental solution. Hence, we get in�nitelymany solutions (xm; ym) of (�):
xm + ymp3 = (1 +
p3)(2 +
p3)m; for m = 0; 1; 2; : : :
Simple congruence considerations, mod 4, show that all solutions (x; y) of(�) consist of odd numbers. Then
n =x2m � 1
2; m = 1; 2; 3; : : :
as claimed.
60
If m = 1, x1 = 5 and n = 12. In general, via conjugation,xm � ym
p3 = (1�
p3)(2�
p3)m. Thus,
xm =1
2((1 +
p3)(2 +
p3)m + (1�
p3)(2�
p3)m)
and
n =xm2 � 1
2=
1
4((2 +
p3)2m+1 + (2�
p3)2m+1 � 4)
(so n = 12; 180; 2520; 35112; 489060; 6811740; 94875312; etc.)
Generalization, Part II, byMichael Lambrou, University of Crete, Crete,Greece.
We show that there are no solutions with n odd.
Let n be odd and write n = 2N � 1; N � 1. Then the second part ofPmax is
(2N � 1)2N�12 (2N + 1)
2N�12 3N
(4N � 1)1=2(4N + 1)1=2=
(a rational number)p(2N � 1)(2N + 1)(4N � 1)(4N + 1)
We now show that the integer inside the square root sign is never a perfectsquare. We have
(2N � 1)(2N + 1)(4N � 1)(4N + 1) = 64N4 � 20N2 + 1
which is easily seen to be strictly between the perfect squares
(8N2�2)2 = 64N4�32N2+4 and (8N2�1)2 = 64N4�16N2+1;
so is not itself a perfect square.
2206. [1997: 46] Proposed by Heinz-J �urgen Sei�ert, Berlin, Ger-many.
Let a and b denote distinct positive real numbers.
(a) Show that if 0 < p < 1, p 6= 1
2, then
1
2
�apb1�p + a1�pbp
�< 4p(1� p)
pab+ (1� 4p(1� p))
�a+ b
2
�:
(b) Use (a) to deduce P �olya's Inequality:
a� b
log a� log b<
1
3
�2pab +
a+ b
2
�:
Note: \log" is, of course, the natural logarithm.
61
I. Solution to part (a) by Christopher J. Bradley, Clifton College, Bris-tol, UK.
The symmetry of the inequality to be proved, between a and b, andabout p = 1
2, allows us, without loss of generality, to suppose that b > a
and 0 < p < 1
2. To save writing, set k = 4p(1�p) and note that 0 < k < 1.
We are then required to prove:
1
2
�apb1�p + a1�pbp
�< k
pab+ (1� k)
�a+ b
2
�:
This is equivalent to each of the following three inequalities:
1
2
�apb1�p + a1�pbp
�<
(1� k)
2
�a+ b� 2
pab�+pab ;�
apb1�p + a1�pbp � 2pab�< (1� k)
�a+ b� 2
pab�;
apbp�b1�2p � 2a
12�pb
12�p + a1�2p
�< (1� 2p)2
�a� 2a
12 b
12 + b
�:
It is su�cient to prove the inequality obtained from taking the positive squareroot of this, namely:
b12�
p
2ap
2 � a12�
p
2 bp
2 < (1� 2p)�b12 � a
12
�:
Setting a = c2 and b = d2 (with d > c > 0), this is the same as
2p(d� c) < d� c+ c1�pdp � d1�pcp
or2p(d� c) < (dp � cp)
�d1�p � c1�p
�:
We now prove this inequality when p is rational. Let p = mn, with m
and n co-prime integers and 2m < n (since p < 1
2). Also, we set c = xn,
d = yn with y > x > 0. With these restrictions and substitutions, it is nowsu�cient to prove that
2m (yn � xn) < n (ym� xm)�yn�m + xn�m
�or
(n� 2m) (yn� xn) > nymxm�yn�2m � xn�2m
�:
On division by (y � x), which is valid since it is a positive quantity, itis su�cient to prove that
(n� 2m)�yn�1 + xyn�2 + : : :+ xn�1
�> nxmym
�yn�2m�1 + xyn�2m�2 + : : :+ xn�2m�1
�:
This is true by repeated application of the Power Means Inequality.This result is then extended to irrational values of p by the familiar continuityarguments.
62
II. Solution by Kee-Wai Lau, Hong Kong.
(a) For t > 0, let
f(t) = 1
2
�tp + y1�p
�� 4p(1� p)
pt�
�1� 4p(1� p)
��t+ 1
2
�:
Di�erentiating, we have:
f 0(t) =ptp�1 + (1� p)t�p
2� 2p(1� p)t�
12 � 1� 4p(1� p)
2; (1)
and
f 00(t) = 1
2p(1� p)t�2
�2t
12 � tp � t1�p
�: (2)
It is easy to check that f 0(1) = f 00(1) = 0. Since p 6= 1
2, we have, for t 6= 1,
2t12 � tp � t1�p < 2t
12 � 2
pt1�ptp = 0 ;
and hence that f 00(t) < 0. Thus f(t) � f(1) = 0 and f(t) = 0 if and only ift = 1. Now putting t = a
b, we easily obtain inequality (a).
(b) From part (a), we see that, for t 6= 1 and p 6= 1
2, we have f(t) < 0,
or1
2
�tp + t1�p
�< 4p(1� p)t
12 + (2p� 1)2
�t+ 1
2
�:
By integrating this inequality with respect to p from p = 1
2to p = 1, we
obtain1
2
�t� 1
log t
�<
1
3
pt+
t+ 1
12:
P �olya's Inequality follows by substituting t = a
b.
Part (a) was also solved by PAUL BRACKEN, CRM, Universit �e deMontr �eal, Montr �eal, Qu �ebec; FLORIAN HERZIG, student, Perchtoldsdorf,Austria; JOE HOWARD, New Mexico Highlands University, Las Vegas, NM,USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; MICHAELLAMBROU, University of Crete, Crete, Greece (two solutions); VEDULAN. MURTY, Visakhapatnam, India; ZUN SHAN and EDWARD T.H. WANG,Wilfrid Laurier University, Waterloo, Ontario; and the proposer.
Part(b) was also solved by PAUL BRACKEN, CRM, Universit �e deMontr �eal, Montr �eal, Qu �ebec; WALTHER JANOUS, Ursulinengymnasium,Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece(two solutions); VEDULA N. MURTY, Visakhapatnam, India; ZUN SHANand EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario;and the proposer.
All solvers, except Bradley, used Calculus in some way or other.
63
2207. [1997: 46] Proposed by Bill Sands, University of Calgary, Cal-gary, Alberta.
Let p be a prime. Find all solutions in positive integers of the equation:
2
a+
3
b=
5
p:
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.Equivalently, we have p(3a + 2b) = 5ab; hence there are three cases
to consider.
First case: p = 5. Then we get
3a+ 2b = ab or (a� 2)(b� 3) = 6 = 1 � 6 = 2 � 3:
Hence all pairs of positive integers (a; b) are (3; 9); (4; 6), (5; 5) and (8;4).
Second case: p divides a. Let a = a1p. Hence
3a1p = b(5a1 � 2):
Then p divides either b or 5a1 � 2. If b = b1p, then 3a1 = b1(5a1 � 2) or
(5a1 � 2)(5b1� 3) = 6
which has only one solution: (a1; b1) = (1;1), whence (a; b) = (p; p). Oth-erwise, 5a1 � 2 = a2p. Therefore,
3a2p+ 2
5= ba2 or 3a2p+ 6 = 5ba2:
Hence a2 divides 6.
If a2 = 1, then (a; b) =�p(p+2)
5;3(p+2)
5
�, but only if p � 3 (mod 5).
If a2 = 2, then (a; b) =�2p(p+1)
5;3(p+1)
5
�, but only if p � 4 (mod 5).
If a2 = 3, then (a; b) =�p(3p+2)
5; 3p+2
5
�, but only if p � 1 (mod 5).
If a2 = 6, then (a; b) =�2p(3p+1)
5;3p+1)
5
�, but only if p � 3 (mod 5).
Third case: p divides b. Let b = b1p. Hence
2b1p = a(5b1 � 3):
Then p divides either a or 5b1 � 3. If p divides a, then we have the samecase (p divides a and b) as already considered above. Again (p; p) is the (only)solution. Otherwise, 5b1 � 3 = b2p. Therefore,
2b2p+ 3
5= ab2 or 2b2p+ 6 = 5ab2:
64
Hence b2 divides 6.
If b2 = 1, then (a; b) =�2(p+3)
5;p(p+3)
5
�, but only if p � 2 (mod 5).
If b2 = 2, then (a; b) =�2p+3
5;p(2p+3)
5
�, but only if p � 1 (mod 5).
If b2 = 3, then (a; b) =�2(p+1)
5;3p(p+1)
5
�, but only if p � 4 (mod 5).
If b2 = 6, then (a; b) =�2p+1
5;3p(2p+1)
5
�, but only if p � 2 (mod 5).
[Note that if p = 2 or 3, this generates only two distinct solutions:(2; 2), (1;6) for p = 2 and (3;3), (12; 2) for p = 3. If p > 5, then the threesolutions are all distinct.]
Also solved by GERALD ALLEN, CHARLES DIMINNIE, TREY SMITHand ROGER ZARNOWSKI (jointly), Angelo State University, San Angelo,TX, USA; �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; MICHEL BATAILLE, Rouen, France; CHRISTOPHER J.BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGEL CABEZ �ON OCHOA,Logro ~no, Spain; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Con-necticut, USA; KEITH EKBLAW, Walla Walla, Washington, USA; RICHARDI. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursul-inengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University ofCrete, Crete, Greece; GERRY LEVERSHA, St Paul's School, London, England;HEINZ-J�URGEN SEIFFERT, Berlin, Germany; ZUN SHAN and EDWARD T.H.WANG, Wilfrid Laurier University, Waterloo, Ontario; DAVID R. STONE,Georgia Southern University, Statesboro, Georgia, USA; and the proposer.There were six incorrect and two incomplete solutions.
2153. [1996:217; 1997:313]The conjectured inequality should read jxnp(1=x)j � 2n�1.
2167. [1996:274; 1997:381]The equation in the last line should read 2(n+4)=2n+ n = 2.
Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell
Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer
Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil
Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,
J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia
65
THE ACADEMY CORNERNo. 17
Bruce Shawyer
All communications about this column should be sent to Bruce
Shawyer, Department of Mathematics and Statistics, Memorial University
of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7
This month, we present another solution to a university entrance schol-arship examination paper from the 1940's, which appeared in the April 1997issue of CRUX with MAYHEM [1997: 129].
10. Two unequal circles of radii R and r touch externally, and P and Q arethe points of contact of a common tangent to the circles, respectively.Find the volume of the frustrum of a cone generated by rotating PQabout the line joining the centres of the circles.
Solution.
OP P 0 OQ Q0
T
P
Q
S
��
�
�
R r
R� r
First, we note that triangle 4TOPOQ has a right angle at T , with
OPOQ = R+ r and OPT = R� r. Hence OQT = 2
pRr.
Because of parallel and perpendicular lines, all of angles \PSOP ,\TOQOP , \OPPP
0 and \OQQQ0 are equal { we denote the common
value by �. From triangle 4OPOQT , we note that
sin� =R� r
R+ rand cos � =
2
pRr
R+ r:
66
Using various right triangles, we obtain:
OPP0
= R sin � =R(R� r)
r + r;
PP 0 = R cos � =2RpRr
R+ r;
P 0S = PP 0 cot �
=2RpRr
R+ r
2
pRr
R� r
=4rR2
R2 � r2;
OQQ0
= r sin� =r(R� r)
r + r;
QQ0 = r cos � =2rpRr
R+ r;
Q0S = QQ0 cot �
=2rpRr
R+ r
2pRr
R� r
=4r2R
R2 � r2:
Hence
V =�
3
h(PP 0)2 P 0S � (QQ0)2Q0S
i
=�
3
�4R3r
(R+ r)2
4rR2
R2 � r2�
4Rr3
(R+ r)2
4r2R
R2 � r2
�
=16�R2r2(R3 � r3)
3(R+ r)2(R2 � r2)
=16�R2r2(R2
+Rr + r2)
3(R+ r)3;
which is the required volume.
67
11. Prove that
sin2(�+ �) + sin
2(� + �)� 2 cos(�� �) sin(�+ �) sin(� + �)
= sin2(�� �): (1)
Solution.
2�sin
2(� + �) + sin
2(� + �)
�= 2� (cos(2� + 2�) + cos(2� + 2�))
= 2� 2 cos(2�+ �+ �) cos(�� �);
�4 cos(�� �) sin(� + �) sin(� + �)
= �2 cos(�� �) (cos(�� �)� cos(2� + �+ �)) ;
so that
2 � LHS of (1)
= 2�1� cos(2�+ �+ �) cos(�� �)
� cos2(�� �) + cos(�� �) cos(2�+ �+ �)
�= 2
�1� cos
2(�� �)
�= 2sin
2(�� �):
What is the equation of this curve?
68
THE OLYMPIAD CORNERNo. 188
R.E. Woodrow
All communications about this column should be sent to Professor R.E.
Woodrow, Department of Mathematics and Statistics, University of Calgary,
Calgary, Alberta, Canada. T2N 1N4.
This number we begin with the problems of The Final Round of theJapan Mathematical Olympiad. My thanks go to Richard Nowakowski forcollecting these when he was Canadian Team Leader at the 35th IMO in HongKong.
JAPANMATHEMATICAL OLYMPIAD
Final RoundFebruary 1994 | 4 hours
1. For positive integer n, let an be the nearest positive integer topn,
and let bn = n + an. Dropping all bn (n = 1; 2; : : : ) from the set of allpositive integersN , we get a sequence of positive integers in ascending orderfcng. Represent cn by n.
2. Five points are located on the plane. Any three of those pointsare not collinear. Let l1; : : : ; l10 be the length of the ten segments obtainedby joining every two points of the �ve points. Assume that l21; : : : ; l
29 are
rational numbers. Prove that l210 is also a rational number.
3. There is a triangle A0A1A2 and seven points P0; : : : ; P6 on theplane. Assume that any Pi and Pi+1 are symmetric with center Ak, where kis the remainder of i divided by 3.
(a) Prove that P0 = P6.(b) Describe the possible position of P0 under the additional assump-
tion that every segment connecting Pi and Pi+1 does not intersect with theinterior of the triangle A0A1A2.
4. We consider a triangle ABC such that \MAC = 15�, where M is
the midpoint of BC. Determine the possible maximum value of \B.
5. There are N persons and N pieces of lot cards. Each number 1through N is written on a card. When theN persons draw these cards, theirorder is determined by the numbers on their cards. After repeating this drawtwo times, we give gifts by the following rule.
Rule: A person X gets the gift, if there is no person Y such thatY is prior to X both times. Otherwise X cannot get the gift.
69
For example, if X is at the top in the �rst lot, X always gets thegift whatever he draws in the second lot.
Then determine the expected number of persons who get gifts.
As a second Olympiad for your puzzling pleasure we give problems ofthe 30
th Spanish Mathematical Olympiad, First Round, November 26{27,1993. My thanks go to Richard Nowakowski, Canadian Team Leader to the35
th IMO in Hong Kong and to Francisco Bellot Rosado, I.B. Emilio Ferrari,Valladolid, Spain, for copies of the contest.
30th SPANISH MATHEMATICAL OLYMPIAD
First Round | November 26{27, 1993Proposed by the Royal Spanish Mathematical Society
Time allowed: 4 hours each day. Each problem carries 10 points.
First Day
1. Show that, for all n 2 N, the fractions
n� 1
n;
n
2n+ 1;
2n+ 1
2n2 + 2n
are irreducible.
2. A sphere of radius R, and a right cone with base a meridian of thesphere and vertex external to the sphere, are given. Find the radius of thecircle of intersection of the sphere and the cone, given that the volume of thecone is half of the volume of the sphere.
3. Solve the following system of equations:
x � jxj + y � jyj = 1; [x] + [y] = 1;
in which jtj and [t] represent the absolute value and the integer part of thereal number t.
4. Let AD the internal bisector of the triangleABC (D 2 BC),E thepoint symmetric to D with respect to the midpoint of BC, and F the point
of BC such that \BAF = \EAC. Show that BFFC
=c3
b3.
Second Day
5. Find all the natural numbers n such that the number
n(n+ 1)(n+ 2)(n+ 3)
has exactly three prime divisors.
70
6. An ellipse is drawn taking as major axis the biggest of the sides ofa given rectangle, such that the ellipse passes through the intersection pointof the diagonals of the rectangle.
Show that, if a point of the ellipse, external to the rectangle, is joinedto the extreme points of the opposite side, then three segments in geometricprogression are determined on the major axis.
7. Let a 2 R given. Find the real numbers x1; : : : ; xn which satisfythe system of equations
x21 + ax1 +�a�12
�2= x2
x22 + ax2 +�a�12
�2= x3
.......................... ...
x2n�1 + axn�1 +�a�12
�2= xn
x2n + axn +�a�12
�2= x1:
8. (The Sisyphus's myth) There are 1001 steps going up, with rocks onsome of them (no more than 1 rock on each step). Sisyphus may pick anyrock and raise it one or more steps up to the nearest empty step. Then hisopponent Hades rolls a rock (with an empty step directly below it) down onestep. There are 500 rocks, originally located on the �rst 500 steps. Sisyphusand Hades move rocks in turn, Sisyphus making the �rst move. His goal isto place a rock on the top step. Can Hades stop him?
We now turn to readers' solutions received before January 1st, to prob-lems for consideration by the International Jury at the 36
th IMO in Canada[1996: 299{301].
4. Let A, B and C be non-collinear points. Prove that there is aunique point X in the plane of ABC such that XA2
+ XB2+ AB2
=
XB2+XC2
+ BC2= XC2
+XA2+CA2.
Solutions by D.J. Smeenk, Zaltbommel, the Netherlands; and by
Vasiliou Meletis, Elefsis, Greece. We give Meletis' solutions.First Solution.
A C
B
A1
B1
C1
X
71
From the hypothesis we have
AX2+AB2
= CX2+ CB2: (1)
If B1 is the midpoint of BX, applying the �rst theorem of the median in thetriangles4ABX, 4CBX we get
2AB21 + 2BB2
1 = 2CB21 + 2BB2
1 or
AB1 = CB1: (2)
This indicates that the perpendicular bisector of the side AC passes throughthe point B1. Let A1, C1 be the midpoints of AX and CX, respectively.
Similarly, we obtain that the perpendicular bisectors of BC and ABpass through the midpoints A1 and C1, respectively. (3)
Furthermore we obtain ABkA1B1, ACkA1C1 and BCkB1C1. (4)From (3) and (4) we get that the circumcentre O of ABC is the ortho-
centre H1 of A1B1C1. (5)Also from (4) the triangles ABC and A1B1C1 are similar with X the
centre of similarity and ratio 1
2. (6)
So, their orthocentres H and H1 lie in the same straight line with thepoint X andHH1 = H1X. (7)
Combining (5) and (7) we getHO = OX; that is the point X is known(constant), because X is symmetric to H with respect to the orthocentre Oof ABC.
Second Solution.The conditions of the problem are equivalent to the system of equations
B
A
A1
q
M
X
CA2
E1 E2
XB2 �XC2= AC2 � AB2 (1)
XC2 �XA2= BA2 � BC2: (2)
Now, taking equation (1) gives a locus of points X satisfying the condition.The relation reminds us of the second theorem of the median in a triangle.
Let AA1, XA2 be the altitudes of the triangles ABC and XBC re-spectively on side BC (extended). Let M be the midpoint of the side BC.
If we suppose
AB � AC � BC; (3)
for illustration, we getXC � XB � XA;
72
and furthermore the point M lies between the points A1 and A2 (4)But
XB2 �XC2= 2BC �MA2; and
AC2 � AB2= 2BC �MA1:
Hence MA1 = MA2 and A2 is a constant point on BC because it is sym-metric to A1 with respect to the midpointM .
Consequently, if (1) holds, the point X lies on the line E1 perpendic-ular to BC at A2. Similarly, if (2) holds, the point X lies on the line E2perpendicular to AC at B2 (where BB1 ? AC and AB1 = CB2).
Hence, the required point X lies at the intersection of E1 and E2.5. The incircle of ABC touches BC, CA and AB at D, E and F
respectively. X is a point insideABC such that the incircle ofXBC touchesBC at D also, and touches CX and XB at Y and Z, respectively. Provethat EFZY is a cyclic quadrilateral.
Solutionsby Toshio Seimiya, Kawasaki, Japan; and by VasiliouMeletis,
Elefsis, Greece. We give Seimiya's argument.
B PD C
A
E
F
XYZ
Let P be the intersection ofEF withBC. Then byMenelaus' Theoremwe have
BP
PC�CE
EA�AF
FB= 1: (1)
Since CE = CD, EA = AF , and FB = BD, we get
BP
PC�CD
BD= 1
so that
BP
PC=BD
CD: (2)
73
Since XZ = XY , BZ = BD and CY = CD, we have from (2)
BP
PC�CY
YX�XZ
ZB=BD
CD�CD
Y X�XY
BD= 1:
Hence by Menelaus' Theorem P , Z and Y are collinear. Since PF � PE =
PD2 and PZ � PY = PD2 we have PF � PE = PZ � PY .Hence EFZY is a cyclic quadrialteral.
Comment. If AB = AC then BD = DC and then it can easily beproved that AD is the perpendicular bisector of EF and Y Z so that EFZYis an isosceles trapezoid, and is a cyclic trapezoid.
6. An acute triangle ABC is given. Points A1 andA2 are taken on theside BC (with A2 between A1 and C), B1 and B2 on the side AC (with B2
between B1 and A) and C1 and C2 on the side AB (with C2 between C1
and B) so that
\AA1A2 = \AA2A1 = \BB1B2 = \BB2B1 = \CC1C2 = \CC2C1:
The lines AA1, BB1, and CC1 bound a triangle, and the lines AA2, BB2
and CC2 bound a second triangle. Prove that all six vertices of these twotriangles lie on a single circle.
Solutions by Toshio Seimiya, Kawasaki, Japan; by D.J. Smeenk, Zalt-
bommel, the Netherlands; and by Vasiliou Meletis, Elefsis, Greece. We give
Meletis' solution.
B
A
CA1 A2
B1
C2
H I
x x
x x
Let AA1, BB1 meet at the point E; AA1, CC2 meet at the point F ;and BB1, CC1 meet at the point I. Also
\A1AA2 = \B1BB2 = \C1CC2 = 2x: (1)
The bisectors of the angles at A1, B1 and C in triangles 4A1AA2,4B1BB2 and 4C1CC2 respectively are perpendicular to their respective
74
bases. Hence they are the altitudes of4ABC. Let H be the orthocentre of4ABC.
Since \A1AH = \B1BH = x and \A1AH = \C1CH = x each oneof the quadrilaterals AHEB, AHDC is inscribable in a circle.
These two circles have a common chord, the segment AH and since\ABH = \ACH = 90
� � \BAC, then the circles have equal radii.Thus, since the inscribed angles \EAH, \DAH are equal, the corre-
sponding chords HE andHD are equal.Therefore HE = HD. Similarly, we prove that HD = HI, and so on
for all six vertices of these two triangles of the problem.Thus, all six vertices lie at the same distance from the pointH, and the
points are concyclic.
Next we give a rather novel solution by Meletis to the second problemof the 37
th IMO itself [1996: 303].
2. Let P be a point inside triangle ABC such that
\APB � \ACB = \APC � \ABC:
Let D, E be the incentres of triangles APB, APC respectively. Show thatAP , BD and CE meet at a point.
Solution by Meletis Vasiliou, Elefsis, Greece.
D0
C DB
A
P
Ey
x
3 4
1 2
Figure 1.
We will �rst prove the converse of the proposition and then apply it toprove the stated problem.
Stage I. The equality \APB�\APC = \ACB�\ABC is equivalentto
\A1 � \A2 = \C3 � \B4 (1)
(see �gure 1).Stage II. Assume the conclusion of the problem; that is. assume that
the bisectors of the angles \PCA and \PBA concur at a point I on AP .Then we have
AC
PC=AI
PI=AB
PB
orAC
AB=PC
PB: (2)
75
This ratio indicates that if AD is the bisector of the angle \BAC, then PDis the bisector of the angle \BPC, or equivalently
\the points A;P belong to the \circle of Apollonius" whosediameter DD0 lies on the lineCB, withD;D0 harmonic con-jugates to the points C;B"
(3)
Stage III. If (3) holds then (1) holds. Since AD bisects \CAB, then
\PAD = x =\A1 � \A2
2: (4)
Since PD bisects \CPB then
\PDB � \PDC = \C3 � \B4
and if we draw the bisector DE of the straight angle \BDC we get
\PDE = y =\C3 � \B4
2: (5)
Because of (3), and since DE ? CD we obtain that DE is tangent at D tothe circumcircle of the triangle4APD.
Hence x = y. (6)Combining (4), (5) and (6) we get that
\A1 � A2 = \C3 � \B4
so the converse of the proposition is true.Stage IV. If (1) holds then (3) holds.
1
6
2 7
3 45
C D B
A
P
P1
Figure 2.
Consider the circle of Apollonius with respect to the angle A of thetriangle ABC withDD0 as diameter. Let P1 be the point at which the circleintersects line CP . We want to show that P � P1 to complete the proof.
Suppose P 6� P1. We distinguish two cases:Case a. P1 is external to the segment CP (see �gure 2). Denote
\A1 = \CAP; \A2 = \BAP;
\A6 = \CAP1; \A7 = \BAP1;
76
\B4 = \CBP and \B5 = \CBP1:
Then we have
\A1 < \A6
\A2 > \A7 (7)
\B4 < \B5: (8)
So from the hypothesis and the conclusion of Stage III
\C3 � \B4 = \A1 � \A2
< \A6 � \A7
= \C3 � \B5
< \C3 � \B4:
or
\C3 � \B4 < \C3 � \B4 ;
a contradiction.
Case b. P1 is internal to the segment CD. The proof is similar. Thiscompletes the proof.
To complete this number of the corner and our �le of solutions submit-ted by the readers to problems from 1996 numbers of the Corner, we giveone solution to a problem proposed to the jury but not used at the 36th IMO.
5. [1996: 348] 36th IMO Problems proposed to the jury and not used.
Let ABC be a triangle. A circle passing throughB andC intersects thesides AB and AC again at C0 and B0, respectively. Prove that BB0, CC0
and HH0 are concurrent, where H andH0 are the orthocentres of trianglesABC and AB0C0 respectively.
Solution by Toshio Seimiya, Kawasaki, Japan.
(See diagram on the next page.) Let BH, CH meet AC, AB at D, Erespectively. Since \BDC = \BEC = 90
�, B, C, D, E are concyclic andthus \ADE = \ABC.
Now, since B, C, B0, C0 are concyclic we have
\AB0C0 = \ABC:
Thus \ADE = \AB0C0, so that DEkB0C0. Let Q be the intersection ofDC0 with EB0. As HE ? AB, B0H0 ? AB, we get HEkH0B0. Similarlywe have HDkH0C0.
77
B C
A
E
C0
D
B0
P
Q
H0
H
Since DEkC0B0, HDkH0C0 and HEkH0B0, DC0, EB0 andHH0 areconcurrent, so that H, Q, H0 are collinear.
By Pappus' Theorem H, P , Q are collinear.
Hence H, P , Q,H0 are collinear. So BB0, CC0, andHH0 are concur-rent.
That is the Corner for this issue! Send me your National and RegionalOlympiad Contest materials for use in the Corner.
78
BOOK REVIEWS
Edited by ANDY LIU
From Erd }os to Kiev: Problems of Olympiad Caliber, by Ross Hons-berger,published by the Mathematical Association of America, 1996,ISBN# 0-88385-324-8, softcover, 257+ pages, $31.00.Reviewed by Bill Sands, University of Calgary, Calgary, Alberta.
This book is number 17 in the Dolciani Mathematical Expositions se-ries of the MAA, and is the seventh in this series by Ross Honsberger. Hisprevious book in this series, More Mathematical Morsels (reviewed in Crux
on [1991: 235{236] by Andy Liu), contained mostly problems from Crux.The same is true of From Erd �os to Kiev | and this time, all problems takenfrom Crux are accompanied by references (year and page number) to wherein Crux the problem (and sometimes solution) appeared. Most of the Crux
problems used here are from Rob Woodrow's Olympiad Corner columns from1987 and 1988; thus they are mostly contest problems, some easy to �nd else-where, like the AIME and IMO contests, others quite obscure. The authorhas sometimes used solutions taken from Crux, duly crediting the originalsolvers, and other times has come up with his own, but in all cases he haswritten (or rewritten) the problems and solutions as he liked: a privilegenot usually available to Crux editors! The result is a readable and enjoyablebook, that every Crux fan will want to own.
Altogether the book has 89 problems collected into 46 unnumbered\chapters", with no discernible reason for the order they are put in. Insteadof a traditional index, at the end the problems are classi�ed under three sub-jects (roughly: combinatorics, algebra/number theory, and geometry) with aone{line description of each problem. This reviewer can't single out manyof the problems in this book for special mention | as the general editor ofCrux, I was not as close to the material in the Olympiad Corners as I was tothe regular Problems and Solutions columns. But Honsberger includes a fewitems from here too, and I recognized with pleasure one of Hidetosi Fuka-gawa's Sangaku problems on page 223 | one of several he submitted to Crux(and not the most striking one in my opinion) before publication of his bookJapanese Temple Geometry Problems with Dan Pedoe in 1989.
Of course, none of you need to be sold on the virtues of Crux! Onehopes, though, that other readers of this book would thus be drawn to sub-scribe to Crux, and maybe some of them will, but wouldn't it have beenappropriate, given the book's debt to Crux, if the words \Crux Mathemati-
corum" had been more visible? Yes, Crux is acknowledged, and praised, inthe Preface, and the book is even dedicated to Crux's late founders L �eo Sauv �eand Fred Maskell, which is a thoughtful touch and is appreciated. But whycouldn't Crux get some mention in the title or at least somewhere on the
79
cover? And the MAA's advertising for this book, as far as I have seen, doesnot mention Crux either. Come on, MAA | fair's fair!
As for other criticisms | well, we could start with the accent on the\o" of Erd }os, which is wrong (I've used the incorrect umlaut till now in thisreview!), in the title and elsewhere in the book too. (As an aside, Imight notethat the title, though catchy, isn't especially appropriate, in that the bookneither begins nor ends with a problem involving Erd }os or Kiev; however,there are Erd }os problems, and a problem from the Kiev Olympiad, in thebook. Picky, picky : : : .)
I didn't notice many misprints or weaknesses of exposition liable toslow readers down. On the bottom of page 242 and the top of page 243,there are two displayed equations in which most of the terms have beenleft out, but the reader can probably reconstruct these with little trouble.On pages 116{117, to show that cn is not a multiple root of the polynomialPn(x) = 0, it is much simpler just to di�erentiate the polynomial Qn(x),which is a factor of Pn(x); we get
(n+ 1)xn + nxn�1 + � � �+ 1
which is obviously positive for x = cn > 0. Thus cn is a single root ofQn(x)
and so also of Pn(x).
On page 107, Daniel Ropp's university should be Washington Univer-sity, not Washington State University (the correct name is given in Crux). Inoddly similar typos, the zeds in Bruce Reznick's last name and ZviMargaliot's�rst name are replaced by s's, on pages 179 and 187 respectively.
In the second half of the book there is a rash of minor errors involvingCrux references; for the bene�t of readers rather than as criticism, I'll listthem here. On page 153, the reference given [1987: 120] is to the originalpublication of the problem; the solution actually appeared on [1988: 182].On page 159, the year for this reference should be 1988, not 1987. Also,the solution for this problem appeared on [1994: 191{193] with a furthercomment on [1995: 82] (both probably too recent to be picked up by Hons-berger's searches). On page 167, the solution for this problem appeared on[1988: 199]. On page 177, again the year of publication of the original Cruxproblem should be 1988 not 1987, and the solution appeared on page 267 ofthe 1989 volume, not page 269. And by the way, these references all refer toproblem 1 of this chapter, which contains two problems. On page 239, thepage reference given as 491 should be just 49.
Having a whole chapter on Olympiad Corner solutions by GeorgeEvagelopoulos, the �rst chapter in fact, was, I think, a mistake. Some ofthese solutions were equally due to other readers, as reported in Crux at thetime. In fact, problem 1 of this chapter is listed as coming from the 1983 Aus-tralian Olympiad, which indeed it does, but the Crux reference given [1985:71] is to the same problem as proposed by Brazil (but not used) at someIMO. The correct reference for the Australian problem is [1983: 173], with
80
the o�cial solution as supplied by Peter O'Halloran published on [1986:22]. This is the same solution that Honsberger gives in his book and at-tributes to Evagelopoulos. Only on [1987: 43] is the Brazil version of theproblem wrapped up, and here no solution is published, only the solversare listed. (And, as Honsberger mentions, these include two others as wellas Evagelopoulos.) Strangely, in two other cases Honsberger fails to men-tion solvers given equal credit in Crux for solutions he uses and ascribes toEvagelopoulos alone. The solution to problem 4 was credited (on [1989:230]) also to Zun Shan and Ed Wang; what Honsberger calls a \brilliant ob-servation" could just as well be attributed to them. And for problem 7, inthe published solution in Crux [1990: 105] Duane Broline is also listed as asolver. There are also remarkable similarities between some of Evagelopou-los's solutions and earlier solutions published elsewhere. For example, forproblem 3, a problem from the Russian journal Kvant and �rst publishedin Crux in 1988, Evagelopoulos's original solution in Crux [1990: 104] con-tained the same square diagram, complete with the same cells labelled \A"and \B", as is present in the Kvant solution (see page 24 of issue 12 of the1987Kvant); no doubt the Kvant solution is a \beautiful gem", as Honsbergercalls Evagelopoulos's solution on page 5. And for problem 6, also from Kvant,Evagelopoulos's solution in Crux [1990: 102] contained the same terminol-ogy (\representative", translated from the Russian) and the same notation(\�") as in the solution published on page 25 of issue 9 of the 1987 Kvant, tomention only two of the amazing resemblances between these two solutions.
The last chapter contains an exposition with proof of the power meaninequality, and so is of a di�erent character from the rest of the book. Whilethis is a useful inequality to know, it's a bit jarring to have it suddenly appearhere, especially when there is only one small, brief, unattributed examplegiven of a problem that can be solved with it. There must be lots of prob-lems from Crux that could have been used. However, let none of the abovereservations prevent anyone from buying this book.
Did you know?
Members of the CanadianMathematical Society are entitled to receive CRUXwith MAYHEM at half price!
With membership, the cost is only $30.00 for hard copy and on-lineaccess (including handling charges and delivery by surface mail).
Contact
81
ONE PROBLEM { SIX SOLUTIONS
Georg Gunther
Sir Wilfred Grenfell College, Corner Brook, Newfoundland
The three ingredients necessary to solve a problem are insight, persistenceand technique. In the following, all three of these play a role as six di�erentsolutions to the following problem are presented.
The Problem:
On the three sides of triangleABC, construct squares facing outwards.LetA0,B0,C0 be the centres of the squares constructed on sidesBC,CA andAB, respectively. Prove that dist(A;A0) = dist(B0; C0) and AA0 ? B0C0.
Solution 1: (Analytic Geometry)
A0
B0
C0
A
B C
6
-x
y
(2b; 2c)
(2a; 0)
A00
B00C00
A000
B000
C000
It is important to make the coordinates work for you. Place the originat B, and the x{axis along BC. Let C = (2a; 0) and A = (2b;2c). Thenlabel the vertices of the three squares as shown.
It is easy to show thatB00=(2a+2c; 2a�2b),A00 = (0;�2a) andC000 =(�2c; 2b). Now we quickly obtain A0 = (a;�a), B0 = (a+ b+ c; a� b+ c)
and C0 = (b� c; b+ c).
Copyright c 1998 Canadian Mathematical Society
82
Now, we have that dist(A;A0) =
p(2b� a)2 + (2c+ a)2 = dist(B0; C0),
slopeAA0 =2c+a2b�a and slopeB0C0 = �2b�a
2c+a, showing that the lines are
perpendicular.
Solution 2: (Geometry)
A0
B0
C0
A
B CM
T
P
Q
c
a
b
Let M be the midpoint of BC. Consider triangles ABA0 and C0BM . Wehave A bBA0 = bB + 45
�= C0 bBM , AB = c, C0B =
cp2, BA0 = ap
2and
BM =a2.
So these triangles are similar, and 4C0BM is obtained from 4ABA0 byrotating through 45
� about B, and scaling by a factor of 1p2.
This means that C0M =AA0p
2=
pp2, and that A bPC0 = 45
�. Likewise, trian-gles ACA0 and B0CM are similar, and 4B0CM is obtained from 4ACA0by rotating through an angle of �45� about C and dilating by 1p
2.
Hence B0M =pp2andM bQP = 45
�.
But this now tells us that B0cMC0 = 90�, and so
C0B02
= C0M2+ B0M
2=
p2
2+p2
2= p2;
hence C0B0 = p = AA0.
Also, since 4C0MB0 is a right-angled isosceles triangle, we note that
McC0B0 = 45�.
83
In 4TC0P , we have bP = cC0 = 45� and so bT = 90
�, telling us thatC0B0 ? AA0.
Solution 3: (Trigonometry)
A0
B0
C0
A
B C
c
a
b
T
q = 45�
q
q
q q
xy
z
x0
y0
z0
Consider the hexagon AB0CA0BC0 and all six of its diagonals.
Now we use the cosine law:
B0C02
= AC02+ AB0
2 � 2AC0 � AB0 cos( bA+ 90�)
=c2
2+b2
2+ bc sin( bA):
Also
AA02
= AC2+ CA0
2 � 2AC � CA0 cos( bC + 45�)
= b2 +a2
2� ab
�cos( bC)� sin( bC)
�
=b2
2+c2
2+ ab sin( bC):
But recall that
bc sin( bA) = ac sin( bC) = 2� (area4ABC) :
Hence B0C0 = AA0.
Likewise, C0A0 = BB0 and A0B0 = CC0.
Now, consider triangles B0C0B and A0AC0. These triangles are congruentsince B0C0 = AA0, C0B = AC0 and BB0 = C0A0. Hence x = x0. Likewise,
84
y = y0 and z = z0. But then 2x+2y+2z = 180� and hence x+y+z = 90
�.It follows, looking at triangle A0B0T , that A0A ? B0C0.
Solution 4: (Vectors)
A0
B0
C0
A =�!a
B =�!b C =
�!c
�����
PPPP
PPi
������ @@@@@R
?
��������7
A00
B00C00
Here, we let the vertices of the triangle be represented by vectors�!a ,�!b and�!
c .
Let�!n be the unit vector which is perpendicular to the plane containing the
diagram.
Recall that for any two vectors�!u ,
�!v , the vector
�!u �
�!v is perpendicular
to both�!u and
�!v , and points in the direction given by the right-hand rule.
It follows that��!AC00 =
�!n �
�!BA =
�!n �
��!a �
�!b�;
��!BA00 =
�!n �
�!CB =
�!n �
��!b �
�!c�;
��!CB00 =
�!n �
�!AC =
�!n �
��!c �
�!a�:
We next compute:
��!AC0 = 1
2
h�!b �
�!a +
�!n �
��!a �
�!b�i;
��!AB0 = 1
2
h�!c �
�!a +
�!n �
��!c �
�!a�i;
and
��!BA0 = 1
2
h�!c �
�!b +
�!n �
��!b �
�!c�i:
85
Let�!u be the vector from
�!C0 to
�!B0 ; let
�!v be the vector
��!AA0.
Now:
�!u =
��!AB0 �
��!AC0 =
1
2
h�!c �
�!b +
�!n �
��!c +
�!b � 2
�!a�i;
and
�!v =
1
2
h�!c �
�!b +
�!n �
��!b �
�!c�i���!a �
�!b�
=1
2
h�!c +
�!b � 2
�!a +
�!n �
��!b �
�!c�i:
But now:
�!n �
�!v =
1
2
h�!n �
��!c +
�!b � 2
�!a�+
��!c �
�!b�i
=�!u :
And since����!n ��� = 1, we conclude that
����!v ��� = ����!u ��� and that�!v ?
�!u .
Solution 5: (Complex Numbers)
A0
B0
C0
A
B C
A00
B00
C000
Place the diagram in the complex plane, with the origin at B. Let A and Crepresent the complex numbers at vertices A and C.
Recall that for any complex number z, the product z0 = zei� is the complexnumber which is obtained by rotating z about the origin through the angle �.
If instead, we wish to rotate w about z through an angle �, thenw0 = z + (w� z)ei�.
86
w
w0
z�
Using this, we obtain the following:
B00 = C + (A� C)(�i) = C(1 + i)� Ai;
C000 = Ai;
A00 = �Ci:
Also, A0 = 1
2C(1� i), B0 = 1
2A(1� i) +C(1 + i) and C0 = 1
2A(1 + i).
Let�!u be the vector from C0 to B0; let
�!v be the vector from A0 to A.
Now�!u =
1
2C(1 + i)�Ai and
�!v = A� 1
2C(1� i).
But now note that�!u i = A� 1
2C(1� i) =
�!v .
So A0A is obtained by rotating C0B0 through 90�, which proves both parts
of the result that we seek.
Solution 6: (Transformations)
A0
B0
C0
A
B C
B000
q = 45�
q
q
Rotate triangle ACA0 about C through �45� and dilate byp2 to obtain
triangle B000CB.
This tells us that B000B is inclined at 45� with respect to AA0.
87
Now rotate triangleAB000B aboutA through�45� and dilate by 1p2to obtain
triangle AB0C0.
This tells us that B0C0 =1p2B000B, and that B0C0 is inclined at 45� with
respect to B000B.
But this proves our result.
So, there it is | one problem, six di�erent approaches. Which solutionis the best? The late Professor Paul Erd }os used to talk about \God's LittleBlack Book", in which could be found the perfect solution to every problem.Of the solutions presented here, perhaps the last one is the closest candidatefor inclusion in this \Little Black Book": it is economical, to the point andcarries with it a wonderful element of surprise. Those are the properties ofbeautiful mathematics.
Have you heard about ATOM?
ATOM is \A Taste Of Mathematics" (Aime{T{On les Math �ematiques).
The ATOM series
The booklets in the series, A Taste of Mathematics, are published bythe Canadian Mathematical Society (CMS). They are designed as enrichmentmaterials for high school students with an interest in and aptitude for math-ematics. Some booklets in the series will also cover the materials useful formathematical competitions at national and international levels.
La collection ATOM
Publi �es par la Soci �et �e math �ematique du Canada (SMC), les livrets de lacollection Aime-t-on lesmath �ematiques (ATOM) sont destin �es au perfection-nement des �etudiants du cycle secondaire qui manifestent un int �eret et desaptitudes pour les math �ematiques. Certains livrets de la collection ATOMservent �egalement de mat �eriel de pr �eparation aux concours de math �emati-ques sur l' �echiquier national et international.
This volume contains the problems and solutions from the 1995{1996Mathematical Olympiads' Correspondence Program. This program has sev-eral purposes. It provides students with practice at solving and writing upsolutions to Olympiad-level problems, it helps to prepare student for theCanadian Mathematical Olympiad and it is a partial criterion for the selec-tion of the Canadian IMO team.
For more information, contact the Canadian Mathematical Society.
88
THE SKOLIAD CORNERNo. 28
R.E. Woodrow
Last number we gave the Preliminary Round of the British ColumbiaColleges Junior High School Mathematics Contest for 1997. To keep in the ow of the contest we give Part A and Part B of the Final Round. Studentswhose performance in class on problems of the Preliminary Round was ex-emplary were invited to nearby colleges to attempt the Final Round as partof a larger day-long mathematical event. Again my thanks go to John GrantMcLoughlin, now of the Faculty of Education, Memorial University of New-foundland, who participated in formulating the exams while he was at Okana-gan College.
BRITISH COLUMBIA COLLEGES
Junior High School Mathematics Contest
Final Round 1997 | Part A
1. The buttons of a phone are arranged as shown below. If the buttonsare one centimeter apart, centre-to-centre, when you dial the number 592-7018 the distance, in centimetres, traveled by your �nger is:
(a)p5(3 +
p2) + 2
p2 (b) 2
p5 +
p2(3 +
p5)
(c) 2p5(1 +
p2) + 2
p2 (d)
p5 +
p2(2 + 3
p5)
(e)p5(1 +
p2) + 4
p10
0
8
5
2
7
4
1
9
6
3
2. What is the total number of ones digits needed in order to write theintegers from 1 to 100?
(a) 11 (b) 18 (c) 20 (d) 21 (e) 100
3. The number of solutions (x; y; z) in positive integers for the equa-tion 3x+ y+ z = 23 is:
(a) 86 (b) 50 (c) 60 (d) 70 (e) 92
89
4. In the diagram below the upper scale AB has ten 1 centimetre divi-sions. The lower scale CD also has ten divisions, but it is only 9 centimetreslong. If the right hand end of the fourth division of scale CD coincides ex-actly with the right hand end of the seventh division of scale AB, what is thedistance, in centimetres, from A to C?
A B
C D
(a) 3:2 (b) 3:25 (c) 3:3 (d) 3:35 (e) 3:4
5. Triangle ABC is equilateral with sides tangent to the circle withcenter at O and radius
p3. The area of the quadrilateral AOCB, in square
units is:
O
C
A
B
(a) 3p3 (b) 3 (c) 6
p3 (d) 3� (e) 2
p3
6. Times such as 1:01, 1:11,: : : are called palindromic times becausetheir digits read the same forwards and backwards. The number of palin-dromic times on a digital clock between 1:00 a.m. and 11:59 a.m. is:
(a) 47 (b) 48 (c) 55 (d) 56 (e) 66
7. Ted's television has channels 2 through 42. If Ted starts on channel15 and surfs, pushing the channel up button 518 times, when he stops he willbe on channel:
(a) 11 (b) 13 (c) 15 (d) 38 (e) 41
8. Consider a three-digit number with the following properties:
1. If its tens and ones digits are switched, its value would increaseby 36.
2. Instead, if its hundreds and ones digits are switched its valuewould decrease by 198.
Suppose that only the hundreds and tens digits are switched. Its valuewould:
(a) increase by 6 (b) decrease by 540 (c) increase by 540
(d) decrease by 6 (e) increase by 90
90
9. Speedy Sammy Seamstress sews seventy-seven stitches in sixty-sixseconds. The time, in seconds, it takes Sammy to stitch �fty-�ve stiches is:
(a) 431
3(b) 441
4(c) 451
5(d) 461
6(e) 471
7
10. How many positive integers less than or equal to 60 are divisibleby 3, 4, or 5?
(a) 25 (b) 35 (c) 36 (d) 37 (e) 44
Final Round 1997 | Part B
1. (a) The pages of a thick telephone directory are numbered from 1
to N . A total of 522 digits are required to print the pages. FindN .(b) There are 26 pages in the local newspaper. Suppose that you pull a
sheet out and drop it on the oor. One of the pages facing you is numbered19. What are the other page numbers on the sheet?
2. Using each of the digits 1, 9, 9, and 7 create expressions for thenumbers 1; 2; 3; : : : ; 10. Note that the digits must appear separately; thatis, numbers like 17 are not allowed. Only the basic operations +, �, �, �and brackets (if necessary) may be used. Other mathematical symbols suchas p are not allowed. Every expression must include one 1, two 9's, andone 7, in any order.
3. (a) Decide which is greater:p6 +
p8 or
p5 +
p9.
(b) Show that x2+1
x� 2 for any real number x > 0.
4. In the plane �gure shown on the right, ABCD is a square withAB = 12. If A0, B0, C0, and D0 are the mid-points of AO, BO, CO, andDO, respectively, then:
D C
BA
A0
C0D0
B0
O
(a) Find the area of the square A0B0C0D0.(b) Find the area of the shaded region.(c) Find the area of the trapezoid AA0B0B.
91
5. The �gure below shows the �rst three in a sequence of square arraysof dots. The number of dots in the three arrays is 1, 5, and 13.
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
(a) Find the number of dots in the next array in the sequence.(b) Find the number of dots in the sixth array in the sequence.(c) Find an expression for an, the number of dots in the nth array in the
sequence, in terms of n alone.(d) Find a relation between an+1, an, and n.
Last issuewe gave the problems of the Preliminary Round of the BritishColumbia Colleges Junior High School Mathematics Contest 1997. Here arethe answers:
1. E 2. A 3. D 4. A 5. C
6. A 7. D 8. E 9. B 10. D
11. C 12. B 13. D 14. C 15. E
That completes this number of the Corner. Send me your comments,suggestions, and suitable materials for use in the Skoliad Corner.
Did you know?
Members of the CanadianMathematical Society are entitled to receive CRUXwith MAYHEM at half price!
With membership, the cost is only $30.00 for hard copy and on-lineaccess (including handling charges and delivery by surface mail).
Contact
92
MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by
High School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.
All material intended for inclusion in this section should be sent to the
Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,
PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronicaddress is still
The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).The rest of the sta� consists of Adrian Chan (Upper Canada College), JimmyChui (Earl Haig Secondary School), Richard Hoshino (University ofWaterloo),David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).
Shreds and Slices
The Fibonacci Triangle
We present an intriguing con�guration of numbers, which might be
called the Fibonacci triangle. Recall that the golden ratio is � =1+p5
2, and
that bxc is the greatest integer less than or equal to x. Starting with a 2, forany entry n, write b�nc and b� 2nc under it. The �rst few rows we obtainare as follows:
9 15 16 26 17 28 29 47 19 31 32 52 33 54 55 89
6 10 11 18 12 20 21 34
4 7 8 13
3 5
2
A general portion of the triangle is the following:
n
b�nc b� 2ncb�b�ncc b� 2b�ncc b�b� 2ncc b� 2b� 2ncc
93
Some patterns are immediately apparent in the Fibonacci triangle; wepresent some now.
Proposition 1. n+ b�nc = b� 2nc.Proof. b� 2nc = b(� + 1)nc = b�n+ nc = b�nc+ n.
Proposition 2. b�nc+ b� 2nc = b�b� 2ncc.Proof. We �rst make a general observation: bxc = a if and only if a
is an integer and a � x < a + 1. Most problems of this form will use thisfact. Let a = b�nc, so a � �n < a + 1. Note b�nc + b� 2nc = 2a + n,and b�b� 2ncc = b� (a + n)c = b�a + �nc. Therefore, we must prove2a+ n � �a+ �n < 2a+ n+ 1. For the left inequality, we have
2a+ n � �a+ �n
if and only if(2� � )a � (� � 1)n
if and only ifa � ��1
2�� n = �n;
which is true; and for right inequality,
�a+ �n < 2a+ n+ 1
if and only if(2� � )a > (� � 1)n� 1
if and only ifa > ��1
2�� n�1
2�� = �n� � � 1;
which is also true, and the statement is proved.
These results, for one, verify that the last two entries of each row doindeed form the Fibonacci sequence. We �nally show the one conspicuousresult in the triangle.
Proposition 3. Each positive integer greater than 1 appears exactly oncein the table.
Proof. If a and b are distinct positive integers greater than 1, then soare b�ac and b�bc. This is because � > 1, so �a and �b must di�er by atleast 1, and hence round down to di�erent integers. The same argumentworks for � 2. Hence, the same integer cannot appear as a left leg or a rightleg.
Now we invoke a classic result:
Beatty's Theorem. If� and � are positive, irrational numbers such that
1
�+
1
�= 1;
then the sequences fb�c; b2�c; b3�c; : : : g, fb�c; b2�c; b3�c; : : : g containeach positive integer exactly once.
94
We can take � = � and � = � 2. By Beatty's theorem, no integer canappear both in the form b�ac and b� 2bc. Hence, no integer appears morethan once. But Beatty's also tells us that each integer greater than 1 mustappear at least once. Hence, each integer greater than 1 appears exactlyonce.
Note that in this proof, we used only the fact that the sum of the recip-rocals of � and � 2 is 1; they can be replaced with any numbers satisfyingthe conditions of Beatty's Theorem (the fact that each is greater than 1 alsofollows from this relationship).
Problems
1. (a) Prove that b� 2nc � b�b�ncc = 1.
(b) Prove that b�b� 2ncc � b� 2b�ncc = 1.
What do these say about the entries in the triangle?
2. An F-sequence is a sequence in which, except for the �rst and secondterms, every term is the sum of the two previous terms.
(a) Show that the positive integers cannot be partitioned into a �nitenumber of F-sequences.
(b) Show that the positive integers can be partitioned into an in�nitenumber of F-sequences.
Mayhem Problems
The Mayhem Problems editors are:
Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.
Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only solutions | thenext issue will feature only problems.
We warmly welcome proposals for problems and solutions. With thenew schedule of eight issues per year, we request that solutions from theprevious issue be submitted by 1 April 1998, for publication in the issue 5months ahead; that is, issue 6 of 1998. We also request that only students
submit solutions (see editorial [1997: 30]), but we will consider particularlyelegant or insightful solutions from others.
95
High School Solutions
Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,Canada. L3S 3E8 <[email protected]>
H217. Let a1, a2, a3, a4, a5 be a �ve-term geometric sequence satis-fying the inequality 0 < a1 < a2 < a3 < a4 < a5 < 100, where each termis an integer. How many of these �ve-term geometric sequences are there?(For example, the sequence 3, 6, 12, 24, 48 is a sequence of this type).
Solutionby Joel Schlosberg, student, Hunter College High School, New
York NY, USA.
Let nm
be the common ratio of the geometric sequence, where n and
m are relatively prime integers, with n > m. Now a5 = a1 � n4
m4 , so leta1 = km4, where k is a positive integer. Thus, our geometric series becomeskm4, km3n, km2n2, kmn3, kn4, and kn4 < 100. If n > 4, then kn4 �n4 > 256 > 100. So n � 3. Hence, there are three cases to consider:
If n = 3 and m = 2, then 81k < 100, so k = 1. The only solution is(16; 24; 36; 54; 81).
If n = 3 and m = 1, then 81k < 100, so k = 1. The only solution is(1; 3; 9; 27; 81).
If n = 2 and m = 1, then 16k < 100, so k = 1, 2, : : : , 6. Thereare six solutions, namely (1; 2; 4; 8; 16), (2; 4; 8; 16; 32), (3; 6; 12; 24; 48),(4; 8; 16; 32; 64), (5;10; 20; 40; 80) and (6;12; 24; 48; 96). In total, there areeight sequences.
There was one incorrect submission received, where the solver incor-
rectly assumed that the common ratio had to be an integer. Hence, the
solution (16; 24;36; 54; 81) was missed.
H218. A Star Trek logo is inscribed inside a circle with centre O andradius 1, as shown. Points A, B, and C are selected on the circle so thatAB = AC and arc BC is minor (that is, ABOC is not a convex quadrilat-eral). The area of �gureABOC is equal to sinm�, where 0 < m < 90 andmis an integer. Furthermore, the length of arc AB (shaded as shown) is equalto a�=b, where a and b are relatively prime integers. Let p = a+ b+m.
A
O
���������������SSSSSBBBBBBBBBB
CB
��������������������������������������������������������������
96
(i) If p = 360, andm is composite, determine all possible values for m.
(ii) Ifm and p are both prime, determine the value of p.
Solution.
(i) Let \BOC = 2x�, where 0 < x < 90. Otherwise, \BOC wouldexceed 180
� and ABOC would be a convex quadrilateral (and not the �gureof a Star Trek logo). Draw OA. Since AB = AC, OAB and OAC arecongruent triangles, and so \AOB = \AOC. Since \AOB + \AOC =
(360 � 2x)�, we have \AOB = (180 � x)�. Thus, the area of ABOC istwice that of triangle OAB, so sinm�
= OA �OB � sin(180� x)� = sinx�.Since 0 < m;x < 90, sinm�
= sinx� implies thatm = x. Hence \AOB =
(180�m)� and the length of arc AB is
2�(180�m)
360=�(180�m)
180:
Thus, 180�m180
=ab, where a and b are relatively prime integers. Hence,
a � 180�m and b � 180 with equality occurring i� 180�m180
is irreducible.In this case, a + b+m = 180 �m+ 180 +m = 360, and so if p = 360,we must have gcd(m;180) = 1. But m is composite and cannot have anycommon divisors with 180 = 2
2�32�5. Sommust be a product of primes
greater than 5. But m < 90, so we only have two possible choices for m:7 � 7 and 7 � 11. Hence, if p = 360 andm is composite, then m = 49 or 77.
(ii) Ifm is a prime larger than 5, wewill have p = 360, since gcd(m;180)
= 1. Since 360 is not prime, we must restrict our choices of m to 2, 3 and5. Ifm = 3 or m = 5, we will �nd that p is composite. However, ifm = 2,we have a
b=
178
180=
89
90, so p = a + b+m = 89 + 90 + 2 = 181, which is
prime. Hence, m = 2 and p = 181.
H219. Consider the in�nite sum
S =a0
100+
a1
102+
a2
104+
a3
106+ � � � ;
where the sequence fang is de�ned by a0 = a1 = 1, and the recurrence
relation an = 20an�1 + 12an�2 for all positive integers n � 2. IfpS can
be expressed in the form apb, where a and b are relatively prime positive
integers, determine the ordered pair (a; b).
Solution.
We have
97
S �20S
102�
12S
104=
�a0
100+
a1
102+
a2
104+
a3
106+ � � �
�
��20a0
102+
20a1
104+
20a2
106+
20a3
108+ � � �
�
��12a0
104+
12a1
106+
12a2
108+
12a3
1010+ � � �
�
=a0
100+
a1
102�
20a0
102+a2 � 20a1 � 12a0
104
+a3 � 20a2 � 12a1
106+a4 � 20a3 � 12a2
108+ � � �
Since an � 20an�1 � 12an�2 = 0 for all positive integers n � 2, we have
S �20S
102�
12S
104=
a0
100+
a1
102�
20a0
102;
and substituting in a0 = a1 = 1, we have 7988S10000
=81
100, so S =
2025
1997. Hence,p
S =45p1997
, and so the desired ordered pair is (a; b) = (45; 1997).
H220. Let S be the sum of the elements of the set f1; 2; 3; : : : ;(2p)n � 1g. Let T be the sum of the elements of this set whose repre-sentation in base 2p consists only of digits from 0 to p � 1. Prove that2n � T
S= (p� 1)=(2p� 1).
Solution.
We have S =[(2p)n�1](2p)n
2. Let R denote the set of numbers that have
at most n digits in base 2p and contain only the digits from 0 to p� 1. ThusT is the sum of the elements of R. For example, when p = 2 and n = 3, wehave R = f0; 1; 10; 11; 100; 101; 110;111g and T = 11104 = 84. Note Rwill have pn elements, because each of the n digits (including leading zeros)can be any one of p di�erent numbers. Since each number in f0; 1; : : : ; p�1gappears the same number of times as a digit in R, there will be exactly pn�1
elements in R that have k as their tth digit, for k = 0; 1; : : : ; p � 1 andt = 1; 2; : : : ; n. Thus,
T =
n�1Xi=0
(2p)i � [pn�1 � 0 + pn�1 � 1 + � � �+ pn�1 � (p� 1)]
=
n�1Xi=0
(2p)i � pn�1 �p(p� 1)
2=
pn(p� 1)
2�n�1Xi=0
(2p)i
=pn(p� 1)
2�(2p)n� 1
2p� 1:
98
Hence,
2n �
T
S=
2n � p
n(p�1)2
� (2p)n�1
2p�1[(2p)n�1](2p)n
2
=p� 1
2p� 1;
as required.
This can also be solved by induction { that is an exercise left for the
reader.
Advanced Solutions
Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.M1G 1C3 <[email protected]>
A193. If f(x; y) is a convex function in x for each �xed y, and a convexfunction in y for each �xed x, is f(x; y) necessarily a convex function in xand y?
Solution.
No; for example, take
f(x; y) =
�exy � 1 if x; y � 0;
0 otherwise,
Then f satis�es the given conditions. However, f(0;2) = f(2;0) = 0, andf(1;1) = e, so f is not convex.
A194. Let H be the orthocentre (point where the altitudes meet) of atriangle ABC. Show that if AH : BH : CH = BC : CA : AB then thetriangle is equilateral.
Solution by Deepee Khosla, Ottawa, ON.
Let a = BC, b = AC, c = AB, as usual, and set A0 = AH \ BC,B0 = BH \AC, C0 = CH \AB. Since \AA0B = \BB0A, it follows thatquadrilateral ABA0B0 is cyclic (with diameter AB). Thus AH � HA0 =
BH �HB0. Together with the givens, we have
HB0
HA0=AH
BH=a
b;
so that
a� BH = b�AH; (1)
a�HA0 = b�HB0: (2)
99
Also, 2K = a�AA0 = a(AH+HA0), whereK is the area of triangleABC. This implies a�HA0 = 2K � a� AH, and similarly for b�HB0.Putting these into equation (2) gives
a� AH = b�BH: (3)
Multiplying (1) and (3) and cancelling yields a2 = b2 or a = b since a; b > 0.A similar argument shows that a = c, so triangle ABC is equilateral.
A195. Compute tan 20�tan 40
�tan 60
�tan 80
�.
Solution I by D.J. Smeenk, Zaltbommel, the Netherlands.Let A = tan20
�tan40
�tan60
�tan80
�. Then A =BC, where B =
8 sin20�sin40
�sin60
�sin 80
� and C = 8 cos 20�cos 40
�cos 60
�cos 80
�.Then
B = 8sin20�sin40
�sin 60
�sin80
�
= 2(cos 60� � cos 100
�)(cos 20
� � cos 100�)
= (1 + 2 sin10�)(cos 20
�+ sin10
�)
= cos 20�+ sin10
�+ 2 sin10
�cos 20
�+ 2sin
210�
= cos20�+ sin10
�+ sin30
� � sin 10�+ (1� cos 20
�) =
3
2;
C = 8cos 20�cos 40
�cos 60
�cos 80
�
= 2(cos 60�+ cos 100
�)(cos 20
�+ cos 100
�)
= (1� 2 sin 10�)(cos 20
� � sin10�)
= cos 20� � sin10
� � 2 sin 10�cos 20
�+ 2sin
210�
= cos20� � sin10
� � sin 30�+ sin10
�+ (1� cos 20
�) =
1
2:
Therefore, A =BC
= 3.
Solution II.By a well-known identity,
tan9� =
�91
�tan � �
�93
�tan
3 � +�95
�tan
5 � ��97
�tan
7 � +�99
�tan
9 ��9
0
���9
2
�tan
2 � +�9
4
�tan
4 � ��9
6
�tan
6 � +�9
8
�tan
8 �:
The LHS is zero for the nine values � = 0�, 20�, 40�, : : : , 160�, so the roots
of �9
1
�x�
�9
3
�x+
�9
5
�x5 �
�9
7
�x7 +
�9
9
�x9 = 0
are the nine distinct values tan0�
= 0, tan20�, tan 40�, : : : , tan160�.
Note that tan 100�= � tan80
�, : : : , tan 160�= � tan20
�. By taking outa factor of x, we see the product of the roots is
tan220�tan
240�tan
260�tan
280�=
�9
1
�= 9;
100
so that tan20� tan40� tan60� tan 80�= 3.
Also solved by Deepee Khosla, Ottawa, ON, and Bob Prielipp, Univer-
sity of Wisconsin{Oshkosh, Wisconsin, USA.
A196. Show that r2 + r2a + r2b + r2c � 4K, where r, ra, rb, rc, andKare respectively the inradius, exradii and area of a triangle ABC.
Solution by Deepee Khosla, Ottawa, ON, and Bob Prielipp, University
of Wisconsin{Oshkosh, Wisconsin, USA.We have
r =K
s; ra =
K
s� a; rb =
K
s� b; and rc =
K
s� c:
Thus, by the AM-GM inequality,
r2 + r2a + r2b + r2c =K2
s2+
K2
(s� a)2+
K2
(s� b)2+
K2
(s� c)2
� 44
sK8
s2(s� a)2(s� b)2(s� c)2= 4
K2
K= 4K:
Challenge Board Solutions
Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <[email protected]>
C73. Proposed by Matt Szczesny, University of Toronto.The sequence fang consists of positive reals such that the sum of the an
diverges. Show that the sum of an=sn diverges, where sn is the nth partialsum.
Solution by the proposer.Since the an are positive, the sequence fsng is increasing. Putting
s0 = 0, for any non-negative integer N and positive integer k, we thereforehave
N+kXn=N+1
an
sn�
N+kXn=N+1
an
sN+k
=1
sN+k
(sN+k � sN) = 1�sN
sN+k
:
For every non-negative integer N there exists, by the divergence of fsng,some k such that sN
sN+k< 1
2. It follows that for each such N we can obtain
a k so thataN+1
sN+1
+ � � � +aN+k
sN+k
�1
2; and hence the sum
a1
s1+a2
s2+ � � �
must diverge.
101
The Order of a Zero
Naoki Satograduate student, Yale University
Recall that a root r of a polynomial p(x) has multiplicitym if (x� r)m
divides p(x) but (x�r)m+1 does not. In other words, p(x) = (x�r)mq(x)for some polynomial q(x), and q(r) 6= 0. This means that p(x) behavesmuch like q(r)(x � r)m around x = r, for example in graphing p(x). Infact, in many applications, the values q(r) andm are all the data we have toknow. We extend these ideas analytically to calculus, where they can play auseful and powerful role. Keep in mind that the notion of multiplicity of aroot will be our main motivation.
First, we give the analogue of multiplicity in calculus.
De�nition. We say f(x) has order r at x = a if
limx!a
f(x)
(x� a)r= c
for some non-zero constant c. Let us call c the r-value of f(x) at x = a.
First, it should be clear that the values of r and c are unique. To seethis, assume that f(x) has order r1 and r2 at a, so
limx!a
f(x)
(x� a)r1= c1 ; and lim
x!a
f(x)
(x� a)r2= c2:
Then
limx!a
(x� a)r1�r2 =c2
c1:
Since c1 and c2 are non-zero, we must have r1 = r2 (and of course, c1 = c2).Thus, we can now speak of the order of f(x) at a. However, the order
at a need not exist (see Example 4 below).
From a numerical viewpoint, f(x) behaves like c(x � a)r for x closeto a, so the order gives a measure of how quickly f(x) approaches (or recedesfrom) 0 as x approaches a. The greater r is, the faster f(x) decreases. Notethat if r is negative, then r must be an integer (why?), and f(x) goes to �1as x approaches a.
Examples.1. If a root r of polynomial p(x) has multiplicity m, and p(x) =
(x � r)mq(x), then p(x) has order m and r-value q(r) at r. In particu-lar, the order is a non-negative integer which is at most the degree of p(x).For example, the polynomial (x� 1)
2x(x+ 3) has order 2 and r-value 4 atx = 1.
Copyright c 1998 Canadian Mathematical Society
102
2. If f is continuous at a and f(a) 6= 0, then f(x) has order 0 andr-value f(a) at a. It neither approaches nor recedes from 0 at x = a.
3. It is well known that limx!0
sinx
x= 1: Hence, sinx has order 1 and
r-value 1 at x = 0.
4. As mentioned, the order need not exist. For example, the somewhattrivial f(x) = 0 has no order at any point, and f(x) = jxj has no orderat x = 0. Also, f(x) = xjxj is di�erentiable at x = 0, but has no orderat x = 0. This shows that even regular behaviour such as di�erentiabilitycannot guarantee the existence of order!
We do have the following su�cient conditions for continuity and dif-ferentiability.
Proposition. If f(x)�f(a)has order greater than 0 at a, then f is con-tinuous at a. If f(x)�f(a) has order at least 1 at a, then f is di�erentiableat a.
Proof. Let r be the order of f(x)�f(a)at a. Assume that r > 0. Recallthat f is continuous at a if and only if f(a) = limx!a f(x), or equivalentlylimx!a(f(x)� f(a)) = 0. But we see
limx!a
(f(x)� f(a)) = limx!a
f(x)� f(a)
(x� a)r� limx!a
(x� a)r = 0:
In this product, the �rst limit exists by de�nition of order, and the secondlimit exists and is 0 since r > 0. Hence, f is continuous at a.
Similarly, recall that f is di�erentiable at a if and only if
limx!a
f(x)� f(a)
x� a
exists, and the derivative f 0(a) is the value of this limit.
Assume that r � 1. Then
limx!a
f(x)� f(a)
x� a= lim
x!a
f(x)� f(a)
(x� a)r� limx!a
(x� a)r�1;
and in this product, both limits exist, so f is di�erentiable at a.
Remarks. Example 4 shows that the converse does not hold. Also, thefunction f(x), which is 0 at x = 0 and 1 everywhere else, has order 1 atx = 0 but is not continuous at x = 0; hence, we do require that r > 0 forcontinuity.
So order can give some useful information, but this seems academic,since continuity and di�erentiability are easy to determine anyway. We nowpresent a result which uses orders and r-values to compute limits.
Proposition. Let f1(x), f2(x) have orders r1, r2 and r-value c1, c2 at arespectively. Then f1(x)f2(x) has order r1 + r2 and r-value c1c2 at a, and
103
f1(x)=f2(x) has order r1 � r2 and r-value c1=c2 at a. Also,
limx!a
f1(x)
f2(x)=
�0 if r1 > r2
c1=c2 if r1 = r2:
If r1 < r2, then
limx!a+
f1(x)
f2(x)= sgn(c1=c2) � 1;
and if r1 � r2 is an integer, then
limx!a�
f1(x)
f2(x)= sgn(c1=c2)(�1)r2�r1 � 1:
Else, if r2 � r1 is not an integer, then this last limit does not exist.
A proof follows almost directly from the de�nition, and we will not in-clude one here. Note that the evaluation of the limit only requires knowledgeof r1, r2, c1, and c2, justifying the claim that these are the only values oneusually needs to know. Actually �nding these values may vary from trivialto di�cult, which we will address in a moment.
The philosophy behind this concept is, as pointed out before, that iff(x) has order r and r-value c at a, then f(x) behaves like c(x� a)r at a.So, in evaluating the limit in the former proposition, the method that shouldwork, and the one we should always pursue is to substitute c(x � a)r forf(x) (since they behave similarly at x = a), cancel all factors of x � a, andsee what is left, because what is left is what precisely determines the valueof the limit.
For example, in taking the limit of a rational function as x approaches a,what we do is precisely as described above; that is, factor and cancel powersof x� a. Other limits are not so straightforward. For example, a recent �nalexam in a �rst year calculus course at the University of Toronto asked thefollowing: Evaluate
limx!0
sinx� arctanx
x2 ln(1 + x):
Analogously, what we would like to do is \factor" out the powers of x, ormore precisely, �nd the orders and r-values of the numerator and denom-inator at 0. For other similar limits, we must also calculate the orders andr-values for fairly general functions. As of now, there does not seem to be aclear way of doing this. We give a systematic method for a reasonable class offunctions, and we return to the polynomial case for motivation of a di�erentdescription of order.
Lemma. Let p(x) = anxn+ an�1xn�1 + � � �+ a1x + a0. Then there
exist unique constants cn, cn�1, : : : , c1, c0 such that p(x) = cn(x� a)n +
cn�1(x� a)n�1 + � � �+ c1(x� a) + c0.
Proof. The ci are determined by the following translation:
p(x+ a) = cnxn+ cn�1xn�1 + � � �+ c1x+ c0
() p(x) = cn(x� a)n + cn�1(x� a)n�1 + c1(x� a) + c0:
104
In particular, we see that by di�erentiating both sides k times,ck = p(k)(a)=k!.
Corollary. (x � a)r divides p(x) if and only if p(k)(a) = 0 fork = 0; 1; : : : ; r � 1.
Proof. Both conditions are equivalent to c0 = c1 = � � � = cr�1 = 0.
Proposition. Let p(x) be a non-zero polynomial. Then the order ofp(x) at a is the value of r such that p(k)(a) = 0 for k = 0; 1; : : : ; r� 1, andp(r)(a) 6= 0, and the r-value at a is p(r)(a)=r!.
Proof. Left as an exercise for the reader.
We now state the generalization we seek.
Theorem. Let f(x) be a function which has a Taylor expansion aroundx = a, say
f(x) = c0 + c1(x� a) + c2(x� a)2 + � � � ;
such that not all the coe�cients ci are zero, or in other words, f(x) is notidentical to zero around x = a. Then the order of f(x) at a exists, and is theunique non-negative integer r such that f (k)(a) = 0 for k = 0; 1; : : : ; r�1,and f (r)(a) 6= 0, and the r-value at a is f (r)(a)=r!.
Proof. For all k � 0, f (k)(a) = ck. Let cr be the �rst coe�cient whichis non-zero. Then r is the order of f(x) at a, as described, and
limx!a
f(x)
(x� a)r= cr = f (r)(a)=r! 6= 0:
Let us now apply this to the example above. Let f1(x) = sinx � arctanx,f2(x) = ln(1 + x). Then
f 01(x) = cosx�
1
x2 + 1; f 0
1(0) = 0;
f 001 (x) = � sinx+2x
(x2 + 1)2; f 001 (0) = 0;
f 0001 (x) = � cosx+2(x2 + 1)
2 � 8x2(x2 + 1)
(x2 + 1)4; f 0001 (0) = 1:
Hence, f1(x) has order 3 and r-value 1=6 at 0, and
f 02(x) =1
1 + x; f 02(0) = 1;
so f2(x) has order 1 and r-value 1 at 0. By a previous proposition,
limx!0
sinx� arctanx
x2 ln(1 + x)=
1
6:
The orders cancel, and we obtain a non-zero number.
105
Remark. An approach using L'Hopital's Rule does work, but one hasto break the limit up as follows:
limx!0
sinx� arctanx
x2 ln(1 + x)= lim
x!0
sinx� arctanx
x3� limx!0
x
ln(1 + x);
and then, the calculations actually amount to the same thing we have done(and the reader, if not sure, should check this).
Lastly, we present a method for determining partial fraction expan-sions, using the ideas so far.
Looking at one root, we wish to �nd constants A1, A2, : : : , An suchthat
p(x)
(x� a)nq(x)=
A1
x� a+
A2
(x� a)2+ � � �+
An
(x� a)n+ � � � : (1)
We examine two cases.
Case I. n = 1
If the exponent is 1, then there is a very simple way of determiningthe coe�cient. In fact, it is so easy one can usually do it in one's head. Weillustrate with an example, where the exponent is 1 for all the roots.
We know
x� 1
(x� 2)x(x+ 1)=
A
x� 2+B
x+
C
x+ 1
for some constants A, B, and C.
Multiplying by x� 2, we obtain
x� 1
x(x+ 1)= A+
B(x� 2)
x+C(x� 2)
x+ 1:
Substituting x = 2, we automatically get A = 1=6. Similarly, we also obtainB = 1=2, and C = �2=3, and so
x� 1
(x� 2)x(x+ 1)=
1
6(x� 2)+
1
2x�
2
3(x+ 1):
One must agree this is much easier and faster than the usual method bysolving a system of linear equations.
Case II. n > 1
When there are multiple roots, the situation becomes much more com-plicated, and the method in Case I no longer works. For example, we know
x+ 1
x2(x� 1)3=A
x+
B
x2+
C
x� 1+
D
(x� 1)2+
E
(x� 1)3
for some constants A, B, C, D, and E. The values of B and E can be foundby the method in Case I, but there is no way to directly isolate the other
106
constants in the same way. The key here, as in a previous proposition, is tomake a smart substitution, to make these terms where the function goes to�1 into nicer terms.
Going back to (1), if we substitute x = 1=t+ a, we obtain
p(1=t+ a)tn
q(1=t+ a)= A1t+ A2t
2+ � � �+ Ant
n+ � � � :
Amazing! We get a garden variety polynomial (terms that blow up at a havebecome powers of t, with this substitution; why?). So, we make the substi-tution, and take the polynomial part. Furthermore, the Ai are simply thecoe�cients of the polynomial. But how do we know these are the right coef-�cients? Could there be more terms to the right, in that last equation, thatcontain polynomial terms?
No, not if we make sure that the expression remaining on the rightstays bounded as t approaches �1. Then it can't contain any powers of t.We illustrate with an example.
Let
r(x) =1
x(x� 1)2=A
x+
B
x� 1+
C
(x� 1)2
for some constants A, B, and C. Then
r(1
t) =
1
1
t(1
t� 1)2
=t3
(1� t)2=
(�y+ 1)3
y2(sub. y = 1� t)
= �y + 3�3
y+
1
y2= 2+ t�
3
1� t+
1
(1� t)2:
The polynomial part of this expression is 2 + t; we can be sure of this sincethe rest of the expression goes to 0 as t approaches �1. Similarly,
r(1 +1
t) =
1
(1 +1
t)(
1
t)2
=t3
t+ 1= t2 � t+ 1�
1
t+ 1:
Hence, the polynomial part is 1� t+ t2.
Therefore, A = 1 (the coe�cient of t in 2+ t), and B = �1 and C = 1
(the coe�cients of t, t2 in 1� t+ t2). Hence,
1
x(x� 1)2=
1
x�
1
x� 1+
1
(x� 1)2:
We will leave it the reader to consider what happens in the case that thedenominator of the fractions has non-real roots.
107
PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, De-
partment ofMathematics and Statistics,Memorial University of Newfound-
land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-
companied by a solution, together with references and other insights which
are likely to be of help to the editor. When a submission is submitted with-
out a solution, the proposer must include su�cient information on why a
solution is likely. An asterisk (?) after a number indicates that a problem
was submitted without a solution.
In particular, original problems are solicited. However, other inter-
esting problems may also be acceptable provided that they are not too well
known, and references are given as to their provenance. Ordinarily, if the
originator of a problem can be located, it should not be submitted without
the originator's permission.
To facilitate their consideration, please send your proposals and so-
lutions on signed and separate standard 81
2"�11" or A4 sheets of paper.
These may be typewritten or neatly hand-written, and should be mailed to
the Editor-in-Chief, to arrive no later than 1 October 1998. They may also
be sent by email to [email protected]. (It would be appreciated if
email proposals and solutions were written in LATEX). Graphics �les should
be in epic format, or encapsulated postscript. Solutions received after the
above date will also be considered if there is su�cient time before the date
of publication. Please note that we do not accept submissions sent by FAX.
2314. Proposed by Toshio Seimiya, Kawasaki, Japan.
Given triangle ABC with AB < AC. The bisectors of angles B and Cmeet AC and AB at D and E respectively, and DE intersects BC at F .
Suppose that \DFC =1
2(\DBC � \ECB). Determine angle A.
2315. Proposed by V�aclav Kone �cn �y, Ferris State University, Big
Rapids, Michigan, USA.
Prove or disprove that F (n) =
sn
�1�
1
n
�n�1, where F (n) is the
maximum value of
f(x1; x2; : : : ; xn) = sinx1 cosx2 : : : cosxn + cosx1 sinx2 : : : cosxn
+ : : :+ cosx1 cosx2 : : : sinxn;
xk 2 [0; �=2], k = 1; 2; : : : ; n, and n > 1 is a natural number.
108
2316. Proposed by Toshio Seimiya, Kawasaki, Japan.
Given triangle ABC with angles B and C satisfying C = 90�+
1
2B.
Suppose that M is the mid-point of BC, and that the circle with centre Aand radius AM meets BC again at D. Prove that MD = AB.
2317. Proposed by Richard I. Hess, Rancho Palos Verdes, California,
USA.
� 2�4� 3�
b a
c
de
�
The quadrilateral shown at theleft has integer elements a throughe. The angles as shown are integermultiples of the smallest.
(a) What is the smallest possiblevalue of c?
(b) What is the smallest possiblevalue of c if � must be obtuse?
2318. Proposed by V�aclav Kone �cn �y, Ferris State University, Big
Rapids, Michigan, USA.
Suppose that ABC is a triangle with circumcentre O and circumradiusR.
Consider the bisector (`) of any side (say AC), and let P (the \pedalpoint") be any point on ` inside the circumcircle.
Let K, L, M denote the feet of the perpendiculars from P to the linesAB, BC, CA respectively.
Show that [KLM ] (the area of the pedal triangleKLM ) is a decreasingfunction of � = OP , � 2 (0; R).
2319. Proposed by Florian Herzig, student, Perchtoldsdorf, Austria.
Suppose that UV is a diameter of a semicircle, and that P , Q are twopoints on the semicircle with UP < UQ. The tangents to the semicircle atP and Q meet at R. Suppose that S is the point of intersection of UP andV Q.
Prove that RS is perpendicular to UV .
2320. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.Two circles on the same side of the line ` are tangent to it at D. The
tangents to the smaller circle from a variable point A on the larger circleintersect ` at B and C. If b and c are the radii of the incircles of trianglesABD and ACD, prove that b+ c is independent of the choice of A.
109
2321. Proposed by David Doster, Choate Rosemary Hall, Wall-
ingford, Connecticut, USA.Suppose that n � 2. Prove that
nXk=2
�n2
k
�=
n2Xk=n+1
�n2
k
�:
Here, as usual, bxc means the greatest integer less than or equal to x.
2322. Proposed by K.R.S. Sastry, Dodballapur, India.
Suppose that the ellipse E has equationx2
a2+y2
b2= 1. Suppose that
� is any circle concentric with E. Suppose that A is a point on E and B is apoint of � such that AB is tangent to both E and �.
Find the maximum length of AB.
2323. Proposed by K.R.S. Sastry, Dodballapur, India.
Determine a positive constant c so that the Diophantine equation
uv2 � v2 � uv � u = c
has exactly four solutions in positive integers u and v.
2324. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,
Madrid, Spain.
Find the exact value of1Xn=1
1
un, where un is given by the recurrence
un = n!
�n� 1
n
�un�1 ;
with the initial condition u1 = 2.
2325?. Proposed by Walther Janous, Ursulinengymnasium, Inns-
bruck, Austria.
Suppose that q is a prime and n is a positive integer. Suppose thatfakg (0 � k � n) is given by
nXk=0
akxk
=1
qn
nXk=0
�qn
qk
�(qx� 1)
k :
Prove that each ak is an integer.
110
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to
consider for publication new solutions or new insights on past problems.
2208. [1997: 47] Proposed by Christopher J. Bradley, Clifton Col-
lege, Bristol, UK.
1. Find a set of positive integers fx; y; z; a; b; c; kg such that
y2z2 = a2 + k2
z2x2 = b2 + k2
x2y2 = c2 + k2
2. Show how to obtain an in�nite number of distinct sets of positive inte-gers satisfying these equations.
I. Solution by C. Festraets-Hamoir, Brussels, Belgium.
1. Partons de triples pythagoriciens �el �ementaires
52 � 12 = 3
2+ 4
2
12 � 52 = 4
2+ 3
2
52 � 52 = 24
2+ 7
2
Multiplions ces �egalit �es respectivement par 32 �72, 72 �42 et 42 �32; on obtient
152 � 72 = 63
2+ 84
2
72 � 202 = 112
2+ 84
2
202 � 152 = 288
2+ 84
2
Donc fx; y; z; a; b; c; kg = f20; 15; 7; 63; 112; 288; 84g.2. Etant donn �ee une solution:
y2z2 = a2 + k2
z2x2 = b2 + k2
x2y2 = c2 + k2
il su�t de multiplier ces �egalit �es respectivement par b2c2, c2a2, a2b2 pouren obtenir une novelle:
b2c2y2z2 = a2b2c2 + k2b2c2
c2a2z2x2 = a2b2c2 + k2c2a2
a2b2x2y2 = a2b2c2 + k2a2b2
111
On poseb2y2 = y21 a2b2c2 = k21 k2b2c2 = a21c2z2 = z21 k2c2a2 = b21a2x2 = x21 k2a2b2 = c21
ce qui donne
y21z21 = k21 + a21
z21x21 = k21 + b21
x21y21 = k1 + c21
II. Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Uni-
versity, Waterloo, Ontario.
We solve part 2 �rst. From our construction we can easily obtain par-ticular solutions. Let q be any even natural number with the property thatq=2 can be written as the product of two distinct factors in at least threedi�erent ways so that q = 2u1v1 = 2u2v2 = 2u3v3 where ui 6= uj ,vi 6= vj , and ui 6= vj for i; j = 1; 2; 3, i 6= j. (For example, we couldlet q = 2p1p2p3 where p1; p2, and p3 are three distinct odd primes and letui = pi, vi = q=2ui.) Let a1 = ju21 � v21j, b1 = ju22 � v22j, c1 = ju23 � v23j.Then
a21 + q2 = (u21 + v21)2= l2 where l = u21 + v21
a22 + q2 = (u22 + v22)2= m2 where m = u22 + v22
a23 + q2 = (u23 + v23)2= n2 where n = u23 + v23
Now, let d = lmn, a = a1d, b = b1d, c = c1d, k = qd, x = mn, y = nl,and z = lm. Then
a2 + k2 = d2(a21 + q2) = d2l2 = l4m2n2 = y2z2
b2 + k2 = d2(b21 + q2) = d2m2= l2m4n2 = z2x2
c2 + k2 = d2(c21 + q2) = d2n2 = l2m2n4 = x2y2
To get a particular solution, we could take q = 24 = 2�1�12 = 2�2�6 =
2 � 3 � 4 so u1 = 1, v1 = 12, u2 = 2, v2 = 6, u3 = 3, v3 = 4. Thena1 = 143, b1 = 32, c1 = 7, l = 145, m = 40, n = 25, d = 2
3 � 54 � 29,
which leads to a = 23�5
4�11�13�29, b = 28�5
4�29, c = 23�5
4�7�29,x = 2
3 � 53, y = 5
3 � 29, z = 23 � 5
2 � 29 and k = 26 � 3� 5
4 � 29.
Also solved by GERALD ALLEN and CHARLES R. DIMINNIE, Angelo
State University, San Angelo, TX, USA; FLORIAN HERZIG, student, Per-
chtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California,
USA; ROBERT B. ISRAEL, University of British Columbia, Vancouver, BC;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL
LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St Paul's
School, London, England; DIGBY SMITH,Mount Royal College, Calgary, Al-
berta; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. There
was one incomplete solution.
112
2209. [1997: 47] Proposed byMiguel Amengual Covas, Cala Figuera,
Mallorca, Spain.
Let ABCD be a cyclic quadrilateral having perpendicular diagonalscrossing at P . Project P onto the sides of the quadrilateral.
1. Prove that the quadrilateral obtained by joining these four projectionsis inscribable and circumscribable.
2. Prove that the circle which passes through these four projections alsopasses through the mid-points of the sides of the given quadrilateral.
Comments. The recent solution to Crux with Mayhem 2194 [1997:530-532] discussed the ambiguity of the word circumscribable. It is perhapspreferable to use the terminology cyclic and circumscribing for our bicentricquadrilateral; cf. Crux with Mayhem 2203.
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.Let the projections of P be K, L, M , N withK 2 AB, etc. Further-
more, let the midpoints of AB, BC, CD, DA be Q, R, S and T . Fromone of Brahmagupta's theorems ([1] theorem 3.23), P is collinear with eachprojection (such as K) and the midpoint opposite that projection (namelyS). Furthermore QRST is a rectangle since its sides are parallel to the per-pendicular diagonals (AC and BD) of the given quadrilateral. The diagonalsQS and RT of the rectangle are diameters of its circumcircle, and this cir-cle contains the projection points (such asK) since they form right triangles(namely QKS) that have a diameter as hypotenuse. In other words, K, L,M , N , Q, R, S, T are concyclic.
Next we have four cyclic quadrilaterals (each having an opposite pair ofright angles): KBLP , LCMP , MDNP , and NAKP . Thus
\PKN = \PAN = \CAD = \CBD = \LBP = \LKP;
and hence, PK is an interior angle bisector ofKLMN . Likewise PL, PM ,PN are interior angle bisectors, whence KLMN has an inscribed circlewith incentre at P . [Note that the existence of an incircle does not requireAC ? BD.]
Reference.
[1] H.S.M. Coxeter and S.L. Greitzer, Geometry Revisited. MAA, 1967.
Comments. The many references supplied by our solvers indicate thateach part of problem 2209 can easily be found elsewhere; for example, seeCrux 1836 [1993: 113; 1994: 84-85], Crux 1866 [1993: 203; 1994: 176],and problem 3 on the 1990 Canadian Mathematical Olympiad [1990: 198-199]. Seimiya found the problem itself as th �eor �eme 159, p. 319, in F.G.-M.,Exercices de g �eom�etrie. (For the information of newcomers to Crux with
Mayhem, F.G.-M. was a favourite reference of founding editor L �eo Sauv �e.
113
The �fth edition of 1912 (Cours de mathematiques No. 267, Tours: MaisonA.Mame et Fils, etc.) contains 2000 theorems with their proofs | nearly everyresult of elementary geometry known at that time. Because of his religiousa�liation, the author had to remain anonymous.)
Lambrou mentions a converse: given a bicentric quadrilateralKLMN
there exists an appropriate cyclic quadrilateral ABCD with orthogonal di-agonals. (The sides of ABCD are the perpendiculars at K; L;M;N to thelines from the incentre P .)
Also solved by CLAUDIO ARCONCHER, Jundia��, Brazil; FRANCISCO
BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; MANSUR BOASE,
student, St. Paul's School, London, England; CHRISTOPHER J. BRADLEY,
Clifton College, Bristol, UK; ADRIAN CHAN, student, Upper Canada Col-
lege, Toronto, Ontario; JORDI DOU, Barcelona, Spain; C. FESTRAETS-
HAMOIR, Brussels, Belgium; CYRUS HSIA, student, University of Toronto,
Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-
tria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,USA;
MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY
LEVERSHA, St Paul's School, London, England; GOTTFRIED PERZ, Pesta-
lozzigymnasium, Graz, Austria; ISTV �AN REIMAN, Budapest, Hungary;
TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Neth-
erlands; YEO KENG HEE, Hwa Chong Junior College, Singapore; and the
proposer.
2211. [1997: 47] Proposed by Bill Sands, University of Calgary, Cal-
gary, Alberta.
Several people go to a pizza restaurant. Each person who is \hungry" wantsto eat either 6 or 7 slices of pizza. Everyone else wants to eat only 2 or 3slices of pizza each. Each pizza in the restaurant has 12 slices.
It turns out that four pizzas are not su�cient to satisfy everyone, but thatwith �ve pizzas, there would be some pizza left over.
How many people went to the restaurant, and howmany of these were \hun-gry"?
Solution by Kathleen E. Lewis, SUNY Oswego. Oswego, NY, USA.
I am interpreting the problem to mean, for example, that the \hungry"people would each be satis�ed with 6 pieces, but would eat 7 if they wereavailable.
If x represents the number of \hungry" people, and y the number ofnon-\hungry" people, then we know that 6x+ 2y > 48, but 7x+ 3y < 60.Because x and y must be integers, the quantities 6x + 2y and 7x + 3y arealso integers and the �rst of these must be even. Therefore we can rewritethe inequalities as
6x+ 2y � 50 and 7x+ 3y � 59:
114
Taking the di�erence, we see that x + y, the total number of people, is atmost 9. If x is seven or less, then 6x + 2y cannot be greater than 48, butif x = 9, then 7x > 60, so x must be 8. If y were zero, then four pizzaswould su�ce, so y must be 1. Thus there are eight \hungry" people and one
non-\hungry" person.
Also solved by MANSUR BOASE, student, St. Paul's School, London,
England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; KEITH
EKBLAW, Walla Walla, Washington, USA; ROBERT GERETSCHL �AGER, Bun-
desrealgymnasium, Graz, Austria; FLORIAN HERZIG, student, Perchtolds-
dorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;
ROBERT B. ISRAEL, University of British Columbia, Vancouver, BC;
EDWARD J. KOSLOWSKA and ROSE MARIE SAENZ, students, Angelo State
University, San Angelo, Texas, USA; MICHAEL LAMBROU, University of
Crete, Crete, Greece; GERRY LEVERSHA, St Paul's School, London,
England; SEAN MCILROY, student, University of British Columbia, Van-
couver, BC; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; JOEL
SCHLOSBERG, student, Hunter College High School, New York NY, USA;
HEINZ-J�URGEN SEIFFERT, Berlin, Germany; ZUN SHAN and EDWARD T.H.
WANG, Wilfrid Laurier University, Waterloo, Ontario; DAVID R. STONE,
Georgia Southern University, Statesboro, Georgia, USA; and the proposer.
Also, one correct but anonymous solution was sent in. Two other readers
misinterpreted the problem.
2212. [1997: 48] Proposed by Edward T.H. Wang, Wilfrid Laurier
University, Waterloo, Ontario.
Let S = f1; 2; : : : ; ng where n � 3.
(a) In how many ways can three integers x, y, z (not necessarily distinct)be chosen from S such that x + y = z? (Note that x + y = z andy + x = z are considered to be the same solution.)
(b) What is the answer to (a) if x, y, z must be distinct?
Solution by William Moser, McGill University, Montreal, Quebec.
The answers to (a) and (b) are the cardinalities, s and t, of the sets:
S = f(x; y) j 1 � x � y; x+ y � ng and
T = f(x; y) j 1 � x < y; x+ y � ng;
where x and y denote integers.
Note �rst that the set
U = f(x; y) j 1 � x; 1 � y; x+ y � ng
has cardinality u =�n2
�. Since the set
V = f(x; y) j 1 � x = y; x+ y � ng
115
has cardinality v =�n2
�, it follows that the set
W = f(x; y) j 1 � x; 1 � y; x 6= y; x+ y � ng = U � V
has cardinality w = u� v =�n2
���n2
�.
From t = w2and s = t+ v, we then obtain that s =
1
2
��n2
�+�n2
��and
t = 1
2
��n2
���n2
��.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,
UK; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; GORAN CONAR,
student, Gymnasium Vara�zdin, Vara�zdin, Croatia; KEITH EKBLAW, Walla
Walla, Washington, USA; IAN JUNE L. GARCES, Manila, The Philippines,
and GIOVANNI MAZZARELLO, Firenze, Italy; ROBERT GERETSCHL �AGER,
Bundesrealgymnasium, Graz, Austria; FLORIAN HERZIG, student, Perch-
toldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;
CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER
JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Fer-
ris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, Uni-
versity of Crete, Crete, Greece; GERRY LEVERSHA, St Paul's School, London,
England; WILLIAMMOSER, McGill University, Montreal, Quebec (a second
solution); GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; HEINZ-
J�URGEN SEIFFERT, Berlin, Germany; ZUN SHAN, Nanjing Normal Univer-
sity, Nanjing, China; DIGBY SMITH,Mount Royal College, Calgary, Alberta;
and the proposer. One incorrect solution was sent in.
As expected, the answers can be given in many di�erent forms. If we
let f(n) and g(n) denote the numbers of ways for (a) and (b), respectively,then the most common answers are:
f(2n) = n2; f(2n+1) = n(n+1); g(2n) = n(n�1) and g(2n+1) = n2;
which, as pointed out by many solvers, are equivalent to the following for-
mulae:
f(n) =
(n2
4if n is even
n2�14
if n is odd; g(n) =
(n(n�2)
4if n is even
(n�1)24
if n is odd:
Janous and Sei�ert observed that g(n) = f(n� 1) for n � 4 and Lambrou
remarked that f(n) = 1
4
�n2 � 1
2(1 + (�1)n)
�.
Besides the solution given above (which was also obtained by Shan),
there is a variety of other interesting \single expression" formulae involving
the oor and/or the ceiling functions. These include:
1. f(n) =
�n2
4
�; g(n) =
�(n� 1)
2
4
�=
�n(n� 2)
4
�(Garces and Mazzarello; Geretschl�ager; and the proposer),
116
2. f(n) =
�n
2
� �n+ 1
2
�; g(n) =
�n� 1
2
� �n
2
�(Perz; and Sei�ert),
3. f(n) =
�n
2
� �n
2
�; g(n) =
�n
2
� �n
2
���n
2
�(Hsia),
4. f(n) =
�n
2
� �n�
�n
2
��; g(n) = (n� 1)
�n
2
����
n
2
��2(Cabez �on).
2213. [1997: 48] Proposed by Victor Oxman, University of Haifa,
Haifa, Israel.
A generalization of problem 2095 [1995: 344, 1996: 373].
Suppose that the function f(u) has a second derivative in the interval(a; b), and that f(u) � 0 for all u 2 (a; b). Prove that
1. (y � z)f(x) + (z � x)f(y) + (x� y)f(z) > 0 for all x; y; z 2 (a; b),z < y < x
if and only if f 00(u) > 0 for all u 2 (a; b);
2. (y � z)f(x) + (z � x)f(y) + (x� y)f(z) = 0 for all x; y; z 2 (a; b),z < y < x
if and only if f(u) is a linear function on (a; b).
Solution by Robert B. Israel, University of British Columbia, Vancou-
ver, BC.
Part 1 is incorrect. The �rst statement is actually equivalent to \f isstrictly convex", not \f 00(u) > 0 for all u 2 (a; b)". Strictly convex meansthat f(tx+ (1� t)y) < tf(x) + (1� t)f(y) for x; y 2 (a; b) and 0 < t <
1. If f is twice di�erentiable this is equivalent to f 00(u) � 0 everywherewith f 00(u) > 0 on a dense set. For example, f(x) = x4 is strictly convexeverywhere but f 00(0) = 0, and in this case
(y� z)x4 + (z � x)y4 + (x� y)z4
= (x� z)(y� z)(x� y)
�(x+ y)2 + (y + z)2 + (x+ z)2
2
�> 0
if z < y < x.
117
More generally, let f be any continuous function on (a; b), and n � 3
an integer. Given x1; x2; : : : ; xn let
L(f) =
nXi=1
(xi�1 � xi+1)f(xi);
where we take x0 = xn and xn+1 = x1. Then the following are equivalent:
(i) L(f) > 0 whenever a < x1 < x2 < � � � < xn < b.
(ii) f is strictly convex on (a; b).
Similarly, for part 2, we will show that f is linear on (a; b) if and onlyif L(f) = 0 whenever a < x1 < x2 < � � � < xn < b.
Proof of (i) =) (ii): If f is not strictly convex, then either it is notconvex at all or its graph includes a straight line segment. In the �rst case,there exist x, z, and t with a < z < x < b and 0 < t < 1 such that
f(tz+ (1� t)x) > tf(z) + (1� t)f(x):
Taking y = tz+(1� t)x and noting that t = (x�y)=(x� z)we can rewritethis as
f(y)>x� y
x� zf(z) +
y � z
x� zf(x)
or (y � z)f(x) + (z � x)f(y) + (x � y)f(z) < 0. Let x1 = z, x2 = y,and x3 = � � � = xn = x, and we have L(f) < 0 (if n > 3, note thatthe terms in f(xj) for 3 < j < n are 0 and the terms in f(x3) and f(xn)add to (y � x)f(x)). If n > 3 we must move x3; : : : ; xn slightly so thatx3 < � � � < xn, but by continuity the inequality will still be true if thechanges are small enough. In the second case, we can take anyx1 < � � � < xnin the interval over which the graph is a straight line f(x) = cx + d, andthen
L(f) = c
nXi=1
xi�1xi �nXi=1
xixi+1
!+ d
nXi=1
xi�1 �nXi=1
xi+1
!= 0
Proof of (ii)=) (i): As noted above, if g is linear on (a; b) then L(g) =0. Thus we can add a linear function g to f without changing either (i) or (ii).With an appropriate choice of g we can make f(x1) = f(xn) = 0, and bystrict convexity we then have f(xj) < 0 for 1 < j < n. Now write
L(f) =
nXi=1
(xi�1 � xi)f(xi) +
nXi=1
(xi � xi+1)f(xi)
=
nXi=1
(xi�1 � xi)(f(xi) + f(xi�1))
118
Note that the i = 1 term is 0, and each of the other terms is strictlypositive, so the sum is strictly positive.
As noted above, if f is linear on (a; b) then L(f) = 0. Conversely,suppose L(f) = 0 whenever a < x1 < x2 < � � � < xn < b. By part 1,L(h) > 0 where h(x) = x2, and so since L is a linear operator L(f + ch) =
cL(h) > 0 for any c > 0. Again using part 1, f+ch is strictly convex, and sof = lim
c7!0+(f + ch) is convex on (a; b). By the same argument, �f is convex,
so f is concave on (a; b). Any function that is both convex and concave on(a; b) is linear on (a; b).
Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria;
MICHAEL LAMBROU, University of Crete, Crete, Greece, (part 1 only).
There were seven incorrect and two incomplete solutions. All of the incorrect
solutions failed to notice the aw in the statement of the problem, and then
proceeded to treat as equivalent the two statements: \f is strictly convex on
(a; b)" and \f 00(u) > 0 for all u 2 (a; b)".
Several solvers pointed out that the condition f(u) � 0 for all
u 2 (a; b) is unnecessary.
Herzig remarks: If the interval considered is the set of real numbers
instead, then it is su�cient to assume y =x+z2
and f is a continuous function
for part 2. The given equation reduces to Jensen's equation
f
�x+ y
2
�=f(x) + f(y)
2
that can be solvedusing elementary methods (tricky substitutionand Cauchy's
method for solving functional equations).
Lambrou lists a series of six equivalent statements for part 1, where he
assumes the existence of derivatives only when needed:
1. (y � z)f(x) + (z � x)f(y) + (x � y)f(z) > 0 for all x; y; z 2 (a; b)
with z < y < x.
2. The �rst di�erence quotientf(u)�f(v)
u�v (u 6= v) is strictly increasing in
each variable (separately).
3. The second di�erence quotient
�f(u)� f(w)
u� w�f(v)� f(w)
v� w
�,(u� v)
(u 6= w, v 6= w, u 6= v) is strictly positive for all u; v;w 2 (a; b).
4. f 0 is strictly increasing on (a; b).
5. f is strictly convex on (a; b).
119
6. f 00(x) � 0 for all x 2 (a; b) and there is no interval I � (a; b) such
that f 00(x) = 0 for all x 2 I. (In the language of topology, this says
that f 00(x) > 0 on a dense subset of (a; b).)
2214. [1997: 109] Proposed by Walther Janous, Ursulinengymnas-
ium, Innsbruck, Austria.Let n � 2 be a natural number. Show that there exists a constant
C = C(n) such that for all real x1; : : : ; xn � 0 we have
nXk=1
pxk �
vuut nYk=1
(xk + C):
Determine the minimum C(n) for some values of n.[For example, C(2) = 1.]
Solution by Michael Lambrou, University of Crete, Crete, Greece.
We show that the inequality is valid for an aggregate of values of C ofwhich the least is
C(n) =n� 1
n�1pnn�2
; n � 2:
Let us �rst do the easier task of proving the existence of C's whichmake the inequality valid. Of course this part will be redundant as soon aswe improve the technique to �nd the least C.
Setting xi = y2i where yi � 0 (i = 1; : : : ; n), we are to show, equiva-lently, that for some C we have
nXi=1
yi
!2
�nYi=1
(y2i + C): (1)
Treating the right hand side of (1) as a polynomial in C, we observe that allcoe�cients are non-negative and that the coe�cient of Cn�1 is
Py2i . Thus
nYi=1
(y2i + C) �
nXi=1
y2i
!Cn�1:
But by the Cauchy{Schwarz inequality we have
nXi=1
yi
!2
� n
nXi=1
y2i
!;
so inequality (1) will be valid if we choose C = n1=(n�1) or larger. Thiscompletes the easier task.
120
It turns out that n1=(n�1) is only a slight overestimate of the mini-mum C, which we now seek. For any C for which (1) is valid, set wi =
yipn� 1=
pC, so that (1) becomes
nXi=1
wi
!2
�Cn�1
(n� 1)n�1
nYi=1
(w2i + n� 1)
or equivalently nXi=1
wi
!2
�Cn�1nn
(n� 1)n�1
nYi=1
�w2i � 1
n+ 1
�: (2)
To �nd the minimumC we shall �rst show that the following inequalityis valid:
nXi=1
wi
!2
� n2nYi=1
�w2i � 1
n+ 1
�: (3)
We shall use the Weierstrass inequality
mYi=1
(1 + ai) � 1 +
mXi=1
ai;
which holds if all ai � 0 or if �1 < ai < 0 for all i [for example, seeitem 3.2.37, page 210 of D. S. Mitrinovi�c, Analytic Inequalities]. Withoutloss of generality let w1; : : : ; wt � 1 and 0 � wt+1; : : : ; wn < 1, wheret 2 f0; 1; : : : ; ng. Then
nYi=1
�w2i � 1
n+ 1
�=
tYi=1
�w2i � 1
n+ 1
��
nYi=t+1
�w2i � 1
n+ 1
�
� 1 +
tXi=1
w2i � 1
n
!0@1 +
nXi=t+1
w2i � 1
n
1A
=1
n2
n� t+
tXi=1
w2i
!0@t+ nXi=t+1
w2i
1A
=1
n2
0@ tXi=1
w2i +
nXi=t+1
12
1A0@ tXi=1
12+
nXi=t+1
w2i
1A
�1
n2
nXi=1
wi
!2
(the last inequality by the Cauchy{Schwarz inequality), which proves (3).Note that equality occurs for w1 = � � � = wn = 1. We conclude that (2)
121
is valid for any C with Cn�1nn=(n� 1)n�1 � n2, i.e., with
C �n� 1
n�1pnn�2
; n � 2:
The minimum value C(n) we seek is then as stated at the beginning, sincefor
xi = y2i = C
�wipn� 1
�2=
C � 1n� 1
the original inequality reduces to equality.
Remark. The above shows C(2) = 1,
C(3) =2p3� 1:1547; C(4) =
3
3p42� 1:1905; C(5) =
4
4p53� 1:1963;
and generally C(n)� n1=(n�1) which approaches 1 in the limit.
The minimum C(n) for all n � 2 was also found by CHRISTOPHER
J. BRADLEY, Clifton College, Bristol, UK; KEE-WAI LAU, Hong Kong; and
HEINZ-J�URGEN SEIFFERT, Berlin, Germany. The existence of C(n) was
proved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria; JOE
HOWARD, NewMexicoHighlandsUniversity, Las Vegas, NM, USA; V �ACLAV
KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; YEO KENG
HEE, student, Hwa Chong Junior College, Singapore; and the proposer.
Of those readers who found the minimum C(n), only Lambrou man-
aged to do it without multivariable calculus. He also sent in a second solu-
tion, using calculus.
Herzig found the minimum C(n) for n = 2 and 3, and Howard found
it for n = 2. Kone�cn �y conjectured the correct minimum value of C(n) for
all n � 2, but proved it only for n = 2 and 3.
The proposer showed more generally that, for all real � > 0 and inte-
gers n � 2, there is a constant C = C�(n) such that
nXi=1
x�i �
nYi=1
(xi + C)
!�
for all real x1; : : : ; xn � 0.
2215?. [1997: 109] Proposed by Theodore Chronis, student, Aris-
totle University of Thessaloniki, Greece.Let P be a point inside a triangle ABC. It is known how to determine
P such that PA+PB+PC is a minimum (known as Fermat's Problem forTorricelli).
Determine P such that PA+ PB + PC is a maximum.
122
We received two types of solutions to this problem:
I. P is considered to be an interior point of triangleABC (this is the problemas it was posed);
II. P is considered to be inside or on the boundary of triangle ABC.
Solution by Michael Lambrou, University of Crete, Greece, [slightlymodi�ed by the editor].
(I) LetP be an internal point. We show the existence of another internalpoint Q such that QA+ QB +QC > PA+ PB + PC.
Consider the ellipse E through P with foci B and C and l the tangentto this ellipse at P . Then the circle centred at A with radius AP meets lat P . Therefore, the points on l to at least one side of P are outside thecircle. Take Q to be a point on l that is outside the circle (hence QA > PA)
but close to P so that it is inside triangle ABC. Since Q is on l, Q is outsideE and QB + QC > PB + PC. Adding we get the required result.
Therefore, PA+ PB + PC has no maximum.
(II) Assume that AC � AB � BC and P is an interior or boundarypoint of triangle ABC; P 6= A. Let the line parallel to BC through P meetAB at D and AC at E. By similarity AE � AD � DE and AE � AP .Hence
PA+ PB + PC < PA+ (BD+DP ) + (PE +EC)
= PA+DE + BD+ EC
� PA+ AD + BD +EC
= PA+ AB +EC
� AE +AB + EC
= AC + AB
[Seimiya and Tsaoussoglou both noted that this inequality, for an interiorpoint P , is due to Visschers (1902) [see F.G.M. Exercises de G �eom�etrie,p. 228, Th. 25 { IV]].
However, if P = A then PA+ PB + PC = AB +AC.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,
UK (I and II); JORDI DOU, Barcelona, Spain (I and II); RICHARD I. HESS,
Rancho PalosVerdes, California, USA (I and II); PETERHURTHIG, Columbia
College, Burnaby, BC (II); ROBERT B. ISRAEL, University of British Columbia,
Vancouver, BC (II); V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,
Michigan, USA (II); GERRY LEVERSHA, St Paul's School, London, England
(I); ISTV �AN REIMAN, Budapest, Hungary (I); TOSHIO SEIMIYA, Kawasaki,
Japan (II); PANOS E. TSAOUSSOGLOU, Athens, Greece (II).
123
2216. [1997: 110] Proposed by Walther Janous, Ursulinengymnas-
ium, Innsbruck, Austria.Suppose that � � 1 is a natural number.
1. Determine the set of all �'s such that the diophantine equationx� + y2 = z2 has in�nitely many solutions.
2.?For any such �, determine all solutions of this equation.
Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-
sity, Waterloo, Ontario.
We show that x�+y2 = z2 has in�nitely many solutions for all naturalnumbers � by �nding all solutions for any �.
Suppose �rst that x� + y2 = z2. Let
x = p�11 p�22 : : : p�kk (1)
where p1 < p2 < � � � < pk are primes and �i 2 N for i = 1; 2; : : : ; k. Then
x�+y2 = z2 is equivalent to (z+y)(z�y) =kYi=1
p�ii and so z+y =
kYi=1
p�ii
and z � y =
kYi=1
p ii where �i, i are nonnegative integers satisfying
�i + i = ��i (2)
for i = 1; 2; : : : ; k and
kYi=1
p�ii >
kYi=1
p ii : (3)
It follows that
y =1
2
kYi=1
p�ii �
kYi=1
p ii
!; (4)
z =1
2
kYi=1
p�ii +
kYi=1
p ii
!: (5)
Conversely, if p1 6= 2 and x, y, z are given by (1), (4), (5) where �i 2 N and�i, i are nonnegative integers satisfying (2) and (3) for i = 1; 2; : : : ; k, thenit is easy to see that x� + y2 = z2 holds. If p1 = 2, then �1 and 1 mustboth be positive integers, since x� even implies that z+ y and z� y are botheven. (In this case, �1 � 1 and 1 � 1 and so ��1 � 2 which is true for all� � 2, while if � = 1 we must assume �1 � 2.)
Note that condition (3) could be replaced by the condition that
�i 6= i (6)
124
for at least one i if we replace (4) by
y =1
2
�����kYi=1
p�ii �
kYi=1
p ii
����� : (7)
In summary, if p1 6= 2 then all solutions to x� + y2 = z2 are given by (1),(5), and (7) where �i 2 N and �i, i are nonnegative integers satisfying (2)and (6). If p1 = 2 we must impose the condition that �1 � 1, 1 � 1.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,
UK; YEO KENG HEE, Hwa Chong Junior College, Singapore; and the pro-
poser. Four solvers solved part 1 correctly: FLORIAN HERZIG, student,
Perchtoldsdorf, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big
Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete,
Greece; and GERRY LEVERSHA, St Paul's School, London, England. There
was one incorrect solution, and one incomplete solution.
2217. [1997: 110] Proposed by Bill Sands, University of Calgary,
Calgary, Alberta.
(a) Prove that for every su�ciently large positive integer n, there are arith-metic progressions a1; a2; a3 and b1; b2; b3 of positive integers such thatn = a1b1 + a2b2 + a3b3.
(b) What happens if we require a1 = b1 = 1?
(This is a variation of problem 3 of the 1995/96 Alberta High School Mathe-matics Competition, Part II, which may appear in a future Skoliad Corner.)
I. Solution to part (a) by Joel Schlosberg, student, Bayside, New York,
USA.
Let the progressions be 2; 3; 4 and a; a + b; a + 2b. We have to showthat
n = 2a+ 3(a+ b) + 4(a+ 2b) = 9a+ 11b
for su�ciently large n. There is a theorem of Euclid that any su�ciently largenumber can be represented in the form ka + lb [for some positive integersa and b] if k and l are relatively prime. Since gcd(9; 11) = 1, the answerfollows immediately.
[Editorial note. Coincidentally, in a recent issue of Mathematics and
Informatics Quarterly (Vol. 7, No. 4, 1997, p. 185), Crux with Mayhem reg-ular K. R. S. Sastry gives a proof of the above \theorem of Euclid", in factshowing more generally the known result that, if k and l are relatively prime,every integer greater than kl can be expressed as ka+ lb for some positiveintegers a and b. Thus Schlosberg's proof shows that the statement of theproblem holds for every integer n � 100. Actually, we can allow b = 0 inthis problem, which (it can easily be checked) lowers the bound to n � 89.For more, see the editorial remarks below.]
125
II. Solution to part (b) by Christopher J. Bradley, Clifton College, Bris-
tol, UK.
[Bradley �rst solved part (a). | Ed.]
If one requires a1 = 1 and b1 = 1 then only those n are representablefor which non-negative x; y exist such that
1 + (1 + x)(1 + y) + (1 + 2x)(1 + 2y) = n; (1)
that is, n = 3 + 3x+ 3y+ 5xy or
5n� 6 = (5x+ 3)(5y+ 3):
Since neither 5x + 3 nor 5y + 3 can equal 1 this implies 5n � 6 has tobe composite as a necessary condition. When 5n � 6 is prime, n = 5 orn = 17 for example, no x and y can possibly be found. Now, by Dirichlet'sTheorem, since 5 and 6 are coprime there are inde�nitely large primes of theform 5n � 6. It follows that there are in�nitely many n which cannot beexpressed in the form (1).
III. Editorial remarks.
Here are extensions of this problem observed by various solvers. Forthe complete list of solvers, see below.
Shan and Wang calculated that the largest integer n not expressible asin part (a) is n = 17. Some other solvers gave larger \impossible" values forn because they did not allow constant arithmetic progressions.
Israel proved part (a) under the stronger condition that a1 is any �xedpositive integer. He also strengthened part (b) by showing that (a) failswhenever a1 and b1 are �xed positive integers.
Lambrou generalized (a) in another direction, namely he found all pos-itive integers m so that every su�ciently large integer n can be written inthe form
n = a1b1 + a2b2 + � � �+ ambm
where a1; : : : ; am and b1; : : : ; bm are m-term arithmetic progressions ofpositive integers. His answer: m = 1; 2; 3 or 6.
Most solvers noted that one of the arithmetic progressions in part (a),say a1; a2; a3, can be �xed while the other varies, and the result is still true;for example see solution I, which uses (a1; a2; a3) = (2;3; 4). Lambrou alsoinvestigated exactly which �xed (a1; a2; a3) allow this. Letting (a1; a2; a3) =(a� h; a; a + h) where a > h > 0 are integers, he proved that every su�-ciently large positive integer n can be expressed as n = a1b1 + a2b2 + a3b3for some positive integer arithmetic progression b1; b2; b3 if and only if 3aand 2h are relatively prime.
As a variation on part (b), Leversha noted that if we impose a1 = 1 =
b3, then no integer of the form p � 6, where p is prime, can be written asin (a).
126
Readers may like to verify the above results for themselves! And thereare still many related problems that await investigation. For example, somereaders' solutions to (a) use arithmetic progressions in which the commondi�erences are bounded (in absolute value). Bradley's solution seems to bethe best in this respect, with his common di�erences always at most 7. Canthis number be lowered? That is, can every su�ciently large positive integerbe written as a1b1+a2b2+a3b3, where a1; a2; a3 and b1; b2; b3 are arithmeticprogressions of positive integers with ja1� a2j and jb1� b2j both at most 6,say? Or what if the common di�erences of the arithmetic progressions in (a)are required to be bounded below by some constant? Finally, in (b), what ifwe only require a1 = b1?
Both parts also solved by FLORIAN HERZIG, student, Perchtoldsdorf,
Austria; ROBERT B. ISRAEL, University of British Columbia, Vancouver, BC;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL
LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St Paul's
School, London, England; ZUN SHAN and EDWARD T.H. WANG, Wilfrid
Laurier University, Waterloo, Ontario; and the proposer. Part (a) only solvedby DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut,
USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; V �ACLAV
KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; and YEO
KENG HEE, student, Hwa Chong Junior College, Singapore.
All proofs of part (b) were similar, though Lambrou observed that the
full statement of Dirichlet's Theorem is not needed, only that there are in-
�nitely many primes of the form 10k � 1 (as can be seen from Solution II
above), and that this special case is a little easier to prove: for example,
see T. Nagell's Introduction to Number Theory, Chapter 5, x50. Is there a
solution to (b) that does not need Dirichlet's Theorem at all?
The original Alberta High School Mathematics Competition problem
was to show that every integer n can be written in the form a1b1 + a2b2 +
a3b3, where a1; a2; a3 and b1; b2; b3 are arbitrary arithmetic progressions of
integers.
2218. [1997: 110] Proposed by Victor Oxman, University of Haifa,
Haifa, Israel.
Suppose that a, b, c are positive real numbers and that
abc = (a+ b� c)(b+ c� a)(c+ a� b):
Clearly a = b = c is a solution. Determine all others.
I Solution by Goran Conar, student, Gymnasium Vara�zdin, Vara�zdin,
Croatia.If any of the three quantities, a + b � c, b + c � a and c + a � b is
negative, say a+ b� c < 0, then we have c > a+ b, so that b+ c� a > 0,
127
and c + a � b > 0. This implies that abc < 0, a contradiction. Hencea+ b� c > 0, b+ c� a > 0 and c+ a� b > 0.
Note that
(a+ b� c)(a� b+ c) = a2 � (b� c)2 � a2; (1)
(b+ c� a)(b� c+ a) = b2 � (c� a)2 � b2; (2)
(c+ a� b)(c� a+ b) = c2 � (a� b)2 � c2: (3)
Multiply the three inequalities and take non-negative square roots. We get
(a+ b� c)(b+ c� a)(c+ a� b) � abc: (4)
Since equality holds in (4), it must also hold in (1), (2) and (3). Therefore,a = b = c is the only solution.
II Solution by Robert B. Israel, University of British Columbia, Van-
couver, BC.Since the given equation is invariant under any permutation of a, b, c,
we may assume that 0 < a � b � c.
The equation can be re-written as
(b+ c� a)(b� c)2 + a(b� c)(c� a) = 0:
Since both terms on the left are non-negative, the only way to obtainequality is with a = b = c. Hence there are no other solutions in positivereals.
Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sara-
jevo, Bosnia and Herzegovina; MANSUR BOASE, student, St. Paul's School,
London, England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;
THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece;
CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA;
FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,
Rancho Palos Verdes, California, USA; JOE HOWARD, New Mexico High-
lands University, Las Vegas, NM, USA; CYRUS HSIA, student, University
of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium,
Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,
Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece;
KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St Paul's School, London,
England; CAN ANH MINH, University of California, Berkeley, CA, USA;
BOB PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; ISTV �AN
REIMAN, Budapest, Hungary; JUAN-BOSCO ROMEROM �ARQUEZ, Univer-
sidad de Valladolid, Valladolid, Spain; ZUN SHAN and EDWARD T.H.WANG,
Wilfrid Laurier University, Waterloo, Ontario; DIGBY SMITH, Mount Royal
College, Calgary, Alberta; YEO KENG HEE, Hwa Chong Junior College, Sin-
gapore; and the proposer. There was also one incorrect solution.
128
A few solvers remarked that if a, b and c are non-negative, thenwe have
the additional solutions: (a; b; c) = (0; t; t); (t; 0; t); (t; t; 0) where t � 0 is
arbitrary.
Howard pointed out that (a + b � c)(b + c � a)(c + a � b) � abc
is problem #3 of the 1981 British Mathematical Olympiad, and referred to
Rabinowitz's Index to Mathematical Problems, 1980{1984, p. 49.[Ed.| Actually, it is much \older" than that and dates back to at least 1925.
See, for example, x 1.3 on p. 12 of Geometric Inequalities by O. Bottema
et al.]
Many solvers, after establishing the fact that a, b, c are the sides of
a triangle, say 4, showed that the given equality is equivalent to the fact
that 2r = R, where r and R denote the inradius and circumradius of4 re-
spectively. Then the conclusion follows from a celebrated theorem of Euler,
which states that 2r � R, with equality if and only if4 is equilateral.
[Ed.| This can be found, for example, in x 5.1 on p. 48 of the bookmentioned
in the previous paragraph.]
2210?. [1997: 47] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,
Alcorc �on, Madrid, Spain.
Given a0 = 1, the sequence fang (n = 1; 2; : : : ) is given recursivelyby �
n
n
�an �
�n
n� 1
�an�1 +
�n
n� 2
�an�2 � : : :�
�n
bn2c
�abn2 c = 0:
Which terms have value 0?
No solutions have been received to date. The problem remains open.
Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell
Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer
Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil
Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,
J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia
129
THE ACADEMY CORNERNo. 18
Bruce Shawyer
All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7
Memorial University Undergraduate
Mathematics Competition
September 25, 1997
We present in this issue, two solutions from Bob Prielipp, University ofWisconsin{Oshkosh, Wisconsin, USA, to the above competition, which wasprinted in Academy Corner No. 15 [1997: 449].
2. The surface area of a closed cylinder is twice the volume. Determinethe radius and height of the cylinder given that the radius and heightare both integers.
Solution.Let r be the radius and h be the height of the closed cylinder, where r
and h are both positive integers.
If the surface area of the closed cylinder is twice the volume, then
2�rh+ 2�r2 = 2�r2h:
It follows that h+ r = rh. Thus
(r � 1)(h� 1) = rh� r � h+ 1 = 0 + 1 = 1:
Hence r� 1 = 1 and h� 1 = 1 (since r and h are both positive integers), sothat r = 2 and h = 2.
It is easily checked that if r = 2 and h = 2, then the surface area ofthe closed cylinder is twice the volume.
130
3. Prove that
1 +1
4+
1
9+ : : :+
1
n2< 2:
Solution. [Slightly shortened by the editor.]We use mathematical induction to prove, for each positive integer n,
1 +1
4+
1
9+ : : :+
1
n2< 2�
1
n:
The result clearly holds for n = 1.
Assume that the result holds for n = k. Then
1 +1
4+
1
9+ : : :+
1
k2+
1
(k+ 1)2
<
�2�
1
k
�+
1
(k+ 1)2
= 2�(k+ 1)2 � k
k(k+ 1)2= 2�
k2 + k+ 1
k(k+ 1)2
< 2�k(k+ 1)
k(k+ 1)2= 2�
1
k + 1:
Hence, by induction, the result holds.
Equality holds if and only if n = 1.
Advance Notice
At the summer 1999 meeting of the Canadian Mathematical Society, tobe held in St. John's, Newfoundland, there will be a Mathematics EducationSession on the topic \What Mathematics Competitions do for Mathematics".
Invited speakers include Edward Barbeau, Toronto; Tony Gardner, Birm-ingham, England; Ron Dunkley, Waterloo; and Rita Janes, St. John's. Any-one interested in givinga paper at this session should contact one of the orga-nizers, Bruce Shawyer or Ed Williams, at the Department ofMathematics andStatistics, Memorial University of Newfoundland, St. John's, Newfoundland,Canada.
email addresses:
131
THE OLYMPIAD CORNERNo. 189
R.E. Woodrow
All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.
We begin this issue with an exam set sent in by one of our newer cor-respondents. My thanks go to Enrique Valeriano, National University ofEngineering, Lima, Peru.
PERU'S SELECTION TEST FOR
THE XII IBEROAMERICAN OLYMPIADAugust 1997 | 3.5 hours
1. Given an integer a0 > 2, the sequence a0; a1; a2; : : : is de�ned asfollows:
ak+1 = ak(1 + ak); if ak is an odd number;ak+1 = ak
2; if ak is an even number:
Prove that there is a nonnegative integer p such that ap > ap+1 > ap+2.
2. A positive integer is called \almost-triangular" if the number isitself triangular or is the sum of di�erent triangular numbers. How manyalmost-triangular numbers are there in the set f1; 2; 3; : : : ; 1997g?
Note: The triangular numbers are a1; a2; a3; : : : ; ak; : : : ; wherea1 = 1, and ak = k + ak�1, for all k � 2.
3. An n� n chessboard (n � 2) is numbered with n2 non-zero num-bers. This chessboard is called an \Incaican Board" if, for each square thenumber written on the square is the di�erence between two of the numberswritten on two of the neighbouring squares (sharing a common edge). Forwhich values of n can one obtain Incaican Boards?
4. Let ABC be a given acute triangle. Give a ruler and compass con-struction of an equilateral triangle DEF with D on BC, E on AC, and Fon AB such that the perpendiculars to BC at D, to AC at E, and to ABat F , respectively, are concurrent.
Next we give the Third and Fourth Grade problems of the 38th Mathe-matics Competition for Secondary School Students of the Republic of Slove-nia. My thanks go to Richard Nowakowski, Canadian Team Leader to theIMO in Hong Kong, for collecting this contest and forwarding it to me.
132
REPUBLIC OF SLOVENIA
38th Mathematics Competition
for Secondary School StudentsApril, 1994
Third Grade1. Let n be a natural number. Prove: if 2n+1 and 3n+ 1 are perfect
squares, then n is divisible by 40.
2. Show that cos(sinx) > sin(cosx) holds for every real number x.
3. The polynomial p(x) = x3+ax2+bx+c has only real roots. Showthat the polynomial q(x) = x3�bx2+acx�c2 has at least one nonnegativeroot.
4. Let the point D on the hypotenuse AC of the right triangle ABCbe such that jABj = jCDj. Prove that the bisector of \BAC, the medianthroughB, and the altitude throughD, of the triangleABD have a commonpoint.
Fourth Grade1. Prove that there does not exist a function f : Z! Z, for which
f(f(x)) = x+ 1 for every x 2Z.
2. Put a natural number in every empty �eld of the table so that youget an arithmetic sequence in every row and every column.
0
74
103
186
3. Prove that every number of the sequence
49; 4489; 444889; 44448889; : : :
is a perfect square (in every number there are n fours, n � 1 eights and anine).
4. Let Q be the midpoint of the side AB of an inscribed quadrilat-eral ABCD and S the intersection of its diagonals. Denote by P and R
the orthogonal projections of S on AD and BC respectively. Prove thatjPQj = jQRj.
133
As a �nal contest for your puzzling pleasure in this number, we give theVIII Nordic Mathematical Contest. Againmy thanks go to Richard Nowakowski,Canadian Team leader to the IMO in Hong Kong, for collecting this contestand forwarding it to me.
VIII NORDIC MATHEMATICAL CONTEST
March 17th, 1994
Time: 4 hours
1. Let O be a point in the interior of an equilateral triangle ABC withside length a. The lines AO, BO and CO intersect the sides of the triangleat the points A1, B1 and C1 respectively. Prove that
jOA1j+ jOB1j+ jOC1j < a:
2. A �nite set S of points in the plane with integer coordinates is calleda two-neighbour set, if for each (p; q) inS exactly two of the points (p+1; q),(p; q+1), (p�1; q), (p; q� 1) are in S. For which n does there exist a two-neighbour set which contains exactly n points?
3. A square piece of paper ABCD is folded by placing the corner Dat some point D0 on BC (see �gure). Suppose AD is carried into A0D0,crossing AB at E. Prove that the perimeter of triangle EBD0 is half as longas the perimeter of the square.
D C
A B
D0
E
A0
4. Determine all positive integers n < 200 such that n2 + (n+ 1)2 isa perfect square.
Turning now to comments and solutions related to the February 1997
number of the corner, we welcome two alternate solutions to problems ofthe Sixth Irish Mathematical Olympiad sent in by another new contributorwhom we also welcome.
3. [1995: 151{152; 1997: 9{13] Sixth Irish Mathematical Olympiad.
For nonnegative integers n, r the binomial coe�cient�n
r
�denotes the
number of combinations of n objects chosen r at a time, with the conventionthat
�n
0
�= 1 and
�n
r
�= 0 if n < r. Prove the identity
1Xd=1
�n� r + 1
d
��r � 1
d� 1
�=
�n
r
�
134
for all integers n, r with 1 � r � n.
Alternate Solution by Mohammed Aassila, UFR de Math �ematique etd'Informatique, L'Universit �e Louis Pasteur, Strasbourg, France.
We have
(1 + x)n = (1 + x)n�r+1(1 + x)r�1
=
"m�r+1Xi=0
�n� r + 1
i
�xi
# 24r�1Xj=0
�r � 1
j
�xj
35
=
m�r+1Xi=0
r�1Xj=0
�n� r + 1
i
��r � 1
j
�xi+j
The coe�cient of xr is
rXd=0
�n� r + 1
d
��r � 1
r � d
�=
rXd=1
�n� r + 1
d
��r � 1
d� 1
�
4. [1995:151{152; 1997: 9{13] Sixth Irish Mathematical Olympiad.
Let x be a real number with 0 < x < �. Prove that, for all naturalnumbers n, the sum
sinx+sin3x
3+
sin5x
5+ � � �+
sin(2n� 1)x
2n� 1
is positive.
Alternate solution by Mohammed Aassila, UFR de Math �ematique etd'Informatique, L'Universit �e Louis Pasteur, Strasbourg, France.
We know that
2 sinx sin(2k� 1)x = cos(2k� 2)x� cos 2kx:
Hence
2 sinx
�sinx+
sin3x
3+ � � �+
sin(2n� 1)x
2n� 1
�
= 1� cos 2x+cos 2x� cos 4x
3+ � � �+
cos(2n� 2)x� cos 2nx
2n� 1
= 1� cos 2x
�1�
1
3
�� cos 4x
�1
3�
1
5
�� � � � �
cos 2nx
2n� 1
� 1���
1�1
3
�+
�1
3�
1
5
�+ � � �+
1
2n� 1
�= 0:
135
Next we turn to solutions to problems of the Latvian 44 MathematicalOlympiad given in the February number of the Corner last year [1997: 78].
LATVIAN 44MATHEMATICAL OLYMPIAD
Final Grade, 3rd RoundRiga, 1994
1. It is given that cosx = cos y and sinx = � siny. Prove thatsin 1994x+ sin1994y = 0.
Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;by Mansur Boase, student, St. Paul's School, London, England; by PanosE. Tsaoussoglou, Athens, Greece; and by Zun Shan and Edward T.H. Wang,Wilfrid Laurier University, Waterloo, Ontario. We give the solutionof Covas.
More generally, we prove that sinmx + sinmy = 0 where m is aninteger.
Since sinmx+sinmy = 2 sin�mx+y
2
�cos
�m�x�y2
��, it is su�cient
to show that sin�x+y
2
�= 0 since sin
�mx+y
2
�= 0 follows easily by induc-
tion from
sin
�(m+ 1)
x+ y
2
�= sin
�m
(x+ y)
2
�cos
�x+ y
2
�
+cos
�m
(x+ y)
2
�sin
�x+ y
2
�:
Now,
cosx = cos y () cosx� cos y = 0
() �2 sinx+ y
2sin
x� y
2= 0 (1)
and
sinx = � siny () sinx+ siny = 0
() 2 sinx+ y
2cos
x� y
2= 0 : (2)
Squaring each of (1) and (2) and adding, we �nd
4 sin2
�x+ y
2
��sin
2 (x� y)
2+ cos2
(x� y)
2
�| {z }
=1
= 0:
Hence sin x+y
2= 0.
3. It is given that a > 0, b > 0, c > 0, a+ b+ c = abc. Prove that atleast one of the numbers a, b, c exceeds 17=10.
136
SolutionsbyMiguel Amengual Covas, Cala Figuera,Mallorca, Spain; byHeinz-J �urgen Sei�ert, Berlin, Germany; by Panos E. Tsaoussoglou, Athens,Greece; and by Zun Shan and Edward T.H. Wang, Wilfrid Laurier University,Waterloo, Ontario. We give two solutions of Shan and Wang.
First Solution.We show, in general, that if xi > 0 for i = 1; 2; : : : ; n, such thatPn
i=1 xi =Qn
i=1 xi, then maxfxi : i = 1; 2; : : : ; ng � n1=(n�1). In partic-ular, when n = 3 we get
maxfx1; x2; x3g �p3 > 1:7 =
17
10:
By the arithmetic-geometric mean inequality we have 1
n
nXi=1
xi
!n�
nYi=1
xi =
nXi=1
xi;
and thusPn
i=1 xi � nn=(n�1).
Without loss of generality, wemay assume thatmaxfxi : i = 1; 2; : : : ; ng =xn. Then nxn �
Pni=1 xi � nn=(n�1) from which xn � n1=(n�1) follows.
Second Solution (without the AM-GM inequality).Suppose xi > 0 (i = 1; 2; : : : ; n) such that
Pn
i=1 xi =Qn
i=1 xi. Thendividing both sides by
Qn
i=1 xi we get
nXi=1
1
x1x2 : : : xi : : : xn= 1;
where xi indicates that the factor xi is missing. Hence for some j we have
1
x1x2 : : : xj : : : xn�
1
nor n � x1x2 : : : xj : : : xn:
Without loss we may assume that xn = maxfxi : i = 1; 2; : : : ; ng.Then xn�1n � x1x2 : : : xj : : : xn � n, from which xn � n1=(n�1) follows. In
particular, for n = 3, we get x3 �p3 > 1:7 = 17
10.
4. Solve the equation 1!+2!+3!+ � � �+n! = m3 in natural numbers.
Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-sity, Waterloo, Ontario.
We solve the more general problem of �nding all solutions of the Dio-phantine equation 1! + 2! + 3! + � � �+ n! = mk in natural numbers, n, m,and k. For convenience, let Sn = 1! + 2! + 3! + � � �+ n!.
When k = 1 clearly m = Sn is the only solution for any n.
When k = 2 we claim that the equation Sn = m2 has exactly twosolutions: n = m = 1 and n = m = 3. Note �rst that d! � 0(mod 10)
137
for all d � 5 and S4 = 1 + 2 + 6 + 24 = 33 � 3(mod 10). HenceSn � 3(mod 10) for all n � 4. However, it is easy to see that the last digitof a perfect square can never be 3 and so there are no solutions if n � 4.Checking the cases when n = 1; 2; 3 directly reveals that there are preciselytwo solutions, as given above.
When k � 3 we show that n = m = 1 is the only solution. Ifn � 2 then clearly Sn � 0(mod 3). But mk � 0(mod 3) impliesm � 0(mod 3) and so mk � 0(mod 27) as k � 3. Since d! � 0(mod 27)
for all d � 9 and since
S8 = 1+ 2 + 6 + 24 + 120 + 720 + 5040 + 40320 = 46233 6= 0 (mod 27)
there are no solutions ifn � 8. On the other hand, direct checking shows thatfor n = 2; 3; 4; 5; 6; 7, Sn = 3; 9; 33; 153;873, and 5913, none of which is aperfect kth power for any k � 3. Finally, it is trivial to see that n = m = 1
is a solution.
Remark: The special case of this problem when k = 2 was proposed byE.T.H. Wang as Quicky Q657 in theMathematics Magazine, 52 (1979), p. 47.The general case was also proposed by him as problem #4203 inMathmedia(in Chinese) 4(2), 1980; p. 64 with solution in 4(3), 1980, p. 49. This is ajournal published by the Institute of Mathematics, Academia Sinica, Taipei,Taiwan.
5. There are 1994 employees in the o�ce. Each of them knows 1600others of them. Prove that we can �nd 6 employees, each of them knowingall 5 others.
Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-sity, Waterloo, Ontario.
LetE denote the set of these 1994 employees. For each x 2 E, let S(x)denote the set of all employees whom x does not know. Then by assumption,jS(x)j = 393 for all x 2 E. Let a and b be any two employees who knoweach other. Since
jS(a) [ S(b)j � 2� 393 = 786 < 1992;
9c 2 E such that a, b, and c form a triple of mutual acquaintances. Since
jS(a) [ S(b)[ S(c)j � 3� 393 = 1179 < 1991;
9d 2 E such that a, b, c, and d form a quadruple of mutual acquaintances.Since
jS(a) [ S(b) [ S(c) [ S(d)j � 4� 393 = 1572 < 1990;
9e 2 E such that a, b, c, d, and e form a quintuple of mutual acquaintances.Finally, since
jS(a) [ S(b) [ S(c) [ S(d) [ S(e)j � 5� 393 = 1965 < 1989;
138
9f 2 E such that a, b, c, d, e, and f form a sextuple of mutual acquaintances.
1st SELECTION ROUND
1. It is given that x and y are positive integers and 3x2+x = 4y2+y.Prove that x� y, 3x+ 3y + 1 and 4x+ 4y+ 1 are squares of integers.
Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;by Panos E. Tsaoussoglou, Athens, Greece; and by Zun Shan and EdwardT.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give the solu-tion by Shan and Wang and their comment.
Note �rst that the given equation implies the following two equations:
(x� y)(3x+ 3y+ 1) = y2 (1)
(x� y)(4x+ 4y+ 1) = x2 (2)
(1) � (2) yields (x � y)2(3x + 3y + 1)(4x+ 4y + 1) = (xy)2 whichimplies that (3x + 3y + 1)(4x + 4y + 1) is a perfect square. But clearlygcd(3x+3y+1;4x+4y+1) = 1 since 4(3x+3y+1)�3(4x+4y+1) = 1.Therefore, 3x + 3y + 1 and 4x + 4y + 1 are both squares, which, togetherwith (1) (or (2)), implies that x� y is also a square.
Comment: This would be a much better problem had it asked to showonly that x�y is a square. This is an example of a case when asking to provetoo many things actually gives the solution away, in some sense.
Shan andWang also proposed the followingproblem inspired by this one.
The Diophantine equation 3x2 +x = 4y2+ y is satis�ed when x = 30
and y = 26.
(a) Find another solution in positive integers.
(b) Are there in�nitely many solutions in positive integers? Is so, de-scribe all of them.
2nd SELECTION ROUND
1. It is given that 0 � xi � 1, i = 1; 2; : : : ; n. Find the maximum ofthe expression
x1
x2x3 : : : xn + 1+
x2
x1x3x4 : : : xn + 1+ � � �+
xn
x1x2 : : : xn�1 + 1:
Solutions by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta; and by Zun Shan and Edward T.H. Wang, Wilfrid Laurier University,Waterloo, Ontario. We �rst give the solution of Shan and Wang.
Clearly n � 2 for the question to make sense. Let Sn denote the givensum. We prove that Sn � n � 1. For n = 2, equality holds if and only ifx1 = 0, x2 = 1 or x1 = 1, x2 = 0 or x1 = x2 = 1. For n � 3, equalityholds if and only if one of the xi's is 0 and the others are all equal to 1. We�rst establish a lemma:
139
Lemma. Suppose that the xi's are reals such that 0 � xi � 1 for alli = 1; 2; : : : ; n, where n � 2. Then x1+x2+ � � �+xn � n�1+x1x2 � � �xnwith equality holding if and only if xi 6= 1 for at most one i, i = 1; 2; : : : ; n.
Proof. For n = 2, x1+x2 � x1x2+1, (1� x1)(1�x2) � 0, whichis clearly true. Equality holds if and only if x1 = 1 or x2 = 1. Suppose theassertion holds for some n � 2. Then
x1 + x2 + � � �+ xn + xn+1 � n� 1 + x1x2 � � �xn + xn+1
� n� 1 + x1x2 � � �xnxn+1 + 1
= n+ x1x2 � � �xn+1:
(The 2nd inequality is by the n = 2 case.)
If equality holds, then it must hold in both inequalities above. By theinduction hypotheses, we then have
(i) at most one of x1; x2; : : : ; xn is di�erent from 1 and
(ii) either xn+1 = 1 or x1x2 � � �xn = 1.
Since x1x2 � � �xn = 1 clearly impliesx1 = x2 = � � � = xn = 1, our assertionabout the equality follows.
Now we proceed to prove the claim about Sn. For n = 2,
S2 =x1
x2 + 1+
x2
x1 + 1�
x1
x2 + x1+
x2
x1 + x2= 1:
It is readily seen that equality holds if and only if x1 = 0, x2 = 1 or x1 = 1,x2 = 0, or x1 = x2 = 1. For n � 3 we apply the lemma above and obtain
Sn �x1 + x2 + � � �+ xn
x1x2 � � �xn + 1�
(n� 1) + x1x2 � � �xnx1x2 � � �xn + 1
� n� 1:
If equality holds, then the 2nd inequality impliesxj 6= 1 for at most one j andthe 3rd inequality implies (n� 1)x1x2 � � �xn = x1x2 � � � xn, which impliesthat xi = 0 for at least one i. Hence xi = 0 for some i and xj = 1 for allj 6= i. It is obvious that this condition is also su�cient. This completes theproof.
And here is Klamkin's somewhat more abbreviated solution with gen-eralization.
Since the expression is convex in each of the variables, the maximumvalue is achieved when each variable takes on 0 or 1. Clearly this occurs whenone variable is 0 and the rest are 1 giving the maximum value of n� 1. Thesame maximum occurs if any of the numerators xi are replaced by x�ii where�i � 1.
A similar result, using convexity, that
X xui
1 + s� xi+Y
(1� xi)v � 1;
140
where 0 � xi � 1, u; v � 1, s =Pxi and the sum and product are over
i = 1; 2; : : : ; n, is given in [1].
Reference:
[1] M.S. Klamkin, USA Mathematical Olympiads 1972{1986, M.A.A.,Washington D.C., 1988, p. 82.
3. A triangle ABC is given. From the vertex B, n rays are constructedintersecting the sideAC. For each of the n+1 triangles obtained, an incirclewith radius ri and excircle (which touches the side AC) with radius Ri isconstructed. Prove that the expression
r1r2 : : : rn+1
R1R2 : : : Rn+1
depends on neither n nor on which rays are constructed.
Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain.If A, B, C are the angles of a triangle, then
r = s tanA
2tan
B
2tan
C
2and ra = s tan
A
2;
where r, ra are the inradius and the radius of the excircle opposite A, and sis the semiperimeter.
It follows thatr
ra= tan
B
2tan
C
2:
Next we apply this result to each of n + 1 triangles obtained (see �gure atthe top of the next page). This yields
r1R1
= tan A2tan �1
2r2R2
= tan 180���12
tan �22
� � � � � � � � �� � � � � � � � �
rnRn
= tan180���n�1
2tan �n
2rn+1Rn+1
= tan 180���n2
tan C2:
Multiplying these equalities, we observe that the product of all theright hand members is
tanA
2��tan
�1
2cot
�1
2
���tan
�2
2cot
�2
2
�� � ��tan
�n
2cot
�n
2
�� tan
C
2
= tanA
2� tan
C
2;
and we getr1r2 � � � rn�1R1R2 � � �Rn+1
= tanA
2� tan
C
2
which depends on neither n nor on which rays are constructed.
141
A �1 �2 �n C
B
3rd SELECTION ROUND
3. Let ABCD be an inscribed quadrilateral. Its diagonals intersectat O. Let the midpoints of AB and CD be U and V . Prove that the linesthrough O, U and V , perpendicular to AD, BD and AC respectively, areconcurrent.
Solution by Toshio Seimiya, Kawasaki, Japan.Case 1. Neither is AC orthogonal to BD, nor is AD a diameter.
A
B
C
D
N
Y
X
M
O
V
U
HS
Let M , N be the feet of the perpendiculars from U , V to BD, ACrespectively, and let S be the intersection of UM and V N . Let X, Y bethe feet of the perpendiculars from A, D to BD, AC respectively, and letH be the intersection of AX and DY . Since U is the midpoint of AB, andUMkAX, M is the midpoint of BX. SimilarlyN is the midpoint of CY .
Since \AXD = \AYD = 90�, A, X, Y , D are concyclic. Therefore\Y XD = \Y AD = \CAD = \CBD. Thus we have XY kBC.
SinceM , N are the midpoints ofBX, CY respectively, we haveMNkXY .
Since SMkHX, XNkHY , andMNkXY , MX, NY and SH are concur-rent at O. Therefore S, H, O are collinear.
Since AH ? OD and DH ? OA, H is the orthocentre of �OAD, so thatHO ? AD. Thus we have SO ? AD.
Thus the lines through O, U and V , perpendicular to AD, BD andACrespectively, are concurrent at S.
142
Case 2. AC is orthogonal to BD.
A
D
B
C
M
V
O
U
LetM be the midpoint of BC. Since U is the midpoint of AB, we getUM ? AC, so that UM ? BD. Similarly we have VM ? AC.
Since AC ? BD and M is the midpoint of BC, by Brahmagupta'sTheorem we have MO ? AD. (See: H.S.M. Coxeter and S.L. Greitzer,Geometry Revisited, p. 59).
Thus the lines through O, U and V , perpendicular to AD, BD andACrespectively, are concurrent at M .
Case 3. AD is a diameter.
A
B
C
D
S
O
q
q
U
V
Let S be the intersection of AB and CD. Since AB ? BD andCD ? AC, S is the orthocentre of �OAD. Thus SO ? AD.
Hence the lines through O, U and V , perpendicular to AD, BD andAC respectively, are concurrent at S.
That completes this number of the Olympiad Corner. Send me yournice solutions and suggestions for future issues.
143
BOOK REVIEWS
Edited by ANDY LIU
Vita Mathematicaedited by Ronald Calinger,published by the MAA, Notes and Reports Series, 1996,ISBN# 0-88385-097-4, softcover, 350+ pages, $34.95.Reviewed byMaria de Losada, Bogot �a, Colombia.
This collection of papers read at the August 1992 Quadrennial Meet-ing of the International Study Group on the Relations between History andPedagogy ofMathematics at the University of Toronto and the Seventh Inter-national Congress onMathematical Education at Universit �e Laval in Quebec,edited and refereed, begins with a somewhat ponderous re ection on the dif-ferent tendencies in research in the history of mathematics, contrasting theapproaches of cultural and mathematical historians.
Looking at the role of problems in the history and teaching of mathe-matics, Evelyne Barbin reminds us that \one of the perverse e�ects of edu-cation : : : [is] that answers are given to questions that have not been asked.The history of mathematics shows us that questions must come �rst, and[that] it is through questions that we make sense of mathematical concepts."
In the historical studies (from antiquity to the Scienti�c Revolution)can be found intriguing material that is not of easy access. Swetz's paper onthe enigmas of Chinese mathematics is informative as it strives to provide abalanced presentation and Katz's treatment of medieval Hebrew and Islamicmathematics provides useful and little known information in a concise andwell ordered manner.
Noteworthy among the more recent historical studies is JudithGrabiner's paper which contrasts historical perspectives of the calculus, thegeometric (McLaurin) and algebraic (Lagrange), linking these naturally witheducation and culture and suggesting that \progress in mathematics is madeby those who sharpen their thinking by exercising the courage of their some-times idiosyncratic convictions". On an entirely di�erent note, the detailedhistory given by Ronald Calinger of the University of Berlin MathematicsSeminar examines every facet of that paradigmatically successful structuringof a research tradition, in which many of the recurrent themes of this book,such as starting from problems and learning from the masters, were put intopractice.
The third section of the collection, devoted to the integration of historywith mathematics teaching recounts many valuable experiences, two titles ofnote being Mathematical Masterpieces: Teaching with Original Sources andA History of Mathematics Course for Teachers, Based on Great Quotations.
144
Although they are presented as summaries and have catchy titles, these ar-ticles refer to teaching experiences that are both substantive and well struc-tured. In the �rst of these, Reinhard Laubenbacher and David Pengelleyreveal not only a polished list of original sources that can be used with un-dergraduates, but also aspects of their pedagogical approach. In the second,Israel Kleiner begins with quotations addressing the question \What ismath-ematics?" that are arranged in \antagonistic" pairs to bring across the un-derlying message of mathematics as an activity whose history is susceptibleto chronological, thematical, topical and biographical study, as long as sightis not lost of Lakatos' remark: \The history of mathematics, lacking the guid-ance of philosophy [is] blind, while the philosophy of mathematics, turningits back on the most intriguing phenomena in the history of mathematics,[is] empty."
It is unfortunate that the book has so many typos! Not only are foot-note numbers routinely formatted incorrectly, but there are also plenty oferrors in the text.
In general, this is good formative reading for those with an appetitefor historical material, but especially useful in its treatment of interrelationsbetween history and pedagogy.
The Canadian Mathematical Society has initiated a new series of book-lets of enrichment material for interested and mathematically talented highschool students.
La Soci �et �e math �ematique du Canada vient de lancer une s �erie de livretsd'enrichissement pour les �el �eves du secondaire int �eress �es et forts en math �e-matiques.
A Taste OfMathematics
Aime{T{On les Math �ematiques
See the enclosed yer for more details of the ATOM series.
Voir le d �epliant ci-inclus pour plus de renseignements sur la collection ATOM .
145
Sum of powers of a �nite sequence:
a geometric approach
William O.J. Moser
In this note we give a geometric-combinatoric derivation of a formulafor the sum of rth powers of a �nite positive integer sequence:
ar1 + ar2 + � � �+ arn =
rX`=1
nXi=1
�ai
`
��(r; `) ; (1)
where �n
k
�=
� n!k!(n�k)! ; if 0 � k � n ,
0; if k > n � 1 ,
and the numbers �(r; `) are determined by the recurrence
�(r; 1) = 1; r = 1; 2; 3; � � � ; �(1; `) = 0; ` = 2; 3; � � � ;�(r; `) = `(�(r� 1; `� 1) + �(r � 1; `)); r � 2; ` � 2 :
(2)
The derivation parallels, simpli�es and generalizes the proof given in[1], and seems to be more elementary than proofs given in [2] and [3].
For �xed (but arbitrary) positive integers m; r, let L(r)m denote ther-dimensional integer lattice points
f(x1; x2; � � � ; xr) jxi integers, 0 � x1; x2; � � � ; xr � mg ;
and forw � 1 let S(r)m (w) denote the set of r-dimensional \cubes" with facesparallel to the coordinate planes, vertices in L(r)m and width w.Of course S(r)m (w) = ; if w > m. A cube in S(r)m (w) is identi�ed by itsvertex (�) = (�1; �2; � � � ; �r) closest to (0; � � � ; 0). Clearly
w+�1 � m; w+�2 � m; � � � ; w+�r � m; �1; �2; � � � ; �r � 0 ;
or
0 � �1 � m� w; 0 � �2 � m� w; � � � ;
� � � 0 � �r � m� w; 1 � w � m; (3)
so the number of cubes in S(r)m (w) is
jS(r)m (w)j =�
(m�w + 1)r ; if 1 � w � m ,0 ; if w > m � 1 .
Copyright c 1998 Canadian Mathematical Society
146
For given 1 � w � m we can also �nd the number of (�)'s satisfying (3) asfollows. Any r-tuple (�) satisfying (3) determines the sequence (�) consist-ing of the distinct values, in increasing order, of the entries in (�):
(�) : 0 � �1 < �2 < � � � < �` � m�w; ` � r : (4)
Given a sequence (�) satisfying (4) there are many r-tuples (�) satisfying(3) which have entries with values (�). How many? Let �(r; `) denote thisnumber. It depends on ` (and not on the particular numbers �1; � � � ; �`)and satis�es the recurrence (2). For, given (�), there are ` choices (namely�1; �2; � � � ; �`) for the �rst entry �1 of (�); (�) can then be completed in�(r � 1; `) ways if the integer chosen for �1 appears among the entries�2; �3; � � � ; �r, and in �(r� 1; `� 1) ways otherwise; (2) then follows.
The numbers �(r; `) are exhibited in the display below:
rn` 1 2 3 4 5 6
1 1 0 0 0 0 0
2 1 2! 0 0 0 0
3 1 6 3! 0 0 0
4 1 14 36 4! 0 0
5 1 30 150 240 5! 0
.
.
....
.
.
....
.
.
....
.
.
.
Since there are�m�w+1
`
�ways of choosing the sequence (�) satisfying
(4) we �nd that
(m� w+ 1)r =
rX`=1
�m� w+ 1
`
��(r; `); 1 � w � m; 1 � ` � r;
and setting a = m� w+ 1 � 1
ar =
rX`=1
�a
`
��(r; `); a � 1; r � 1 :
Summing over a set of a's we have (1). Taking ai = a + (i � 1)d,i = 1; 2; � � � ; n in (1), we have
ar + (a+ d)r + � � �+ (a+ (n� 1)d)r =
rX`=1
nXi=1
�a+ (i� 1)d
`
��(r; `) ;
the sum of the rth powers of an arithmetic progression. When a = d = 1
s(r)(n) = 1r + 2r + � � �+ nr =
rX`=1
�n+ 1
`+ 1
��(r; `) ;
147
since
nXi=1
�i
`
�=
�n+ 1
`+ 1
�.
The display below exhibits s(r)(n) for r = 1; 2; 3; 4; 5:
s(1)(n) 1�n+1
2
�s(2)(n) 1
�n+1
2
�+ 2!
�n+1
3
�s(3)(n) 1
�n+1
2
�+ 6
�n+1
3
�+ 3!
�n+1
4
�s(4)(n) 1
�n+1
2
�+ 14
�n+1
3
�+ 36
�n+1
4
�+ 4!
�n+1
5
�s(5)(n)1
�n+1
2
�+ 30
�n+1
3
�+ 150
�n+1
4
�+ 240
�n+1
5
�+ 5!
�n+1
6
�References
1. [1] Moser, W., Sums of dth powers. Math. Gazette 75 (1991) 332.
2. [2] Paul, J.L., On the sum of the kth powers of the �rst n integers.Amer. Math. Monthly 78 (1971) 271{273. MR 43 #4092.
3. [3] Wagner, C., Combinatorial proofs of formulas for power sums. Arch.Math. (Basel) 68 (1997), no. 6, 464{467.
148
THE SKOLIAD CORNERNo. 29
R.E. Woodrow
In this number we give the �rst round of the 1996{97 Alberta HighSchool Mathematics Competition written in November of 1996. The top stu-dents from Round 1 are invited to take part in the second round competitionwritten in February. There are book prizes for individuals and teams basedon the Round 1 results, and scholarships and cash prizes for Round 2. (See[1997:410-411, 479-481] where we gave Round 2 of the 1996{97 contest andthe answers.) More information about the contest can be found at the web-site: www.math.ualberta.ca/ ahsmc/. My thanks go to the organizers of thecommittee (chaired by T. Lewis, University of Alberta) for permission to usethese materials.
ALBERTA HIGH SCHOOL
MATHEMATICS COMPETITION
Part I | November, 1996
1. An eight-inch pizza is cut into 3 equal slices. A ten-inch pizza is cutinto 4 equal slices. A twelve-inch pizza is cut into 6 equal slices. A fourteeninch pizza is cut into 8 equal slices. From which pizza would you take a sliceif you want as much pizza as possible?
(a) 8-inch (b) 10-inch (c) 12-inch (d) 14-inch (e) does not matter
2. One store sold red plums at four for a dollar and yellow plums atthree for a dollar. A second store sold red plums at four for a dollar andyellow plums at six for a dollar. You bought m red plums and n yellowplums from each store, spending a total of ten dollars. How many plums inall did you buy?
(a) 10 (b) 20 (c) 30 (d) 40 (e) not enough information
3.
A
B
C
D
EF
Six identical cardboard pieces arepiled on top of one another, andthe result is shown in the dia-gram.
The third piece to be placed is:
(a) A (b) B (c) C (d)D (e) E
149
4. A store o�ered triple the GST in savings. A sales clerk calculated theselling price by �rst reducing the original price by 21% and then adding the 7%GST based on the reduced price. A customer protested, saying that the storeshould �rst add the 7%GST and then reduce that total by 21%. They agreed ona compromise: the clerk just reduced the original price by the 14% di�erence.How do the three ways compare with one another from the customer's pointof view?
(a) The clerk's way is the best. (c) The compromise is the best.(b) The customer's way is the best. (d) All three ways are the same.(e) The compromise is the worst while the other two are the same.
5. Ifm andn are integers such that 2m�n = 3, then what willm�2n
equal?
(a) �3 only (b) 0 only (c) only multiples of 3(d) any integer (e) none of these
6. If x is x% of y, and y is y% of z, where x, y and z are positive realnumbers, what is z?
(a) 100 (b) 200 (c) 10; 000 (d) does not exist (e) cannot be determined
7. About how many lines can one rotate a regular hexagon throughsome angle x, 0� < x < 360�, so that the hexagon again occupies its originalposition?
(a) 1 (b) 3 (c) 4 (d) 6 (e) 7
8. AB is a diameter of a circle of radius 1 unit. CD is a chord perpen-dicular to AB that cuts AB at E. If the arc CAD is 2=3 of the circumferenceof the circle, what is the length of the segment AE?
(a) 23
(b) 32
(c) �2
(d)p32
(e) none of these
9. One of Kerry and Kelly lies onMondays, Tuesdays and Wednesdays,and tells the truth on the other days of the week. The other lies on Thursdays,Fridays and Saturdays, and tells the truth on the other days of the week. Atnoon, the two had the following conversation:
Kerry : I lie on Saturdays.Kelly : I will lie tomorrow.Kerry : I lie on Sundays.On which day of the week did this conversation take place?
(a) Monday (b)Wednesday (c) Thursday (d) Saturday (e) Sunday
150
10. How many integer pairs (m;n) satisfy the equation
m(m+ 1) = 2n ?
(a) 0 (b) 1 (c) 2 (d) 3 (e)more than 3
11. Of the following triangles given by the lengths of their sides, whichone has the greatest area?
(a) 5; 12; 12 (b) 5; 12; 13 (c) 5; 12; 14 (d) 5; 12; 15 (e) 5; 12; 16
12. If x < y and x < 0, which of the following numbers is nevergreater than any of the others?
(a) x+ y (b) x� y (c) x+ jyj (d) x� jyj (e)�jx + yj
13. An x by y ag, with x < y, consists of two perpendicular whitestripes of equal width and four congruent blue rectangles at the corners. Ifthe total area of the blue rectangles is half that of the ag, what is the lengthof the shorter side of each blue rectangle?
(a)x�y+
px2+y2
4(b)
x�y+px2+y2
2(c)
3x+y+px2+y2
4
(d)3x+y+
px2+y2
2(e) none of these
14. A game is played with a deck of ten cards numbered from 1 to 10.Shu�e the deck thoroughly.
i) Take the top card. If it is numbered 1, you win. If it is numbered k,where k > 1, go to (ii).
ii) If this is the third time you have taken a card, you lose. Otherwise,put the card back into the deck at the kth position from the top and go to (i).
What is the probability of winning?
(a) 15
(b) 518
(c) 1345
(d) 310
(e) none of these
15. Five of the angles of a convex polygon are each equal to 108�.In which of the following �ve intervals does the maximum angle of all suchpolygons lie?
(a) (105�; 120�) (b) (120�; 135�) (c) (135�; 150�)(d) (150�; 165�) (e) (165�; 180�)
16. Which one of the following numbers cannot be expressed as thedi�erence of the squares of two integers?
(a) 314159265 (b) 314159266 (c) 314159267(d) 314150268 (e) 314159269
151
In the last number we gave the problems of the Final Round of theBritish Columbia Colleges Junior High School Mathematics Contest 1997.Here are the o�cial solutions. My thanks for sending the materials to me goto John Grant McLoughlin, now of the Faculty of Education, Memorial Uni-versity of Newfoundland, who participated in formulating the exams whilehe was at Okanagan College.
BRITISH COLUMBIA COLLEGES
Junior High School Mathematics Contest
Final Round 1997 | Part A
1. The buttons of a phone are arranged as shown at the right. If thebuttons are one centimetre apart, centre-to-centre, when you dial the num-ber 592-7018 the distance, in centimetres, travelled by your �nger is:
0
8
5
2
7
4
1
9
6
3
Solution. Each of the distances between successive digits in the phonenumber is the hypotenuse of a right-angled triangle with integer sides. The6 lengths can be easily computed as
p2;p5;p5;p2;p10; and
p5, for a
total of 2p2 + 3
p5 +
p2p5. This can be rearranged into
p5(3 +
p2) +
2p2. Answer is (A)
2. What is the total number of ones digits needed in order to write theintegers from 1 to 100?
Solution. Clearly we need only one 1 for all the single digit numbers,and (since we are only interested in one 3-digit number, namely 100) we needonly one 1 for all the 3-digit numbers under consideration. For the 2-digitnumbers there are nine which have a 1 in the units position (11;21; : : : ; 91),and 10 which have a 1 in the tens position (10; 11; : : : ; 19), for a grand totalof 21 ones needed. Answer is (D)
3. The number of solutions (x; y; z) in positive integers for the equa-tion 3x+ y+ z = 23 is:
Solution. Clearly, there are only 7 possible values for x, namely 1
through 7. When x = 7, we have y + z = 2, which has a single solutiony = z = 1; when x = 6, we have y + z = 5, which has 4 solutions:(y; z) = (1; 4); (2;3); (3; 2); (4; 1). In this way we see that by reducing thevalue of x by 1 we increase the number of solutions by 3. Thus the totalnumber of solutions are 1 + 4 + 7 + 10 + 13 + 16 + 19 = 70.Answer is (D)
152
4. In the diagram below the upper scale AB has ten 1 centimetre divi-sions. The lower scale CD also has ten divisions but it is only 9 centimetreslong. If the right hand end of the fourth division of scale CD coincides ex-actly with the right hand end of the seventh division of scale AB, what is thedistance, in centimetres, from A to C?
A B
C D
Solution. Let E be the right hand end of the fourth division of scaleCD (which is also the right hand end of the seventh division of scale AB).The distance AE is 7 cm. The distance CE is 4 � 0:9 = 3:6 cm. Thus thedistance AC = AE � CE = 3:4 cm. Answer is (E)
5. Triangle ABC is equilateral with sides tangent to the circle withcentre at O and radius
p3. The area of the quadrilateral AOCB, in square
units, is:
O
C
A
B
Solution. First join the points B and O. Since an intersecting tangentand radius of a circle are perpendicular, this produces 2 congruent right an-gled triangles, namely BOA and BOC, each of which is a 30� � 60� � 90�
triangle, which means that side BO is twice the length of side AO. Usingthe Theorem of Pythagoras we have
BA2 = (2p3)2 � (
p3)2 = 12� 3 = 9:
Thus BA = BC = 3. The area of triangle BOA is now 12�p3 � 3. Since this
is half the area we seek, the answer is 3p3. Answer is (A)
6. Times such as 1:01, 1:11,: : : are called palindromic times becausetheir digits read the same forwards and backwards. The number of palin-dromic times on a digital clock between 1:00 a.m. and 11:59 a.m. is:
Solution. Let us �rst consider single digit hours. The �rst and last digitsmust be the same, but the middle digit can be anything. Thus there are 6
such palindromic times each hour (since there are 6 possible middle digits),for a total of 54 such palindromic times between 1:00 am and 9:59 am. Fortwo digit hours (that is, 10 and 11), in order to be a palindromic time theminutes are completely determined by the hour; there are only two suchtimes, namely 10:01 and 11:11. So the grand total number of palindromictimes between 1:00 a.m. and 11:59 a.m. is 56. Answer is (D)
153
7. Ted's television has channels 2 through 42. If Ted starts on channel15 and surfs, pushing the channel up button 518 times, when he stops he willbe on channel:
Solution. Ted's television set has a total of 41 channels. Thus everytime he pushes the channel up button 41 times he returns to the channel hestarts with. Since
518 = 12 � 41 + 26
we see that he has e�ectively gone up 26 channels from his starting point atchannel 15. He should thus be at channel 15+26=41. Answer is (E)
8. Consider a three-digit number with the following properties:
1. If its tens and ones digits are switched, its value would increaseby 36.
2. Instead, if its hundreds and ones digits are switched its valuewould decrease by 198.
Suppose that only the hundreds and tens digits are switched. Its valuewould:
Solution. Let n be a three-digit number with the given properties. Leta, b, and c represent its hundreds digit, its tens digit and its ones digit, re-spectively. Then n = 100a+ 10b+ c. By property 1 we see that
100a+ 10c+ b = n+ 36 = (100a+ 10b+ c) + 36;
or 9(c� b) = 36;
that is, c� b = 4:
By property 2 we have
100c+ 10b+ a = n� 198 = (100a+ 10b+ c)� 198;
or 99(a� c) = 198;
that is, a� c = 2:
Note that the two relations established above (when added) yield a� b = 6.Now let us consider the switch of the hundreds and the tens digits of n; thenew number is
100b+ 10a+ c = 100a+ 10b+ c+ 100(b� a) + 10(a� b)
= n� 90(a� b) = n� 540:
Answer is (B)
9. Speedy Sammy Seamstress sews seventy-seven stitches in sixty-sixseconds. The time, in seconds, it takes Sammy to stitch �fty-�ve stitches is:
Solution. It takes Sammy 67seconds to sew a single stitch. For 55
stitches it will take 55 � 67= 471
7seconds. Answer is (E)
154
10. How many positive integers less than or equal to 60 are divisibleby 3, 4, or 5?
Solution. There are 20 integers between 1 and 60which are divisible by3, 15 divisible by 4, and 12 divisible by 5. If we simply add up these numbersto get 20+15+12 = 47, we will have counted those integers divisible by atleast two of 3, 4, or 5more than once. Thus we must subtract from this totalthe number of integers between 1 and 60 divisible by both 3 and 4 (that is,divisible by 12), namely 5; the number divisible by both 3 and 5, namely 4;and the number divisible by both 4 and 5, namely 3. Thus our accumulatedtotal is now 47� (5 + 4 + 3) = 35. However, the integer 60, which is theonly positive integer in our range divisible by all of 3, 4, and 5 was at �rstcounted 3 times, and now is not counted at all. Thus we must add 1 backinto the total. So our total is 36.
Alternate method: Simply enumerate all the numbers between 1 and60 and test each one. Those that are divisible by 3, 4, or 5 are: 3, 4, 5, 6, 8,9, 10, 12, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 33, 35, 36, 39, 40, 42,44, 45, 48, 50, 51, 52, 54, 55, 56, 57 and 60. There are 36 numbers in thislist. Answer is (C)
Final Round 1997 | Part B
1. (a) The pages of a thick telephone directory are numbered from 1
to N . A total of 522 digits is required to print the pages. FindN .
Solution. From 1 to 9 inclusive there are 9 single digit numbers, whichuses up 9 digits. From 10 to 99 inclusive there are 90 2-digit numbers, whichuses up a further 180 digits. So far we have used up 189 digits before weget to 3-digit numbers leaving us with 522 � 189 = 333 digits still to beaccounted for. With this many digits we can make up 111 3-digit numbers.Starting at 100 and proceeding for 111 numbers brings us to page 210. ThusN = 210.
(b) There are 26 pages in the local newspaper. Suppose that you pull asheet out and drop it on the oor. One of the pages facing you is numbered19. What are the other page numbers on the sheet?
Solution. Since the newspaper has 26 pages, the outer sheet has pages1, 2, 25, and 26. The next outermost sheet has pages 3, 4, 23, and 24. Thenext sheet has 5, 6, 21, and 22. Finally the sheet we are interested in haspages 7, 8, 19, and 20. So the other page numbers are 7, 8, and 20.
2. Create expressions for the numbers 1; 2; 3; : : : ; 10 by using each ofthe digits 1, 9, 9, and 7. Note that the digits must appear separately; that is,numbers like 17 are not allowed. Only the basic operations +, �, �, � andbrackets (if necessary) may be used. Other mathematical symbols such asp
are not allowed. Every expression must include one 1, two 9's, and one 7, inany order.
155
Solution. There are several acceptable answers for this question. Hereare some:
1 = 7� (9� 9) + 1
2 = (9 + 7)� (9� 1)
3 = 9� (9 + 1� 7)
4 = (9� 1)� (9� 7)
5 = (9 + 1)� (9� 7) = 7� (9� 9)� 1
6 = 9� 9 + 7� 1
7 = 9� 9 + 7� 1 = (9� 9)� 1 + 7
8 = 9� 9 + 7 + 1
9 = 9� 9 + 7 + 1
10 = 9 + 9� 7� 1
3. (a) Decide which is greater:p6 +
p8 or
p5 +
p9.
Solution. Let x =p6 +
p8 and y =
p5 +
p9. Then
x2 = 6+ 8 + 2p48 = 14 + 2
p48
y2 = 5 + 9 + 2p45 = 14 + 2
p45
Sincep48 >
p45, we see that x2 > y2 and thus it follows that x > y since
both are positive real numbers. Therefore, the larger value isp6 +
p8.
(b) Show that x2+1
x� 2 for any real number x > 0.
Solution. Note that (x � 1)2 � 0 for all real numbers x. This can berewritten as x2 � 2x+ 1 � 0, and thus we have x2 + 1 � 2x. Since we aregiven x > 0 we can divide both sides of the inequality by x and preserve thedirection of the inequality to get
x2 + 1
x� 2:
4. In the plane �gure shown below, ABCD is a square with AB =
12. If A0, B0, C0, and D0 are the midpoints of AO, BO, CO, and DO,respectively, then:
D C
BA
A0
C0D0
B0
O
156
(a) Find the area of the square A0B0C0D0.
Solution. Triangles AOB and A0OB0 are similar since both are isosce-les right angled triangles. SinceA0O is one half the length of AO we see thatA0B0 is also one half the length of AB; that is, A0B0 is 6 units in length,which makes the square A0B0C0D0 have area 36 square units.
(b) Find the area of the shaded region.
Solution. Triangle BA0B0 has a base A0B0 of 6 units, and an altitudewhich is one half of the altitude of triangle AOB (from its base AB); that is,triangle BA0B0 has altitude 3 from its base A0B0. Thus the area of triangleBA0B0 is 1
2� 6 � 3 = 9 square units. Since the shaded area comprises 4 such
triangles, the total shaded area is 36 square units.
Alternate method for (b): Triangles A0OB and A0B0B have the samealtitude from the baseOB0B, and they have the same length for a base, sinceB0 is the midpoint of OB. Thus their areas must be the same. But the areaof triangle A0OB0 is clearly 1
4of the area of the square A0B0C0D0 computed
in part (a); that is, 9 square units. Thus, triangle BA0B0 has area 9 and we�nish as in the �rst method.
(c) Find the area of the trapezoid AA0B0B.
Solution. Since the parallel sides of the trapezoid have lengths12 and 6,and since its altitude is 3 (as seen in part (b) above), we see that the area ofthe trapezoid is:
1
23(6 + 12) = 27:
Alternate method for (c): The area we seek is 14of the di�erence be-
tween the areas of the squares ABCD and A0B0C0D0, which is:
1
4(122� 62) =
1
4(144� 36) = 27:
5. The �gure below shows the �rst three in a sequence of square arraysof dots. The numbers of dots in the three arrays are 1, 5, and 13.
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
r
(a) Find the number of dots in the next array in the sequence.
Solution. There are several ways of looking at the arrays. One way is tonotice that in going from one array to the next we build around the outsideof the array a new set of dots. That is, we add 4 dots to the single dot to get
157
array 2. We then add 4 � 2 = 8 dots around array 2 to get array 3. We wouldthen add 4 � 3 = 12 dots around array 3 to get array 4. This leaves us with13 + 12 = 25 dots in the next array.
(b) Find the number of dots in the sixth array in the sequence.
Solution. To carry on to array 6 we need �rst to get array 5 from array4. Using the process described in part (a) we would get 25 + 4 � 4 = 41 dotsin array 5 and 41 + 4 � 5 = 61 dots in array 6.
(c) Find an expression for an in terms of n alone.
Solution. By observing the dot pattern, the number of dots seems tobe in the order of n2. If we subtract n2 from an for the �rst few values of nwe see that what remains is (n�1)2. Since an = n2+(n�1)2 satis�es therelation from part (d) above, and since it coincides with the �rst few valuesof an, this must be the solution for an.
(d) If an is the number of dots in the nth array in the sequence, �nd arelation between an+1, an, and n.
Solution. From the above analysis, if we denote by anthe number ofdots in the nth array, then
an+1 = an + 4n:
That completes the Skoliad Corner for this number. I need contestmaterials suited to this feature, so please send me your materials, as well ascomments and suggestions for the future of the Skoliad Corner.
158
MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and byHigh School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.
All material intended for inclusion in this section should be sent to theMayhem Editor, Naoki Sato, Department of Mathematics, Yale University,PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronicaddress is still
The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).The rest of the sta� consists of Adrian Chan (Upper Canada College), JimmyChui (Earl Haig Secondary School), Richard Hoshino (University ofWaterloo),David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).
Mayhem Problems
The Mayhem Problems editors are:
Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.
Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only problems | thenext issue will feature only solutions.
We warmly welcome proposals for problems and solutions. We requestthat solutions from this issue be submitted by 1 February 1999, for publica-tion in issue 4 of 1999. Also, starting with this issue, we would like to re-openthe problems to all CRUX with MAYHEM readers, not just students, so nowall solutions will be considered for publication.
Erratum
We regret to report that the High School porblems in volume 24, issue 1[1998: 42], were mislabelled. Instead of H223, H224, H225 and H226,they shouldhave been labelledH233, H234, H235 andH236 respectively.We kindly ask readers to respect the new labellingwhen submitting solutions.
159
High School Problems
Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,Canada. L3S 3E8 <[email protected]>
H237. The letters of the word MATHEMATICAL are arranged at ran-dom. What is the probability that the resulting arrangement contains noadjacent A's?
H238. Johnny is dazed and confused. Starting atA(0; 0) in the Carte-sian grid, he moves 1 unit to the right, then r units up, r2 units left, r3 unitsdown, r4 units right, r5 units up, and continues the same pattern inde�-nitely. If r is a positive number less than 1, he will be approaching a pointB(x; y). Show that the length of the line segment AB is greater than 7
10.
H239. Find all pairs of integers (x; y) which satisfy the equationy2(x2 + 1) + x2(y2 + 16) = 448.
H240. Proposed by Alexandre Trichtchenko, Brook�eld High School,Ottawa, Ontario.
A Pythagorean triple (a; b; c) is a triple of integers satisfying the equa-tion a2 + b2 = c2. We say that such a triple is primitive if gcd(a; b; c) = 1.Let p be an odd integer with exactly n prime divisors. Show that there existexactly 2n�1 primitive Pythagorean triples where p is the �rst element ofthe triple. For example, if p = 15, then (15;8; 17) and (15;112; 113) are theprimitive Pythagorean triples with �rst element 15.
Advanced Problems
Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.M1G 1C3 <[email protected]>
A213. Show that the number of non-negative integer solutions to theequation a+ b+ c+ d = 98, where a � b � c � d, is equal to the numberof non-negative integer solutions to the equation p+ 2q + 3r+ 4s = 98.
A214. Show that any rational number can be written as the sum ofa �nite number of distinct unit fractions. A unit fraction is of the form 1
n,
where n is an integer.
A215. For a �xed integer n � 2, determine the maximum value ofk1+� � �+kn, where k1, : : : , kn are positive integers with k31+� � �+k
3n � 7n.
(Polish Mathematical Olympiad)
160
A216. Given a continuous function f : R! R satisfying the condi-tions:
f(1000) = 999;
f(x) � f(f(x)) = 1 for all x 2 R:
Determine f(500).(Polish Mathematical Olympiad)
Challenge Board Problems
Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <[email protected]>
Editorial Notes: Part (b) is the interesting part of C77. Part (a) is acommonly asked problem, but I think it is better to ask it again than simplyto state it for use in (b).
C77. Let Fn denote the nth Fibonacci number, with F0 = 1 andF1 = 1. (Then F2 = 2, F3 = 3, F4 = 5, etc.)
(a) Prove that each positive integer is uniquely expressible in the formFa1 + � � �+ Fak , where the subscripts form a strictly increasing sequence ofpositive integers, no pair of which are consecutive.
(b) Let � = 12(1 +
p5), and for any positive integer n, let f(n) equal
the integer nearest to �n. If n = Fa1 + � � � + Fak is the expression for nfrom part (a), prove that f(n) = Fa1+1 + � � �+ Fak+1.
C78. Let n be a positive integer. An n�nmatrix A is amagic matrixof order m if each entry is a non-negative integer and each row and columnsum ism. (That is, for all i and j,
PkAik =
PkAkj = m.)
Let A be a magic matrix of order m. Show that A can be expressed asthe sum of m magic matrices of order 1.
161
Tips on Inequalities
Naoki Satograduate student, Yale University
Inequalities can be di�cult to solve because there are few systematicmethods for tackling even the most simple formulations. Indeed, solvingusually involves a trial and error of di�erent approaches, before one hitsthe right combination of estimations and manipulations. In this article, weexpose some useful standard approaches and techniques. We recall two basicand fundamental inequalities:
AM-GM Inequality. For all x1, x2, : : : , xn � 0,
x1 + x2 + � � �+ xn
n� npx1x2 � � �xn;
with equality if and only if x1 = x2 = � � � = xn.
Cauchy-Schwarz Inequality (CSB). For all real xi, yi, i = 1, 2, : : : , n,
(x1y1 + x2y2 + � � �+ xnyn)2 � (x21 + x22 + � � �+ x2n)(y
21 + y22 + � � �+ y2n);
with equality if and only if the vectors (x1; x2; : : : ; xn) and (y1; y2; : : : ; yn)
are proportional.
Manipulating the ExpressionsA typical approach in proving an inequality of the form A � B is to
�nd intermediate expressions, so we have a chain
A � P1 � P2 � � � � � Pn � B:
These are usually found through using the classic inequalities, and by manip-ulating terms until we get what we want. As anyone who has worked withinequalities knows, this takes great care; one constantly has to make surethat the estimates are not too crude, and that the inequality signs are goingthe right way. In this kind of approach, there are several things one shouldkeep in mind:
1. Is the inequality sharp or strict?An inequality is sharp if equality occurs at a point, and strict if equal-
ity never occurs. It is always a good idea to check which type it is, thoughit is usually given or obvious. A strict inequality may allow for generousestimates, but not always. A sharp inequality leaves no such allowance.
2. If equality does occur, when/where does it occur?
Copyright c 1998 Canadian Mathematical Society
162
The points where equality occurs are points you must work around. Inthe chain above, each intermediate inequality must become an equality atthese points. This is a good check of whether your intermediate expressionsare the right ones.
Pairing and Grouping
In inequalities where several terms are involved, it might be possibleto group terms together and prove \smaller" inequalities.
Problem 1.
(a) Prove that x2 + y2 + z2 � xy + yz + zx for all x, y, z 2 R.(b) Prove that x3 + y3 + z3 � x2y + y2z + z2x for all x, y, z � 0.
Solution. (a) No grouping is immediately obvious. We know x2+y2 �2xy, but how can we incorporate this? By adding x2 + y2 � 2xy, y2+ z2 �2yz, and z2+x2 � 2zx, and dividing by 2, we obtain the desired inequality.
(b) Here, we know 2x3 + y3 = x3 + x3 + y3 � 3x2y by AM-GM. Wethen add the other two corresponding inequalities, and then divide by 3.
Problem 2. For all a, b, c, d > 0, show that
a3 + b3 + c3
a + b+ c+a3 + b3 + d3
a+ b+ d+a3 + c3 + d3
a+ c+ d+b3 + c3 + d3
b+ c+ d� a
2+b
2+c
2+d
2:
Solution. We claim that
a3 + b3 + c3
a+ b+ c�
a2 + b2 + c2
3:
Then, the problem follows by symmetry. Our inequality is equivalent to
3a3 + 3b3 + 3c3 � (a2 + b2 + c2)(a+ b+ c) ;
which, in turn, is equivalent to
2a3+ 2b
3+ 2c
3 � a2b� ab
2 � a2c� ac
2 � b2c� bc
2
= (a3 � a
2b� ab
2+ b
3) + (a
3 � a2c� ac
2+ c
3) + (b
3 � b2c� bc
2+ c
3)
= (a� b)2(a + b) + (a� c)
2(a+ c) + (b� c)
2(b+ c) � 0 ;
which is true. Note that this inequality also follows from Chebyshev's in-equality.
The Cauchy-Schwarz Inequality
The Cauchy-Schwarz inequality is a good way to deal with squares andespecially fractions.
Problem 3. For x1, x2, : : : , xn > 0, show that
x21
x1 + x2+
x22
x2 + x3+ � � �+
x2n
xn + x1�
x1 + x2 + � � �+ xn
2:
163
Solution. By CSB,
[(x1 + x2) + (x2 + x3) + � � �+ (xn + x1)]
��
x21
x1 + x2+
x22
x2 + x3+ � � �+
x2n
xn + x1
�� (x1 + x2 + � � �+ xn)
2 :
The result then follows by dividing each side by 2 (x1 + x2 + � � �+ xn).
Problem 4. Prove that for a1, a2, : : : , an > 0,
(a1 + a2 + � � �+ an)2
2(a21 + a22 + � � �+ a2n)�
a1
a2 + a3+
a2
a3 + a4+ � � �+
an
a1 + a2:
(1990{1991 IMO Correspondence)
Solution. Recall that CSB states that for all real xi, yi, i = 1, 2, : : : , n,
(x1y1 + x2y2 + � � �+ xnyn)2 �
�x2
1 + x2
2 + � � �+ x2
n
� �y2
1 + y2
2 + � � �+ y2
n
�:
Setting xi = 4pai, yi =
4pa3i , we obtain
(a1 + a2 + � � �+ an)2
��p
a1 +pa2 + � � �+
pan� �a1pa1 + a2
pa2 + � � �+ an
pan�:
Setting xi =pai, yi = ai, we obtain�
a1pa1 + a2
pa2 + � � �+ an
pan�2
� (a1 + a2 + � � �+ an)�a21 + a22 + � � �+ a2n
�:
Combining these two, we obtain
(a1 + a2 + � � �+ an)3
��p
a1 +pa2 + � � �+
pan�2 �
a21 + a22 + � � �+ a2n�:
Finally, setting xi =pai+1 + ai+2, yi =
qai
ai+1+ai+2, we obtain
�pa1 +
pa2 + � � �+
pan�2
� 2 (a1 + a2 + � � �+ an)
�a1
a2 + a3+
a2
a3 + a4+ � � �+
an
a1 + a2
�:
The last two inequalities give the desired result.
Elementary Symmetric Polynomials
Given a set of variablesX, such asX = fx; y; zg, a polynomial is sym-metric inX if it is invariant under any permutation of the variables underX,and homogeneous of degree k if every term in the polynomial has degree k.
164
For example, x2 + y2 + z2 � xyz is symmetric but not homogeneous in X,and x3 + x2y � xyz is homogeneous of degree 3 but not symmetric in X.
The elementary symmetric polynomials in X are the polynomials ob-tained as the sum of the products of the variables in X, taken k at a time.For X = fx; y; zg, these would be x+ y+ z, xy+xz+ yz, and xyz. Notethat each of these is homogeneous. A theorem of Gauss states that any sym-metric, homogeneous polynomial in X can be expressed as a polynomial inthese elementary polynomials.
After all these tedious de�nitions, we �nally get to the point that it canbe useful to use these elementary polynomials.
Problem 5. For non-negative reals x, y, and z satisfyingx+y+z = 1,show that �
1
x+ 1
��1
y+ 1
��1
z+ 1
�� 64 :
Proof. Expanding, we �rst must show
1 +1
x+
1
y+
1
z+
1
xy+
1
xz+
1
yz+
1
xyz� 64 :
By AM-GM,
xyz ��x+ y + z
3
�3=
1
27so that
1
xyz� 27 :
Therefore,
1 +1
x+
1
y+
1
z+
1
xy+
1
xz+
1
yz+
1
xyz
� 1 +3
3pxyz
+3
3px2y2z2
+1
xyz
=
�1 +
1
3pxyz
�3� 43 = 64 :
Problem 6. Prove that
0 � yz + zx+ xy � 2xyz �7
27;
where x, y, and z are non-negative real numbers for which x+ y + z = 1.(1984 IMO, #1)
Proof. Let S = xy+ xz+ yz� 2xyz, P = (1� 2x)(1� 2y)(1� 2z).Then
P = 1� 2(x+ y+ z) + 4(xy+ xz + yz)� 8xyz = 4S � 1 :
165
We must show that 0 � S � 727, or equivalently, �1 � P � 1
27. Since
0 � x; y; z � 1, �1 � 2x � 1; 2y � 1; 2z � 1 � 1, so �1 � P . Now, ifone of the variables, say x, was greater than 1=2, then the other two wouldbe less than 1=2, and we would have 1� 2x < 0, 1 � 2y, 1 � 2z > 0, andP < 0 < 1
27. Otherwise, x, y, and z are at most 1=2, and all factors of P
are non-negative, so by AM-GM,
P ��1� 2x+ 1� 2y + 1� 2z
3
�3=
1
27:
Introducing and Removing Constraints
Inequalities often come with constraints on the variables. Removingthese constraints can simplify the problem. Alternately, introducing themmay help as well. The most common way of removing a constraint is to\homogenize" the given inequality. For example, suppose we are given theexpression x3 + xy � 2, where xyz = 1. Then
x3 + xy � 2 = x3 + xy 3pxyz � 2xyz:
This new expression is homogeneous of degree 3. It's not pretty, but forthe expressions given in problems, most of the time it will be nice and mucheasier to work with.
At this point, we introduce an inequality that is not well-known, butthat seems to pop up from time to time.
Schur's Inequality. For all x, y, z � 0 and non-negative integers n,
xn(x� y)(x� z) + yn(y� z)(y� x) + zn(z � x)(z � y) � 0 :
For n = 1, this becomes
x3 + y3 + z3 + 3xyz � x2y + xy2 + x2z + xz2 + y2z + yz2 ;
or in shorthand,Px3 +3xyz �
Px2y. This is a useful inequality to know,
with which we present another solution.
Problem 6. Prove that
0 � yz + zx+ xy � 2xyz �7
27;
where x, y, z are non-negative real numbers for which x+ y + z = 1.(1984 IMO, #1)
Solution. This is equivalent to the following problem: For x, y, z � 0,prove that
0 � (yz+ zx+ xy)(x+ y+ z)� 2xyz �7
27(x+ y + z)3 :
166
This is the homogeneous version of the original inequality. The expressionin the middle expands to
Px2y + xyz, which is clearly non-negative. We
focus on the right inequality, which becomes
Px2y+ xyz �
7
27
Px3 +
7
9
Px2y +
14
9
Pxyz ;
which implies6Px2y � 7
Px3 + 15xyz :
This is where we must think backwards. What results do we know thatwe can use to prove this? By Schur's, 5
Px2y � 5
Px3 + 15xyz (we try
to eliminate the xyz term). Hence, to prove the above inequality, we mustshow that
Px2y � 2
Px3, which is left as an exercise for the reader (it has
been virtually done already in this article).
We stated early in the solution that the modi�ed problem was equiva-lent to the original problem. It is easy to see that the modi�ed problem im-plies the original problem (which is the only direction we actually needed),but what about the converse? What if x+ y+ z 6= 1? In such a case, we cannormalize.
A property of homogeneous polynomials, and an alternate de�nition,is the following: p(x1; x2; : : : ; xn) is homogeneous of degree k if
p(�x1; �x2; : : : ; �xn) = �kp(x1; x2; : : : ; xn)
for all � 2 R.Going back to the original problem,
p(x; y; z) = (yz+ zx+ xy)(x+ y+ z)� 2xyz :
If x + y + z = 0, then all three variables must be 0, and the inequalityfollows. Otherwise, we can set � = 1
x+y+z, and so we must show
0 � p
�x
x+ y + z;
y
x+ y+ z;
z
x+ y + z
��
7
27:
This is the original problem. Thus, we can set x + y + z equal to 1, orindeed anything we want to (except 0). It can be vital to exploit this degreeof freedom. We perform a similar setting in the next problem.
Problem 7. Given 0 < a � b, and x1, x2, : : : , xn � 0, show that
(xa1 + xa2 + � � �+ xan)1=a � (xb1 + xb2 + � � �+ xbn)
1=b:
Solution. If xb1 + xb2 + � � � + xbn = 0, then the problem is solved.Otherwise, by the arguments above, we can assume xb1+xb2+ � � �+xbn = 1.Then
xbi � 1 =) xi � 1 =) xb�ai � 1 =) xbi � xai
=) xa1 + xa2 + � � �+ xan � xb1 + xb2 + � � �+ xbn = 1
=) (xa1 + xa2 + � � �+ xan)1=a � 1 = (xb1 + xb2 + � � �+ xbn)
1=b:
167
Problems
1. Prove that if x, y, and z are non-negative real numbers such thatx+ y+ z = 1, then
2(x2 + y2 + z2) + 9xyz � 1 :
2. For any real numbers a, b, and c, show that
min�(a� b)2; (b� c)2; (c� a)2
��
a2 + b2 + c2
2:
3. Let a, b, and c be the sides of a triangle with perimeter 2. Prove thata2 + b2 + c2 + 2abc < 2.
4. For non-negative reals x, y, and z satisfying 2xyz+xy+xz+yz = 1,prove that x+ y + z � 3
2.
5. For all positive integers n, show that
1
2�3
4�5
6� � �
2n� 1
2n�
1p3n+ 1
:
(Hint: The inequality is almost certainly not sharp, so there is someroom for approximation. The RHS suggests squaring.)
6. Show that if x, y, and z are non-negative reals such that x+y+z = 1,then �
1
x� 1
��1
y� 1
��1
z� 1
�� 8 :
(Note: The solution in Problem 5 does not work!)
7. Given a, b, c, d, e > 0, abcde = 1, show that
a4 + b4 + c4 + d4 + e4 � a+ b+ c+ d+ e :
(A favourite of Ravi Vakil's. Find as many di�erent solutions as youcan.)
8. Show that if non-negative reals a, b, and c satisfy
1
1 + a+
1
1 + b+
1
1 + c= 1 ;
then abc � 8.
168
Riveting Properties of Pascal's Triangle
Richard Hoshinostudent, University of Waterloo
Consider the following table of integers, known as Pascal's Triangle:
11 1
1 2 11 3 3 1
1 4 6 4 11 5 10 10 5 1
.
.
.
This table is named after the French mathematician Blaise Pascal, whoconceived of it at the age of thirteen (although the Chinese discovered someproperties of this triangle in the early 14th century, centuries before Pascalwas even born!).
To generate a row of Pascal's Triangle, look at the row immediatelyabove it. Each element of the triangle is the sum of the two elements aboveit (for example, the second element of the fourth row is 4 because 4 = 1+3).By convention, we denote the top row as the 0th row, and we denote the leftmost entry of each row as the 0th entry, even though it may seem a littleawkward at �rst.
Let us show that this table is the same as the following table:�0
0
��1
0
� �1
1
��2
0
� �2
1
� �2
2
��3
0
� �3
1
� �3
2
� �3
3
��4
0
� �4
1
� �4
2
� �4
3
� �4
4
��5
0
� �5
1
� �5
2
� �5
3
� �5
4
� �5
5
�...
In this table,�n
k
�is determined by the formula n!
k!(n�k)!. Thus, we will
show that the kth element of the nth row of Pascal's Triangle equals�n
k
�for
all integers n and k. Each of these elements in the triangle, namely�0
0
�,�1
0
�,�
1
1
�,�2
0
�,�2
1
�, : : : , is called a binomial coe�cient, since the coe�cients of the
expansion of the binomial (1 + x)n correspond to the entries of the nth row
Copyright c 1998 Canadian Mathematical Society
169
of Pascal's Triangle (for example, (1 + x)4 = x4 +4x3 +6x2 + 4x+1). Letus prove these statements, as they will be fundamental to our analysis of theproperties of Pascal's Triangle.
Now�n
k
�denotes the number of ways we may select k objects from a
set of n objects. By convention, we let�n
0
�= 1, since there is technically
only \one" way we may select nothing from a set of n objects. Let us showthat �
n
k
�+
�n
k� 1
�=
�n+ 1
k
�
for all n and k. Note that the right side denotes the number of ways we mayselect a k-member committee from a class of n girls and one boy (we choosekmembers from a set of (n+1)people). Now, if the boy is on the committee,then we have
� n
k�1�ways of selecting the remaining k�1members. And if he
is not on the committee, then there are�n
k
�ways of selecting the committee.
Hence, �n
k
�+
�n
k � 1
�=
�n+ 1
k
�:
This is known as Pascal's Identity.
Since�0
0
�= 1, the table of binomial coe�cients corresponds directly
to Pascal's Triangle { since the initial element is the same and, like Pascal'sTriangle, each element is the sum of the two directly above it. Thus, we candetermine any element in Pascal's Triangle with this formula. For example,
the thirty-�fth element in the seventy-ninth row of Pascal's Triangle is�79
35
�.
We now prove that the entries in the nth row of Pascal's Triangle arethe coe�cients in the expansion of (1 + x)n. We proceed by induction. Thecase is trivial for n = 1. Suppose that
(1 + x)n =
�n
0
�+
�n
1
�x+
�n
2
�x2 + � � �+
�n
k
�xk + � � �+
�n
n
�xn
for some n. Then
(1 + x)n+1 = (1 + x)n(1 + x)
=
��n
0
�+
�n
1
�x+
�n
2
�x2 + � � �+
�n
n
�xn�(1 + x)
=
�n
0
�+
��n
1
�+
�n
0
��x+ � � �+
��n
n
�+
�n
n� 1
��xn +
�n
n
�xn+1:
Since�n
0
�=�n+1
0
�=�n
n
�=�n+1
n+1
�= 1, and using Pascal's Identity for
all the other terms, we immediately arrive at the case for n+ 1. Hence, thecoe�cient of xk in (1 + x)n is equal to
�nk
�.
Now we illustrate some of the really neat properties of Pascal'sTriangle.
170
Theorem 1.
(i) The sum of the coe�cients in the nth row of Pascal's Triangle is 2n.
(ii) If we alternately add and subtract the digits in the nth row of Pascal'sTriangle, we always arrive at zero. For example, 1� 4+ 6� 4+ 1 = 0
for n = 4.
Proof. (i) Since
(1 + x)n =
�n
0
�+
�n
1
�x+ � � �+
�n
k
�xk + � � �+
�n
n
�xn;
substituting x = 1 into this expression yields
2n =
�n
0
�+
�n
1
�+ � � �+
�n
n
�:
(ii) The second part of the theorem is left as an exercise. It is the sametechnique as above; just substitute a di�erent value for x. Can you guesswhich value of x we ought to substitute?
Theorem 2. If the binary representation ofn contains p ones, then thereare 2p odd numbers in the nth row of Pascal's Triangle. For example, since9 = 10012, there are 22 = 4 odd numbers in the ninth row.
Proof. We shall analyze everything in modulo 2. For those of you notfamiliar with modular arithmetic, when we say that x is congruent to y mod-ulom, which is written x � y (mod m), we mean that x and y are numberssuch that when both are divided by m, they give the same remainder. Forexample, 1998 is congruent to 2 modulo 4. Also, if r � 0(mod 2), then r iseven.
We �rst show that
(1 + x)2n
� 1 + x2n
(mod 2)
for all non-negative integers n. We proceed by induction. If n = 0, the claimis immediate. Suppose the claim is true for some n = k. Then
(1+x)2k+1
� ((1+x)2k
)2 � (1+x2k
)2 � 1+2x2k
+x2k+1
� 1+x2k+1
(mod 2);
and so it is also true for n = k+1. Hence, by induction, the claim has beenveri�ed.
Here is an example for a speci�c case. Let n = 50. Then 50 = 21 +
24 + 25 = 2+ 16 + 32. Hence,
(1 + x)50 = (1 + x)2(1 + x)16(1 + x)32
� (1 + x2)(1 + x16)(1 + x32)
� 1 + x2 + x16 + x18 + x32 + x34 + x48 + x50(mod 2):
171
Hence, there are eight odd entries in the 50th row of Pascal's Triangle,
namely�50
0
�,�50
2
�,�50
16
�,�50
18
�,�50
32
�,�50
34
�,�50
48
�, and
�50
50
�. Since 50 = 110102,
this veri�es that there are exactly 23 = 8 odd terms in this row.
For a general n, suppose there are p ones in the binary representationof n. Then
(1 + x)n = (1 + x)2a1(1 + x)2
a2 � � � (1 + x)2ap
;
where 0 � a1 < a2 < � � � < ap and the athi digit of the binary representation
of n is 1, starting from 0 at the right. The right side is congruent to
(1 + x2a1)(1 + x2
a2) � � � (1 + x2
ap
)
modulo 2, and when we expand the right side, we will arrive at an expressionwith 2p terms. Note that there must be exactly 2p terms, since each exponentxk can be formed in only one way by multiplying coe�cients from the setfx2
a1; x2
a2; : : : ; x2
apg. This is a direct result from the binary representationof k. Hence, if n has p ones in its binary representation, (1 + x)n modulo 2
has 2p terms, and hence there are 2p odd entries in the nth row of Pascal'sTriangle.
Theorem 3. (The Fibonacci sequence is the sequence, 1, 1, 2, 3, 5, 8,13, 21, 34, : : : , where each element in the sequence is the sum of the twobefore it. More formally, we say that the Fibonacci sequence fFng satis�esthe conditions F0 = 1, F1 = 1, and Fn = Fn�1 + Fn�2 for all n > 1.)We can derive the Fibonacci sequence from Pascal's Triangle in the followingmanner.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
.
.
.
�����:
�����:
-F5 = 8
�����:
�����:
-F4 = 5
�����:
-F3 = 3�����:
-F2 = 2
-F1 = 1
-F0 = 1
In other words, for all n, we have
F2n =
�2n
0
�+
�2n� 1
1
�+
�2n� 2
2
�+ � � �+
�n
n
�
and
F2n+1 =
�2n+ 1
0
�+
�2n
1
�+
�2n� 1
2
�+ � � �+
�n+ 1
n
�:
172
For example,
F7 =
�7
0
�+
�6
1
�+
�5
2
�+
�4
3
�= 1+ 6 + 10 + 4 = 21:
Proof. We prove this theorem using double induction: we know thatthe claim is true for F0 and F1. Suppose the claim is true for some F2k andF2k+1. Then we want to show that
F2k+2 =
�2k+ 2
0
�+
�2k+ 1
1
�+
�2k
2
�+ � � �+
�k + 1
k + 1
�;
that is, the right side is equal to the (2k+ 2)th Fibonacci number. But
F2k+2 = F2k + F2k+1
=
�2k+ 1
0
�+
��2k
0
�+
�2k
1
��+ � � �+
��k+ 1
k� 1
�+
�k + 1
k
��+
�k
k
�
=
�2k+ 1
0
�+
�2k+ 1
1
�+
�2k
2
�+ � � �+
�k + 2
k
�+
�k
k
�;
by Pascal's Identity. Since�2k+1
0
�= 1 =
�2k+2
0
�and
�k
k
�= 1 =
�k+1
k+1
�, we
have
F2k+2 =
�2k+ 2
0
�+
�2k+ 1
1
�+
�2k
2
�+ � � �+
�k + 1
k + 1
�;
as required.
The second part of the induction follows similarly { assume the claim istrue for F2k�1 and F2k, and show that the claim is also true for F2k+1. Theproof is left as an exercise. Now we are done { since the proposition is truefor k = 0 and k = 1, it is true for k = 2. Since it is true for k = 1 and k = 2,it is true for k = 3, etc. Then by double induction, the claim is true for allnon-negative integers n.
Exercises
1. Show that�n
k
�=�n
n�k�, and use this fact to show that Pascal's Triangle
is symmetric about the vertical line that separates the triangle into twoequal halves.
2. For which n is�2n
n
�odd?
3. Let an represent the number of elements in the nth row of Pascal'sTriangle that are congruent to 1modulo 3. Let bn represent the numberof elements in the nth row of Pascal's Triangle that are congruent to 2
modulo 3. Prove that for all n, an � bn is a power of two.
(This problem was on the IMO short list one year { it is very tough!)
173
Swedish Mathematics Olympiad
1985 Qualifying Round
1. The real numbers a, b, and c satisfy the equations
ab+ b = �1bc+ c = �1ca+ a = �1
Calculate the product abc.
2. 1985 runners reported for a marathon. They were assigned the num-bers 1, 2, : : : , 1985. However, a number of runners dropped out be-fore the race. In fact, among the starting runners, there were no twofor whom one runner's number was 10 times the other. What is thegreatest number of runners that could have participated?
3. In the system of equations
a1x+ b1y + c1z + d1u = 0
a2x+ b2y + c2z + d2u = 0
a3x+ b3y + c3z + d3u = 0
a4x+ b4y + c4z + d4u = 0
the coe�cients a1, b2, c3, and d4 are even integers and the other co-e�cients are odd integers. Prove that the only solution in integers isx = y = z = u = 0.
4. The non-negative integers p, q, r, and s satisfy the equality
(p+ q)2 + p = (r+ s)2 + r:
Show that p = r and q = s.
5. Let f be de�ned by
f(x) =4x2 sin2 x+ 9
x sinx:
Find the least value of f over the interval 0 < x < �.
6. The point P lies on the perimeter or inside a given triangle T . The pointP 0, in the plane of the triangle, lies at a distance d from P . Let r and r0
be the radii of the smallest circles, with centres P and P 0 respectively,which contain T . Show that
r + d � 3r0:
Give an example where equality holds.
174
1985 Final Round
1. Let a > b > 0. Prove that
(a� b)2
8a<a+ b
2�pab <
(a� b)2
8b:
2. Find the least natural number such that if the �rst digit is placed last,the new number is 7=2 times as large as the original number. (Thenumbers are written in the decimal system.)
3. A, B, and C are three points on a circle with radius r, and AB = BC.D is a point inside the circle such that the triangle BCD is equilateral.The line through A and D meets the circle at the point E. Show thatDE = r.
4. The polynomial p(x) of degree n has real coe�cients, and p(x) � 0 forall x. Show that
p(x) + p0(x) + p00(x) + � � �+ p(n)(x) � 0:
5. In a right-angled coordinate system, a triangle has vertices A(a; 0),B(0; b), and C(c; d), where the numbers a, b, c, and d are positive.Show that if we denote the origin by O,
AB + BC + CA � 2CO:
6. X-wich has a vibrant club-life. For every pair of inhabitants there isexactly one club to which they both belong. For every pair of clubs thereis exactly one person who is a member of both. No club has fewer than3 members. At least one club has 17 members. How many people livein X-wich?
175
PROBLEMSProblem proposals and solutions should be sent to Bruce Shawyer, De-
partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.
In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.
To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81
2"�11" or A4 sheets of paper.
These may be typewritten or neatly hand-written, and should be mailed tothe Editor-in-Chief, to arrive no later than 1 November 1998. They may alsobe sent by email to [email protected]. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication. Please note that we do not accept submissions sent by FAX.
2306. Proposed by Vedula N. Murty, Visakhapatnam, India.CORRECTION to (a) Give an elementary proof of the inequality:�
sin
��x
2
��2>
2x2
1 + x2; (0 < x < 1): (1)
2326?. Proposed by Walther Janous, Ursulinengymnasium, Inns-
bruck, Austria.Prove or disprove that if A, B and C are the angles of a triangle, then
2
�<
Xcyclic
�1� sin A
2
� �1 + 2 sin A
2
�� � A
�9
�:
2327. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.
The sequence fang is de�ned by a1 = 1, a2 = 2, a3 = 3, and
an+1 = an � an�1 +a2n
an�2; n � 3 :
Prove that each an 2 N, and that no an is divisible by 4.
176
2328?. Proposed by Walther Janous, Ursulinengymnasium, Inns-
bruck, Austria.It is known from Wilson's Theorem that the sequence fyn : n � 0g,
with yn =n! + 1
n+ 1, contains in�nitely many integers; namely, yn 2 N if and
only if n+ 1 is prime.
(a) Determine all integer members of the sequences fyn(a) : n � 0g, with
yn =n! + a
n+ a, in the cases a = 2, 3, 4.
(b) Determine all integer members of the sequences fyn(a) : n � 0g, with
yn =n! + a
n+ a, in the cases a � 5.
2329?. Proposed by Walther Janous, Ursulinengymnasium, Inns-
bruck, Austria.Suppose that p and t > 0 are real numbers. De�ne
�p(t) := tp + t�p + pp and �p(t) :=�t+ t�1
�p+ 2 :
(a) Show that �p(t) � �p(t) for p � 2.
(b) Determine the sets of p: A, B and C, such that
1. �p(t) � �p(t),
2. �p(t) = �p(t),
3. �p(t) � �p(t).
2330. Proposed by Florian Herzig, student, Perchtoldsdorf, Austria.Prove that
e = 3�1!
1 � 3+
2!
3 � 11�
3!
11 � 53+
4!
53 � 309�
5!
309 � 2119+ : : : ;
where11 = 3 � 3 + 2 � 1;53 = 4 � 11 + 3 � 3;
309 = 5 � 53 + 4 � 11;2119 = 6 � 309 + 5 � 53;
..
.
[Ed: There is enough information here to deduce the general term.]
2331. Proposed by Paul Yiu, Florida Atlantic University, Boca Ra-ton, Florida, USA.
Let p be an odd prime. Show that there is at most one non-degenerateinteger triangle with perimeter 4p and integer area. Characterize those primesfor which such triangles exist.
177
2332. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.Suppose x and y are integers. Solve the equation
x2y2 � 7x2y+ 12x2 � 21xy � 4y2 + 63x+ 70y � 174 = 0:
2333. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.You are given that D and E are points on the sides AC and AB re-
spectively of4ABC. Also,DE is not parallel to CB. Suppose F andG arepoints of BC and ED respectively such that
BF : FC = EG : GD = BE : CD:
Show that GF is parallel to the angle bisector of \BAC.
2334. Proposed by Toshio Seimiya, Kawasaki, Japan.Suppose that ABC is a triangle with incentre I, and that BI, CI meet
AC, AB atD, E respectively. Suppose that P is the intersection ofAI withDE. Suppose that PD = PI.
Find angle ACB.
2335. Proposed by Toshio Seimiya, Kawasaki, Japan.Triangle ABC has circumcircle �. A circle �0 is internally tangent to �
at P , and touches sidesAB, AC at D, E respectively. Let X, Y be the feetof the perpendiculars from P to BC, DE respectively.
Prove that PX = PY sin A2.
2336. Proposed by Toshio Seimiya, Kawasaki, Japan.The bisector of angleA of a triangleABC meetsBC atD. Let � and �0
be the circumcircles of triangles ABD and ACD respectively, and let P , Qbe the intersections of AD with the common tangents to �, �0 respectively.
Prove that PQ2 = AB � AC.
2337. Proposed by Iliya Bluskov, Simon Fraser University, Burnaby,BC.
Let F (1) =
�n2 + 2n+ 2
n2 + n+ 1
�, and, for each i > 1, let
F (i) =
�n2 + 2n+ i+ 1
n2 + n+ iF (i� 1)
�.
Find F (n).
Some readers have pointed out that problem 2287 [1997: 501] is thesame as problem 2234 [1997: 168], and that problem 2288 [1997: 501] isthe same as problem 2251 [1997: 299]. Also part (a) of problem 2306 [1998:46; 175] is the same as the �rst part of 2296 [1997; 503]. The editors missedthese duplications. Proposers are asked not to submit the same problemmore than once.
178
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.
2219. [1997: 110] Proposed by Christopher J. Bradley, Clifton Col-lege, Bristol, UK.
Show that there are an in�nite number of solutions of the simultaneousequations:
x2 � 1 = (u+ 1)(v� 1)
y2� 1 = (u� 1)(v+ 1)
with x; y; u; v positive integers and x 6= y.
I. Solution by Charles Ashbacher, Cedar Rapids, Iowa, USA; EdwardJ. Barbeau, University of Toronto, Toronto, Ontario; Charles R. Diminnie,Angelo State University, San Angelo, TX, USA; Florian Herzig, student, Per-chtoldsdorf, Austria; Cyrus Hsia, student, University of Toronto, Toron-to, Ontario; Walther Janous, Ursulinengymnasium, Innsbruck, Austria; andV�aclav Kone �cn �y, Ferris State University, Big Rapids, Michigan, USA.
For positive integers n, the quadruples
(x; y; u; v) = (1;2n+ 1; 2n2 + 2n+ 1; 1)
give an in�nite set of solutions in which x 6= y, since
x2 � 1 = (u+ 1)(v� 1) = 0
andy2� 1 = 4n2 + 4n = (u� 1)(v+ 1):
II. Solution by Edward J. Barbeau, University of Toronto, Toronto, On-tario; Digby Smith, Mount Royal College, Calgary, Alberta; and the pro-poser.
For any positive integer n, the quadruple
(x; y; u; v) = (2n2� n; 2n2 + n; 4n3 + n; n)
is a solution in which x 6= y, since
x2 � 1 = (u+ 1)(v� 1) = 4n4 � 4n3 + n2 � 1
andy2� 1 = (u� 1)(v+ 1) = 4n4 + 4n3 + n2 � 1:
III. Solution by Charles R. Diminnie, Angelo State University, San An-gelo, TX, USA; and Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-sity, Waterloo, Ontario.
179
It is well known that the Pell equation s2�2t2 = 1 has in�nitely manysolutions in positive integers s, t. Clearly, s > t > 1.
If we set u = s+ t, v = s� t, x = t� 1 and y = t+ 1, then
(u+ 1)(v� 1) = s2 � (t+ 1)2 = t2 � 2t = (t� 1)2� 1 = x2 � 1, and
(u� 1)(v+ 1) = s2 � (t� 1)2 = t2 + 2t = (t+ 1)2 � 1 = y2� 1.
Also solved by MANSUR BOASE, student, St. Paul's School, London,England; FLORIAN HERZIG, student, Perchtoldsdorf, Austria (a secondsolution); RICHARD I. HESS, Rancho Palos Verdes, California, USA;MICHAEL LAMBROU, University of Crete, Crete, Greece (two solutions);GERRY LEVERSHA, St Paul's School, London, England; and PANOS E.TSAOUSSOGLOU, Athens, Greece.
The families given by I, II and III above, do not exhaust all the possi-ble solutions. It is interesting to note that the \smallest" solution producedby all these families is (1; 3; 5; 1). The next smallest ones are (1; 5; 13; 1),(6; 10; 34; 2) and (11;13; 29; 5), respectively. Both Herzig and Leversha ob-tained another in�nite set of solutions in which v = 2, by considering thePell equation 3x2 � y2 = 8. Their \smallest" solution is (6; 10; 34; 2) listedabove. However, the next solution, (22; 38; 482; 2) is not obtainable fromany of the families given in I, II and III.
2220. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Madrid, Spain.
Let V be the set of an icosahedron's twelve vertices, which can be par-titioned into four classes of three vertices, each one in such a way that thethree selected vertices of each class belong to the same face.
How many ways can this be done?
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.I will prove that there are 10 di�erent ways of partitioning the vertices
of an icosahedron in the described manner. Since only the topological prop-erties of the icosahedron are important, consider the graph of the vertices:see �gure on next page.
We want to �nd four triangles in this graph such that each vertex isused exactly once. Consider the vertex A. There are �ve triangles havingA as vertex. Because of (spatial) symmetry we may assume that triangleABM is chosen. Thus for the triangle containing point C only two choicesremain: 4CFL and4CFK. Without loss of generality we may assume thattriangle CFL is chosen. Now notice that the only \free" triangle containingD is4DZY and �nally4EKX remains.
We have covered all possibilities already. For this case there are twodi�erent ways of �nding \disjoint" triangles because we may choose fromtwo equivalent triangles for vertex C. If all �ve possible triangles at vertex
180
X
YZ
K
F
LM
E
D
A
B C
r
rr
r
r
r
r
r
r
r
r r
A are considered, we get a total of 10 di�erent con�gurations as claimed.
Also solved by MANSUR BOASE, student, St. Paul's School, London,England; JORDI DOU, Barcelona, Spain; RICHARD I. HESS, Rancho PalosVerdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece;and the proposer. There were two incorrect solutions.
Several solvers noted that the centres of the faces used in the partitionare the vertices of a tetrahedron.
2221. [1997: 111] Proposed by �Sefket Arslanagi�c, University of Sara-jevo, Sarajevo, Bosnia and Herzegovina.
Find all members of the sequence an = 32n�1 + 2n�1 (n 2 N) whichare the squares of any positive integer.
Solution by David Doster, Choate Rosemary Hall, Wallingford, Con-necticut, USA.
We have a1 = 4 and a2 = 29. For n � 3, 32n�1 � 3(mod 4) and2n�1 � 0(mod 4). Thus, an � 3(mod 4). But a positive integer is asquare only if it is congruent to 0 or 1(mod 4). Hence, a1 = 4 is the onlysquare in the sequence.
Also solved by SAM BAETHGE, Nordheim, Texas, USA; FRANCISCOBELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain and MAR�IAASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain;MANSUR BOASE, student, St. Paul's School, London, England;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGELCABEZ �ON OCHOA, Logro ~no, Spain; ADRIAN CHAN, student, Upper CanadaCollege, Toronto, Ontario; GORAN CONAR, student, Gymnasium Vara�zdin,Vara�zdin, Croatia; YEO KENG HEE, Hwa Chong Junior College, Singapore;
181
FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,Rancho Palos Verdes, California, USA; CYRUS HSIA, student, University ofToronto, Toronto, Ontario; ROBERT B. ISRAEL, University of BritishColumbia, Vancouver, BC; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece;KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St Paul's School, London,England; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; BOBPRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; ISTV �ANREIMAN, Budapest, Hungary; HEINZ-J�URGEN SEIFFERT, Berlin, Germany;D.J. SMEENK, Zaltbommel, the Netherlands; DAVID R. STONE, GeorgiaSouthern University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU,Athens, Greece; ZUN SHAN and EDWARD T.H. WANG, Wilfrid Laurier Uni-versity, Waterloo, Ontario; and the proposer.
2222. [1997: 111] Proposed by Shawn Godin, St. Joseph ScollardHall, North Bay, Ontario.
Find the value of the continued root:vuut4 + 27
s4 + 29
r4 + 31
q4 + 33
p� � � :
NOTE: This was inspired by the problems in chapter 26 \Ramanujan, In�nityand the Majesty of the Quattuordecillion", pp. 193{195, in \Keys to In�nity"by Cli�ord A. Pickover, John Wiley and Sons, 1995.
I. Solution by Robert B. Israel, University of British Columbia, Van-couver, BC.
The answer is 29. More generally, for any positive integer n, we claimthat s
4 + n
r4 + (n+ 2)
q4 + (n+ 4)
p� � � = n+ 2 ;
where the left side is de�ned as the limit of
F (n;m) =
vuuut4 + n
vuut4 + (n+ 2)
s4 + (n+ 4)
r� � �q4 +m
p4
as m!1 (where m is an integer and (m� n) is even).
If g(n;m) = F (n;m)� (n+ 2), we have
F (n;m)2 � (n+ 2)2 = (4 + nF (n+ 2;m))� (4 + n(n+ 4))
= n(F (n+ 2;m)� (n+ 4)) ;
sog(n;m) =
n
F (n;m) + n+ 2g(n+ 2;m) :
182
Clearly F (n;m) > 2, so
jg(n;m)j <n
n+ 4jg(n+ 2;m)j :
By iterating this, we obtain
jg(n;m)j <n(n+ 2)
m(m+ 2)jg(m;m)j <
n(n+ 2)
m:
Therefore g(n;m)! 0 asm!1.
II. Solution by Efstratios Rappos, Girton College, University of Cam-bridge, England
Let
Sn =
s4 + (2n� 1)
r4 + (2n+ 1)
q4 + (2n+ 3)
p� � �
Sn satis�es the recurrence relation
Sn =
q4 + (2n� 1)Sn+1
if and only if(Sn � 2)(Sn+ 2) = (2n� 1)Sn+1 :
By inspection, this admits Sn = 2n+1 as a solution. We only have to provethat S1 = 3 to make this induction complete. Let
Tn =
vuut4 +
s4 + 3
r� � � (2n� 3)
q4 + (2n� 1)
p2n+ 3)
and
Un =
s4 +
r4 + 3
q� � � (2n� 3)
p4 + (2n� 1)(2n+ 3) = 3 :
Clearly Tn � Un and the latter is identically equal to 3. Therefore, using the
fact that B � A > 0 implies thatp(4 +A)=(4 + B) �
pA=B,
1 �Tn
3=
Tn
Un=
r4 +
q� � �+ (2n� 1)
p2n+ 3q
4 +p� � � (2n� 1))(2n+ 3)
�
rq� � �+ (2n� 1)
p2n+ 3qp
� � �+ (2n� 1)(2n+ 3)
� � � � � 2n+1
s1
2n+ 3
=1
(2n+ 3)(12 )n+1
�! 1
183
as n!1 [for example, by rewriting as expf� ln(2n+3)=2n+1g and usingL'Hopital's rule]. This proves that S1 = limn!1 Tn = 3. The requiredexpression is precisely S14 and hence its value is 29.
Also solved or answered byMANSUR BOASE, student, St. Paul's School,London, England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA;RICHARD I.HESS, Rancho PalosVerdes, California, USA;WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris StateUniversity, Big Rapids, Michigan, USA; MICHAEL LAMBROU, Universityof Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, StPaul's School, London, England; J.A. MCCALLUM, Medicine Hat, Alberta;HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the proposer.
Several readers made reference to the July 1996 issue of the Mathe-matical Gazette, which contained a similar problem, Problem 80E posed byTony Ward: evaluate: s
1 + 1
r1 + 2
q1 + 3
p� � � :
Bradley notes that the solution appeared in the March 1997 issue where thevalue of the continued root was proved to be 2, and to be \well-de�ned".He also adds that the editor of the Problems section of the Gazette con-cludes \Clearly, the problem raises some deep questions about the meaningof `well-de�ned'." Along the same lines as this editorial comment, MurrayS. Klamkin, University of Alberta, Edmonton, Alberta adds that \the valuecan be anything since there is no de�nition regarding the continuationof theroot". Klamkin refers to A. Herschfeld, On In�nite Radicals, Amer. Math.Monthly, 42 (1935) 419-429 where Herschfeld notes that Ramanujan's so-lution for r
1 + 2
q1 + 3
p1 + � � � = 3
is incomplete since one may write similarly that
4 =p1 + 2 � (15=2) =
q1 + 2
p1 + 3 � (221=12)
=
r1 + 2
q1 + 3
p1 + � � � :
Despite these comments most solvers expressed no di�culty in understand-ing the meaning of the continued root, and for that reason we have decidedto print the above \solutions". Another reference to similar problems givenby several readers was to J.M. Borwein, G. de Barra, Nested Radicals, Amer.Math. Monthly, 98 (1991) 735-739.
184
2224. [1997: 111] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.
Point P lies inside triangle ABC. Triangle BCD is erected outwardlyon sideBC such that \BCD = \ACP and\CBD = \ABC. Prove that ifthe area of quadrilateral PBDC is equal to the area of triangle ABC, thentriangles ACP and BCD are similar.
Solution by Ian June L. Graces, Manila, The Philippines; and GiovanniMazzarello, Firenze, Italy.
Let A0 and P 0 be the respective images of A and P under re ectionin the line BC. Note that B;D; A0 are collinear (by the de�nition of D).Denoting by [XY Z] the area of 4XY Z, we have
[P 0CB] =1
2P 0C � BC sin\P 0CB;
and
[A0CD] =1
2A0C �CD sin\A0CD:
As a consequence of the given conditions ([PBDC] = [ABC]) and the e�ectof the re ection, [P 0CB] = [A0CD] (these are the complements of4BDCin PBDC and in4A0CB) and \A0CD = \P 0CB. Thus
A0C � CD = P 0C � BC:
In other words, (by SAS) we have4A0CP 0 � 4BCD.
From the re ection we have 4A0CP 0 �= ACP , and the desired resultfollows.
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, andMAR�IA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid,Spain; MANSUR BOASE, student, St. Paul's School, London, England;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN,student, Upper Canada College, Toronto, Ontario; FLORIAN HERZIG, stu-dent, Perchtoldsdorf, Austria; WALTHER JANOUS, Ursulinengymnasium,Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece;GERRY LEVERSHA, St Paul's School, London, England; ISTV �AN REIMAN,Budapest, Hungary; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK,Zaltbommel, the Netherlands; and the proposer.
185
2225. [1997: 111] Proposed by Kenneth Kam Chiu Ko, Mississauga,Ontario.
(a) For any positive integer n, prove that there exists a unique n-digitnumber N such that:
(i) N is formed with only digits 1 and 2; and(ii) N is divisible by 2n.
(b) Can digits \1" and \2" in (a) be replaced by any other digits?
Solution by Robert B. Israel, University of British Columbia, Vancou-ver, BC.
We can use any two non-zero digits whose di�erence is odd. Let thedigits be a and b, where a; b 2 f1;2; : : : ; 9g and a� b is odd.
There are 2n di�erent n{digit numbers formed with these two digits,and 2n residue classes modulo 2n. I claim that the 2n numbers are all indistinct residue classes. By the Pigeonhole Principle, exactly one of thesenumbers must be in the residue class 0.
Consider two distinct n{digit numbers, N1 and N2, formed with thedigits a and b. Suppose that the �rst digit, counting from right to left, wherethey di�er, is in the 10k position, 0 � k � n� 1, where N1 has a and N2
has b. Then N1 �N2 � (a� b)10k = (a � b)5k2k (mod 10k+1), and thusmodulo 2k+1. Since (a� b)5k is odd, we have N1 6� N2 (mod 2n).
Also solved by SAM BAETHGE, Nordheim, Texas, USA; CARL BOSLEY,student,Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHERJ. BRADLEY, Clifton College, Bristol, UK; CURTIS COOPER, Central Mis-souri State University, Warrensburg, Missouri, USA; CHARLES R.DIMINNIE, Angelo State University, San Angelo, TX, USA; FLORIANHERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Pa-los Verdes, California, USA; CYRUS HSIA, student, University of Toronto,Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-tria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,USA;MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRYLEVERSHA, St Paul's School, London, England; KATHLEEN E. LEWIS, SUNYOswego, Oswego, NY, USA; GOTTFRIED PERZ, Pestalozzigymnasium, Graz,Austria; JOEL SCHLOSBERG, student, Hunter College High School, NewYork, NY, USA; ZUN SHAN and EDWARD T.H. WANG, Wilfrid Laurier Uni-versity, Waterloo, Ontario; DAVID R. STONE, Georgia Southern University,Statesboro, Georgia, USA; KENNETH M. WILKE, Topeka, Kansas, USA; YEOKENG HEE, student, Hwa Chong Junior College, Singapore; and the pro-poser.
Besides Israel, only Lambrou, Leversha and Schlosberg proved themoregeneral result presented above. Shan and Wang pointed out that if fa; bg =f2; 4g, f2; 8g, f4; 6g, f6; 8g or f4; 8g, then the \existence" claim is still trueand the \uniqueness" claim would be true if one strengthens condition (ii)to 2n+1jN for the �rst four pairs, and to 2n+2jN for the pair f4; 8g.
186
Diminnie asked whether any similar results are possible if 2n is re-placed by kn for 3 � k � 9.
2226. [1997: 166] Proposed by K.R.S. Sastry, Dodballapur, India.
An old man willed that, upon his death, his three sons would receivethe uth, vth, wth parts of his herd of camels respectively. He had uvw � 1
camels in the herd when he died. Obviously, their sophisticated calculatorcould not divide uvw � 1 exactly into u, v or w parts. They approached adistinguished CRUX problem solver for help, who rode over on his camel,which he added to the herd and then ful�lled the old man's wishes, and tookthe one camel that remained, which was, of course, his own.
Dear CRUX reader, how many camels were there in the herd?
I. Solution by Robert B. Israel, University of British Columbia, Van-couver, BC.
There were 41 camels. We may assume u � v � w. Of course u � 2,and vw+ uw+ uv = uvw� 1.
If u = 2, the equation becomes (v � 2)(w � 2) = 5. Since the onlyfactorization of 5 is 1� 5, this means v = 3, w = 7, and uvw� 1 = 41.
If u = 3, the equation becomes (2v� 3)(2w� 3) = 11. Again there isonly one factorization, and v = 2 (which violates u � v).
Finally, if u � 4 we have 1� 1=(uvw) = 1=u+ 1=v + 1=w � 3=4 souvw � 4, which is impossible.
II. Solution by Michael Lambrou, University of Crete, Crete, Greece.This solution is so much in keeping with the spirit of the problem that theeditor felt a need to share it with all CRUX readers.
\In the holy name of the Almighty," said the distinguished problemsolver who saw the CRUX of the matter, \if you add my humble camel toyour herd then there will be uvw camels to share. Your kind selves, whoserenowned hospitality o�ered me your well to quench my thirst, will receiveuv, vw, and wu camels respectively. This amply ful�ls the deceased's will,whose soul may rest in tranquility."
\So," interrupted the sophisticated calculator, a student of logistics (theart of the practical arithmetician), perpetuator of the Pythagorean doctrinethat whole numbers are the essence of nature, \we must then have that uv+vw+wu+1 equals uvw, since numbers are the balance of ideas, the epitomeof fairness, and we must take into account the camel of our distinguishedguest. We cannot let him leave our oasis for his long journey to redeem hispilgrimage pledge, without a fair chance to cross the desert."
The sophisticated calculator, well versed in the new art of Al-jabr, could
187
easily re-write the condition as
u+ v + w = (u� 1)(v� 1)(w� 1):
\What?" he exclaimed. \If any of u, v, w is 1, then the right hand side is aproduct of numbers, one of which is nothing!" He was perplexed althoughhe had seen this `nothing' number in Ptolemy's Almagest, the monumentalastronomical work, in the Table of Chords in Chapter One. \I do not un-derstand," he continued. \How can nothing exist? It is contrary to nature.Nature abhors void because it wouldmake motion impossible, as falling bod-ies would have to have in�nite speed."
\On the contrary," said the distinguished problem solver, \in my longtravels I have heard that the wise men of the east have discovered a nothingnumber. They call it `as-sifr', and it comes from the Sanskrit `sunya'. It hasthe property that when multiplied by anything it gives as-sifr. So, mathe-matically speaking, we have to exclude equality of any of u, v,w to 1 becausethis would contradict the equation: u+ v + w = (u� 1)(v� 1)(w� 1)."
\Perhaps mathematically we have to exclude this case but we have aninheritance problem," answered the calculator, trying to gain time, \ and thisnothing is not, philosophically speaking, on solid ground." He then turned tothe respectful cadi, the assessor of values and conservator of culture, whosejudgment had a Rhadamanthean wisdom. \What do you say, esteemed rev-erend?"
The cadi replied that \if any of u, v, w was 1, then one of the sonswould take the entire herd, which is contrary to our sacred traditions. Surelyit was not the intention of their late father to incite hatred in the thoughts ofthe two losers. Surely he did not want to upset the good values and bondsof his family." This answer satis�ed the calculator.
\Fine! May your shadow never be less, but let us continue the analysis.We may assume that w � v � u > 1 since birth rites allow shares to belarger or lesser. But could u be 4 or more?"
\I hope not," continued the calculator, \because the uth part would betoo small, unworthy of the respect showed to his late father. Ah yes, if u � 4
then
3w � u+ v + w = (u� 1)(v� 1)(w� 1) � 3 � 3(w� 1)
giving 9 � 6w, which cannot be, since then w = 1, but then he wouldtake the entire herd, inciting hatred in the thoughts of the other two, asforewarned by the incontestable cadi."
\So we must have u � 3. Let us then see what happens if u = 3. Herew � v � u = 3 gives
3 + 2w � u+ v+ w = (u� 1)(v� 1)(w� 1) � 2 � 2(w� 1) ;
188
that is, 7 � 2w, giving w = 1, 2, or 3. The cases w = 1; 2 are excludedsince w � u = 3, leaving 3 = w � u = 3; that is, all shares equal,u = v = w = 3. But this does not satisfy the original equation and must bedropped."
The dropping of the equal shares possibility came as a relief to all. Itmust have been Almighty's wish since not all three sons deserved an equalshare. One of the three was certainly more praiseworthy spiritually, as heattended prayers and consulted often the holy book.
\Last but not least we have to analyze the possibility u = 2." Every-body listened carefully, especially potential brides, because u = 2 meantthat one son would take half the herd. That is, as much as the other twotogether. Wise is the Lord!
\So we have2 + v+ w = 1(v� 1)(w� 1)
which, after some Al-jabr, gives
(v� 2)w = 2v + 1."
At this point the problem solver interrupted again. \Observe," he said,\if v = 2 then the left hand side gives as-sifr, which is incompatible with theright, so this case must be dropped."
\I will do as you say," replied the calculator, \although I think there isa deeper reason for that. Harmony with nature does not allow u = v = 2
because the original equation then becomes
4 + w = w � 1
and, if anything is added to 4, be it something of substance or void, you getat least 4 more than the addend, and not one less than the addend."
\We, therefore, have," he continued:
\w =2v+ 1
v� 2= 2 +
5
v� 2
This is a di�cult situation. How can an integer equal a fraction? Only whenwhat seems a fraction is not really a fraction but an integer concealed. We arerescued from this di�cult situation by appealing to the ideas of Diophantus.I am so glad I have a recent manuscript with a translation of his eternal book,because the original Greek is too di�cult."
\This is how he approaches such problems: The denominator v � 2
must be a divisor of 5, a sacred number, the number of Platonic solids andthe length of the hypotenuse of the eternal triangle. Divine wisdom arrangedthat 5 has the prime property that it possesses precisely two divisors, unityand itself. So v is either 3 or 7. It cannot be 7 because w would then be 3, asmaller number. This leaves v = 3 and w = 7."
189
\Thus the total number of camels in the original herd is 41. One camelis left over for our guest, which he can have back, as it is not counted inthe 41. The shares are 21, 14, and 6 respectively," concluded the calculatorboastfully.
Everybody applauded the sagacity of this artful manipulator of numberswhose eurhythmic mind interpreted the inheritance laws with the infallibleways of the mathematician.
The distinguished problem solver smiled to himself. He had succeededagain. In a true Socratic manner he led the dialogue by giving imperceptiblehints, the CRUX of the matter, to his counterpart who then discovered thetruth that had been known to the problem solver from the very beginning.He saddled his camel, thanked for the hospitality and the knowledge he ac-quired, savoured a sip of water and left for the next stage of his Prometheanjourney.
Also solved or answered by SAM BAETHGE, Nordheim, Texas, USA;FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MAR�IAASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain;MANSUR BOASE, student, St. Paul's School, London, England;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN,student, Upper Canada College, Toronto, Ontario; CHARLES R. DIMINNIE,Angelo State University, San Angelo, TX, USA; IAN JUNE L. GARCES, Ateneode Manila University, The Philippines and GIOVANNI MAZZARELLO, Fer-rovie dello Stato, Florence, Italy; DAVID HANKIN, Hunter College CampusSchools, New York, NY, USA; FLORIAN HERZIG, student, Perchtoldsdorf,Austria; JOHN G. HEUVER, Grande Prairie Composite High School, GrandePrairie, Alberta; D. KIPP JOHNSON, Beaverton, Oregon, USA; GERRYLEVERSHA, St Paul's School, London, England; GOTTFRIED PERZ, Pesta-lozzigymnasium, Graz, Austria; REZA SHAHIDI, student, University of Wa-terloo, Waterloo, Ontario; D.J. SMEENK, Zaltbommel, the Netherlands;DAVID STONE and VREJ ZARIKIAN, Georgia Southern University, States-boro, Georgia; PANOS E. TSAOUSSOGLOU, Athens, Greece; PAUL YIU,Florida Atlantic University, Boca Raton, Florida, USA; and the proposer.There were two incomplete solutions.
Lambrou also considers the general problem where the number of camelsis p � 1, and u, v, w all divide evenly into p. In addition to the solutiongiven above he �nds 13 other solutions (u; v;w; p), where w � v � w:(2; 3; 8; 24), (2; 3; 9; 18), (2;3; 10; 15), (2; 3; 12; 12), (2; 4; 5; 20),(2; 4; 6; 12), (2; 4; 8; 8), (2;5; 5; 10), (2; 6; 6; 6), (3; 3; 4; 12),(3; 3; 6; 6), (3; 4; 4; 6), (4; 4; 4; 4):
190
2229. [1997: 167] Proposed by Kenneth Kam Chiu Ko, Mississauga,Ontario.
(a) Letm be any positive integer greater than 2, such that x2 � 1 (mod m)
whenever (x;m) = 1.
Let n be a positive integer. Ifmjn+1, prove that the sum of all divisorsof n is divisible bym.
(b)?
Find all possible values ofm.
Solution by Kee-Wai Lau, Hong Kong (modi�ed by the editor).(a) We �rst show that n cannot be a perfect square.
Suppose that n = k2. Then k2 � �1(mod m). But kjn,mjn+ 1 and(n; n + 1) = 1 together imply that (k;m) = 1, and so, k2 � 1(mod m).Thus 1 � �1(mod m), which is false since m > 2. Therefore, all the di-visors of n can be grouped into pairs (s; t), where st = n and s 6= t. Itthen su�ces to show that mj(s + t). As above, (s;m) = 1 impliesthat s2 � 1(mod m). Adding st � �1(mod m), we have thats(s+ t) � 0(mod m), or s+ t � 0(mod m).
(b) We show that the possible values ofm are precisely 3, 4, 6, 8, 12, and 24.
For each m such that 3 � m � 24, direct checking of those x with1 < x < m and (x;m) = 1 reveals that these values are indeed the onlyones that satisfy the described condition.
Assume then that m > 24 and let p1, p2, p3, : : : , denote the se-quence of prime numbers. Then 2jm, for otherwise (2;m) = 1 implies that22 � 1(mod m), which is false. Similarly, 3jm and 5jm. If (7;m) = 1,then 72 � 1(mod m), or mj48, which is impossible since 5jm. Thus 7jm.
Suppose that pijm for all i = 1, 2, : : : , k, for some k � 4. If
(m; pk+1) = 1, then p2k+1 � 1(mod m), which implies that p2k+1 >Qk
i=1 pi.
However, this contradicts the Bonse Inequality, which states that for allk � 4, p2k+1 <
Qk
i=1 pi. (See, for example, chapter 27 of The Enjoyment
of Mathematics by H. Rademacher and O. Toeplitz; Dover, 1990.)
It follows that pk+1jm and so m is divisible by any prime, whichis clearly impossible. This shows that if m > 24, we cannot havex2 � 1(mod m) whenever (x;m) = 1, and the proof is complete.
Also solved by ADRIAN BIRKA, student, Lakeshore Catholic HighSchool, Port Colbourne,Ontario;MANSUR BOASE, student, St. Paul's School,London, England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,Rancho Palos Verdes, California, USA; D. KIPP JOHNSON, Beaverton, Ore-gon, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; andHEINZ-J�URGEN SEIFFERT, Berlin, Germany.
191
Part (a) only was solved by JIMMY CHUI, student, Earl Haig Sec-ondary School, North York, Ontario; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; SEAN MCILROY, student, University of BritishColumbia, Vancouver, BC; JOEL SCHLOSBERG, student, Hunter CollegeHighSchool, New York, NY, USA; and the proposer.
Regarding the solution to (b), the Bonse Inequality was also used, ex-plicitly or implicitly, by Boase, Bradley, Herzig and Sei�ert. This inequalityis an easy consequence of Bertrand's Postulate as shown by Herzig and Seif-fert. Johnson gave a solution using Bertrand's Postulate directly.
2230. [1997: 167] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.
Triangles BCD and ACE are constructed outwardly on sides BC andCA of triangle ABC such that AE = BD and \BDC + \AEC = 180�.The point F is chosen to lie on the segment AB so that
AF
FB=DC
CE:
Prove thatDE
CD+ CE=
EF
BC=
FD
AC:
Solution by Toshio Seimiya, Kawasaki, Japan.
LetG be a point onAE produced beyondE such that\ECG = \DCB.Since \CEG = 180� � \AEC = \CDB, we have 4CEG � 4CDB,(directly similar) from which we have 4CBG � 4CDE. Thus
\BGC = \DEC: (1)
Since AF
FB= CD
CE= BD
EG= AE
EG; we get FE k BG, so that
\AGB = \AEF: (2)
Hence we have from (1) and (2)
\FED = \AEC � (\AEF + \DEC)
= \AEC � (\AGB + \BGC)
= \AEC � \AGC
= \ECG
= \BCD: (3)
Similarly we have
\FDE = \ACE: (4)
192
Let H be a point on CD produced beyond D such that DH = EC. SinceBD = AE and \BDH = 180� � \BDC = \AEC, we have
4BDH = 4AEC;
so that BH = AC, and \BHD = \ACE:
As \FED = \BCD = \BCH, and \FDE = \ACE = \BHD =
\BHC, we have4FDE � 4BHC.
Thus we getDE
HC=EF
BC=
FD
BH:
Since HC = CD +DH = CD +CE, and BH = AC, we have
DE
CD + CE=
EF
BC=FD
AC:
Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria;ISTV �AN REIMAN, Budapest, Hungary; and the proposer.
Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell
Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer
Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil
Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,
J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia
193
THE ACADEMY CORNERNo. 19
Bruce Shawyer
All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7
Memorial University Undergraduate
Mathematics Competition
September 25, 1997
Solutions by Solomon W. Golomb, USC, Los Angeles, CA, USA, whowrites: \It's reassuring to know that I can still do freshman/high school
mathematics, after all these years".
1. Determine whether or not the following system has any real solutions.
If so, state how many real solutions exist.
x+1
x= y; y +
1
y= z; z +
1
z= x: (1)
Solution. If (1) is to have real solutions, then xyz 6= 0. Hence
x2 + 1 = xy; y2 + 1 = yz; z2 + 1 = zx:
Multiplying gives�x2 + 1
� �y2 + 1
� �z2 + 1
�= x2y2z2.
But, for real x, y, z, we have x2 + 1 > x2, y2 + 1 > y2, z2 + 1 > z2,
so there are no real solutions.
2. The surface area of a closed cylinder is twice the volume. Determine
the radius and height of the cylinder given that the radius and height
are both integers.
Solution. \The surface area of a closed cylinder is twice the volume" is
absurd. It is dimensionally incorrect. One has units of area, the other
has units of volume.
What was no doubt intended was 2�rh+ 2�r2 = 2�r2h, h+ r = rh,1
r+ 1
h= 1, with no integer solution except r = h = 2. But what are
the units in which h+ r = rh? (No physical object corresponds to this
solution, independent of the arbitrary choice of units.)
194
3. Prove that
1 +1
4+
1
9+ � � �+ 1
n2< 2:
Solution (a).
1 +1
4+
1
9+ � � �+ 1
n2<
1Xk=1
1
k2
=�2
6<
10
6
=5
3< 2:
Solution (b). (For those who do not know the value of �(2).)
1 +1
4+
1
9+ � � �+ 1
n2< 1 +
Z n+1
1
dt
t2
< 1 +
Z 11
dt
t2= 1� 1
t
����1
1
= 1+ 1 = 2:
4. Describe the set of points (x; y) in the plane for which
sin(x+ y) = sinx+ siny:
Solution. Since sin(x + y) = sinx cos y + siny cosx, it follows that
sin(x + y) = sinx + siny has no solutions in the �rst quadrant, in
which cosx < 1, cosy < 1, so that sin(x+ y)< sinx+ siny.
There are no solutions in the third quadrant, in which
sin(x+ y) = sinx cos y + siny cosx > 0 > sinx+ siny:
The line y = �x, which bisects the second and fourth quadrants, is
clearly a solution line: sin(x+ (�x)) = 0 and sinx+ sin(�x) = 0.
To show that there are no other solutions, observe:
any solution (x; y) in the second or fourth quadrant corresponds to a
�rst-quadrant solution of sin(x� y) = sinx� siny.
If x 6= y, suppose (without loss of generality) that x > y, so that
sinx > siny while cos y > cosx. But then
sin(x� y) = sinx cos y � siny cosx > sinx� siny:
195
5. In a parallelogram ABCD, the bisector of angle ABC intersects AD
at P . If PD = 5, BP = 6 and CP = 6, �nd AB.
Solution.
A D
CB
P
z
y 5
y 6 6
�
� �
� �
We label the parallelogram as
shown.
All angles� are equal, becauseBCP
is an isosceles triangle, BP bisects
\ABC, and alternate interior angles
are equal.
Thus BPA is also an isosceles tri-
angle, similar to triangle BCP , and
z = 5+ y because BC = AD.
From similar triangles, we see that 6
z= y
6, 36 = yz = y(5 + y), and
y2+5y�36 = 0. By the quadratic formula, the positive root is y = 4,
which is the length of side AB. (Triangle CDP is a 4{5{6 triangle!)
6. Show that, where k+ n � m,
nXi=0
�n
i
��m
k+ i
�=
�m+ n
n+ k
�:
Solution. With k+n � m, we represent
�m+ n
n+ k
�by taking j elements
from n, and the remaining (n+ k)� j elements from m, for every j,
0 � j � n. Thus
�m+ n
n+ k
�=
nXj=0
�n
j
��m
n+ k � j
�
=
0Xi=n
�n
n� i
��m
k+ i
�
=
nXi=0
�n
i
��m
k+ i
�;
where we substituted i = n� j, and used
�n
a
�=
�n
n� a
�.
Solutions were also received from D. KIPP JOHNSON, Beaverton, Ore-
gon, USA and D.J. SMEENK, Zaltbommel, the Netherlands.
196
THE OLYMPIAD CORNERNo. 190
R.E. Woodrow
All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.
Since the \summer break" is coming up, we give three Olympiad con-
tests from three di�erent parts of the world. My thanks go to Bill Sands for
collecting the contest materials for me when he was helping to coordinate
marking of the IMO held in Toronto in 1995.
We �rst give the problems of the Grade XI and Grade XII versions of
the Lithuanian Mathematical Olympiad.
44th LITHUANIANMATHEMATICALOLYMPIAD (1995)
GRADE XI
1. You are given a set of 10 positive integers. Summing nine of them
in ten possible ways we only get nine di�erent sums: 86, 87, 88, 89, 90, 91,
93, 94, 95. Find those numbers.
2. What is the least possible number of positive integers such that the
sum of their squares equals 1995?
3. � � � � � � � � �� � � � � � � �� � � � � � �� � � � � �� � � � �� � � �� � �� ��
Replace the asterisks in the
\equilateral triangle" by the
numbers 1, 2, 3, 4, 5, 6, 7, 8, 9
so that, starting from the second
line, each number is equal to the
absolute value of the di�erence
of the nearest two numbers in
the line above.
Is it always possible to inscribe the numbers 1; 2; : : : ; n, in the way
required, into the equilateral triangle with the sides having n asterisks?
4. A function f : N! N is such that
f(f(m)+ f(n)) = m+ n
for all m;n 2 N (N = f1; 2; : : : g denotes the set of all positive integers).
Find all such functions.
197
5. In a trapezium ABCD, the bases are AB = a, CD = b, and the
diagonals meet at the point O. Find the ratio of the areas of the triangle
ABO and trapezium.
GRADE XII
1. Consider all pairs (x; y) of real numbers satisfying the inequalities
�1 � x+ y � 1; �1 � xy + x+ y � 1:
Let M denote the largest possible value of x.
(a) Prove thatM � 3.
(b) Prove thatM � 2.
(c) FindM .
2. A positive integer n is called an ambitious number if it possesses
the following property: writing it down (in decimal representation) on the
right of any positive integer gives a number that is divisible by n. Find:
(a) the �rst 10 ambitious numbers;
(b) all the ambitious numbers.
3. The area of a trapezium equals 2; the sum of its diagonals equals 4.
Prove that the diagonals are mutually orthogonal.
4. 100 numbers are written around a circle. Their sum equals 100. The
sum of any 6 neighbouring numbers does not exceed 6. The �rst number is
6. Find the remaining numbers.
5. Show that, at any time, moving both the hour-hand and the minute-
hand of the clock symmetrically with respect to the vertical (6 � 12) axis
results in a possible position of the clock-hands. How many straight lines
containing the centre of the clock-face possess the same property?
Next we give the problems of the Korean Mathematical Olympiad.
8th KOREAN MATHEMATICAL OLYMPIADFirst Round
Morning Session | 2.5 hours
1. Consider �nitely many points in a plane such that, if we choose any
three points A, B, C among them, the area of4ABC is always less than 1.
Show that all of these �nitely many points lie within the interior or on the
boundary of a triangle with area less than 4.
2. For a given positive integer m, �nd all pairs (n; x; y) of positive
integers such thatm, n are relatively prime and (x2+y2)m = (xy)n, where
n, x, y can be represented by functions ofm.
198
3. Let A, B, C be three points lying on a circle, and let P , Q, R be the
midpoints of arcs BC, CA, AB, respectively. AP , BQ, CR intersect BC,
CA, AB at L,M , N , respectively. Show that
AL
PL+BM
QM+CN
RN� 9:
For which triangle ABC does equality hold?
4. A partition of a positive integer n is a sequence (�1; �2; : : : ; �k) of
positive integers such that �1 + �2 + � � � + �k = n and �1 � �2 � � � � ��k � 1. Each �i is called a summand. For example, (4; 3; 1) is a partition
of 8 whose summands are distinct. Show that, for a positive integer m with
n > 1
2m(m+1), the number of all partitions of n into distinctm summands
is equal to the number of all partitions of n� 1
2m(m+1) into r summands
(r � m).
Afternoon Session | 2.5 hours
5. If we select at random three points on a given circle, �nd the prob-
ability that these three points lie on a semicircle.
6. Show that any positive integer n > 1 can be expressed by a �nite
sum of numbers satisfying the following conditions:
(i) they do not have factors except 2 or 3;
(ii) any two of them are neither a factor nor a multiple of each other.
That is,
n =
NXi=1
2�i3�i ;
where �i, �i (i = 1; 2; : : : ; N ) are nonnegative integers and, whenever
i 6= j, the condition (�i� �j)(�i � �j) < 0 is satis�ed.
7. Find all real valued functions f de�ned on real numbers except 0
such that1
xf(�x) + f
�1
x
�= x; x 6= 0:
8. Two circles O1, O2 of radii r1; r2 (r1 < r2), respectively, intersect
at two points A and B. P is any point on circle O1. Lines PA, PB and
circle O2 intersect at Q and R, respectively.
(i) Express y = QR in terms of r1, r2, and � = \APB.
(ii) Show that y = 2r2 is a necessary and su�cient condition that
circle O1 be orthogonal to circle O2.
199
Final Round
First Day | 4.5 hours
1. For any positive integer m, show that there exist integers a, b sat-
isfying
jaj � m; jbj � m; 0 < a+ bp2 � 1 +
p2
m+ 2:
2. Let A be the set of all non-negative integers. Find all functions
f : A! A satisfying the following two conditions:
(i) for anym;n 2 A,
2f(m2 + n2) = ff(m)g2 + ff(n)g2;
(ii) for any m;n 2 A withm � n,
f(m2) � f(n2):
3. Let 4ABC be an equilateral triangle of side length 1, let D be a
point on BC, and let r1, r2 be inradii of trianglesABD, ADC, respectively.
Express r1r2 in terms of p = BD, and �nd the maximum of r1r2.
Second Day | 4.5 hours
4. Let O and R be the circumcentre and the circumradius of 4ABC,respectively, and let P be any point on the plane ofABC. Let perpendiculars
PA1, PB1, PC1, be dropped to the three sides BC, CA, AB. Express
[A1B1C1]
[ABC]
in terms of R and d = OP , where [ABC] denotes the area of4ABC.5. Let p be a prime number such that
(i) p is the greatest common divisor of a and b;
(ii) p2 is a divisor of a. Prove that the polynomial xn+2 + axn+1 +
bxn+a+b cannot be decomposed into the product of two polynomials with
integral coe�cients, whose degrees are greater than one.
6. Let m, n be positive integers with 1 � n � m� 1. A box is locked
with several padlocks, all of which must be opened to open the box, and all
of which have di�erent keys. A total ofm people each have keys to some of
the locks. No n people of them can open the box but any n+ 1 people can
open the box. Find the smallest number l of locks and in that case �nd the
number of keys that each person has.
200
Now, problems selected from the 1995 IsraelMathematical Olympiads.
Selected Problems FromISRAEL MATHEMATICAL OLYMPIADS 1995
1. n positive integers d1; d2; : : : ; dn divide 1995. Prove that there ex-
ist di and dj among them, such that the denominator of the reduced fractiondidj
is at least n.
2. Two players play a game on an in�nite board that consists of 1� 1
squares. Player I chooses a square and marks it with O. Then, Player II
chooses another square and marks it with X. They play until one of the
players marks a whole row or a whole column of 5 consecutive squares, and
this player wins the game. If no player can achieve this, the result of the
game is a tie. Show that Player II can prevent Player I from winning.
3. Two thieves stole an open chain with 2k white beads and 2m black
beads. They want to share the loot equally, by cutting the chain to pieces in
such a way that each one gets k white beads andm black beads. What is the
minimal number of cuts that is always su�cient?
4. � and � are two given circles that intersect each other at two points.
Find the geometric locus of the centres of all circles that are orthogonal to
both � and �.
5. Four points are given in space, in a general position (that is, they arenot contained in a single plane). A plane � is called \an equalizing plane" if
all four points have the same distance from �. Find the number of equalizing
planes.
6. n is a given positive integer. An is the set of all points in the plane,
whose x and y coordinate are positive integers between 0 and n. A point
(i; j) is called \internal" if 0 < i; j < n. A real function f , de�ned on An,
is called a \good function" if it has the following property: for every internal
point x, the value of f(x) is the mean of its values on the four neighbouring
points (the neighbouring points of x are the four points whose distance from
x equals 1).
If f and g are two given good functions and f(a) = g(a) for every point
a in An which is not internal (that is, a boundary point), prove that f � g.
7. Solve the system
x+ log(x+px2 + 1) = y
y + log(y+py2 + 1) = z
z + log(z +pz2 + 1) = x
201
8. Prove the inequality
1
kn+
1
kn+ 1+
1
kn+ 2+ � � �+ 1
(k+ 1)n� 1� n
n
rk+ 1
k� 1
!
for any positive integers k, n.
9. PQ is a diameter of a half circle H. The circle O is tangent to H
from the inside and touches diameter PQ at the point C. A is a point on
H and B is a point on PQ such that AB is orthogonal to PQ and is also
tangent to the circle O. Prove that AC bisects the angle PAB.
10. � is a given real number. Find all functions f : (0;1) 7! (0;1)
such that the equality
�x2f
�1
x
�+ f(x) =
x
x+ 1
holds for all real x > 0.
Next we turn to solutions to problems posed in the February 1997 num-
ber of the Corner. Some new solutions arrived after we went to press last
issue, and at least one other batch of solutions was incorrectly �led and just
turned up! Pavlos Maragoudakis, Pireas, Greece sent in solutions to prob-
lems 1, 3 and 4 of the Final Grade, Third Round, and also to Problem 1 of the
1st Selection Round. The misplaced solutions were from D.J. Smeenk, who
gave solutions to Problem 3 of the Final Grade, 3rd Round and to Problem 3
of each of the second and third Selection Rounds. Because they are interest-
ing and di�erent, we give his two solutions to Problem 3 of the Final Grade.
(Look at all the 3's in the above!)
3. [1997: 78, 1998: 13{14] Latvian 44Mathematical Olympiad.
It is given that a > 0, b > 0, c > 0, a + b + c = abc. Prove that at
least one of the numbers a, b, c exceeds 17=10.
Solution by D.J. Smeenk, Zaltbommel, the Netherlands.In any triangle ABC the following identity holds
tan�+ tan� + tan = tan� tan� tan :
Hence, in this problem a, b, c can be considered to be the tangents of angles
of an acute angled triangle. At least one of the angles is at least �3, and
tan��3
�=p3 > 17
10.
202
Second solution.
We may suppose a � b � c,
abc = a+ b+ c � 3c:
So ab � 3, a � b and a � p3 > 17
10.
Now we turn to readers' solutions of problems from the March 1997
Corner and the 3rdMathematical Olympiad of the Republic of China (Taiwan)
[1997: 66].
3rd MATHEMATICAL OLYMPIAD OF THEREPUBLIC OF CHINA (Taiwan)
First Day | April 14, 1994
1. Let ABCD be a quadrilateral with AD = BC and let
\A+\B = 120�. Three equilateral triangles4ACP ,4DCQ and4DBRare drawn on AC, DC and DB away from AB. Prove that the three new
vertices P , Q and R are collinear.
Solution by Toshio Seimiya, Kawasaki, Japan.
A
D
PQ O
R
C
B
`
Let O be the intersection of AD and BC. Since \A + \B = 120�,we get \AOB = 60�. Let l be the exterior bisector of angle AOB. Since
\APC = 60� = \AOC, we have that O, P , A, C are concyclic. Hence
\POA = \PCA = 60�. The exterior angle of \AOB is 120�, showing thatPO bisects the exterior angle of \AOB. Thus P lies on l. Similarly Q and
R lie on l. Hence, P , Q and R are collinear.
Comment. As is shown in the proof, the condition AD = BC is not
necessary.
2. Let a, b, c be positive real numbers, � be a real number. Suppose
that
f(�) = abc(a� + b� + c�)
g(�) = a�+2(b+ c� a) + b�+2(a� b+ c) + c�+2(a+ b� c)
Determine the magnitude between f(�) and g(�).
203
Solution by Panos E. Tsaoussoglou, Athens, Greece.
bca�+1 + acb�+1 + abc�+1 � a�+2(b+ c� a)
�b�+2(a� b+ c)� c�+2(a+ b� c)
= a�+1(bc� a(b+ c) + a2) + b�+1(ac� b(a+ c) + b2)
+c�+1(ab� c(a+ b) + c2)
= a�+1(a� b)(a� c) + b�+1(b� a)(b� c) + c�+1(c� a)(c� b)
� 0
which is an inequality of Schur.
Next we move to solutions of Selected Problems from the Israel Math-
ematical Olympiads, 1994 [1997: 131].
SELECTED PROBLEMS FROM THE ISRAELMATHEMATICAL OLYMPIADS, 1994
1. p and q are positive integers. f is a function de�ned for positive
numbers and attains only positive values, such that f(xf(y)) = xpyq. Prove
that q = p2.
Solutions by Pavlos Maragoudakis, Pireas, Greece; and Michael Selby,University of Windsor, Windsor, Ontario. We give the solution ofMaragoudakis.
For x =1
f(y), we get f(y) =
yq=p
(f(1))1=p.
For y = 1, we get f(1) = 1, so f(y) = yq=p. Hence f(x � yq=p) = xp � yq.For y = zp=q we get f(x � z) = xpzp or f(x) = xp.
Thus q
p= p, whence q = p2.
2. The sides of a polygon with 1994 sides are ai =p4 + i2,
i = 1; 2; : : : ; 1994. Prove that its vertices are not all on integer mesh points.
Solution by Michael Selby, University of Windsor, Windsor, Ontario.One assumes integer mesh points are lattice points.
Assume that Pi, the ith vertex, has coordinates (xi; yi) where xi, yi are
integers. Let�!di = (xi+1 � xi; yi+1 � yi) = (�i; �i) be a vector represen-
tation of the ith side. (The indices are read cyclically.) Then j�!di j2 = a2i =
4 + i2. Also��!d1994 = (x1 � x1994; y1 � y1994).
We know that
1994Xi=1
(�!di )
2 =
1994Xi=1
(4 + i2); or
204
1994Xi=1
(�2
i + �2i ) = 4 � 1994 + (1994)(1995)(3989)
6
= 4 � 1994 + (997)(665)(3989);
which is an odd integer.
However, we know
1994Xi=1
�i =
1994Xi=1
�i = 0, since
1994Xi=1
�!di =
�!0 .
Therefore
1994Xi=1
(�i+ �i)
!2
= 0.
Thus
1994Xi=1
(�2
i + �2i ) + 2
0@X
ij
�i�j +Xi<j
�i�j +Xi<j
�i�j
1A = 0
and
2
0@X
i;j
�i�j +Xi<j
�i�j +Xi<j
�i�j
1A = �
Xi
��2
i + �2i�!
;
giving an odd integer on the right, and an even one on the left, a contradic-
tion.
Therefore, not all the vertices can be lattice points.
5. Find all real coe�cients polynomials p(x) satisfying
(x� 1)2p(x) = (x� 3)2p(x+ 2)
for all x.
Solutions by F.J. Flanigan, San Jose State University, San Jose, Califor-nia, USA; and Michael Selby, University of Windsor, Windsor, Ontario. Wegive Flanigan's solution.
We consider polynomials p(x) with coe�cients in a �eld F of arbitrarycharacteristic and �nd as follows:
(i) If char(F) = 0, (in particular, if F = R), then p(x) = a(x � 3)2,
where a is any scalar (possibly 0) in F;
(ii) If char(F) = 2, then every p(x) satis�es the equation (clear);
(iii) If char(F) = an odd prime, l, then there are in�nitely many so-
lutions, including all p(x) = a(x � 3)2(xl� � x + c) with a; c 2 F, and
� = 0; 1; 2; : : : . (Note that p(x) has the form a(x� 3)2 if � = 0.)
To prove this, observe that if char(F) 6= 2, then x � 1 and x � 3 are
coprime, whence p(x) = (x� 3)2q(x) in F[x].
Thus our equation becomes
(x� 1)2(x� 3)2q(x) = (x� 3)2(x� 1)2q(x+ 2) (�)
205
whence q(x) = q(x+ 2), as polynomials; that is, elements of F[x].
Now if char(F) = 0, then (�) has only constant solutions.(The most elementary proof of this: without loss of generality,
q(x) = xn+ axn�1 + � � � . Then q(x+2)� q(x) = 2nxn�1 + � � � , and thisis non-zero if n � 1. Another proof: (�) implies that q(x) is periodic, which
forces equations q(x) = c to have in�nitely many roots x, a contradiction).
This establishes the assertion (i).
Re: assertion (iii). Let char(F) = l and q(x) = xl� � x+ c.
Then for x = 0; 1; : : : ; l � 1, (that is for each element of the prime
�eld), we have q(x) = c and so q(x) = q(x + 1) = q(x + 2) = : : : ,
yielding polynomials of degree greater than or equal to l which satisfy (�).This establishes the assertion (iii).
We next turn to solutions to Problems From the Bi-National Israel{
Hungary Competition, 1994 [1997: 132].
PROBLEMS FROM THE BI-NATIONALISRAEL-HUNGARY COMPETITION, 1994
1. a1; : : : ; ak; ak+1; : : : ; an are positive numbers (k < n). Suppose
that the values of ak+1; : : : ; an are �xed. How should one choose the values
of a1; : : : ; an in order to minimizePi;j;i6=j
aiaj?
Solutions by F.J. Flanigan, San Jose State University, San Jose, Califor-nia, USA; and Michael Selby, University of Windsor, Windsor, Ontario. Wegive Flanigan's solution.
To minimize the given rational function, choose
ai =
ak+1 + � � �+ an
1
ak+1+ � � �+ 1
an
!1=2
= (A � H)1=2; i = 1; 2; : : : ; k
where A is the arithmetic and H the harmonic mean of ak+1; : : : ; an.
To prove this, we will be forgiven if we change notation: let xi = ai,
i = 1; 2; : : : ; k and br = ak+r, r = 1; : : : ; m with k +m = n, and denote
the given rational function F (x1; : : : ; xk). Then we have F (x1; : : : ; xk) =
X + Y + B, where
X =X
1�i<j�k
�xi
xj+xj
xi
�;
Y =X
1�i�k
X1�r�m
�xi
br+br
xi
�;
B =X
1�r<s�m
�br
bs+bs
br
�:
206
Note that B is �xed and Y can be improved to
Y =X
1�i�k
0@0@ X1�r�m
1
br
1Axi +
0@ X1�r�m
bi
1A 1
xi
1A
=Xi
�m
Hxi +
mA
xi
�
where A is the arithmetic mean and H is the harmonic mean of the br.
Now we recall that the simple function�x+ �
x(with �;�; x all positive)
assumes its minimum when �x = �
x; that is x =
p�=�. Thus each of
the terms in Y (and so Y itself) assumes its minimum when we choose, for
i = 1; 2; : : : ; k,
xi =
smA
(m=H)=pAH;
as asserted.
But there ismore. It is also known that each term inX, (and soX itself)
assumes its minimum when xi = xj , with 1 � i < j � k. Thus choosing
all xi =pAH minimizes both X and Y and, since B is �xed, minimizes
F (x1; : : : ; xk) as claimed.
Comments:
(1) It is unusual to minimize the sum of two terms in the same variables by
minimizing each term simultaneously.
(2) Whenm = 2, thenpAH = G, the geometric mean of b1, b2.
(3) Fmin = k(k� 1) + 2k(n� k)2pA=H+ B.
3. m, n are 2 di�erent natural numbers. Show that there exists a real
number x, such that 1
3� fxng � 2
3and 1
3� fxmg � 2
3, where fag is the
fractional part of a.
Solution by F.J. Flanigan, San Jose State University, San Jose, Califor-nia, USA.
We work in the �rst quadrant of the standard uv{plane, studying the
ray
R = f(u; v) = (xm; xn) : x > 0g:The key is to observe that the problem is equivalent to showing that the ray
R contacts at least one of the \small" 1
3by 1
3squares in the centres of the
large standard 1 by 1 lattice squares (considered as a 3 � 3 checkerboard).
(For if (xm; xn) lies in one of these small squares then (fxmg; fxng) lies inthe small square f(u; v) : 1
3� u; v � 2
3g closest to the origin, as desired.)
207
To establish this contact, we assume, without loss of generality, that
0 < n < m, so that R is given by v = n
mu, u > 0.
Consider the sequence of rays v = 1
2u, v = 1
5u, v = 1
8u, v = 1
11u; : : : ,
with u > 0. These rays are determined by the lower right corners of the 1st,
2nd, 3rd, 4th; : : : , small central square.
v = u
v = 1
2u
typicalR
v = 1
5uv = 1
8u v = 1
11u
1
1 2 3 4
It is now apparent that our ray R lies between, (or on) the ray v = u
and v = 1
2u, or v = 1
2u and v = 1
5u, or : : : , and hence R will contact the
�rst or the second or : : : , small square, as required.
Comment: We can now estimate the least x for given m, n in the se-
quence 1; 12; 15; 18; : : : of slopes and use this to determine which small square
is the �rst to be contacted by the ray R. From this one can estimate the
coordinates of (xm; xn) in various ways.
4. An \n-m society" is a group of n girls andm boys. Show that there
exist numbers n0 andm0 such that every n0-m0 society contains a subgroup
of 5 boys and 5 girls in which all of the boys know all of the girls or none of
the boys knows none of the girls.
Solution by Michael Lebedinsky, student, Henry Wise Wood School,Calgary, Alberta.
We will show that we can take n0 = 9. For n0 � 9, observe that for
each girl there must be at least 5 boys whom she knows, or 5 boys whom she
does not know. We associate to each girl an ordered pair, the �rst element
of which is a subset of 5 of the boys all of whom she knows or all of whom
she does not know, and the second element of which is 0 or 1 according as
she knows the boys or not. There are�9
5
��2 = 252 such pairs. Invoking the
Pigeonhole Principle, if m0 � 4� 252 + 1 = 1009, at least 5 girls must be
assigned the same ordered pair, producing 5 girls and 5 boys for which each
girl knows each boy, or no girl knows any of the boys.
That completes this Corner for this issue. Enjoy solving problems over
the next weeks | and send me your nice solutions as well as your Olympiad
materials.
208
BOOK REVIEWS
Edited by ANDY LIU
Mathematical Challenge, by Tony Gardiner,
published by Cambridge University Press, 1996,
ISBN# 0-521-55875-1, softcover, 138+ pages.
More Mathematical Challenges, by Tony Gardiner,
published by Cambridge University Press, 1997,
ISBN# 0-521-58568-6, softcover, 140+ pages.
Reviewed by Ted Lewis, University of Alberta.
These two books are companion volumes which cover the recent phenome-
nal development in mathematics competitions in the United Kingdom. Although the
country has a long and distinguished history in this endeavour, featuring the famed
Cambridge Tripos, it was not until the late 1980's when a popularization movement,
under the leadership of the author Tony Gardiner, made it a truly national event.
This grassroots approach has nurtured the young talents into a force to be reckoned
with consistently in the International Mathematical Olympiad.
The �rst book contains the problems for the UK Schools Mathematical Chal-
lenge Papers from 1988 to 1993. The target audience are children aged 12 to 14. Each
paper consists of 25 multiple-choice questions, of which the �rst 15 are relatively
straightforward. The paper is to be attempted in one hour, and students are not ex-
pected to �nish it. Random guessing is discouraged, and calculators are forbidden.
Answers, statistics and brief comments are given, but the reader has to work out the
solutions.
The second book contains the problems for the UK Junior Mathematical
Olympiad from 1989 to 1995. The target audience is children aged 11 to 15. Except
in 1989, each paper consists of 10 problems in Section A and 6 problems in Section B.
The 1989 paper consists of 13 problems, and may be regarded as a long Section B.
The following paragraph is quoted from page 2 of the book.
Section A problems are direct, \closed" problems, each requiring aspeci�c calculation and having a single numerical answer. Section B prob-lems are longer and more \open". Thus, while the �nal mathematicalsolution is often quite short, and should involve a clear claim, followedby a direct deductive calculation or proof, there will generally be a pre-liminary phase of exploration and conjecture, in which one tries to sortout how to tackle the problem.
There is a section titled \Comments and hints" and another titled \Outline solutions".
Thus a student who is frustrated by a particular problem has several recourses for
assistance. The outline solutions are precisely that, with gaps for the reader to �ll in,
but revealing enough details to guide a diligent student to the full solution.
We now present some sample problems.
1988/14.
Weighing the baby at the clinic was a problem. The baby would not keep still and
209
caused the scales to wobble. So I held the baby and stood on the scales while the
nurse read o� 79 kg. Then the nurse held the baby while I read o� 69 kg. Finally I
held the nurse while the baby read o� 137 kg. What is the combined weight of all
three (in kg)?
(a) 142 (b) 147 (c) 206 (d) 215 (e) 284
1993/18.
Sam the super snail is climbing a vertical gravestone 1 metre high. She climbed at
a steady speed of 30 cm per hour, but each time the church clock strikes, the shock
causes her to slip down 1 cm. The clock only strikes the hours, so at 1 o'clock she
would slip back 1 cm, at 2 o'clock she would slip back 2 cm and so on. If she starts
to climb just after the clock strikes 3 pm, when will she reach the top?
(a) 3:50 pm (b) 6:20 pm (c) 6:50 pm (d) 7:04 pm (e) 7:20 pm
1994/A6.
A normal duck has two legs. A lame duck has one leg. A sitting duck has no legs.
Ninety nine ducks have a total of 100 legs. Given that there are half as many sitting
ducks as normal ducks and lame ducks put together, �nd the number of lame ducks.
1995/B6.
I write out the numbers 1, 2, 3, 4 in a circle. Starting at 1, I cross out every second
integer till just one number remains: 2 goes �rst, then 4, leaving 1 and 3; 3 goes next,
leaving 1 | so \1" is the last number left. Suppose I start with 1, 2, 3, : : : , n in a
circle. For which values of n will the number \1" be the last number left?
Comments and hints for 1995/B6.
The question \For which values of n : : : ?" is more complicated than it looks. It is
not enough to answer \n = 4" (even though n = 4 works), since we clearly want to
know all possible values of n that work. Hence a solution must not only show that
certain values of n work, but must also somehow prove that no other values of ncould possibly work.
Outline solution of 1995/B6.
The easy step is to realize that, if n is odd (and � 3) then 1 will be crossed out at
the start of the second circuit. Hence, for 1 to survive, either n = 1 or n must be
e?e? (say n = 2m). Suppose n = 2m. Then at the start of the second circuit, there
remain exactly m numbers 1, 3, 5, : : : , 2m � 1. For 1 to survive next time it is
essential that eitherm = 1 (so n = 2) ormmust be e?e? (saym = 2p). Continuing
in this way we see that 1 will be the last uncrossed number if and only if n is a ?o?e?
of ??o. [Ed. here ? indicates some number.]
As a bonus, the �rst book provides 420 additional questions, and the second 40
Section A problems and 20 Section B problems. For one reason or another, these did
not make it into the competition papers, but are nevertheless excellent questions.
A lot of thought has gone into the planning of these two books. They are to be
done actively rather than read passively. The problems are the important features,
not the solutions. Particularly valuable are the author's comments on how things are
done and why. This dynamic package is a must for anyone interested in mathematics
competitions for youngsters.
210
A Note on Special Numerals
in Arbitrary Bases
Glenn Appleby, Peter Hilton and Jean Pedersen
1 Introduction
In [1], at the end of a section devoted to the Pigeonhole Principle, the author
posed the following problem:
\4.2.28 Show that, for any integer n, there exists a multiple of
n that contains only the digits 7 and 0."
Plainly the author is referring to the representation of the multiple as a
numeral in base 10. Thus the reader of the problem is encouraged to believe
that the validity of the statement depends in some way on the numbers 7 and
10 and their mutual relationship. Our analysis of the problem shows that,
in fact, we may replace 10 by any base b � 2, and 7 by any digit t in base b.
(Of course, we assume t 6= 0 to avoid absurd triviality.) Moreover, we prefer
to restrict ourselves to numerals consisting of a sequence of k t's followed
by l 0's; and, if we assume, as we may without real loss of generality, that
n > 0, then we have k � 1, l � 0. We then give two arguments for the
conclusion. The �rst does employ the Pigeonhole Principle but does not give
us a means of calculating all pairs (k; l) from the data. Our second approach
yields the minimal pair (k; l) and shows, as expected, that an arbitrary pair
(K; L) in the set of solutions is obtained from the minimal pair by taking K
to be an arbitrary multiple of k, and L an arbitrary integer satisfying L � l.
We prefer to replace the modulusn of the problem as stated bym, thus
freeing n to stand for the multiple we are seeking. Also we point out that,
of course, the analysis we carry out does not depend on the condition t < b
imposed by the requirement that t be a digit in base b. Thus the number n
we seek may be represented, for arbitrary t and base b, by the expression
n = t(bk � 1)bl
b� 1: (1.1)
2 The two main arguments
Let b be an arbitrary base and let t, 0 < t � b� 1, be an arbitrary non-zero
digit in base b.
Copyright c 1998 Canadian Mathematical Society
211
Theorem 1 For any integerm > 0, there exists a positive integer n, writtenas a numeral in base b consisting of a sequence of t's followed by a (possiblyempty) sequence of 0's, such that n � 0 mod m.
We give two proofs, the �rst uses the Pigeonhole Principle, the second
being an exercise in modular arithmetic.
Proof A.
Consider the sequence of numerals 0, t, tt, ttt, : : : . The remainders
mod m of these integers lie in a set containing m elements. Thus after at
most (m+ 1) such numerals, some remainder must have been repeated. If
ttt : : : (k + l occurrences), and ttt : : : , (l occurrences), k > 0 yield the same
remainders, then the numeral ttt : : : 000 : : : (k occurrences of t, l zeros) ful�ls
the conclusion of the theorem.
This elegant proof has the single defect that it gives no clue as to how
one �nds the minimal values of k and l as functions of b, t andm. Our second
proof is not so neat, but gives more information.
Proof B.
Let z = b� 1. We show �rst how to �nd a smallest number n = nz,
whose numeral in base b consists of a sequence of z's followed by a sequence
of 0's, such that n � 0 mod m.
Write m = vw, where v is prime to b and, for every prime p, pjwimplies that pjb. Note that this factorization of m is unique. Let k be the
order of b mod v and let l be the smallest non-negative integer such that
wjbl. Then the number n, represented in base b by a sequence of k z's
followed by a sequence of l 0's, is obviously a positive integer nz of the
form
nz = zzz : : : 000 : : : (2.1)
divisible by m. We will show below that nz, given by (2.1), is minimal for
this property.
It is now a trivial matter to complete the proof of the theorem. For
if we replace m by mz in the argument above, we �nd a number
nz = zzz : : : 000 : : : such that nz � 0 mod mz, so that n1, given by
zn1 = nz gives us
n1 = 111 : : : 000 : : : � 0 mod m: (2.2)
But then
n = tn1 = ttt : : : 000 : : : � 0 mod m: (2.3)
However, if we want to make sure we have the smallest number n of
the required form, we should proceed more cautiously.
212
We �rst prove that if n = nz is chosen as in (2.1), then it is the smallest
integer of the given form to satisfy n � 0 mod m.
Now if n is of the form zzz : : : 000 : : : , withK z's and L 0's, then [see
(1.1)] n = (bK � 1)bL. Moreover, since v, w are coprime, we have
n � 0 mod m () n � 0 mod v and n � 0 mod w:
Since v is prime to b,
n � 0 mod v () bK � 1 � 0 mod v () kjK:
Further, since, for all primes p, pjw implies that pjb, we have that w is
prime to bK � 1, so that
n � 0 mod w () bL � 0 mod w () L � l:
Now suppose that n is chosen to be the smallest integer of the form
zzz : : : 000 : : : such that n � 0 mod mz, and let n = zn1. Then since
zn1 � 0 mod mz () n1 � 0 mod m; (2.4)
it follows that n1, given by (2.2), is the smallest integer of the form
111 : : : 000 : : : to satisfy n1 � 0 mod m.
We want now to �nd the smallest integer nt represented in base b by
a sequence of t's followed by a sequence of 0's [compare (2.3)] to satisfy
nt � 0 mod m.
Our recipe for constructing nt is as follows. Let d = gcd(m; t),
m = m0d, t = t0d. Then if n1 has the form 111 : : : 000 : : : ,
tn1 � 0 mod m () t0n1 � 0 mod m0 () n1 � 0 mod m0: (2.5)
Thuswe construct n as in (2.1) to be minimal satisfyingn � 0 mod m0z.Then n = zn1, and, by (2.4), we have that n1 is minimal of the required form
to satisfy n1 � 0 mod m0, so that �nally, by (2.5), we have that nt = tn1 is
minimal of the form ttt : : : 000 : : : to satisfy nt � 0 mod m.
Example 1.
Let us work in base b = 10 and look for the smallest number n6 of the
form
n6 = 666 : : : 000 : : :
divisible by 99. We have b = 10, z = 9, m = 99, t = 6, so d = 3, m0 = 33,
m0z = 297. To �nd n, minimal of the form 999 : : : 000 : : : , to satisfy
n � 0 mod 297, we factorize 297, as in the construction above,
as 297 = 297 � 1, since 297 is prime to 10. We �nd k = 6 (that is, the
order of 10 mod 297 is 6) and, of course, l = 0. Thus n = 999999, so
n1 = 111111, n6 = 666666.
213
Example 1 brings out the important practical point that, to �nd
nt, we simply �nd the minimal n of the form zzz : : : 000 : : : to satisfy
n � 0 mod m0z, and then replace z by t in the numeral for n. Indeed,
all we have to do is to �nd the minimal values of k and l which yield n (see
Section 3).
It is also plain that, having obtained our minimal nt, involving k t's
followed by l 0's, we obtain all solutions of the congruence N � 0 mod m
of the required form by taking K t's followed by L 0's, where kjK and
L � l. Notice that this implies that every solution of the congruence
N � 0 mod m, of the required form, is a multiple of the minimum solu-
tion.
3 The algorithm
We extract from the analysis in Section 2 the algorithm for �nding theminimal
pair (k; l) as functions of b, t andm.
Given: b (base), t (digit) andm (modulus).
� Set gcd(m; t) = d, and m = m0d.
� Writem0(b� 1) = vw where v is prime to b and, for all primes p,
pjw =) pjb:
� Finally, let k be the order of b mod v (that is, bk � 1 mod v, with k
positive minimal) and let l be minimal such that wjbl. Then
nt =
k timesz }| {ttt : : :
l timesz }| {000 : : :
is the minimal numeral n, in base b, consisting of a sequence of k t's
followed by a sequence of l 0's, such that mjn.
Example 2.
Given: b = 7, t = 5 andm = 2499.
� gcd(2499;5) = 1, so that m = m0.
� Write 2499� 6 = 306� 49 (thus, v = 306, w = 49).
� Finally, let k be the order of 7 mod 306 (that is, 7k � 1 mod 306,
with k positive minimal), and let l be minimal such that 49j7l. Using
modular arithmetic1 we see that k = 48, and it is clear that l = 2. Thus
n5 =
48 timesz }| {555 : : : 00
1 306 = 2 � 9 � 17. The order of 7 mod 2 is 1; the order of 7 mod 9 is 3; the order of
7 mod 17 is 16. To see the last, observe that, by Fermat's Theorem, 716 � 1 mod 17; but
72� �2 mod 17, so 78 � �1 mod 17. Thus the order of 7 mod 306 is 48.
214
is the minimal numeral, in base 7, consisting of a sequence of 5's fol-
lowed by a sequence of 0's such that 2499jn5.
Example 2 shows that it may sometimes be very tedious to apply the Pigeon-
hole Principle to obtain k and l.
Notice that, in the special case t = b�1, we may simplify the algorithm
by cutting out the �rst step and replacingm0(b�1) bym in the second step.
Acknowledgment The authors would like to thank Professor Robert
Bekes for bringing the original problem in [1] to their attention.
Reference [1] Zeitz, Paul, The Art and Craft of Problem Solving, JohnWiley & Sons, 1998.
Addresses
Glenn Appleby
Department of Mathematics
Santa Clara University
Santa Clara, California 95053-0290
e-mail: [email protected]
Peter Hilton
Department of Mathematics
University of Central Florida
Orlando, Florida 32816-1364
and
Department of Mathematics and Computer Science
University of New England
Armidale, NSW 2351
Australia
Jean Pedersen
Department of Mathematics
Santa Clara, California 95053-0290
e-mail: [email protected]
and
Department of Mathematics and Computer Science
University of New England
Armidale, NSW 2351
Australia
215
THE SKOLIAD CORNERNo. 30
R.E. Woodrow
In the last number, we gave the problems of Part I of the Alberta High
School Mathematics Competition. Here are the \o�cial" solutions.
ALBERTA HIGH SCHOOLMATHEMATICS COMPETITION
Part I | November, 1996
1. An eight-inch pizza is cut into three equal slices. A ten-inch pizza is
cut into four equal slices. A twelve-inch pizza is cut into six equal slices. A
fourteen inch pizza is cut into eight equal slices. From which pizza would you
take a slice if you want as much pizza as possible?
Solution. (b) The area of a circle is proportional to the square of its
radius. It follows that we are comparing 16=3, 25=4, 36=6 and 49=8.
2. One store sold red plums at four for a dollar and yellow plums at
three for a dollar. A second store sold red plums at four for a dollar and
yellow plums at six for a dollar. You bought m red plums and n yellow
plums from each store, spending a total of ten dollars. How many plums in
all did you buy?
Solution. (d) The total expenditure is 10 = m4+ n
3+ m
4+ n
6=
(m+n)
2.
3.
A
B
C
D
EF
Six identical cardboard pieces are
piled on top of one another, and
the result is shown in the dia-
gram.
The third piece to be placed is:
Solution. (b) Clearly, F , E, C, B and A are in that order from top
to bottom. If D is pointing up, it is under A. If it is pointing down, it is
under B.
4. A store o�ered triple the GST in savings. A sales clerk calculated the
selling price by �rst reducing the original price by 21% and then adding the 7%
GST based on the reduced price. A customer protested, saying that the store
should �rst add the 7%GST and then reduce that total by 21%. They agreed on
216
a compromise: the clerk just reduced the original price by the 14% di�erence.
How do the three ways compare with one another from the customer's point
of view?
Solution. (e) Since 1:07 times 0:79 is 0:8103, both the customer's way
and the clerk's way yield a discount approximately 19%.
5. Ifm andn are integers such that 2m�n = 3, then what willm�2n
equal?
Solution. (c)We havem�2n = m�2n+2m�n�3 = 3(m�n�1).
6. If x is x% of y, and y is y% of z, where x, y and z are positive real
numbers, what is z?
Solution. (a) Since y is y% of z, z = 100. The situation is possible if
and only if y = 100 also.
7. About how many lines can one rotate a regular hexagon through
some angle x, 0� < x < 360�, so that the hexagon again occupies its originalposition?
Solution. (e) The axes of rotational symmetry are the three lines joining
opposite vertices, the three lines joining the midpoints of opposite sides, and
the line through the centre and perpendicular to the hexagon.
8. AB is a diameter of a circle of radius 1 unit. CD is a chord perpen-
dicular to AB that cuts AB at E. If the arc CAD is 2=3 of the circumference
of the circle, what is the length of the segment AE?
Solution. (b) By symmetry, ACD is an equilateral triangle. Hence its
centroid is the centre O of the circle. SinceAO = 1,AE = AO+OE = 3=2.
9. One of Kerry and Kelly lies onMondays, Tuesdays and Wednesdays,
and tells the truth on the other days of the week. The other lies on Thursdays,
Fridays and Saturdays, and tells the truth on the other days of the week. At
noon, the two had the following conversation:
Kerry : I lie on Saturdays.
Kelly : I will lie tomorrow.
Kerry : I lie on Sundays.
On which day of the week did this conversation take place?
Solution. (b) Kerry is clearly lying, and is the one who tells the truth onSaturday. Hence the conversation takes place Monday, Tuesday or Wednes-
day, and Kelly's statement is true only on the last day.
10. How many integer pairs (m;n) satisfy the equation:
m(m+ 1) = 2n?
Solution. (c) Either both m and m + 1 are powers of 2, or both are
negatives of powers of 2. The two solutions are (m;n) = (1; 1) and (�2;1).
217
11. Of the following triangles given by the lengths of their sides, whichone has the greatest area?
Solution. (b)Note that 52+122 = 132. With two sides of length 5 and
12, the area is maximum when there is a right angle between them.
12. If x < y and x < 0, which of the following numbers is never
greater than any of the others?
Solution. (d) If y > 0 or y = 0, then x� y = x� jyj is the minimum.
If y < 0, then x+ y = �jx + yj = x� jyj is the minimum.
13. An x by y ag, with x < y, consists of two perpendicular white
stripes of equal width and four congruent blue rectangles at the corners. If
the total area of the blue rectangles is half that of the ag, what is the length
of the shorter side of each blue rectangle?
Solution. (a) Let z be the length of the shorter side of a blue rectangle.
Then the longer side has length z +(y�x)
2. From 8z(z+ (y� x)) = xy, we
have z = 1
4(x�y+fx2+y2g1=2) or z = 1
4(x� y�fx2+y2g1=2). Clearly,
the negative square root is to be rejected.
14. A game is played with a deck of ten cards numbered from 1 to 10.
Shu�e the deck thoroughly.
(i) Take the top card. If it is numbered 1, you win. If it is numbered k,
where k > 1, go to (ii).
(ii) If this is the third time you have taken a card, you lose. Otherwise,
put the card back into the deck at the kth position from the top and go to (i).
What is the probability of winning?
Solution. (c) If card number 1 is initially in the �rst or second position
from the top, you will win. You can also win if it is in the third, and card
number 2 is not in the �rst. Hence the winningprobability is 1
10+ 1
10+ 1
10
�8
9
�.
15. Five of the angles of a convex polygon are each equal to 108�.In which of the following �ve intervals does the maximum angle of all such
polygons lie?
Solution. (a) The sum of the exterior angles of the �ve given angles is
5(180� � 108�) = 360�. Hence these �ve angles are the only angles of the
convex polygon.
16. Which one of the following numbers cannot be expressed as the
di�erence of the squares of two integers?
Solution. (b) Suppose k = m2 � n2 = (m� n)(m+ n). If k is odd,
we can setm�n = 1 andm+n = k. If k = 4l, we can set m�n = 2 and
m + n = 2l. However, if k = 4l + 2, then m� n and m+ n cannot have
the same parity.
218
As a contest this number we give the Fifteenth W.J. Blundon Contest,
which was taken by students in Newfoundland and Labrador. The contest is
supported in part by the Canadian Mathematical Society. My thanks go to
Bruce Shawyer for forwarding me a copy.
15th W.J. BLUNDON CONTESTFebruary 18, 1998
1. (a) Find the exact value of
1
log2 36+
1
log3 36:
(b) If log15 5 = a, �nd log15 9 in terms of a.
2. (a) If the radius of a right circular cylinder is increased by 50% and
the height is decreased by 20%, what is the change in the volume?
(b) How many digits are there in the number 21998 � 51988?3. Solve: 32+x + 32�x = 82.
4. Find all ordered pairs of integers such that x6 = y2 + 53.
5. When one-�fth of the adults left a neighbourhood picnic, the ratio
of adults to children was 2 : 3. Later, when 44 children left, the ratio of
children to adults was 2 : 5. How many people remained at the picnic?
6. Find the area of a rhombus for which one side has length 10 and the
diagonals di�er by 4.
7. In how many ways can 10 dollars be changed into dimes and quar-
ters, with at least one of each coin being used?
8. Solve:p2 + 10 + 4
px+ 10 = 12.
9. Find the remainder when the polynomial x135+x125�x115+x5+1
is divided by the polynomial x3 � x.
10. Quadrilateral ABCD below has the following properties: (1) The
midpoint O of side AB is the centre of a semicircle; (2) sides AD, DC and
CB are tangent to this semicircle. Prove that AB2 = 4AD � BC.
r
r r
r
r r
r
r
A O B
D
C
E
F
G
219
Now a correction to a solution given in the October number of the Sko-
liad Corner.
4. [1997:347] 1995 P.E.I. Mathematics Competition.An autobiographical number is a natural number with ten digits or less
in which the �rst digit of the number (reading from left to right) tells you how
many zeros are in the number, the second digit tells you how many 1's, the
third digit tells you how many 2's, and so on. For example, 6; 210; 001; 000
is an autobiographical number. Find the smallest autobiographical number
and prove that it is the smallest.
Correction by Vedula N. Murty, Visakhapatnam, India. The answer
given is 1201, but the smallest example is 1210.
Editor's Note: Dyslexia strikes again!
That is all for this issue of the Skoliad Corner. I need your suitable
materials and your suggestions for directions for this feature.
Advance Announcement
The 1999 Summer Meeting of the Canadian Mathematical Society will
take place at Memorial University in St. John's, Newfoundland, from Satur-
day, 29 May 1999 to Tuesday, 1 June 1999.
The Special Session on Mathematics Education will feature the topic
What Mathematics Competitions do for Mathematics.The invited speakers are
Ed Barbeau (University of Toronto),
Ron Dunkley (University of Waterloo),
Tony Gardiner (University of Birmingham, UK), and
Rita Janes (Newfoundland and Labrador Senior Mathematics League).
Requests for further information, or to speak in this session, as well as sug-
gestions for further speakers, should be sent to the session organizers:
Bruce Shawyer and Ed Williams
CMS Summer 1999 Meeting, Education Session
Department of Mathematics and Statistics, Memorial University
St. John's, Newfoundland, Canada A1C 5S7
220
MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by
High School and University Students. It continues, with the same emphasis,
as an integral part of Crux Mathematicorum with Mathematical Mayhem.
All material intended for inclusion in this section should be sent to the
Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,
PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic
address is still
The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).
The rest of the sta� consists of Adrian Chan (Upper Canada College), Jimmy
Chui (Earl Haig Secondary School), Richard Hoshino (University ofWaterloo),
David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).
Shreds and Slices
From the Archives
An excerpt from \Wilson's Theorem", by Oliver Johnson, Volume 5,
Issue 2, Mathematical Mayhem:
One of the more beautiful results in number theory is Wilson's
Theorem. [So named because Sir John Wilson didn't discover it,
and Leibniz who did discover it didn't prove it|Lagrange did. This
is of course an example of the lawyers doing the best out of every-
thing; Wilson was a judge, and it is worth considering how many
of Fermat's theorems weren't just won in lawsuits rather than
proved by himself... So sue me Pierre.]
Primitive Roots and Quadratic Residues, Part 1
For the following we will need some de�nitions, none of which should
be too abstruse or unfamiliar. For a positive integer n, the polynomial xn�1
has n distinct roots in the �eld of complex numbers C. These roots are called
the nth roots of unity, and denoted by �n. An element � in �n is called
primitive if �k 6= 1 for 1 � k � n � 1; in other words, n is the smallest
positive exponent k we must raise � to the power of to obtain 1. It can be
shown that there are exactly �(n) primitive nth roots of unity. If n is a
prime p, then for � to be primitive, it su�ces that � 6= 1; in other words, any
pth root of unity not equal to 1 is primitive. This turns out to be the case we
are interested in.
221
Now, recall quadratic residues modulo n: these are the set of non-zero
squares in modulo n. For example, the non-zero squares modulo 10 are
1, 4, 9, 16 � 6, 25 � 5, 36 � 6, and so on, so f1; 4; 5; 6; 9g is the set of
quadratic residues modulo 10. Let An denote the quadratic residues modulo
n, and Bn the remaining numbers, so B10 = f2; 3; 7; 8g. If n is a prime p,
then both Ap and Bp contain exactly (p� 1)=2 elements, and this is where
things �nally get interesting.
Let p be a prime, let � be a primitive pth root of unity, and let
x =Xa2Ap
�a; y =Xb2Bp
�b:
Then both x+ y and xy are always integers. Actually x+ y = �1, and this
is not hard to see. Since �p = 1, or
�p � 1 = (� � 1)��p�1 + �p�2 + : : :+ � + 1
�= 0
and � 6= 1, the latter factor must be 0. Also, Ap and Bp form a partition of
f1; 2; : : : ; p� 1g. Hence,
x+ y = � + �2 + : : :+ �p�1 = �1:
But why xy is an integer is a little deeper. Let us consider an example. For
p = 7, x = � + �2 + �4 and y = �3 + �5 + �6. Then
xy =�� + �2 + �4
� ��3 + �5 + �6
�= �4 + �6 + �7 + �5 + �7 + �8 + �7 + �9 + �10
= 3+ � + �2 + �3 + �4 + �5 + �6 = 2:
In particular, this shows that to calculate xy, we do not need to resort
to any messy cis notation, and that it is quite accessible by just pencil and
paper. Playing around with other primes reveals the following values:
p 3 5 7 11 13
xy 1 �1 2 3 �3
The absolute value of xy always seems to be (p� 1)=4 (the sign is the
one which makes the resulting value an integer). Why is this? And what
determines the sign of xy? We encourage readers to play around with this,
and to send in any interesting results. We will divulge the reasons behind
this phenomenon, as well as some of the deeper theory it will lead to, in the
next issue.
Hint: If indeed x+y and xy are integers, then x and y are the roots of
a quadratic equation with integer coe�cients. What is this quadratic? What
does this quadratic say about x and y?
222
Mayhem Problems
The Mayhem Problems editors are:
Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.
Note that all correspondence should be sent to the appropriate editor |
see the relevant section. In this issue, you will �nd only solutions | the
next issue will feature only problems.
We warmly welcome proposals for problems and solutions. We request
that solutions from the previous issue be submitted by 1 February 1999, for
publication in issue 4 of 1999. Also, starting with this issue, we would like to
re-open the problems to all CRUX withMAYHEM readers, not just students,
so now all solutions will be considered for publication.
High School Solutions
Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,
Canada. L3S 3E8 <[email protected]>
H221. Let P = 195 +6605 +13165. It is known that 25 is one of the
forty-eight positive divisors of P . Determine the largest divisor of P that is
less than 10; 000.
Solution.If k divides a+b, then k divides a5+b5 since the latter is a multiple of
a+b. If k also divides c, then k divides a5+b5+c5, since k divides c5. Notice
that 19 + 660 + 1316 = 1995 = 3 � 5 � 7 � 19. Since 3 divides 660 and 3 alsodivides 19+1316 = 1335, by our result, 3 divides 6605+195+13165 = P .
Similarly, since 5 divides 660, 7 divides 1316 and 19 divides 19, we can show
that P is also divisible by 5, 7 and 19. Therefore, P can be expressed in the
form 3 � 52 � 7 � 19 �R for some positive integer R, since we are given that 25
is a divisor of P .
If R is prime, we know that P will have 2 � 3 � 2 � 2 � 2 = 48 divisors.
(Recall that if p1, p2, : : : , pn are distinct primes, then p1a1p2
a2 � � � pnan has
(a1 + 1)(a2 + 1) � � � (an + 1) positive divisors.)
Thus, R must be prime; otherwise P will have more than 48 divisors,
and so P = 3 � 52 � 7 � 19 � R = 9975 � R, for some large prime R. Since
13165 > 10005 = 1015, R will certainly be larger than 10; 000. (Using
Maple, one can show that R is 408; 255; 160; 853.) Hence, it follows that
the largest divisor of P less than 10; 000 is 9975.
223
H222. McGregor becomes very bored one day and decides to write
down a three digit number ABC, and the six permutations of its digits. To
his surprise, he �nds thatABC is divisible by 2,ACB is divisible by 3,BAC
is divisible by 4, BCA is divisible by 5, CAB is divisible by 6 and CBA is
a divisor of 1995. Determine ABC.
Solution by Evan Borenstein, student, Woodward Academy, CollegePark, Georgia.
Since BCA is divisible by 5, A must be 0 or 5. But we are given that
ABC is a three-digit number, so A = 5. Since ABC is divisible by 2 and
CAB is divisible by 6, both B and C must be even. Since BAC is divisible
by 4, 10A + C = 50 + C must be a multiple of 4. So C can be 2 or 6.
If C = 2, B must be 2 or 8, since ACB is divisible by 3 and B is even.
If C = 6, B must be 4. Hence, there are only three possibilities for ABC:
522, 582 and 546. And of these three, only the second one will make CBA
a divisor of 1995, since 1995 = 285 � 7. Thus we conclude that ABC is 582.
Also solved by Joel Schlosberg, student, Robert Louis Stevenson School,New York, NY, USA.
H223. There are n black marbles and two red marbles in a jar. One by
one, marbles are drawn at random out of the jar. Jeanette wins as soon as
two black marbles are drawn, and Fraserette wins as soon as two red marbles
are drawn. The game continues until one of the two wins. Let J(n) and F (n)
be the two probabilities that Jeanette and Fraserette win, respectively.
1. Determine the value of F (1) + F (2) + � � �+ F (3992).
2. As n approaches in�nity, what does J(2)� J(3)� J(4)� � � � � J(n)
approach?
Solution.
1. If Fraserette wins, the balls must be drawn in one of the following
three ways: red, red; red, black, red; or black, red, red. This must be the
case, as otherwise two black ballswill be drawn and Jeanette will win. Hence,
the probability that Fraserette wins is the sum of the probabilities of each of
three cases above.
Thus,
F (n) =2
n+ 2� 1
n+ 1+
2
n+ 2� n
n+ 1� 1n+
n
n+ 2� 2
n+ 1� 1n
=2
(n+ 1)(n+ 2)+
2
(n+ 1)(n+ 2)+
2
(n+ 1)(n+ 2)
=6
(n+ 1)(n+ 2):
224
Note that F (n) = 6
n+1� 6
n+2, so we can use a telescoping series to
calculate the desired sum. We have
F (1) + F (2)+ � � �+ F (3992) =
�6
2� 6
3
�+
�6
3� 6
4
�+
�6
4� 6
5
�
+ � � �+�
6
3993� 6
3994
�
=6
2� 6
3994
= 3� 3
1997
=5988
1997:
2. Now J(n) = 1�F (n) = 1� 6
(n+1)(n+2)= n2+3n�4
(n+1)(n+2)=
(n�1)(n+4)
(n+1)(n+2).
Thus,
J(2)� J(3)� J(4)� � � � � J(n)
=1 � 63 � 4 �
2 � 74 � 5 �
3 � 85 � 6 �
4 � 96 � 7 � � �
(n� 1) � (n+ 4)
(n+ 1) � (n+ 2)
=1 � 2 � 3 � 4 � � � (n� 2) � (n� 1)
3 � 4 � 5 � 6 � � �n � (n+ 1)
� 6 � 7 � 8 � 9 � (n+ 3) � (n+ 4)
4 � 5 � 6 � 7 � � � (n+ 1) � (n+ 2)
=1 � 2
n � (n+ 1)� (n+ 3) � (n+ 4)
4 � 5
=1
10� (n+ 3)(n+ 4)
n(n+ 1)
=1
10� n+ 3
n� n+ 4
n+ 1:
Thus, as n approaches in�nity, n+3
n= 1+ 3
nand n+4
n+1= 1+ 3
n+1both
approach 1, and so J(2)� J(3)� J(4)� � � � � J(n) approaches 1
10.
H224. Consider square ABCD with side length 1. Select a point M
exterior to the square so that \AMB is 90�. Let a = AM and b = BM .
Now, determine the point N exterior to the square so that CN = a and
DN = b. Find, as a function of a and b, the length of the line segmentMN .
I. Solution by Adrian Chan, student, Upper Canada College, Toronto,Ontario.
Let O be the center of the square. Consider a horizontal re ection
through the line parallel to DA and passing through O. Let the image ofM
andN about this line beM 0 andN 0, respectively. There exists a point where
225
N
C B
M
AD
b
a
rO
N
C B
M
AD
N 0
M 0
MN intersects M 0N 0. Let this point beK. Since K lies on both lines, this
point must lie on this horizontal line of re ection.
Now, the diagram is also symmetrical about the line parallel to DC
passing through O. ThenK must lie on this line as well. This leaves us with
K coinciding with O, since the two lines intersect at O.
Now, because of the symmetry, NO = OM . That is, OM = MN2
.
Now consider quadrilateral OAMB. Since \AOB is right and so is \AMB,
then the quadrilateral is cyclic.
By Ptolemy's Theorem on this quadrilateral, we have OM � AB =
AM � OB +MB � OA = a � 1p2+ b � 1p
2, since OA = OB = 1p
2. Since
OB = 1, we have OM =p2
2(a + b), and so MN =
p2(a + b), since
MN = 2OM . Thus the length of line segmentMN isp2(a+ b).
II. Solutionby Joel Schlosberg, student, Robert Louis StevensonSchool,New York, NY, USA.
Let \BAM = x. Then a = cosx and b = sinx. Let O be the centre
of ABCD and R be the midpoint of AB. Then \BAM = \RAM =
\RMA = x, and so \ORM = \ARO+\ARM = �2+(��2x) = 3�
2�2x.
Now OR = RA = RM = 1
2. By the cosine law on triangle ROM , we have:
OM2 = OR2 +RM2 � 2OR � OM cos\ORM
=1
4+
1
4� 2 � 1
2� 12� cos
�3�
2� 2x
�
=1
2� 1
2� (� sin2x)
=1
2� 1
2� 2 sinx cosx =
1
2+ ab:
Thus using the fact that a2 + b2 = 1, we have 4OM2 = 2(1 + 2ab) =
2(a2 + b2 + 2ab) = 2(a + b)2. Hence, 2OM =p2(a + b), by taking the
square root of both sides (Note: a, b and OM are all positive).
Since 2OM =MN , we haveMN =p2(a+ b).
226
Advanced Solutions
Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.
M1G 1C3 <[email protected]>
A197. Calculate Z �2
��2
sin(2N + 1)�
sin �d�;
where N is a non-negative integer.
Solution by D.J. Smeenk, Zaltbommel, the Netherlands, with minormodi�cations.
Let f(N) denotesin(2N+1)�
sin�. Then
f(N + 1)� f(N) =sin(2N + 3)� � sin(2N + 1)�
sin�:
Now
sin(2N + 3)� � sin(2N + 1)�
= sin(2N + 1)� cos 2� + cos(2N + 1)� sin 2� � sin(2N + 1)�
= sin(2N + 1)�(cos 2� � 1) + cos(2N + 1)� sin2�
= �2 sin(2N + 1)� sin2 � + 2 cos(2N + 1)� sin� cos �
= 2 sin�(cos(2N + 1)� cos � � sin(2N + 1)� sin �)
= 2 sin� cos(2N + 2)�:
These two equations imply f(N + 1)� f(N) = 2 cos(2N + 2)�. So,Z �2
��2
f(N + 1) d� �Z �
2
� �2
f(N) d� = 2
Z �2
� �2
cos(2N + 2)� d� = 0:
This means
f(N) = f(0) =
Z �2
��2
sin�
sin�d� = �:
A198. Given positive real numbers a, b, and c such that a+ b+ c = 1,
show that aabbcc + abbcca + acbacb � 1.
Solution.Using the weighted AM-GM inequality three times, we have the fol-
lowing:
c � a+ a � b+ b � cc+ a+ b
� (acbacb)1
a+b+c ;
b � a+ c � b+ a � cb+ c+ a
� (abbcca)1
a+b+c ;
a � a+ b � b+ c � ca+ b+ c
� (aabbcc)1
a+b+c :
227
Adding these inequalities together gives
1 = a+ b+ c =(a+ b+ c)2
a+ b+ c
=a2 + b2 + c2 + 2ab+ 2ac+ 2bc
a+ b+ c
� aabbcc + abbcca + acbacb:
A199. Let P be a point inside triangle ABC. Let A0, B0, and C0 bethe re ections of P through the sides BC, AC, and AB respectively. For
what points P are the six points A, B, C, A0, B0, and C0 concyclic?
Solution.
r
r
r
r
r r
r
A
B
CB0
A0
C0
P
We show that AP must be perpendicular to BC. Similar arguments
will show the same for BP and CP . Thus, P must be the orthocentre of
triangle ABC.
To see that AP is perpendicular to BC, consider angles C0A0A and
C0A0P as shown in the diagram. Then \C0A0A = \C0BA = \PBA =
\PBF = \FDP = \C0A0P from the quadrilaterals AC0BA0 and BFPDbeing concyclic and the last equality from similar triangles PFD and PC0A0.Thus, P lies on AA0 so AP is perpendicular to BC.
A200. Given positive integers n and k, for 0 � i � k � 1, let
Sn;k;i =X
j�i (mod k)
�n
j
�:
Do there exist positive integers n, k > 2, such that Sn;k;0, Sn;k;1, : : : ,
Sn;k;k�1 are all equal?
Solution.
The answer is NO. To see this, consider the kth roots of unity. In par-
ticular, since k > 2, there is a k such that !k = 1, ! 6= �1. Now consider
228
the expansion of (1 + !)n:
(1 + !)n =
nXj=0
�n
j
�!j
=X
j�0 (mod k)
�n
j
�+ !
Xj�1 (mod k)
�n
j
�
+ � � �+ !k�1X
j�k�1 (mod k)
�n
j
�
= Sn;k;0 + Sn;k;1! + � � �+ Sn;k;k�1!k�1:
Now if all the Sn;k;i are equal, say to A, then we have (1 + !)n =
A(1 + ! + � � �+ !k�1) = A � 0 = 0. Thus ! = �1, contradiction.
Challenge Board Solutions
Editor: David Savitt, Department of Mathematics, Harvard University,
1 Oxford Street, Cambridge, MA, USA 02138 <[email protected]>
C74. Prove that the k-dimensional volume of a parallelepiped in Rn
spanned by vectors ~v1, : : : , ~vk is the square root of the determinant of the
k � k matrix f~vi � ~vjgi;j.Solution.
First, note that by restricting to any k-dimensional subspace of the
n-space which contains ~v1, : : : , ~vk, we may assumewithout loss of generality
that k = n. LetM be the n�nmatrix whose ith column is ~vi, and let P be
the parallelepiped spanned by the ~vi. Under the coordinate transformation
� sending ~x to M~x, the ith elementary basis vector ~ei = (0; : : : ; 1; : : : ; 0)
is sent to ~vi, and so � transforms the unit cube [0; 1]n onto P . We �nd then,
that
Volume(P ) =
ZP
1 dV =
Z[0;1]n
j det�0j dV:
Since�0 =M , it follows thatVolume(P ) = jdetM j = (detMTM)1=2,
and since the i; jth-entry ofMTM is indeed ~vi � ~vj , we are done.Remark. The above proof that Volume(P ) = j detM j is actually a bit
bogus, since use is usually made of that result when deriving the change-of-
variables formula for integration. So, for those of you who are dissatis�ed:
let V (~v1; : : : ; ~vn) be the oriented volume of the parallelepiped spanned by
~v1, : : : , ~vn in that order. One can then verify that:
(1) V (~e1; : : : ; ~en) = 1,
229
(2) V (~v1; : : : ; ~vi; : : : ; ~vj ; : : : ; ~vn) = �V (~v1; : : : ; ~vj ; : : : ; ~vi; : : : ; ~vn), and
(3) V (~v1; : : : ; ~vi + c~v0i; : : : ; ~vn) = V (~v1; : : : ; ~vi; : : : ; ~vn)
+cV (~v1; : : : ; ~v0i; : : : ; ~vn),
where the last assertion follows because V (~v1; : : : ; ~w; : : : ; ~vn) is directly
proportional to the component of ~w which lies perpendicular to the hyper-
plane spanned by ~v1, : : : , ~vi�1, ~vi+1, : : : , ~vn. However, we know from
linear algebra that the determinant is the unique function satisfying condi-
tions (1)-(3), so we conclude that V = det, the oriented volume of P is
detM , and indeed Volume(P ) = j detM j.
In clearing out the Mayhem archives, we also dug up this solution to
an old problem.
S21. Proposed by Colin Springer.
There are n houses situated around a certain lake, in a circle. Each is
painted one of k colours, chosen at random. Find the probability that no two
neighbouring houses are of the same colour.
Solution by Philip Oppenheimer, South Norwalk, CT.
For convenience, we use the following notation: Let P (n; k) denote the
desired probability. Let h(a) denote the colour of the ath house, starting
from a �xed house, going around the circle.
Without loss of generality, we may �x h(1) = 1. Under this condition,
let p(n; a; b) denote the probability that h(a) = b, if indeed no two adjacent
houses are painted the same colour.
With h(1) = 1, there are k� 1 ways to colour the second house (out ofk colours), the third house, etc., until the (n� 1)th house. If h(n� 1) = 1,then there are k�1 ways to colour the nth house. However, if h(n�1) 6= 1,then there are k� 2 ways to colour the nth house. Hence,
P (n; k) =
�k � 1
k
�n�2 �k� 1
kp(n;n� 1; 1) +
k� 2
k(1� p(n;n� 1; 1))
�:
But for all a,
p(n;a; 1) =1
k� 1(1� p(n; a� 1; 1))
=1
k� 1� 1
k � 1p(n;a� 1; 1)
230
so that
p(n; a; 1)� 1
k=
1
k(k� 1)� 1
k � 1p(n;a� 1; 1)
=�1k� 1
�p(n;a� 1; 1)� 1
k
�
=
� �1k� 1
�2 �p(n;a� 2; 1)� 1
k
�
=...
=
� �1k� 1
�a�1 �p(n;1; 1)� 1
k
�
=(�1)a�1(k� 1)a�1
� k� 1
k
=(�1)a�1
k(k� 1)a�2:
Therefore,
p(n; n� 1; 1) =(�1)n
k(k� 1)n�3;
and
P (n; k) =
�k� 1
k
�n�2 �k� 1
k
�1
k+
(�1)nk(k� 1)n�3
�
+k� 2
k
�1� 1
k� (�1)nk(k� 1)n�3
��
=
�k� 1
k
�n�2 �k � 1
k2+
(�1)n(k� 1)
k2(k� 1)n�3
+(k� 1)(k� 2)
k2� (�1)n(k� 2)
k2(k� 1)n�3
�
=
�k� 1
k
�n�2 �(k� 1)2
k2+
(�1)nk2(k� 1)n�3
�
=
�k� 1
k
�n+
(�1)n(k� 1)
kn:
231
Swedish Mathematics Olympiad
1986 Qualifying Round
1. Show that
(1986!)1=1986 < (1987!)1=1987:
2. Show that for t > 0,
t2 +1
t2� 3
�t+
1
t
�+ 4 � 0:
3. A circle C1 with radius 1 is internally tangent to a circle C2 with radius
2. Let ` be a line through the centres of the circles C1 and C2. A circle
C3 is tangent to C1, C2, and `. Find the radius of C3.
4. In how many ways can 11 apples and 9 pears be shared among 4 chil-
dren, so that every child gets 5 fruit? (The apples are identical, as are
the pears.)
5. P is a polynomial of degree greater than 2 with integer coe�cients and
such that P (2) = 13 and P (10) = 5. It is known that P has a root
which is an integer. Find it.
6. The numbers 1, 2, : : : , n are placed in some order at di�erent points
on the circumference of a circle. Form the product of each pair of neigh-
bouring numbers. How should the numbers be placed in order for the
sum of these products to be as large as possible?
1986 Final Round
1. Show that the polynomial
x6 � x5 + x4 � x3 + x2 � x+ 3
4
has no real roots.
2. ABCD is a quadrilateral, and O is the intersection of the diagonals
AC and BD. The triangles AOB and COD have areas S1 and S2respectively, and the area of ABCD is S. Show thatp
S1 +pS2 �
pS:
Show also that equality holds if and only if the lines AB and CD are
parallel.
232
3. LetN be a positive integer,N � 3. Form all pairs (a; b) of consecutive
integers such that 1 � a < b � N and consider the quotient q = b
a
for every such pair. Remove all pairs with q = 2. Show that of the
remaining pairs, there are as many with q < 2 as there are with q > 2.
4. Show that the only positive solution of
x+ y2 + z3 = 3
y + z2 + x3 = 3
z + x2 + y3 = 3
is x = y = z = 1.
5. In the arrangement of pn real numbers below, the di�erence between
the greatest and least numbers in every row is at most d, where d > 0:
a11 a12 � � � a1na21 a22 � � � a2n
.
.
....
. . ....
ap1 ap2 � � � apn
In each column, the numbers are now ordered by size, so that the great-
est appears in the �rst row, the next greatest in the second row, and so
on. Show that the di�erence between the greatest and least numbers
in each of the rows is still at most d.
6. A �nite number of intervals on the real line together cover the inter-
val [0; 1]. Show that one can choose a number of these intervals such
that no two have any points in common and whose total length is at
least 1=2.
J.I.R. McKnight Problems Contest 1982
1. (a) Given the equal positive rationals ab, cd, ef, prove that�
ma4 + nc4 + pe4
mb4 + nd4 + pf4
�1=4is equal to each of the given rationals.
(b) Given that a4+b4 = 7c and that a and b are the roots of x2�5x+3,
�nd c.
2. Consider AB, the major axis of an ellipse centred at the origin with
focus F as shown. Let P be any point on the ellipse. Draw the lines
BP and AP and extend them so that they cross the directrix of F at R
and S respectively. Prove that \RFS is a right angle.
233
BO
A
F
P
R
S
3. Solve the system of equations:
xy + yz + zx = �4y + z � yz = 4
x� y � z = 3
4. If cos�1 x+cos�1 y+cos�1 z = �, prove that x2+y2+z2+2xyz = 1.
5. A shopkeeper orders 19 large and 3 small packets of marbles, all alike.
When they arrive at the shop it is discovered that all the packets have
come open with the marbles loose in the container. If the total number
of marbles is 224, can you help the shopkeeper put up the packets with
the proper number of marbles in each?
6. A radar tracking station is located at ground level vertically below the
path of an approaching aircraft ying at 900 km/h at a constant height
of 10000 m. Find the rate in degree/s at which the radar beam to the
aircraft is turning at the instant when the aircraft is at a horizontal dis-
tance of 3 km from the station.
�x
10 000 m
� 900 km/h
234
PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, De-partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.
In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.
To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81
2"�11" or A4 sheets of paper.
These may be typewritten or neatly hand-written, and should be mailed tothe Editor-in-Chief, to arrive no later than 1 December 1998. They may alsobe sent by email to [email protected]. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication. Please note that we do not accept submissions sent by FAX.
2338. Proposed by Toshio Seimiya, Kawasaki, Japan.Suppose ABCD is a convex cyclic quadrilateral, and P is the intersec-
tion of the diagonals AC and BD. Let I1, I2, I3 and I4 be the incentres of
triangles PAB, PBC, PCD and PDA respectively. Suppose that I1, I2,
I3 and I4 are concyclic.
Prove that ABCD has an incircle.
2339. Proposed by Toshio Seimiya, Kawasaki, Japan.A rhombus ABCD has incircle �, and � touches AB at T . A tangent
to � meets sides AB, AD at P , S respectively, and the line PS meets BC,
CD at Q, R respectively. Prove that
(a)1
PQ+
1
RS=
1
BT,
and
(b)1
PS� 1
QR=
1
AT:
235
2340. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.
Let � > 0 be a real number and a, b, c be the sides of a triangle. Prove
that Ycyclic
s+ �a
s� a� (2�+ 3)3:
[As usual s denotes the semiperimeter.]
2341. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.
Let a, b, c be the sides of a triangle. For real � > 0, put
s(�) :=
������Xcyclic
"�a
b
����b
a
��#������and let4(�) be the supremum of s(�) over all triangles.
1. Show that 4(�) is �nite if � 2 (0; 1] and4(�) is in�nite for � > 1.
2.? What is the exact value of4(�) for � 2 (0;1)?
2342. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.Given A and B are �xed points of circle �. The point C moves on �,
on one side of AB. D and E are points outside 4ABC such that 4ACDand 4BCE are both equilateral.
(a) Show that CD and CE each pass through a �xed point of � when C
moves on �.
(b) Determine the locus of the midpoint of DE.
2343. Proposed by Doru Popescu Anastasiu, Liceul \Radu Gre-ceanu", Slatina, Olt, Romania.
For positive numbers sequences fxngn�1, fyngn�1, fzngn�1 with con-ditions: for n � 1, we have
(n+1)x2n+(n2+1)y2n+(n2+n)z2n = 2pn�nxnyn +
pnxnzn + ynzn
�;
and for n � 2, we have
xn +pnyn � nzn = xn�1 + yn�1 �
pn� 1zn�1:
Find limn!1 xn, limn!1 yn and limn!1 zn.
2344. Proposed by Murali Vajapeyam, student, Campina Grande,Brazil and Florian Herzig, student, Perchtoldsdorf, Austria.
Find all positive integers N that are quadratic residues modulo all
primes greater than N .
236
2345. Proposed by Vedula N. Murty, Visakhapatnam, India.Suppose that x > 1.
(a) Show that ln(x) >3(x2 � 1)
x2 + 4x+ 1.
(b) Show thata� b
ln(a)� ln(b)<
1
3
�2pab+
a+ b
2
�;
where a > 0, b > 0 and a 6= b.
2346. Proposed by Juan-Bosco Romero M�arquez, Universidad deValladolid, Valladolid, Spain.
The angles of4ABC satisfy A > B � C. Suppose that H is the foot
of the perpendicular from A to BC, that D is the foot of the perpendicular
from H to AB, that E is the foot of the perpendicular from H to AC, that
P is the foot of the perpendicular form D to BC, and that Q is the foot of
the perpendicular from E to AB.
Prove that A is acute, right or obtuse according as AH �DP �EQ is
positive, zero or negative.
2347. Proposed by �Sefket Arslanagi�c, University of Sarajevo, Sara-jevo, Bosnia and Herzegovina.
Prove that the equation x2+y2 = z1998 has in�nitely many solutions
in positive integers, x, y and z.
2348. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.Without the use of trigonometrical formulae, prove that
sin (54�) = 1
2+ sin (18�) :
2349. Proposed by V�aclav Kone �cn �y, Ferris State University, BigRapids, Michigan, USA.
Suppose that 4ABC has acute angles such that A < B < C. Prove
that
sin2B sin�A2
�sin
�A+ B
2
�> sin2A sin
�B2
�sin�B + A
2
�:
2350. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.
Suppose that the centroid of 4ABC is G, and that M and N are the
mid-points of AC and AB respectively. Suppose that circles ANC and
AMB meet at (A and) P , and that circle AMN meets AP again at T .
1. Determine AT : AP .
2. Prove that \BAG = \CAT .
237
SOLUTIONSNo problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.
2223. [1997: 111] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,Alcorc �on, Madrid, Spain.
We are given a bag with n identical bolts and n identical nuts, which
are to be used to secure the n holes of a gadget.
The 2n pieces are drawn from the bag at random one by one. Throughout
the draw, bolts and nuts are screwed together in the holes, but if the number
of bolts exceeds the number of available nuts, the bolt is put into a hole until
one obtains a nut, whereas if the number of nuts exceeds the number of bolts,
the nuts are piled up, one on top of the other, until one obtains a bolt.
Let L denote the discrete random variable which measures the height of the
pile of nuts.
Find E[L] +E[L2].
Solution by Gerry Leversha, St. Paul's School, London, England.[Editor's comment: Leversha noticed the typographical error in the
original statement of the problem and recognized correctly that what wasrequired was E[L] + E[L2]. He found the equivalent: E[L2 + L].]
Denote the sequences of bolts and nuts by 0's and 1's respectively;
thus, for n = 2 there are six possible sequences: 1100, 1010, 1001, 0110,
0101, and 0011, and, in general, there are�2n
n
�di�erent sequences. Call
the maximum height of the pile of nuts represented by such a sequence the
value of the sequence. In the sequences above the heights are respectively
2; 1; 1; 1; 0; 0.
We approach this problem by deriving a probability generating function
Gn(x) for the case of n nuts and n bolts. De�ne gn(x) =�2n
n
�Gn(x); then
gn(x) =
nXr=0
anrxr, where anr is the number of sequences of length 2n which
have value r. The �rst few such polynomials are as follows:
g0(x) = 1
g1(x) = x+ 1
g2(x) = x2 + 3x+ 2
g3(x) = x3 + 5x2 + 9x+ 5
g4(x) = x4 + 7x3 + 20x2 + 28x+ 14
Notice in each case that the sum of the coe�cients is gn(1) =
�2n
n
�, as
required. In fact, the polynomials can be calculated from the following in-
ductive de�nition:
238
xgn+1(x) = (x+ 1)2gn(x)� (x+ 1)gn(0) (�)starting with g0(x) = 1. It is easy to check that gn(x) is indeed a polynomial
of degree n.
(�) follows by virtue of the following relationships between the coe�-
cients:
an+1
n+1= ann
an+1
n = 2ann + ann�1
an+1
r = anr+1+ 2anr + anr�1 (1 � r � n� 1)
an+1
0= an
1+ an
0
The �rst of these is immediate, since it merely states that the leading coe�-
cient is 1, and there is obviously only one way to obtain a value of n+ 1 in
an n� 1 sequence, namely by having all the 1's at the front.
The second statement follows because a value of n can be obtained
either by placing 10 in front of the unique sequence of value n, or by placing
01 in front of the sequence of value n, or by placing 1 in a sequence of value
n� 1 in front of that part of the sequence which creates the value.
The third statement is the most di�cult to see. It states that a sequence
of value r, where r is at least 1 and at most n � 1, can be made either by
placing 01 or 10 in front of an existing r sequence, or by devaluing an existing
r + 1 sequence by placing an extra 0 in front of the part which matters, and
an extra 1 at the end, or by adding 1 to an existing r � 1 sequence.
The �nal statement says that a 0 sequence can be created either by
putting 01 in front of an existing 0 sequence or by devaluing an existing 1
sequence by placing an extra 0 in front of the critical part.
These claims are best checked by looking at, say, the sequences for
n = 3 and seeing how the sequences for n = 4 are constructed.
We can now proceed to a calculation ofE[L+L2]; this is done by �nding
the value of G00n(1) + 2G0n(1) =g00n(1) + 2g0n(1)
gn(1). In fact, we can now make
the following claim: E[L2 + L] = n.
This is proved using induction on n. It is trivially true for n = 0, so we
assume that it is true for n = k; that is, we assume that
g00k (1) + 2g0k(1) = kgk(1):
By di�erentiating twice the de�ning equation for gk+1(x), we have
xg0k+1(x) + gk+1(x) = (x+ 1)2g0k(x) + 2(x+ 1)gk(x)� gk(0) ;
xg00k+1(x) + 2g0k+1
(x) = (x+ 1)2g00k (x) + 4(x+ 1)g0k(x) + 2gk(x) :
239
Hence, putting x = 1, we have
g00k+1(1) + 2g0k+1
(1) = 4g00k(1) + 8g0k(1) + 2gk(1)
= (4k+ 2)gk(1) (by the inductive hypothesis)
Hence, for n = k + 1; E[L2 + L] =(4k+ 2)gk(1)
gk+1(1).
However, we know that gk(1) =
�2k
k
�, and so
E[L2 + L] =(4k+ 2)(2k)!(k+ 1)!(k+ 1)!
k!k!(2k+ 2)!;
and it is straightforward to check that this reduces to k+1. This �nishes the
induction step and establishes the claim.
Also solved by the proposer (by a completely di�erent method).
2227. [1997: 166] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,Alcorc �on, Madrid, Spain.
Evaluate Yp
" 1Xk=0
�2k
k
�(2p)2k
#:
where the product is extended over all prime numbers.
Composite solution by David Doster, Choate Rosemary Hall, Wall-ingford, Connecticut, USA and Kee-Wai Lau, Hong Kong.
It is well known that for jxj < 1
4, we have
1Xk=0
�2k
k
�xk = (1� 4x)�1=2; and hence,
1Xk=0
�2k
k
�(2p)2k
=
�1� 1
p2
��1=2:
It is also known that �(s) =Yp
�1� 1
ps
��1for s > 1, where � is the
Riemann Zeta function, and that �(2) = �2
6. (See, for example, Hardy and
Wright, An Introduction to the Theory of Numbers, 5th edition, p. 246.)
Therefore,
Yp
1Xk=0
�2k
k
�(2p)2k
!=Yp
�1� 1
p2
��1=2=��(2)
�1=2=
�p6:
Also solved by MANSUR BOASE, student, St. Paul's School, London, England;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; CON AMORE
240
PROBLEM GROUP, Royal Danish School of Educational Studies, Copenhagen, Den-mark; RICHARD I. HESS, Rancho Palos Verdes, California, USA; ROBERT B. ISRAEL,University of British Columbia, Vancouver, BC; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; MICHAEL LAMBROU,University of Crete, Crete, Greece;HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the proposer.
Most of the submitted solutions are similar to, or virtually the same as, the
one given above. By using the same argument, one can easily show that the value
of the slightly more general product P (�) =Yp
�2k
k
�(4p�)k
!is��(�)
�1=2. This was
pointed out by Janous and the proposer.
2228. [1997: 167] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.
Let A be the set of all real numbers from the interval (0; 1) whose dec-
imal representation consists only of 1's and 7's; that is, let
A =
( 1Xk=1
ak
10k: ak 2 f1; 7g
):
Let B be the set of all reals that cannot be expressed as �nite sums of mem-
bers of A. Find supB.
Solution by the proposer.
We prove that supB = 1.
Fix a 2 [1; 7] and consider b = (a � 1)=6. Since b 2 [0; 1], b can
be easily expressed as a sum of nine reals from [0; 1] each consisting only
of 0's and 1's in decimal representation. Now multiplying each by 6 and
next adding :1111 : : : to the result, we obtain (since 1 = 9 � :1111 : : : ) an
expression for a as a sum of nine reals, each consisting only of 1's and 7's.
Now take any a > 7 and subtract from a as many :1111 : : : 's as needed
to get into the interval [1; 7]. This procedure shows that any a � 1 is a �nite
sum of members of A; that is, supB � 1.
Now let x 2 (8=9;1) be a �nite sum of members of A (so that
x 62 B). Fixing our attention, assume that x is a sum of k members of
A; since x 2 (8=9;1) and each member of A is at least :1111 : : : , we know
that 2 � k � 8. Consider
y =x� k(:1111 : : : )
6:
Then y is a sum of k real numbers, each consisting only of 1's and 0's. This
implies, since k � 8, that y does not contain a digit 9 in its decimal repre-
sentation. Thus we have shown that a number x 2 (8=9; 1) must be in B if
the following property holds:
241
(�): all seven numbers 1
6(x � k(:1111 : : : )); k = 2; 3; : : : ; 8,
contain a 9 in their decimal representation.
Denote the interval �8� k
54;9� k
54
�
by Jk (k = 2; 3; : : : ; 8). The set Ck of all numbers from Jk having a unique
decimal representation and containing at least one 9 is easily seen to be open
and dense in Jk. [Editorial note. For the bene�t of readers not familiar with
these terms, here are a couple of explanations. \Ck is open" means that for
any c 2 Ck, there is some interval (c� �; c+ �) (� > 0) which is completely
contained in Ck; that is, all numbers close enough to c also have a 9 in their
decimal representation. \Ck is dense in Jk" means that for any number
y 2 Jk, and for any � > 0, the interval (c � �; c + �) must contain some
member of Ck; that is, there are numbers with at least one digit 9 as close
as you like to y, even if y itself does not have a 9 in it.]
For each k = 2; 3; : : : ; 8, the map
'k(x) =x� k(:1111 : : : )
6=
9x� k
54
is increasing and continuous and takes (8=9; 1) onto Jk. Therefore the sets
'�1k (Ck), k = 2; 3; : : : ; 8 are open and dense in (8=9;1), and thus so is the
intersection
'�12(C2) \ '�13
(C3) \ � � � \ '�18(C8):
[Editorial note. In other words, the intersection of �nitely many open and
dense sets is open and dense. Proof left to the reader! Or look at problem 16,
Chapter 2 of Rudin's Principles of Mathematical Analysis.] But all elements
x in this intersection have the property (�), and thus lie in B, which means
that B is dense in (8=9;1). This completes the proof that supB = 1.
Remark. In the last part of the proof we have used Baire's Theorem, in
fact its easier version for the intersection of �nitely many open and dense sets
instead of countably many. Also, the argument used in the solution shows
that there are plenty of elements of B in (8=9;1)! So I believe that a reader
not familiar with topology might �nd an explicit example of a sequence of
elements of B which get arbitrarily close to 1.
Also solved by GERALD ALLEN, CHARLES DIMINNIE, TREY SMITH andROGER ZARNOWSKI (jointly), Angelo State University, San Angelo, Texas, USA;MICHAEL LAMBROU, University of Crete, Crete, Greece; and DAVID STONE andVREJ ZARIKIAN, Georgia Southern University, Statesboro, Georgia, USA. Two otherreaders sent in incomplete solutions.
The other solutions to this problem actually contain the explicit examples that
the proposer asks for in his remark above. Allen, Diminnie, Smith and Zarnowski
give bn = :99 : : : 944999, where there are (3n + 1) 9's before the two 4's. Stone
and Zarikian use bn = :99 : : : 9546, where there are (3n � 2) 9's before the 546.
242
Lambrou uses somewhat more complicated bn's. In all cases the solvers then must
do some calculation to show that the bn's all lie in B.
2231. [1997: 167] Proposed by Herbert G �ulicher, WestfalischeWilhelms-Universit�at, M�unster, Germany.
In quadrilateral P1P2P3P4, suppose that the diagonals intersect at the
point M 6= Pi (i = 1; 2; 3; 4). Let \MP1P4 = �1, \MP3P4 = �2,
\MP1P2 = �1 and \MP3P2 = �2.
Prove that
�13 :=jP1M jjMP3j
=cot�1 � cot�1
cot�2 � cot�2;
where the+(�) sign holds if the line segment P1P3 is located inside (outside)
the quadrilateral.
Preliminary comment. No solver made use of directed angles and seg-
ments, which would have reduced the problem to a single case. The pro-
poser's angles are assumed to be positive, and nobody explicitlywent through
all four resulting possibilities: M can lie between P1 and P3 or not, and be-
tween P2 and P4 or not. Since P1P2P3P4 need not have an interior, a more
careful statement of the proposer's two cases would distinguish whether or
not P2 and P4 lie on opposite sides of the line P1P3. We shall feature the
opposite-side case, and leave the other to the reader.
Solution by Michael Lambrou, University of Crete, Crete, Greece(modi�ed by the editor to explicitly distinguish the two subcases).
Assume M to be any point of the line P1P3 except P1; P3, and let
P2P4 be any other line throughM with P2; P4 on opposite sides ofM . Let
! = \P1MP4. Then sin\P1P4M = sin(��(�1+!)) = sin(�1+!), while
sin\P3P4M = sin(!��2) where `+' corresponds to the subcase whereM
is outside the segment P1P3, while `�' corresponds to whereM is between
P1 and P3. From the sine rule applied to trianglesMP1P4 andMP3P4 we
have
P1Msin�1
sin(�1 + !)=MP4 =MP3
sin�2
sin(!� �2):
Hence,
P1M
MP3=
sin�2 sin(�1 + !)
sin(! � �2) sin�1
=sin�2 sin�1 cos! + sin�2 cos�1 sin!
sin! cos�2 sin�1 � cos! sin�2 sin�1
=cot! + cot�1
cot�2 � cot!(1)
(where we divided both the numerator and the denominator by the product
of three sines to get the last equality).
243
Similarly from trianglesMP1P2 andMP3P2 we have
P1M
MP3=
sin�2 sin(! � �1)
sin�1 sin(! � �2)
=cot�1 + cot!
cot�2 � cot!: (2)
The result follows by combining (1) and (2) using the following property of
proportions: ifa
b=
c
d=
e
fthen a
b=
(c+e)
(d+f).
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho
Palos Verdes, California, USA; JOHN G. HEUVER, Grande Prairie Composite High
School, Grande Prairie, Alberta; WALTHER JANOUS, Ursulinengymnasium, Inns-
bruck, Austria; GERRY LEVERSHA, St. Paul's School, London, England; ISTV �AN
REIMAN, Budapest, Hungary; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK,
Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the
proposer.
2232. [1997: 168] Proposed by �Sefket Arslanagi�c, University of Sara-jevo, Sarajevo, Bosnia and Herzegovina.
Find all solutions of the inequality:
n2 + n� 5 <
�n
3
�+
�n+ 1
3
�+
�n+ 2
3
�< n2 + 2n� 2; (n 2 N):
(Note: If x is a real number, then bxc is the largest integer not exceeding x.)Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,
Ontario.The only solution is n = 2. Let
M =
�n
3
�+
�n+ 1
3
�+
�n+ 2
3
�:
Then
M � n
3+n+ 1
3+n+ 2
3= n+ 1:
Hence n2 + n� 5 < n+ 1, or n2 < 6, and so n = 1 or 2. But when n = 1,
M = 1 = n2+2n� 2, and so n = 1 is not a solution. By inspection, n = 2
is a solution.
Also solved by CHARLES ASHBACHER, Cedar Rapids, Iowa, USA; MICHELBATAILLE, Rouen, France; ADRIANCHAN, student, Upper Canada College, Toronto,Ontario; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA;FLORIAN HERZIG, student, Perchtoldsdorf, Austria; JOE HOWARD, New MexicoHighlands University, Las Vegas, NM, USA; WALTHER JANOUS, Ursulinengymnas-ium, Innsbruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA; MICHAEL
244
LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School,London, England; SEAMMCILROY, student, Vancouver, BC; ANNEMARTIN, West-mont College, Santa Barbara, California, USA; CAN ANH MINH, Berkeley, Califor-nia, USA; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; CHRISTOSSARAGOTIS, student, Aristotle University, Thessaloniki, Greece; ROBERT P. SEALY,Mount Allison University, Sackville, New Brunswick; HEINZ-J�URGEN SEIFFERT,Berlin, Germany; REZA SHAHIDI, student, University of Waterloo, Waterloo, On-tario, Canada; D.J. SMEENK, Zaltbommel, the Netherlands; DIGBY SMITH, MountRoyal College, Calgary, Alberta; DAVID R. STONE, Georgia Southern University,Statesboro, Georgia, USA; and the proposer. There were also three incomplete andone incorrect solution submitted.
Some solvers initially prove the equalityjn
3
k+
�n+ 1
3
�+
�n+ 2
3
�= n;
or some more general form of it. The following general form is well-known.
n�1Xi=0
�x+
i
n
�= bnxc for n 2 N; n 6= 0; and x 2 R
(See for example, D.O. Shklarsky, N.N. Chentzov, and I.M. Yaglom, The USSR
Olympiad Problem Book, W.H. Freeman and Company, 1962, p. 24, problem 101,
(3).)
2233. [1997: 168] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.
Let x, y, z be non-negative real numbers such that x+ y+ z = 1, and
let p be a positive real number.
(a) If 0 < p � 1, prove that
xp + yp+ zp � Cp�(xy)p+ (yz)p+ (zx)p
�;
where
Cp =
(3p if p � log 2
log 3�log 2 ;
2p+1 if p � log 2
log 3�log 2 :
(b)?Prove the same inequality for p > 1.
Show that the constant Cp is best possible in all cases.
Solution by G.P. Henderson, Garden Hill, Campbellcroft, Ontario.(a) We are to prove that F (x; y; z) � 0 where
F =X
xp � CX
xpyp
(the sums here and below are cyclic over x; y; z). Here
C = min(3p; 2p+1) = 3p
245
for 0 < p � 1. For any real u; v;w,
uv + vw+ wu � 1
3(u+ v + w)2:
[Editorial note. This follows from the identity
(u+ v +w)2 = 3(uv+ vw+ wu) + 1
2
�(u� v)2 + (v �w)2 + (w� u)2
�;
as other solvers point out.] Therefore
F �X
xp � C
3
�Xxp�2
=�X
xp��
1� 3p�1X
xp�� 0;
the last inequality following since
maxX
xp = 31�p:
[For example, since p � 1,
�xp + yp + zp
3
�1=p� x+ y+ z
3=
1
3
by the power mean inequality. | Ed.] This is the best possible C because
F (1=3;1=3; 1=3) = 0.
(b) With no loss of generality we assume x � y � z, which implies
z � 1
3and x � 1� z
2:
Give z a �xed value in [0; 1=3] and set y = 1�x� z [so that dy=dx = �1].Then F is a function of x only and
dF
dx= pxp�1 � pyp�1 � Cpxp�1yp +Cxppyp�1 + Cpyp�1zp � Cpzpxp�1
= p(xp�1 � yp�1)(1� Czp) + pCxp�1yp�1(x� y):
Since C � 3p and zp � 3�p (and p > 1), we see that dF=dx � 0 and we
only need to prove F � 0 when x has its minimum value, (1�z)=2. That is,we are to prove F (x;x; z) � 0 where x = (1� z)=2 = y and 0 � z � 1=3;
that is,
2xp + zp �C(x2p + 2xpzp) � 0:
This is equivalent to showing that
min0�z�1=3
2xp + zp
x2p + 2xpzp� C:
246
Calling the function to be minimized G(z), we �nd [using dx=dz = �1=2]
sgn
�dG
dz
�= sgn
�(x2p + 2xpzp)(�pxp�1 + pzp�1)
�(2xp + zp)(�px2p�1 � pxp�1zp + 2pxpzp�1)�
= sgn�pxp�1z2p + px2p�1zp � 3px2pzp�1 + px3p�1
�= sgn
"�z
x
�2p+
�z
x
�p� 3
�z
x
�p�1+ 1
#:
Set
t =z
x=
2z
1� z;
that is,
z =t
t+ 2; 0 � t � 1:
As t increases from 0 to 1, z increases from 0 to 1=3. Then sgn(dG=dz) =
sgn H(t) where
H(t) = t2p + tp � 3tp�1 + 1:
We havedH
dt= tp�2(2ptp+1 + pt� 3p+ 3):
The expression in brackets is a continuous, increasing function. It is negative
at t = 0 and positive at t = 1. Therefore it has a unique root, t0, in [0; 1]. It
follows thatH decreases to a minimum at t0 then increases. SinceH(0) = 1
andH(1) = 0,H(t0) < 0 and we see that H has a unique root t1 in (0; t0).
Therefore
dG
dz� 0 for 0 � t � t1 and
dG
dz� 0 for t1 � t � 1:
Hence, putting z1 = t1=(t1 + 2), G increases in 0 � z � z1, is a maximum
at z1 and decreases in z1 � z � 1=3. Therefore
minG = min[G(0);G(1=3)] = min(2p+1; 3p) = C:
Since the minimum is attained in both cases, C is best possible.
Both parts also solved by RICHARD I. HESS, Rancho Palos Verdes, California,USA; and MICHAEL LAMBROU, University of Crete, Crete, Greece. Part (a) onlysolved by HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the proposer. One otherreader sent in a comment.
Sei�ert and the proposer had the same proof for part (a) as Henderson (and
incidentally, the proof seems to hold for negative p as well). For part (b) both Hess
and Lambrou used multivariable calculus.
247
2234. [1997: 168] Proposed by Victor Oxman, University of Haifa,Haifa, Israel.
Given triangle ABC, its centroid G and its incentre I, construct, using
only an unmarked straightedge, its orthocentre H.
Solution by the proposer edited to �t the rewording of the problemfrom his original submission.
We will �rst establish a lemma:
Lemma. Given a line segment and its midpoint, and any other point O
in the plane, a line passing through O and parallel to the given line segment
may be constructed using only an unmarked straightedge.
Proof: LetAD be the given line segment and letK be its midpoint. Let P be
any point on AO extended. Connect P with D and with K. Draw segment
OD. LetM be the intersection ofOD andPK, and letN be the intersection
of AM with PD. We will show that ON is parallel to AD. Suppose instead
that ON1 is parallel to AD with N1 on PD. Let M1 be the intersection
of AN1 and OD, let K1 be the intersection of PM1 and AD, and let L1
be the intersection of ON1 with PK1. Then 4AM1K1 � 4N1M1L1 and
4OM1L1 � 4DM1K1. Therefore,
AK1
L1N1
=K1M1
M1L1
=K1D
OL1
. (1)
Also 4APK1 � 4OPL1 and4DK1P � 4N1L1P , which implies
AK1
OL1
=K1P
L1P=
K1D
L1N1
. (2)
From (1) and (2) it follows that
AK1 � OL1 = L1N1 �K1D; AK1 � L1N1 = OL1 �K1D:
Consequently,
(AK1)2 � OL1 � L1N1 = (K1D)2 � OL1 � L1N1 ;
from which we have (AK1)2 = (K1D)2, and AK1 = K1D. Thus point
K1 coincides with point K, PK1 coincides with PK, pointM1 withM and
point N1 with N . Thus ON is parallel to AD.
AK K1
D
M1
L1MN1
NO
P
r r
r
r
rr
r r
r
r r
248
Using the centroid G draw the medians AA1, BB1, CC1, where A1, B1,
C1 are the midpoints of BC, CA, AB, respectively. Similarly, using the
incentre I draw the angle bisectorsAA2,BB2,CC2, whereA2, B2, C2 lie on
BC, CA,AB, respectively. Draw triangleA1B1C1 and denote C1A1\BB2
by B3, A1B1 \ CC2 by C3, B1C1 \ AA2 by A3. Thus A3, B3, C3 are the
midpoints of AA2, BB2, CC2, respectively.
By applying the lemma twice we may draw a straight line BE through
B parallel to angle bisector CC2 and draw a straight line BF parallel to
angle bisector AA2 (with points E and F located on the line CA). We see
that triangles FAB and BCE are isosceles (\FBA = \BFA, \EBC =
\BEC). Draw a straight lineA1K parallel to CA withK onBE. ThusK is
the midpoint of BE. Similarly get the midpointN of FB. Thus AN ? FB
and CK ? BE.
F A P B2 B1 C E
T
N C1 B3 A1 K
B
Q
R
A2C2 I
A3
C3
r
r r rr
r
rr
r
r
r
r r
r
r r
r r r r r
Now we may draw lines parallel to FB and BE through the points C1 and
A1 respectively, and get the midpoints of AN and CK. Then applying the
lemma we may draw FT parallel to CK and ET parallel to AN . Denote
ET \ FB by R and BE \ FT by Q. Thus FR ? ET , EQ ? FT and
point B is the point of intersection of the altitudes of triangle FTE. Draw
line TB, denote TB \ CA by P and get BP ? CA; that is, BP is an
altitude of triangleABC. We can construct a second altitude similarly. Their
intersection isH, the orthocentre.
Also solved or answered by JORDIDOU, Barcelona, Spain and TOSHIO SEIMIYA,
Kawasaki, Japan. There was one incorrect solution.
249
2235. [1997: 168] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands, not Walther Janous, Ursulinengymnasium, Innsbruck, Austria aswas printed.
Triangle ABC has angle \CAB = 90�. Let �1(O;R) be the circum-
circle and �2(T; r) be the incircle. The tangent to �1 at A and the polar line
of A with respect to �2 intersect at S. The distances from S to AC and AB
are denoted by d1 and d2 respectively.
Show that
(a) STkBC,
(b) jd1 � d2j = r.
[For the bene�t of readers who are not familiar with the term \polar line", we
give the following de�nition as in, for example,Modern Geometries, 4th Edi-tion, by James R. Smart, Brooks/Cole, 1994:
The line through an inverse point and perpendicular to the line joiningthe original point to the centre of the circle of inversion is called the polar ofthe original point, whereas the point itself is called the pole of the line.]
Solution by Istv�an Reiman, Budapest, Hungary.
Assume without loss of generality that AB > AC. [ShouldAB = AC
the polar line and tangent would be parallel so that S would be at in�nity,
while AB < AC would involve only minor changes in notation.] Let P
be the point where �2 touches AB, and Q be where it touches AC. Thus
AQTP is a square whose sides have length r. Next, let U and V be the feet
of the perpendiculars from S to AB and to AC respectively, so that AUSV
is a rectangle.
The polar of A is the line PQ. Since PQT is an isosceles right triangle,
PQmakes a 45o angle withAC, which implies that QSV is also an isosceles
right triangle. Consequently SV = QV = d1, and AV � SV = d2 � d1 =
AV � QV = AQ = r (which proves (b)).
Now, \QAS = \CBA, since both angles are subtended by the chord
AC of the circle �1. Moreover, \QAS = \QTS because these angles are
symmetric about the line PQ. Since the corresponding side vectors�!TQ and
�!BA have the same direction, it follows that so do the vectors
�!TS and
�!BC,
and we conclude that the lines ST and BC are parallel as desired.
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;
FRANCISCO BELLOTROSADO, I.B. Emilio Ferrari, Valladolid, Spain; ADRIANBIRKA,
student, Lakeshore Catholic High School, Niagara Falls, Ontario; MANSUR BOASE,
student, St. Paul's School, London, England;; CHRISTOPHER J. BRADLEY, Clifton
College, Bristol, UK; ADRIAN CHAN, student, Upper Canada College, Toronto, On-
tario; JIMMY CHUI, student, Earl Haig Secondary School, North York, Ontario;
FILIP CRNOGORAC, student, Western Canada High School, Calgary, Alberta; JORDI
250
DOU, Barcelona, Spain; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; JOHN
G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta;
WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria;MICHAEL LAMBROU,
University of Crete, Crete, Greece (2 solutions); JAMES LEE, student, Eric Hamber
Secondary School, Vancouver, BC; GERRY LEVERSHA, St. Paul's School, London,
England; DAVID NICHOLSON, student, Fenelon Falls Secondary School, Fenelon
Falls, Ontario; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; TOSHIO
SEIMIYA, Kawasaki, Japan; CHRISTOPHER SO, student, Franch Liberman Catholic
High School, Scarborough, Ontario; KARENYEATS, student, St. Patrick'sHigh School,
Halifax, Nova Scotia; and the proposer.
2236. [1997: 169] Proposed by Victor Oxman, University of Haifa,Haifa, Israel.
Let ABC be an arbitrary triangle and let P be an arbitrary point in the
interior of the circumcircle of 4ABC. Let K, L, M , denote the feet of the
perpendiculars from P to the lines AB, BC, CA, respectively.
Prove that [KLM ] � [ABC]
4.
Note: [XY Z] denotes the area of 4XY Z.Almost identical solutions were submitted by Niels Bejlegaard, Sta-
vanger, Norway; Mansur Boase, student, St. Paul's School, London, Eng-land; Istv�an Reiman, Budapest, Hungary; Toshio Seimiya, Kawasaki, Japan;and Panos E. Tsaoussoglou, Athens, Greece.
For 4KLM , with O as circumcentre, we have
[KLM ] =R2 � OP 2
4R2[ABC] � [ABC]
4:
[Various di�erent references were given.]
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y,
Ferris State University, Big Rapids, Michigan, USA;MICHAEL LAMBROU, University
of Crete, Crete, Greece; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; D.J. SMEENK,
Zaltbommel, the Netherlands; and the proposer.
251
2237. [1997: 169] Proposed by Meletis D. Vasiliou, Elefsis, Greece.
ABCD is a square with incircle �. Let ` be a tangent to �. Let A0, B0,C0, D0 be points on ` such that AA0, BB0, CC0, DD0 are all perpendicularto `.
Prove that AA0 � CC0 = BB0 �DD0.I. Solution by Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid,
Spain.We give a solution with coordinates. Clearly, without loss of generality,
we can take A(�1;1), B(1;1), C(1;�1), D(�1;�1), and the equation of
the incircle is x2+y2 = 1. Then, if P (cos t; sin t) is any point of the incircle,
the equation of the line `, tangent to the incircle, is
(cos t)x+ (sin t)y = 1:
(The cases in which ` is parallel to either of the axes are trivial.) Then, cal-
culating
AA0 = d(A; `) = j � cos t+ sin t� 1jBB0 = d(B; `) = j cos t+ sin t� 1jCC0 = d(C; `) = j cos t� sin t� 1jDD0 = d(D; `) = j � cos t� sin t� 1j
we easily obtain
AA0 � CC0 = j sin2tj = j � sin2tj = BB0 �DD0;
and we are done.
II. Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria.
r r r
r r r
r
r
r
r
r
r
r
r
A U B X
M
D CY
D0
TA0
C0
B0
`
�
First we note that the equation holds obviously if the point of tangency
of ` and � is the midpoint of a side of the square. Otherwise, by rotational
symmetry about the centreM of ABCD, we may assume that, without loss
of generality, ` touches the arc of � \next to C", say at T . Let X and Y be
252
the points of intersection of l with AB produced and CD, respectively, and
U and V the midpoints of AB and CD, respectively. Note that quadrangles
MUXT and Y VMT are similar, since their respective angles correspond,
and XU = XT andMV =MT . Hence it follows that
AA0
BB0=
AX
BX=
AU + UX
UX � UB=
MU + UX
UX �MU
=Y V + VM
VM � Y V=
Y V + V D
CV � Y V=
Y D
CY=
DD0
CC0:
III. Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.There exist vectors E and F with E � F = 0 and kFk = 1, such that
` = fE + tF j t 2 Rg:We have A = xE + yF and B = uE + vF for some reals x; y; u; v. If
A0 = E + tAF, where tA 2 R, then from (A � A0) � F = 0, it follows that
tA = y, so that A0 = E + yF. Similarly, we �nd B0 = E + vF, and since
C = �A and D = �B, we have C0 = E � yF and D0 = E � vF. From
2r2 = kAk2 = kBk2, where r denotes the radius of �, and A � B = 0, we
obtain
2r2 = x2r2 + y2 = u2r2 + v2 and xur2 + yv = 0:
The �rst equation gives v2 = r2(2�u2), so that, using the second equation,
we have
x2u2r4 = y2v2 = y2r2(2� u2)
or u2(x2r2 + y2) = 2y2:
Using the �rst equation again, we get y2 = u2r2 and then 2 = x2 + u2,
which by AA0 = kA� A0k = jx � 1jr, BB0 = kB� B0k = ju � 1jr,CC0 = kC� C0k = jx+1jr, andDD0 = kD�D0k = ju+1jr, implies the
desired equation.
Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; SAM BAETHGE, Nordheim, Texas, USA; MANSUR BOASE, stu-dent, St. Paul's School, London, England; CHRISTOPHER J. BRADLEY, Clifton Col-lege, Bristol, UK; DENISE CHEUNG, student, Albert Campbell Collegiate Institute,Scarborough, Ontario; JIMMY CHUI, student, Earl Haig Secondary School, NorthYork, Ontario; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; DAVIDDOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; JORDI DOU, Barce-lona, Spain; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I.HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; MICHAELLAMBROU, University of Crete, Crete, Greece (4 methods); KEE-WAI LAU, HongKong; GERRY LEVERSHA, St. Paul's School, London, England; ISTV �AN REIMAN,Budapest, Hungary; K.R.S. SASTRY, Dodballapur, India; TOSHIO SEIMIYA, Kawa-saki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.
253
Most of the solutions were similar to either I or II above. Several solvers madegeneralizations in di�erent directions.
Janous considers a regular 2n-gon P2n := A1A2 � � �A2n with incircle � and`, a tangent line to �; then for the orthogonal projections A0
1; A0
2; : : : ; A0
2n of thevertices of P2n to ` we have
2nYk=1k even
AkA0
k =
2nYk=1k odd
AkA0
k :
Kone �cn �y observes that the condition \AA0, BB0, CC0, DD0 are all perpen-dicular to `" is not necessary; it is su�cient to state that they make the same anglewith `; that is, they are parallel to one another.
Sastry shows that if we start withABCD being a rhombus with incircle �, andwith A0, B0, C0, D0 de�ned as before, we can prove that the equation holds for alllines ` tangent to � if and only if ABCD is a square.
Finally, Seimiya comments that if we start with ABCD any quadrilateral hav-ing an incircle � with centre O, and with ` being a line tangent to �, then when A0,B0, C0, D0 are the feet of the perpendiculars from A, B, C, D, respectively, to `,we have
AA0 � CC0
BB0 �DD0=
AO � CO
BO �DO; a constant:
2238. [1997: 242] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.
A four-digit number abcd is said to be faulty if it has the following
property:
The product of the two last digits c and d equals the two-digit
number ab, while the product of the digits c� 1 and d� 1 equals
the two digit number ba.
Determine all faulty numbers!
Solution by David R. Stone, Georgia Southern University, Statesboro,Georgia, USA.
Implicit in the de�ning properties is that a 6= 0, c � 1, d � 1, and
b 6= 0. Translating the given conditions, if abcd is to be faulty we must have
c � d = ab = 10a+ b; (1)
and
(c� 1) � (d� 1) = ba = 10b+ a:
These imply
10b+ a = cd� d� c+ 1 = 10a+ b� d� c+ 1;
254
so
9(a� b) = c+ d� 1:
Thus 1 � 9(a � b) � 17, forcing 9(a � b) = 9, or a = b + 1. Hence
c+ d = 10. Substituting into (1), we get c(10� c) = 10(b+ 1) + b, or
c2 � 10c+ (11b+ 10) = 0:
By the quadratic formula, c = 5 �p15� 11b, which forces 15� 11b = 4,
or b = 1. Thus a = 2 and c = 7 or c = 3, which forces d = 3 or d = 7,
respectively. That is, the only two faulty numbers are
2137 and 2173:
(Checking, 3� 7 = 21 and (3� 1)� (7� 1) = 12.)
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;�SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina;CHARLESASHBACHER, Cedar Rapids, Iowa, USA; SAMBAETHGE,Nordheim, Texas,USA; FRANK P. BATTLES, Massachusetts Maritime Academy, Buzzards Bay, Massa-chusetts, USA; MANSUR BOASE, student, St. Paul's School, London, England;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;MIGUEL ANGELCABEZ �ONOCHOA, Logro ~no, Spain; CON AMORE PROBLEM GROUP, The Royal Danish Schoolof Educational Studies, Copenhagen, Denmark; MAYUMI DUBREE, student, AngeloState University, San Angelo, Texas, USA; KEITH EKBLAW, Walla Walla, Washing-ton, USA; JEFFREY K. FLOYD, Newnan, Georgia, USA; FLORIAN HERZIG, student,Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;JOHN G. HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON,Beaverton, Oregon, USA; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRYLEVERSHA, St. Paul's School, London, England; KATHLEEN E. LEWIS, SUNY Os-wego, Oswego, NY, USA; J.A. MCCALLUM, Medicine Hat, Alberta; GRADYMYDLAK, student, University College of the Cariboo, Kamloops, BC; CHRISTOSSARAGIOTIS, student, Aristotle University, Thessaloniki, Greece; JOELSCHLOSBERG, student, Robert Louis Stevenson School, New York, NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; REZA SHAHIDI, student, University of Water-loo, Waterloo, Ontario; SKIDMORE COLLEGE PROBLEM GROUP, Saratoga Springs,New York, USA;D.J. SMEENK, Zaltbommel, the Netherlands; DIGBY SMITH,MountRoyal College, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; and theproposer.
Although nobody raised the issue, readers must have wondered why the pro-
poser chose the term \faulty" for numbers satisfying this condition. He reveals his
reason at the end of his solution with the following remark: \Crux problems 2137
and 2173 both have been corrected in Crux, 2137 even twice! They are not faulty
anymore!" (See [1996: 317] and [1997: 48] for 2137, and [1997: 169] for 2173.)
255
2239. [1997: 242] Proposed by Kenneth Kam Chiu Ko, Mississauga,Ontario.
Suppose that 1 � r � n and consider all subsets of r elements of the
set f1, 2, 3, : : : , ng. The elements of these subsets are arranged in ascending
order of magnitude. For i from 1 to r, let ti denote the ith smallest element
in the subset. Let T (n; r; i) denote the arithmetic mean of the elements ti.
Prove that T (n; r; i) = in+ 1
r + 1.
Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,Ontario.
Let X(n) = f1; 2; :::; ng. Since there are�n
r
�subsets of X(n) with
r elements,
T (n; r; i) =
Pti�n
r
� ;where the summation is over all r-subsets. Thus we are to show thatX
ti = in+ 1
r + 1
�n
r
�= i
�n+ 1
r + 1
�: (1)
For each k 2 X(n), let uk denote the number of r-subsets
(a1; a2; :::; ai�1; k; ai+1; :::; ar)
ofX(n) with k being the ith smallest element. Clearly, we must have k � i.
There are�k�1i�1�ways of choosing the i�1 elements before k, and
�n�kr�i
�ways
of choosing the r � i elements after k. Thus
uk =
�k � 1
i� 1
��n� k
r � i
�
from which we get
Xti =
Xk�i
kuk =Xk�i
k
�k� 1
i� 1
��n� k
r� i
�: (2)
From (1) and (2) we see that it remains to prove that
Xk�i
k
i
�k� 1
i� 1
��n� k
r � i
�=
�n+ 1
r + 1
�;
or, equivalently,
Xk�i
�k
i
��n� k
r � i
�=
�n+ 1
r + 1
�: (3)
The equation (3) is well-known (for example, formula (11) on page 207 of
Applied Combinatorics, 2nd edition by Alan Tucker). For completeness we
256
give a combinatorial proof of this identity. Note �rst that�n+1
r+1
�is the num-
ber of (r + 1)-subsets of X(n + 1). Again, we partition the family of all
these subsets according to their (i+ 1)th smallest element, where i is �xed,
1 � i � r. Speci�cally, for each k 2 X(n + 1), we count the number,
vk, of (r + 1)-subsets with k + 1 being the (i + 1)th smallest element.
Clearly, k � i, and, by the same argument as before, with n; r; i; k replaced
by n+ 1; r + 1; i+ 1, and k+ 1, respectively, we have
vk =
�k
i
��n� k
r � i
�:
Summing over k, we get�n+ 1
r + 1
�=Xk�i
vk =Xk�i
�k
i
��n� k
r � i
�
which establishes (3) and completes the proof.Also solved by NIELS BEJLEGAARD, Stavanger, Norway; FRANCISCO BELLOT
ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; ADRIAN BIRKA, student, LakeshoreCatholic High School, Port Colbourne, Ontario; MANSUR BOASE, student, St. Paul'sSchool, London, England; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario; JIMMY CHUI,student, Earl Haig Secondary School, North York, Ontario; CON AMOREPROBLEM GROUP, Royal Danish School of Educational Studies, Copenhagen, Den-mark; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; WALTHER JANOUS, Ur-sulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete,Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's School,London, England; ALAN LING, student, University of Toronto, Toronto, Ontario;CHRISTOPHER SO, student, Francis Liberman Catholic High School, Scarborough,Ontario; and the proposer. There was also one incomplete solution submitted.
Francisco Bellot Rosado noted that this problem is a generalization of the sec-
ond problem from the IMO 1981 (Washington), where the question was to �nd the
arithmetic mean of the smallest elements of the r-subsets. He also pointed out that a
solution to the general question is given in a Romanian Olympiad book: Cuculescu, I.,
Olimpiadele Internationale de Matematica ale elevilor, Ed. Tehnica, Bucarest 1984,
p. 315.
Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell
Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer
Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil
Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,
J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia
257
THE ACADEMY CORNERNo. 20
Bruce Shawyer
All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7
THE BERNOULLI TRIALS 1998
The Bernoulli Trials, an undergraduate mathematics competition, was
held Saturday, March 7 at the University of Waterloo. This is the second
year for this event, which is a double knockout competition with \true" or
\false" as the answers on each round. The participants have 10 minutes for
each question, and drop out after their second incorrect answer.
There were 29 student participants in the competition, which lasted 4
hours and 16 rounds. The winner was third year student Frederic Latour.
Second place went to �rst year student Joel Kamnitzer, and third and fourth
to Richard Hoshino and Derek Kisman. In keeping with the nature of the
answers required, the prizes were awarded in coins: 100 toonies for �rst
place, 100 loonies for second, and quarters for third and fourth.
Ian Goulden and Christopher Small
1. In the �gure below, the two circles are tangent at A. The point C is
the center of the larger circle, and FC is perpendicular to AB. The line
segment DB is of length 9, and the line segment FE is of length 5.
BA
F
C D
E
TRUE OR FALSE? The diameter of the larger circle is less than or equal
to 49.
258
2. One of the participants in the Bernoulli Trials answers by ipping a fair
coin. He chooses \True" when the coin lands heads and \False" when
the coin lands tails.
TRUE OR FALSE? Using this method, on average he will last exactly 4
rounds until he drops out.
3. TRUE OR FALSE? There exist 11 distinct positive integers
a1; a2; : : : ; a11
so that all integers from 8 to 1998 inclusive can be written as sums over
subsets of the ai's.
4. Let ABCD be a rectangle labelled clockwise as shown, and let P be
any point in the plane.
D C
A B
P
TRUE OR FALSE?
jPAj2 � jPBj2 + jPCj2 � jPDj2 = 0 :
(In this expression jEF j denotes the distance from point E to point F .)
5. TRUE OR FALSE? The volume of a regular tetrahedron of side 1 is less
than the volume of a sphere of radius 1=�.
6. TRUE OR FALSE? The smallest perfect square ending in 9009 is
1265032 = 16003009009 :
7. Let f : [0; 1] ! [0; 1] be a continuous strictly increasing function such
that f(0) = 0 and f(1) = 1.
TRUE OR FALSE? The inequality
9Xi=1
f
�i
10
�+
9Xi=1
f�1�i
10
�� 99
10
holds true for any such function.
259
8. Let k be the smallest positive integer such that kt + 1 is a triangular
number whenever t is a triangular number.
TRUE OR FALSE? There exists such a k and k � 12.
(A number t is triangular if it can be written in the form
t = 1+ 2 + 3 + � � �+ n
where n � 1. The �rst 4 triangular numbers are 1, 3, 6, and 10.)
9. TRUE OR FALSE? It is possible to construct a planar quadrilateral with
sides of length 1, 2, 3, and 4, and with area 2p6.
10. Let f : [0;1) ! [0;1) be a continuous strictly increasing function
satisfying the equation
f
�x
1 + x
�=
f(x)
1 + f(x)
for all x � 0.
TRUE OR FALSE? Then f(x) = x for all x � 0.
11. TRUE OR FALSE?
232323
> 323232
:
12. TRUE OR FALSE?4
3<
Z 1
0
x dx
ex � 1<
5
3:
13. A triangle has squares constructed externally on each of its three sides.
Suppose these squares have area 8, 13 and 41 square meters respec-
tively.
TRUE OR FALSE? The area of the triangle is greater than one square
meter.
41
813
14. TRUE OR FALSE?
There exists a continuous function
f : R! R
such that
f [f(x)] = ex � e
ex
260
15. Let A and B be two random points in the interior of a circle C. Let D
be the circle with diameter AB, and let ` be the line tangent to D at
B.
B
A
D
C
`
Let R1 be the area of the region that is interior to both C and D. Let
R2 be the area of the region that is interior to C on the other side of `
from D.
TRUE OR FALSE? The average value ofR1 is strictly less than the average
value of R2.
16. The following game is played between two players using the petals of a
daisy. Each player takes a turn plucking either one petal or two neigh-
bouring petals from a daisy. The player who takes the last petal wins.
TRUE OR FALSE? With best play by both sides for a daisy with thirteen
petals, the �rst player can force a win.
12
3
4
5
6
78
9
10
11
12
13
261
THE OLYMPIAD CORNERNo. 191
R.E. Woodrow
All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.
Welcome back from the \summer break". I hope all of you have been
taking some time to write up your nice solutions to problems given so far.
We start this number with two traditional pieces. First we give the problems
of the 1998 Canadian Mathematical Olympiad which we reproduce with the
permission of the Canadian Mathematical Olympiad Committee of the Cana-
dian Mathematical Society. Thanks go to Daryl Tingley, University of New
Brunswick, and chair of the CMO Committee, for forwarding the questions
to me.
1998 CANADIANMATHEMATICAL OLYMPIADFinal
November 20
1. Determine the number of real solutions a to the equation�1
2a
�+
�1
3a
�+
�1
5a
�= a :
Here, if x is a real number, then [ x ] denotes the greatest integer that
is less than or equal to x.
2. Find all real numbers x such that
x =
�x� 1
x
�1=2+
�1� 1
x
�1=2:
3. Let n be a natural number such that n � 2. Show that
1
n+ 1
�1 +
1
3+ � � �+ 1
2n� 1
�>
1
n
�1
2+
1
4+ � � �+ 1
2n
�:
4. Let ABC be a triangle with \BAC = 40� and \ABC = 60�. LetD andE be the points lying on the sidesAC andAB, respectively, such that
\CBD = 40� and \BCE = 70�. Let F be the point of intersection of the
lines BD and CE. Show that the line AF is perpendicular to the line BC.
262
5. Let m be a positive integer. De�ne the sequence a0; a1; a2; : : : by
a0 = 0; a1 = m; and an+1 = m2an� an�1 for n = 1; 2; 3; : : : . Prove that
an ordered pair (a; b) of non-negative integers, with a � b, gives a solution
to the equationa2 + b
2
ab+ 1= m
2
if and only if (a; b) is of the form (an; an+1) for some n � 0.
The next set of problems is from the twenty-seventh annual United
States of America Mathematical Olympiad written April 28, 1998. These
problems are copyrighted by the committee on the American Mathemati-
cal Competitions of the Mathematical Association of America and may not
be reproduced without permission. Solutions, and additional copies of the
problems may be obtained from Professor Walter E. Mientka, AMC Execu-
tive Director, 917 Oldfather Hall, University of Nebraska, Lincoln, NE, USA
68588{0322. As always, we welcome your original, \nice" solutions and gen-
eralizations which di�er from the published solutions.
27th UNITED STATES OF AMERICAMATHEMATICAL OLYMPIAD
Part I 9 a.m. -12 noonApril 28, 1998
1. Suppose that the set f1; 2; � � � ; 1998g has been partitioned into disjoint
pairs fai; big (1 � i � 999) so that for all i, jai � bij equals 1 or 6. Prove
that the sum
ja1 � b1j+ ja2 � b2j+ � � �+ ja999 � b999j
ends in the digit 9.
2. Let C1 and C2 be concentric circles, with C2 in the interior of C1.From a point A on C1 one draws the tangent AB to C2 (B 2 C2). Let C be
the second point of intersection of AB and C1, and let D be the midpoint of
AB. A line passing through A intersects C2 at E and F in such a way that
the perpendicular bisectors of DE and CF intersect at a point M on AB.
Find, with proof, the ratio AM=MC.
3. Let a0; a1; � � � ; an be numbers from the interval (0; �=2) such that
tan(a0 ��
4) + tan(a1 �
�
4) + � � �+ tan(an �
�
4) � n� 1:
Prove that
tan a0 tana1 � � � tanan � nn+1
:
263
Part II 1 p.m. - 4 p.m.
4. A computer screen shows a 98 � 98 chessboard, coloured in the
usual way. One can select with a mouse any rectangle with sides on the lines
of the chessboard and click the mouse button: as a result, the colours in
the selected rectangle switch (black becomes white, white becomes black).
Find, with proof, the minimum number of mouse clicks needed to make the
chessboard all one colour.
5. Prove that for each n � 2, there is a set S of n integers such that
(a� b)2 divides ab for every distinct a; b 2 S.
6. Let n � 5 be an integer. Find the largest integer k (as a func-
tion of n) such that there exists a convex n-gon A1A2 : : : An for which ex-
actly k of the quadrilateralsAiAi+1Ai+2Ai+3 have an inscribed circle. (Here
An+j = Aj.)
As a third Olympiad for this issuewe give the 11th Grade and 12th Grade
problems of the 1994 Latvian Mathematics Olympiad. My thanks go to Bill
Sands, of the University of Calgary, who collected these problems for me
when he was assisting with the 1995 International Mathematical Olympiad
held in Canada.
45th LATVIANMATHEMATICAL OLYMPIAD, 199411th Grade
1. Prove for each choice of real non-zero numbers a1; a2; : : : ; c3, the
\stars" can be replaced by \<" and \>" so that the system8<:
a1x+ b1y + c1 � 0a2x+ b2y + c2 � 0a3x+ b3y + c3 � 0
has no solution.
2. Solve in natural numbers:
x(x+ 1) = y7
3. Given 4 non-coplanar points, how many parallelepipeds having
these points as vertices can be constructed?
4. Let ABCD be a convex quadrilateral, M 2 AB, N 2 BC, P 2CD, Q 2 DA; AM = BN = CP = DQ, and MNPQ is a square. Prove
that ABCD is a square, too.
5. A square consists of n � n cells, n � 2. A letter is inserted into
each cell. It is given that every two rows di�er. Prove that there is a column
which can be deleted from the square so that all rows are again di�erent after
this deletion.
264
12th Grade
1. Solve the equation cosx � cos 2x � cos 3x = 1.
2. All faces of a convex polytope are triangles. What can be the number
of the faces?
3. Does there exist a polynomial P (x; y) in two variables such that
(a) P (x; y) > 0 for all x; y,
(b) for each c > 0 there exist such x and y that P (x; y) = c?
4. Let S(x) be the digital sum of natural number x. Prove that
S(2n)!1 when n!1, n natural.
5. The centres of four equal circles are the vertices of a square. How
must A, B, C, D be chosen so that each circle contains at least one of them
and the area of ABCD is as big as possible?
As a �nal set of problems for you to puzzle over after the hiatus, we give
the problems of the Dutch Mathematical Olympiad, Second Round, written
16 September 1994. My thanks again go to Bill Sands for collecting these
problems for me while he was helping out at the IMO in Toronto.
DUTCH MATHEMATICAL OLYMPIADSecond Round
16 September, 1994
1. A unit square is divided in two rectangles in such a way that the
smaller rectangle can be put on the greater rectangle with every vertex of the
smaller on exactly one of the edges of the greater.
Calculate the dimensions of the smaller rectangle.
2. Given is a sequence of numbers a1; a2; a3; : : : with the property:
a1 = 2; a2 = 3 and
�an+1 = 2an�1 or
an+1 = 3an � 2an�1for all n � 2:
Prove that no number between 1600 and 2000 can be an element of the
sequence.
265
3. (a) Prove that every multiple of 6 can be written as the sum of four
third powers of integers.
(b) Prove that every integer can be written as the sum of �ve third
powers of integers.
4. Let P be any point on the diagonal BD of a rectangle ABCD.
F is the projection of P on BC. H lies on BC such that BF = FH.
PC intersects AH in Q.
A B
CD
H
FP
Q
Prove: Area �APQ = Area �CHQ.
5. Three real numbers a, b and c satisfy the inequality:��ax2 + bx+ c�� � 1 for all x 2 [�1;+1]:
Prove:��cx2 + bx+ a
�� � 2 for all x 2 [�1;+1].
We now give the solutions to the Canadian Mathematical Olympiad
given earlier this number. (I hope you have already solved them all!) These
\o�cial solutions" were selected from the most interesting student solutions
by Daryl Tingley, University of New Brunswick and Chair of the Canadian
Mathematical Olympiad Committee of the Canadian Mathematical Society.
Of course the original submissions have been somewhat edited.
1998 CANADIANMATHEMATICAL OLYMPIADSolutions of Students
1. Solutionby David Arthur, Upper Canada College, Toronto, Ontario.Let a = 30k+ r, where k is an integer and r is a real number between
0 and 29 inclusive.
Then�12a�=�12(30k+ r)
�= 15k+
�r2
�. Similarly
�13a�= 10k+
�r3
�and
�15a�= 6k+
�r5
�.
266
Now,�12a�+�13a�+�15a�= a, so
�15k+
�r2
��+�10k+
�r3
��+�
6k +�r5
��= 30k + r, and hence k = r �
�r2
���r3
���r5
�:
Clearly, r has to be an integer, or r ��r
2
���r
3
���r
5
�will not be an
integer, and therefore, cannot equal k.
On the other hand, if r is an integer, then r��r2
���r3
���r5
�will also
be an integer, giving exactly one solution for k.
For each r (0 � r � 29); a = 30k + r will have a di�erent remainder
mod30, so no two di�erent values of r give the same result for a.
Since there are 30 possible values for r (0; 1; 2; : : : ; 29), there are then
30 solutions for a.
2. Solution by Jimmy Chui, Earl Haig Secondary School, North York,Ontario.
Since�x� 1
x
�1=2 � 0 and�1� 1
x
�1=2 � 0; then 0 ��x� 1
x
�1=2+�
1� 1x
�1=2= x.
Note that x 6= 0. Else, 1xwould not be de�ned, so x > 0.
Squaring both sides gives,
x2 =
�x� 1
x
�+
�1� 1
x
�+ 2
s�x� 1
x
��1� 1
x
�
x2 = x+ 1� 2
x+ 2
rx� 1� 1
x+
1
x2:
Multiplying both sides by x and rearranging, we get
x3 � x
2 � x+ 2 = 2px3 � x2 � x+ 1
(x3 � x2 � x+ 1)� 2
px3 � x2 � x+ 1 + 1 = 0
(px3 � x2 � x+ 1� 1)2 = 0
px3 � x2 � x+ 1 = 1
x3 � x
2 � x+ 1 = 1
x(x2 � x� 1) = 0
x2 � x� 1 = 0 since x 6= 0 :
Thus x = 1�p5
2. We must check to see if these are indeed solutions.
Let � = 1+p5
2; � = 1�
p5
2. Note that � + � = 1; �� = �1 and
� > 0 > �.
267
Since � < 0; � is not a solution.
Now, if x = �, then
��� 1
�
�1=2+
�1� 1
�
�1=2= (�+ �)1=2 + (1 + �)1=2 (since �� = �1)
= 11=2 + (�2)1=2 (since �+ � = 1 and �2 = � + 1)
= 1� � (since � < 0)
= � (since �+ � = 1):
So x = � is the unique solution to the equation.
3. Solution 1 by Chen He, Columbia International Collegiate,Hamilton, Ontario.
1 +1
3+ : : :+
1
2n� 1=
1
2+
1
2+
1
3+
1
5+ : : :+
1
2n� 1(1)
Since
1
3>
1
4;
1
5>
1
6; : : : ;
1
2n� 1>
1
2n;
(1) gives
1 +1
3+ : : :+
1
2n� 1>
1
2+
1
2+
1
4+
1
6(2)
+ : : :+1
2n=
1
2+
�1
2+
1
4+
1
6+ : : :+
1
2n
�:
Since
1
2>
1
4;
1
2>
1
6;
1
2>
1
8; : : : ;
1
2>
1
2n
then
n
2=
1
2+
1
2+
1
2+ : : :+
1
2| {z }n
>1
2+
1
4+
1
6+ : : :+
1
2n
so that
1
2>
1
n
�1
2+
1
4+
1
6+ : : :+
1
2n
�: (3)
268
Then (1), (2) and (3) show that
1 +1
3+ : : :+
1
2n� 1
>1
n
�1
2+
1
4+
1
6+ : : :+
1
2n
�+
�1
2+
1
4+
1
6+ : : :+
1
2n
�
=
�1 +
1
n
��1
2+
1
4+ : : :+
1
2n
�
=n+ 1
n
�1
2+
1
4+ : : :+
1
2n
�:
Therefore 1n+1
�1 + 1
3+ : : :+ 1
2n�1
�>
1n
�12+ 1
4+ : : :+ 1
2n
�for all
n 2 N and n � 2.
Solution 2 by Yin Lei, Vincent Massey Secondary School, Windsor,Ontario.
Since n � 2; n(n+ 1) � 0. Therefore the given inequality is equiva-
lent to
n
�1 +
1
3+ : : :+
1
2n� 1
�� (n+ 1)
�1
2+
1
4+ : : :+
1
2n
�:
We shall use mathematical induction to prove this.
For n = 2, obviously 13
�1 + 1
3
�= 4
9>
12
�12+ 1
4
�= 3
8.
Suppose that the inequality stands for n = k; that is,
k
�1 +
1
3+ : : :+
1
2k� 1
�> (k+ 1)
�1
2+
1
4+ : : :+
1
2k
�: (1)
Now we have to prove it for n = k+ 1.
We know�1 +
1
3+ : : :+
1
2k� 1
���1
2+
1
4+ : : :+
1
2k
�
=
�1� 1
2
�+
�1
3� 1
4
�+
�1
5� 1
6
�+ : : :+
�1
2k� 1� 1
2k
�
=1
1� 2+
1
3� 4+
1
5� 6+ : : :+
1
(2k� 1)(2k):
Since
1� 2 < 3� 4 < 5� 6 < : : : < (2k� 1)(2k) < (2k+ 1)(2k+ 2)
then
1
1� 2+
1
3� 4+ : : :+
1
(2k� 1)(2k)>
k
(2k+ 1)(2k+ 2)
269
.
Hence
1 +1
3+ : : :+
1
2k � 1>
1
2+
1
4+ : : :+
1
2k+
k
(2k+ 1)(2k+ 2): (2)
Also
k + 1
2k + 1� k+ 2
2k+ 2
=2k2 + 2k + 2k+ 2� 2k2 � 4k� k� 2
(2k+ 1)(2k+ 2)= � k
(2k+ 1)(2k+ 2):
Therefore
k + 1
2k + 1=
k + 2
2k + 2� k
(2k+ 1)(2k+ 2): (3)
Adding 1, 2 and 3 gives
�1 +
1
3+ : : :+
1
2k� 1
�+
�1 +
1
3+ : : :+
1
2k � 1
�+
k+ 1
2k+ 1
> (k+ 1)
�1
2+
1
4+ : : :+
1
2k
�+
�1
2+
1
4+ : : :+
1
2k
�
+k
(2k+ 1)(2k+ 2)+
k+ 2
2k+ 2� k
(2k+ 1)(2k+ 2)
.
Rearranging both sides, we get
(k+ 1)
�1 +
1
3+ : : :+
1
2k + 1
�> (k+ 2)
�1
2+
1
4+ : : :+
1
2k+ 2
�:
This proves the induction.
4. Solution 1 by Keon Choi, A.Y. Jackson Secondary School, NorthYork, Ontario.
SupposeH is the foot of the perpendicular line fromA toBC; construct
equilateral 4ABG, with C on BG. I will prove that if F is the point where
AH meetsBD, then \FCB = 70�. (Because that meansAH, and the given
lines BD and CE meet at one point, this answers the question.) Suppose
BD extended meets AG at I.
270
A
B H C G
F
E
D
I
Now BF = GF and \FBG = \FGB = 40�, so that \IGF = 20�.Also \IFG = \FBG+ \FGB = 80�, so that
\FIG = 180� � \IFG� \IGF
= 180� � 80� � 20�
= 80�:
Therefore 4GIF is an isosceles triangle, so
GI = GF = BF : (1)
But 4BGI and 4ABC are congruent, since BG = AB,
\GBI = \BAC; \BGI = \ABC.
Therefore
GI = BC : (2)
From (1) and (2) we get
BC = BF :
So in4BCF ,
\BCF =180� � \FBC
2=
180� � 40�
2= 70� :
Thus \FCB = 70� and that proves that the given lines CE and BD
and the perpendicular line AH meet at one point.
Solution 2 by Adrian Birka, Lakeshore Catholic High School, PortColborne, Ontario.
First we prove the following lemma:
In4ABC; AA0; BB0; CC0 intersect if and only if
271
sin�1
sin�2
� sin�1sin�2
� sin 1sin 2
= 1 ;
where �1; �2; �1; �2; 1; 2 are as shown in the diagram just below.
[Editor: This is a known variant of Ceva's Theorem.]
A C
B
B0
A0
C0
Da1
a2c1
c2
b1 b2
�1
�2 2
1
�1�2
Proof: Let \BB0C = x; then \BB0A = 180��x. Using the Sine Law
in4BB0C yields
b2
sin�2=
a
sinx: (1)
Similarly using the Sine Law in4BB0A yields
b1
sin�1=
c
sin(180� � x)=
c
sinx: (2)
Hence,
b1 : b2 =c sin�1
a sin�2(3)
(from (1),(2)). [Editor: Do you recognize this when �1 = �2?]
Similarly,
a1 : a2 =b sin�1
c sin�2
; c1 : c2 =a sin 1
b sin 2: (4)
By Ceva's Theorem, the necessary and su�cient condition for AA0,BB
0, CC0 to intersect is: (a1 : a2) � (b1 : b2) � (c1 : c2) = 1. Using (3),
(4) on this yields:
272
b
c� sin�1
sin�2
� ab� sin 1sin 2
� ca� sin�1sin�2
= 1 ;
so that
sin�1
sin�2
� sin�1sin�2
� sin 1sin 2
= 1 : (5)
This is just what we needed to show; therefore the lemma is proved.
Now, in our original question, give \BAC = 40�; \ABC = 60�. Itfollows that \ACB = 80�.
Since \CBD = 40�; \ABD = \ABC � \DBC = 20�. Similarly,
\ECA = 20�.
A
B
CD
KE
F
Now let us show that \FAD = 10�. Suppose otherwise. Let F 0
be such that F; F 0 are in the same side of AC and \DAF 0 = 10�. Then
\BAF0 = \BAC � \DAF
0 = 30�.
Thus
sin\ABD
sin\DBC� sin\BCEsin\ECA
� sin\CAF0
sin\F 0AB=
sin20�
sin 40�� sin 70
�
sin 10�� sin 10
�
sin 30�
=sin20�
2 sin 20� cos 20�� cos 20
�
sin30�
=1
2 sin30�= 1 :
By the lemma above, AF 0 passes through CE \ BD = F . Therefore
AF0 = AF , and \FAD = 10�, contrary to assumption. Thus \FAD must
be 10�. Now let AF \ BC = K. Since \KAC = 10�; \KCA = 80�, itfollows that \AKC = 90�. Therefore AK ? BC ) AF ? BC as needed.
5. Solution by Adrian Chan, Upper Canada College, Toronto, Ontario.
Let us �rst prove by induction thata2n+a
2n+1
an�an+1+1= m
2 for all n � 0:
Proof: Base Case (n = 0) :a20+a
21
a0�a1+1 = 0+m2
0+1= m
2:
273
Now, let us assume that it is true for n = k, k � 0. Then,
a2k + a
2k+1
ak � ak+1 + 1= m
2
a2k + a
2k+1 = m
2 � ak � ak+1 +m2
a2k+1 +m
4a2k+1 � 2m2 � ak � ak+1 + a
2k
= m2 +m
4a2k+1 �m
2 � ak � ak+1a2k+1 + (m2
ak+1 � ak)2 = m
2 +m2ak+1(m
2ak+1 � ak)
a2k+1 + a
2k+2 = m
2 +m2 � ak+1 � ak+2 :
Thereforea2k+1+a
2k+2
ak+1�ak+2+1= m
2, proving the induction. Hence (an; an+1)
is a solution to a2+b2
ab+1= m
2 for all n � 0.
Now, consider the equation a2+b2
ab+1= m
2 and suppose (a; b) = (x; y)
is a solution with 0 � x � y. Then
x2 + y
2
xy + 1= m
2: (1)
If x = 0 then it is easily seen that y = m, so (x; y) = (a0; a1). Since
we are given x � 0, suppose now that x > 0.
Let us show that y � m2x.
Proof by contradiction: Assume that y > m2x. Then y = m
2x + k
where k � 1.
Substituting into (1) we get
x2 + (m2
x+ k)2
(x)(m2x+ k) + 1= m
2
x2 +m
4x2 + 2m2
xk + k2 = m
4x2 +m
2kx+m
2
(x2 + k2) +m
2(kx� 1) = 0:
Now, m2(kx � 1) � 0 since kx � 1 and x2 + k2 � x
2 + 1 � 1 so
(x2 + k2) +m
2(kx� 1) 6= 0.
Thus we have a contradiction, so y � m2x if x > 0.
Now substitute y = m2x� x1, where 0 � x1 < m
2x, into (1).
274
We have
x2 + (m2
x� x1)2
x(m2x� x1) + 1= m
2
x2 +m
4x2 � 2m2
x � x1 + x21 = m
4x2 �m
2x � x1 +m
2
x2 + x
21 = m
2(x � x1 + 1)
x2 + x
21
x � x1 + 1= m
2: (2)
If x1 = 0, then x2 = m2. Hence x = m and (x1; x) = (0;m) =
(a0; a1). But y = m2x � x1 = a2, so (x; y) = (a1; a2). Thus suppose
x1 > 0.
Let us now show that x1 < x.
Proof by contradiction: Assume x1 � x.
Then m2x� y � x since y = m
2x� x1, and
�x2+y2
xy+1
�x� y � x
since (x; y) is a solution to a2+b2
ab+1= m
2.
So x3 + xy2 � x
2y+ xy
2 + x+ y. Hence x3 � x2y+ x+ y, which is
a contradiction since y � x > 0.
With the same proof that y � m2x, we have x � m
2x1. So the sub-
stitution x = m2x1 � x2 with x2 � 0 is valid.
Substituting x = m2x1 � x2 into (2) gives
x21+x22
x1�x2+1 = m2.
If x2 6= 0, then we continue with the substitutionxi = m2xi+1
�xi+2 (*)until we get
x2j+x2j+1
xj �xj+1+1= m
2 and xj+1 = 0. (The sequence xi is decreasing,
non-negative and integer.)
So, if xj+1 = 0, then x2j = m2 so xj = m and (xj+1; xj) = (0;m) =
(a0; a1).
Then (xj; xj�1) = (a1; a2) since xj�1 = m2xj � xj+1 (from (*)).
Continuing, we have (x1; x) = (an�1; an) for some n. Then (x; y) =
(an; an+1).
Hence a2+b2
ab+1= m
2 has solutions (a; b) if and only if (a; b) = (an; an+1) for
some n.
That is all we have room for this issue. Enjoy solving the problems |
and send me your nice solutions as well as Olympiad contests for use in the
corner.
275
BOOK REVIEWSEdited by ANDY LIU
Models that Work, by Alan Tucker, published by Mathematical Association
of America, 1995, ISBN# 0-88385-096-6, softcover, 88+ pages, US$24.00.
Reviewed by Jim Timourian, University of Alberta.
At the University of Alberta, like many other institutions, we have large
numbers of students enrolling in �rst year calculus, but very few (outside of
engineering students) taking more advanced courses. The number of students
choosing a mathematical science as a major is small, given the number of
students who attend the university.
This report documents mathematics programs at a variety of institu-
tions that are very successful in attracting mathematics majors and also stu-
dents for advanced mathematics courses from other programs. In spite of
the fact that the e�ective undergraduate programs studied are at diverse in-
stitutions, from two year community colleges to Ph.D.-granting institutions
with outstanding reputations for graduate education, the investigators found
a lot in common. Most of the report describes that commonality, while the
rest discusses speci�c site visits and answers to a set of questions.
An e�ective program is considered to be one that succeeds in attracting
large numbers of students asmajors, or prepares students to pursue advanced
study in mathematics, or prepares future teachers or attracts and prepares
under-represented groups in mathematics.
The departments are uni�ed by some underlying philosophies, but oth-
erwise they use diverse methods to achieve their goals. Even within depart-
ments there is a variety of approaches used by di�erent faculty members.
The report explicitly lists themes that are part of the department culture
in each of the cases studied. These are respecting students, caring about their
welfare, and enjoying the role of being college instructors. The respect for
the students is characterized by the comment \teaching for the students one
has, not the students one wished one had." This is mirrored in the curriculum
\: : : geared toward the needs of the students, not the values of the faculty."
The most important part of a successful department's culture is not
explicitly mentioned in the report but is obvious throughout. As a whole
a department has to adopt the philosophy that it is a good thing for more
students to take more courses in the mathematical sciences. In many de-
partments this idea is not shared by enough of the sta� members. There are
those who think \there are no jobs" so we should not be attracting students.
Others think that only the very best students, who have the potential to go
on to graduate school in a mathematical science, are worthy and that the rest
are \weak" and can be ignored.
A department that wants to improve its programs needs a sense of what
it is trying to achieve. Read this report to learn what comparable institutions
have done and how they have gone about doing it.
276
Pythagoras Strikes Again!
K.R.S. Sastry
Is it possible that a given triangle is similar to the triangle formed from
its medians? In other words, is it possible that the side lengths of a tri-
angle are proportional in some order to its medians' triangle? For a non-
trivial example note the triangle with side lengths (a; b; c) = (23;7; 17)
and the medians' lengths (ma; mb;mc) =�7p32; 23
p32; 17
p32
�in which
a : mb = b : ma = c : mc is possible. You may use the formulas
4m2a = 2b2 + 2c2 � a
2 (1)
etc. to calculate the lengths of the medians in the above example. The for-
mulas (1) themselves are derived by means of the cosine rule. The type of
triangle as described exists and shall be called a self-median triangle [1].
Is it possible, this time, that a given triangle is similar to the triangle
formed from its altitudes? That is, the lengths of the sides of a triangle are
proportional in some order to the lengths of its altitudes? Again, trivially
it's so in an equilateral triangle but non-trivially we have the triangle with
the lengths of the sides (a; b; c) = (6; 9; 4) and the lengths of the altitudes
(ha; hb; hc) =�246;249;244
�,4 being the area of the triangle. Observe that
a : ha = b : hc = c : hb holds. This type of triangle too exists and matching
the description of self-median triangles we will call the present type self-altitude triangles [2].
Interestingly, as we shall see, with the exception of the equilateral tri-
angle, both of these triangle types are simply reincarnations of appropriate
right triangles. The theorems that follow illustrate this fact.
Theorems on self-median triangles
Theorem 1 tells us how an appropriate right triangle yields a self-median
triangle.
THEOREM 1: Let (a0; b0; c0) be a right triangle in which a20 + b
20 = c
20,
a0 > b0 and c0 > 2b0 hold. Then (a; b; c) = (a0 + b0; a0 � b0; c0) is a
self-median triangle in which a > c > b holds.
Proof: Letm1;m2;m3 denote the lengths of the medians drawn to the sides
a0 + b0; a0 � b0; c0 respectively. Then from (1) we can easily deduce that
4m21 = 3(a0 � b0)
2; 4m2
2 = 3(a0 + b0)2; 4m2
3 = 3c20:
Copyright c 1998 Canadian Mathematical Society
277
Hencea0 + b0
m2
=a0 � b0
m1
=c0
m3
=2p3:
It should be noted that the lengths a0+b0; a0�b0; c0 do not form a triangle
for every given right triangle (a0; b0; c0). For example (a0; b0; c0) = (4;3; 5)
yields (a0+b0; a0�b0; c0) = (7; 1; 5), and there is no triangle with these side
lengths. However (a0; b0; c0) = (12;5; 13) yields the self-median triangle
(a0 + b0; a0 � b0; c0) = (17; 7; 13). In order to assure the formation of
a self-median triangle, the lengths a0 + b0; a0 � b0; c0 have to satisfy the
triangle inequality (a0 � b0) + c0 > a0 + b0. This simpli�es to c0 > 2b0.
In other words the hypotenuse must exceed twice the shorter leg to give
us a self-median triangle. It is trivial to check that the other two triangle
inequalities are satis�ed. Moreover, in the right triangle (a0; b0; c0), we have
a0 + b0 > c0 > a0 � b0 which is the same as a > c > b.
Theorem 2 tells us how to recover the right triangle from a given self-
median triangle.
THEOREM 2: Let (a; b; c) be a self-median triangle in which a > c > b
holds. Then (a0; b0; c0) =�12(a+ b); 1
2(a� b); c
�is a right triangle in which
a20 + b
20 = c
20; a0 > b0 and c0 > 2b0 hold.
Proof: Since a > c, we �nd that r(m2a�m2
c) = 3(c2�a2) < 0, or,ma < mc.
Thus a > c > b() mb > mc > ma. From this and the de�nition of self-
median triangle we have amb
= bma
= cmc
. We square two of these equations,
use the formulas (1) and simplify. This yields (a2� b2)(a2+ b
2� 2c2) = 0.
The hypothesis a > c > b shows that this is equivalent to a2+b2 = 2c2. We
consistently arrive at this same equation (no matter which two equations are
squared) which is the same as�12(a+ b)
�2+�12(a� b)
�2= c
2 or a20+b20 = c
20,
with a0 > b0. Now in triangle (a; b; c), we have c > a� b so that c0 > 2b0holds and the proof is complete.
Remark 1: The proof of Theorem 2 says much more than its statement. Since
the argument is reversible it has given two characterizations of self-median
triangles: Suppose the side lengths of triangle (a; b; c) satisfy a > c > b.
Then the triangle is self-median (i) if and only if a : mb = b : ma = c : mc
or (ii) if and only if a2 + b2 = 2c2. Also, it is easy to see that if triangle
(a; b; c) has a : ma = b : mb = c : mc then it must be equilateral. This
useful observation enables us to give an elegant proof of Theorem 3.
Theorems 1 and 2 considered the determination of self-median trian-
gles in each of which the sides have distinct length measures. Theorem 3
shows that if an isosceles triangle is self-median then it must be equilateral.
THEOREM 3: The equilateral triangle is the only self-median triangle not
covered by the Theorems 1 and 2.
Proof: Any self-median triangle (a; b; c) not covered by the Theorems 1 and 2
must have at least two sides equal, say a = b. Then ma = mb. Let
278
m1;m2;m3 denote the mediansma;mb;mc in some order so that the def-
inition of a self-median triangle enables us to write
a
m1
=b
m2
=c
m3
:
If m1 = mc then m2 = m3 = ma = mb. But then a = b forces
ma = mb = mc and hence the triangle is equilateral. If m1 = m2 =
ma = mb then m3 = mc. This again forces the triangle to be equilateral as
mentioned in Remark 1.
Exercise 1: Find the right triangle that generates the self-median triangle
(23; 7; 17).
Exercise 2: Suppose (a; b; c) is a right triangle with a2 + b2 = c
2. Show that
the triangle (ap2; bp2; c) is self-median.
Exercise 3: Show that there are exactly two self-median triangles in each
of which (i) all the side lengths are integers and (ii) two side lengths are 7
and 17.
Theorems on self-altitude triangles
Theorem 4 tells us how an appropriate right triangle yields a self-altitude
triangle.
THEOREM 4: Let (a0; b0; c0) be a right triangle in which a20 + b20 = c
20 and
a0 > 2b0 hold. Then (a; b; c) = (a0; c0+b0; c0�b0) is a self-altitude trianglein which b > a > c holds.
Proof: Let ha; hb; hc denote the altitudes to the sides a; b; c and 4 denote
the area of this triangle. Then aha = bhb = chc = 24 or
a0ha = (c0 + b0)hb = (c0 � b0)hc = 24 :
Now
a0
ha
=a20
a0ha
=a20
24 =c20 � b
20
(c0 � b0)hc=
c20 � b
20
(c0 + b0)hb
ora0
ha
=c0 + b0
hc
=c0 � b0
hb
() a
ha
=b
hc
=c
hb
implying that the triangle (a; b; c) is self-altitude. Furthermore, in triangle
(a0; b0; c0), we have that c0 + b0 > a0 > c0 � b0 holds. This in triangle
(a; b; c) is equivalent to b > a > c.
Again, not any right triangle (a0; b0; c0) yields a self-altitude triangle.
For example, (a0; b0; c0) = (3; 4; 5) yields (a; b; c) = (3; 9; 1) and there is
no triangle with these sidelengths. However, (a0; b0; c0) = (12; 5; 13) yields
(a; b; c) = (12;18; 8). We may divide these side lengths by their gcd 2 and
279
take the primitive self-altitude triangle (6; 9; 4). Here the triangle inequality
to be satis�ed is a0 + (c0 � b0) > c0 + b0 which simpli�es to a0 > 2b0. It
is trivial to check that the other two triangle inequalities are satis�ed.
Theorem 5 tells us how to recover the right triangle from a given self-
altitude triangle.
THEOREM 5: Let (a; b; c) be a self-altitude triangle in which b > a > c
holds. Then
(a0; b0; c0) =
�a;
1
2(b� c);
1
2(b+ c)
�
is a right triangle in which a20 + b20 = c
20 and a0 > 2b0 hold.
Proof: From the hypothesis b > a > c we have hb < ha < hc. From the
de�nition of self-altitude triangle we have
a
ha
=b
hc
=c
hb
() a2
24 =bc
24 =bc
24 ;
where 4 is the area of triangle (a; b; c). Thus
a2 = bc =
�1
2(b+ c)
�2��1
2(b� c)
�2
which is a20 + b20 = c
20. From a > b� c follows a0 > 2b0 as required.
Remark 2: The proof of Theorem 5 gives two characterizations of self-altitude
triangles: Suppose the side lengths of triangle (a; b; c) satisfy b > a > c.
Then the triangle is self-altitude
(i) if and only if a : ha = b : hc = c : hb, or
(ii) if and only a2 = bc.
It is easy to see that if in triangle (a; b; c) we have a : ha = b : hb = c : hcthen it must be equilateral. This observation enables us to give an elegant
proof of Theorem 6.
Theorems 4 and 5 considered the determination of self-altitude trian-
gles in each of which the sides have distinct length measures. Theorem 6
shows that if an isosceles triangle is self-altitude then it must be equilateral.
THEOREM 6: The equilateral triangle is the only self-altitude triangle not
covered by the Theorems 4 and 5.
Proof: We omit. It is similar to that of Theorem 3.
Exercise 4: Find the right triangle that generates the self-altitude tri-
angle (35; 49; 25).
Exercise 5: Let ABC be a right triangle with right angle at C. CD is drawn
perpendicular to AB with the pointD onAB. Prove that the triangle whose
280
side lengths are AD, DB;CD is self-altitude. In terms of a; b; c give an
answer to the question: when do the lengths AD;DB;CD form a triangle?
Remarks 1 and 2 suggest the following.
OPEN PROBLEM: Suppose AD;BE;CF are three concurrent cevians of tri-
angle ABC. Assume that BC
AD= CA
BE= AB
CFholds. Prove or disprove that
the triangle ABC is equilateral.
Acknowledgement: I thank the referee for his painstaking e�orts to improve
the presentation. He also suggested Theorems 3 and 6.
References
[1] K.R.S. Sastry, Self-Median Triangles, Mathematical Spectrum, 22(1989/90), pp. 58-60.
[2] K.R.S. Sastry, Self-Altitude Triangles, Mathematical Spectrum, 22(1989/90), pp. 88-90.
K.R.S. Sastry
Jeevan Sandhya
Doddakalsandra Post
Raghuvanahalli
Bangalore 560 062, India
Advance Notice
At the summer 1999 meeting of the Canadian Mathematical Society, to
be held in St. John's, Newfoundland, there will be a Mathematics Education
Session on the topic \What Mathematics Competitions do for Mathematics".
Invited speakers include Edward Barbeau, Toronto; Tony Gardner, Birm-
ingham, England; Ron Dunkley, Waterloo; and Rita Janes, St. John's. Any-
one interested in givinga paper at this session should contact one of the orga-
nizers, Bruce Shawyer or Ed Williams, at the Department ofMathematics and
Statistics, Memorial University of Newfoundland, St. John's, Newfoundland,
Canada.
Email addresses:
281
THE SKOLIAD CORNERNo. 31
R.E. Woodrow
As a contest this month we give the U.K. Intermediate Mathematical
Challenge which was written Thursday, 5th February 1998. This contest is
organized from the United Kingdom Mathematics Trust from the School of
Mathematics, the University of Leeds. The contest is open to students in
School year 11 or below. My thanks go to John Grant McLaughlin of the
Faculty of Education, Memorial University of Newfoundland, who sent me a
copy of the questions.
U.K. INTERMEDIATE MATHEMATICAL CHALLENGEFebruary 5, 1998
Time: 1 hour
Five marks are awarded for each correct answer to questions 1{15. Sixmarks are awarded for each correct answer to questions 16{25. Each in-correct answer to questions 16{20 loses 1 mark. Each incorrect answer toquestions 21{25 loses 2 marks.
1. One quarter of a number is 24. What is one third of the original
number?
(a) 6 (b) 8 (c)32 (d) 72 (e) 96
2. 6% of 6 plus 8% of 8 equals
(a) 0:14 (b) 1 (c)1:4 (d) 1:96 (e) 2
3. Starting at A, a point on a �xed circle with centre, O, I �rst move
anticlockwise one quarter of the way round the circle to a point, W , hop
across to X | the opposite end of the diameter throughW , then travel one
�fth of the way round the circle clockwise to the point Y before hopping
across to Z, the point at the opposite end of the diameter through Y . How
big is \AOZ?
(a) 18� (b) 22� (c)162� (d) 198� (e) 270�
4. Which fraction is the odd one out?
(a) 1+47+4
(b) 20140
(c)0:21:4
(d) 1�117�11 (e) 8
56
5. J is the set of High Court judges; K is the set of living things
beginning withK; L is the set of all living creatures; M is the set of brilliant
mathematicians. Kevin is a very ordinary kangaroo. In which of the �ve
regions A-E of the diagram does Kevin belong?
282
A B
E D
C
K
J
L M
(a) A (b) B (c) C (d) D (e) E
6. ABCD is a square with sides of length 9 cm. How many points
(inside or outside the square) are equidistant from B and from C, and are
exactly 6 cm from A?
(a) 0 (b) 1 (c) 2 (d) 3 (e)more than 3
7. Each person's birthday product is obtained by multiplying the day
of the month in which they were born by the number of the month in which
they were born, and then multiplying the answer by the year in which they
were born. Here are �ve English queens and their birthdays. Which of them
has the same birthday product as someone born today? [Ed. \today" is
February 5, 1998.]
(a) Mary I, 18 February 1516 (b) Elizabeth I, 7 September 1533
(c) Anne, 6 February 1665 (d) Victoria, 24 May 1819
(e) Elizabeth II, 21 April 1926
8. How large will an angle of 212
�appear to be if you enlarge it by look-
ing through a stack of �ve magnifying glasses, each one of which magni�es
by a factor of 2?
(a) 212
�(b) 121
2
�(c) 25� (d) 40� (e) 80�
9. What is the total number of letters in all the incorrect options for
this question?
(a) eleven (b) twenty two (c) thirty three (d) forty four (e) �fty �ve
10. On four tests, each marked out of 100, my average was 85. What
is the lowest mark I could have scored on any one test?
(a) 0 (b) 40 (c) 60 (d) 81 (e) 85
11. TheWorldWide Fund For Nature estimates that 54 acres of Brazil-
ian rainforest are destroyed every minute of every day. Approximately how
many acres are lost each week?
(a) 50; 000 (b) 80; 000 (c) 200; 000 (d) 500; 000 (e) 2; 000; 000
283
12. IfC� Celsius is the same temperature as F � Fahrenheit, then F =
(95)C + 32. To avoid working with fractions and awkward numbers, some
people use the approximate formula F 0 = 2C+30. What is the temperature
in degrees Celsius when the approximate formula gives an answer which is
too large by 1?
(a) 5 (b) 9 (c) 10 (d) 12 (e) 15
13. I fold a piece of paper in half, then in half again before cutting a
shape from the folded paper as shown.
XXy
When I unfold the paper, what do I see?
A B C D E
(a) A (b) B (c) C (d) D (e) E
14. In a sponsored \Animal Streak" the cheetah ran at 90 km/hr while
the snail slimed along at 20 hr/km. The cheetah kept going for 18 seconds.
Roughly how long would the snail take to cover the same distance?
(a) 9 months (b) 9 weeks (c) 9 days (d) 9 hours (e) 9 minutes
15. Wallace and Gromit are waiting in a queue. There are x people
behind Wallace, who is y places in front of Gromit. If there are n people in
front of Gromit, how long is the queue?
(a) n� x+ y + 2 (b) n+ x� y (c) n� x+ y � 1
(d) n+ x� y + 1 (e) n� x+ y
16. I made just enough sticky treacle mixture to exactly �ll a square
tin of side 12 inches. But all I could �nd were two 812inch square tins. How
well would the mixture �t?
(a) easily (b) just (with a teeny bit of room to spare) (c) an exact �t
(d) nearly (with a small over ow) (e) no way (major over ow)
284
17. Which of the four trianglesW , X, Y , Z are right-angled?
W
42�
132�
X
4x 5x
6x Y
x 3x
4x
Z6
5
4
(a) onlyW (b)W and X (c) X and Z (d) Y andW (e) Y and Z
18. The integers from 1 to 20 are listed below in such a way that the
sum of each adjacent pair is a prime number. Missing numbers are marked
as �s.20; �; 16; 15; 4; �; 12; �; 10; 7; 6; � ; 2; 17; 14; 9; 8; 5; 18; �:
Which number goes in the place which is underlined?
(a) 1 (b) 3 (c) 11 (d) 13 (e) 19
19. ABCDE is a regular pentagon. FAB is a straight line and
FA = AB. What is the ratio x : y : z?
A B
CE
D
F
zyx
(a) 1 : 2 : 3 (b) 2 : 2 : 3 (c) 2 : 3 : 4 (d) 3 : 4 : 5 (e) 3 : 4 : 6
20. The total length of all the edges of a cube is L cm. If the surface
area of the cube has the same numerical value L cm2, what is its volume
in cm3?
(a) 1 (b) L (c) 2 (d) L3 (e) 8
21. A piece of thin card in the shape of an equilateral triangle with
side 3 cm and a circular piece of thin card of radius 1 cm are glued together so
that their centres coincide. How long is the outer perimeter of the resulting
2{dimensional shape (in cm)?
(a) 2� (b) 6 + � (c) 9 (d) 3� (e) 9 + 2�
22. Shape A is made from 6 unit squares; shape B is made from 8,
C from 4, D from 8 and E from 3 unit squares. For four of these shapes,
four exact copies can be �tted together to make a rectangle. Which is the odd
one out?
285
A B C D E
(a) A (b) B (c) C (d) D (e) E
23. In this unusual game of noughts and crosses the �rst player to
form a line of three Os or three Xs loses. It is X's turn. Where should she
place her cross to make sure that she does not lose?
E
C
A
X
X
O
O
D
B
(a) A (b) B (c) C (d) D (e) E
24. Each of the sides of this regular octagon has length 2 cm. What
is the di�erence between the area of the shaded region and the area of the
unshaded region (in cm2)?
r
(a) 0 (b) 1 (c) 1:5 (d) 2 (e) 2p2
25. A square is inscribed in a 3{4{5 right-angled triangle as shown.
What fraction of the triangle does it occupy?
3
4
5
(a) 1225
(b) 2449
(c) 12
(d) 2549
(e) 1325
That completes this number of the corner. Send me contest materials,
nice solutions, and any suggestions about items you would like see covered
in the Skoliad Corner.
286
MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and byHigh School and University Students. It continues, with the same emphasis,
as an integral part of Crux Mathematicorum with Mathematical Mayhem.
All material intended for inclusion in this section should be sent to theMayhem Editor, Naoki Sato, Department of Mathematics, Yale University,PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronicaddress is still
The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).
The rest of the sta� consists of Adrian Chan (Upper Canada College), Jimmy
Chui (Earl Haig Secondary School), Richard Hoshino (University ofWaterloo),
David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).
Shreds and Slices
Primitive Roots and Quadratic Residues, Part 2
Recall the work in Part 1 [1998: 220]: Let p be an odd prime, and Ap,
Bp the quadratic residues and non-residues respectively. Let � be a primitive
pth root of unity; that is, �p = 1 and � 6= 1. These imply 1 + � + �
2 + � � �+�p�1 = 0, an important fact we will use later. Let
x =Xa2Ap
�a; y =
Xb2Bp
�b:
Then x+ y = �1, and we saw xy always seems to be an integer, but why,
and which one?
We introduce some basic quadratic residue theory. Let a be a non-zero
integer modulo p. We de�ne the Legendre symbol (ap) as follows:
�a
p
�=
�1 if a is a quadratic residue;
�1 if a is a quadratic non-residue:
Although this notation has some deep consequences, we should think of it
as just that: notation. We will now prove a useful relation, known as Euler's
Criterion: �a
p
�� a
(p�1)=2 (mod p) :
If a is a q.r. (quadratic residue), then x2 � a for some x, so a(p�1)=2 �xp�1 � 1 (mod p) by Fermat's Little Theorem. Otherwise, if a is a non-q.r.
287
(quadratic non-residue), then for all x, 1 � x � p� 1, there is a unique y,
1 � y � p�1, y 6= x, such that xy � a. All integers in f1; 2; : : : ; p�1g canbe so paired. Their product is a(p�1)=2 � 1 � 2 � � � (p� 1) � (p� 1)! � �1(mod p), by Wilson's Theorem.
As a corollary, note that��1p
�=
�1 if p � 1 (mod 4);
�1 if p � 3 (mod 4):
Consider the case p � 3 (mod 4), so p = 4k + 3 for some k, and
(p � 1)=2 = 2k + 1. Let a1, a2, : : : , a2k+1 be the q.r.'s and b1, b2, : : : ,
b2k+1 the non-q.r.'s. Since (�1p) = �1, a is a q.r. if and only if�a � p� a
is a non-q.r.. Hence, for all i, there exists a unique j such that ai + bj = p.
The expression we wish to calculate is
xy = (�a1 + �a2 + � � �+ �
a2k+1)(�b1 + �b2 + � � �+ �
b2k+1)
= �a1+b1 + �
a1+b2 + � � �+ �a1+b2k+1
+ �a2+b1 + �
a2+b2 + � � �+ �a2+b2k+1
+ � � �+ �
a2k+1+b1 + �a2k+1+b2 + � � �+ �
a2k+1+b2k+1 :
Note that there are (2k + 1)2 terms, and there is a �p = 1 term in each
row. This leaves (2k+ 1)2 � (2k + 1) = 2k(2k+ 1) terms. We claim that
�, �2, : : : , �p�1 = �4k+2 appear the same number of times among these
2k(2k + 1) terms. Let n be a non-zero integer modulo p. We wish to �nd
the number of ways n can be expressed as the sum of a q.r. u and a non-q.r.
v: n � u+ v.
Assume 1 can be expressed in t distinct such ways: 1 � u1+v1 � u2+
v2 � � � � = ut+ vt. Then n � nu1+nv1 � nu2 +nv2 � � � � � nut+nvt.
If n is a q.r., then so is nui, and nvi remains a non-q.r.. Otherwise, nuiand nvi switch roles, and in either case, there are again t distinct ways of
expressing n as the sum of a q.r. and a non-q.r..
Therefore, among the 2k(2k + 1) terms, each power of � appears
2k(2k+ 1)=(p� 1) = k times. Hence,
xy = 2k+ 1+ k(� + �2 + � � �+ �
4k+2)
= 2k+ 1� k = k + 1 =1 + p
4:
The case p � 1 (mod 4), say p = 4k + 1, is similar, and is only brie y
sketched here. In this case, (�1p) = 1, so a and �a � p � a both occur as
either ai or bj. Hence, when we expand xy to obtain (2k)2 = 4k2 terms,
288
there are no terms of the form �p = 1, and using the same reasoning, the
powers of � appear evenly among the 4k2 terms, so
xy =4k2
p� 1(� + �
2 + � � �+ �4k) = �k =
1� p
4:
In general, we may write
xy =1�
��1p
�p
4:
Hence, x and y are the roots of the quadratic
t2 + t+
1���1p
�p
4
which are namely
1
2
�1�
s��1p
�p
!:
Which is x and which is y? It depends on the choice of �. Note that the
discriminant of this quadratic is
(x� y)2 = x2 � 2xy+ y
2 = (x+ y)2� 4xy =
��1p
�p:
Hence, it is possible to express��1p
�p in terms of any primitive pth root of
unity.
Mayhem Problems
The Mayhem Problems editors are:
Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.
Note that all correspondence should be sent to the appropriate editor |
see the relevant section. In this issue, you will �nd only problems | the
next issue will feature only solutions.
We warmly welcome proposals for problems and solutions.
289
High School Problems
Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,
Canada. L3S 3E8 <[email protected]>
H241. Find the integer n that satis�es the equation
1 � 1998 + 2 � 1997 + 3 � 1996 + � � �+ 1997 � 2 + 1998 � 1 =
�n
3
�:
H242. Let a and b be real numbers that satisfy a2 + b2 = 1. Prove
the inequality
ja2b+ ab2j �
p2
2:
Determine the values of a and b for which equality occurs. (See how many
ways you can solve this!)
H243. For a positive integer n, let f(n) denote the remainder of
n2 +2 when divided by 4. For example, f(3) = 3 and f(4) = 2. Prove that
the equation
x2 + (�1)yf(z) = 10y
has no integer solutions in x, y, and z.
H244. For a positive integer n, let P (n) denote the sum of the digits
of n. For example, P (123) = 1 + 2 + 3 = 6. Find all positive integers n
satisfying the equation P (n) = n74.
Advanced Problems
Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.
M1G 1C3 <[email protected]>
A217. Proposed by Alexandre Trichtchenko, OAC student, Brook�eldHigh School, Ottawa.
Show that for any odd prime p, there exists a positive integer n such
that n, nn, nnn
, : : : all leave the same remainder upon division by p where
n does not leave a remainder of 0 or 1 upon division by p.
A218. Proposed by Mohammed Aassila, Universit �e Louis Pasteur,Strasbourg, France.
(a) Suppose f(x) = xn+qxn�1+t, where q and t are integers, and suppose
there is some prime p such that p divides t but p2 does not divide t.
Show, by imitating the proof of Eisenstein's Theorem, that either f is
irreducible or f can be reduced into two factors, one of which is linear
and the other irreducible.
290
(b) Deduce that if both q and t are odd then f is irreducible.
(Generalization of Question 1, IMO 1993)
A219. Proposed by Mohammed Aassila, Universit �e Louis Pasteur,Strasbourg, France.
Solve the following system:
3
�x+
1
x
�= 4
�y +
1
y
�= 5
�z +
1
z
�;
xy + yz + zx = 1 :
A220. Proposed by Waldemar Pompe, student, University ofWarsaw, Poland.
P is an interior point of triangle ABC. D, E, F are the feet of the
perpendiculars from P to the linesBC, CA, AB, respectively. Let Q be the
interior point of triangle ABC such that
\ACP = \BCQ and \BAQ = \CAP :
Prove that \DEF = 90� if and only ifQ is the orthocentre of triangleBDF .
Challenge Board Problems
Editor: David Savitt, Department of Mathematics, Harvard University,
1 Oxford Street, Cambridge, MA, USA 02138 <[email protected]>
C79. Proposed by Mohammed Aassila, Universit �e Louis Pasteur,Strasbourg, France.
Let f : R+ ! R+ be a non-increasing function, and assume that there
are two constants p > 0 and T > 0 such thatZ 1
t
f(s)p+1 ds � Tf(0)pf(t)
for all t 2 R+. Prove that
f(t) � f(0)
�T + pt
T + pT
�� 1p
for all t � T .
C80. Suppose a1, a2, : : : , am are transpositions inSn (the symmetric
group on n elements) such that a1a2 � � � am = 1. Show that if the ai generate
Sn, thenm � 2n� 2.
291
The Pentagram Theorem
Hiroshi Koterateacher, Todaijigakuen Junior High School, Japan
A pentagram ABCDE is a �ve-pointed star-shaped �gure made by extend-ing the sides of a convex pentagon FGHIJ until they meet, and is sometimes
used as a mystic symbol.
A
B
C
D
E F
G
H
I
J
In my math class, Eiji Konishi, a Todaijigakuen Junior High School stu-
dent in Japan, found a beautiful theorem concerning the pentagram. I would
like to show the theorem he proved by himself. We need two lemmas.
Lemma 1. In triangle ABC, let D and E be points on sides AB and
AC respectively, and let F be the intersection of BE and CD. Then
AC
AB� DBDC
� EBEC
� FCFB
= 1 :
Proof. Let Hi (i = 1; 2; 3; 4) be points on side BC such that DH1,
AH2, FH3, and EH4 are all perpendicular to BC.
A
B C
D
EF
H1 H2 H3 H4
Copyright c 1998 Canadian Mathematical Society
292
Since
FH3 = DH1 �FC
DCand DH1 = AH2 �
DB
AB;
we have
FH3 = AH2 �DB
AB� FCDC
: (1)
Similarly, since
FH3 = EH4 �FB
EBand EH4 = AH2 �
EC
AC;
we have
FH3 = AH2 �EC
AC� FBEB
: (2)
Equations (1) and (2) imply
DB
AB� FCDC
=EC
AC� FBEB
:
It follows thatAC
AB� DBDC
� EBEC
� FCFB
= 1 :
Lemma 2. Let D and E be two points on side AB of triangle ABC,
and let F be a point on side BC. Let G be a point on side BC, extended.
Let H be the intersection of DG and AC, let I be the intersection of DG
and AF , and let J be the intersection of GE and AC. Then
DE
EB� BFFC
� CJJH
� HI
ID= 1 :
Proof. Let L be the point on AC such that BLkDG, and let N be the
point on DG such that NBkAC. Let M be the intersection of AF and BL,
and let P be the intersection of GE and BN .
By Menelaus's Theorem in triangles BCL and BDN , we have
BF
FC� CAAL
� LMMB
= �1 ; (3)
BP
PN� NG
GD� DEEB
= �1 : (4)
By similar triangles,LM
MB=
HI
ID;
293
A
B
L
M
N
P
G
H
I
D
E
F C
J
andAH
AD=
AL
AB;
so by (3),
BF
FC� LMMB
=BF
FC� HI
ID
= �AL
CA= �AH
AD� ABAC
: (5)
Similarly,BP
PN=
CJ
JH;
andGC
GH=
GB
GN;
so by (4),
BP
PN� DEEB
=CJ
JH� DEEB
= �GD
NG= �GD
GH� GCGB
: (6)
Equations (5) and (6), and Lemma 1 imply
BF
FC� HI
ID� CJJH
� DEEB
=AH
AD� ABAC
� GDGH
� GCGB
= 1 :
It follows thatDE
EB� BFFC
� CJJH
� HI
ID= 1 :
294
Theorem. Let A1A2A3A4A5 be a pentagram with pentagon
B1B2B3B4B5 as shown in the following �gure. Let Ck be the intersection
of the line AkBk with the side Bk+2Bk+3, k = 1, 2, 3, 4, 5. Then
B3C1
C1B4
� B4C2
C2B5
� B5C3
C3B1
� B1C4
C4B2
� B2C5
C5B3
= 1 :
A1
A2
A3
A4
A5 B4
B5
B1
B2
B3
C1
Proof. Lemma 2 yields
B3C1
C1B4
� B4C2
C2B5
=B1A4
B5B1
� B2B3
A4B2
;
and
B5C3
C3B1
� B1C4
C4B2
=B3A1
B2B3
� B4B5
A1B4
:
By Menelaus's Theorem in triangle A4B5C5, we have
A4B1
B1B5
� B5A5
A5C5
� C5B2
B2A4
= �1 ;
and in triangle A1B5C5, we have
A1B4
B4B5
� B5A5
A5C5
� C5B3
B3A1
= �1 :
295
It follows that
B3C1
C1B4
� B4C2
C2B5
� B5C3
C3B1
� B1C4
C4B2
� B2C5
C5B3
=B1A4
B5B1
� B2B3
A4B2
� B3A1
B2B3
� B4B5
A1B4
� B2C5
C5B3
=
�B1A4
B5B1
� B2C5
A4B2
���B3A1
A1B4
� B4B5
C5B3
�
=A5C5
B5A5
� B5A5
A5C5
= 1 :
This result may be something already well known, but I presume the
fact that its discovery has been made by a junior high school student is sig-
ni�cant.
Have you heard about ATOM?
ATOM is \A Taste Of Mathematics" (Aime{T{On les Math �ematiques).
The ATOM series
The booklets in the series, A Taste of Mathematics, are published by
the Canadian Mathematical Society (CMS). They are designed as enrichment
materials for high school students with an interest in and aptitude for math-
ematics. Some booklets in the series will also cover the materials useful for
mathematical competitions at national and international levels.
La collection ATOM
Publi �es par la Soci �et �e math �ematique du Canada (SMC), les livrets de la
collection Aime-t-on lesmath �ematiques (ATOM) sont destin �es au perfection-
nement des �etudiants du cycle secondaire qui manifestent un int �eret et des
aptitudes pour les math �ematiques. Certains livrets de la collection ATOM
servent �egalement de mat �eriel de pr �eparation aux concours de math �emati-
ques sur l' �echiquier national et international.
This volume contains the problems and solutions from the 1995{1996
Mathematical Olympiads' Correspondence Program. This program has sev-
eral purposes. It provides students with practice at solving and writing up
solutions to Olympiad-level problems, it helps to prepare student for the
Canadian Mathematical Olympiad and it is a partial criterion for the selec-
tion of the Canadian IMO team.
For more information, contact the Canadian Mathematical Society.
296
A Combinatorial Triad
Cyrus Hsiastudent, University of Toronto
One of the beautiful aspects of mathematics is that solving a problem
can be done in so many ways. In combinatorics, the study of counting and
ways of counting, this is more than evident in its many interesting problems.
See any standard textbook on combinatorics [1] to convince yourself. Con-
sider the following three standard combinatorial problems and see if you can
�nd the link between them.
Problem 1. What is the number of ways to distribute 10 identical apples
into 3 baskets labelled A, B, and C?
Problem 2. How many ways are there to select 10 ice-cream scoops
from 3 di�erent avours on an ice-cream cone?
Problem 3. Find the number of non-negative integer solutions to the
diophantine equation x1 + x2 + x3 = 10.
If you said the answers are all the same, then you are correct. The three
problems above are di�erent ways of counting something equivalent. Each
way of thinking may be more helpful in certain situations than others. The
following are solutions to approaching the apparently di�erent, but similar,
problems.
Solution to Problem 1. The standard approach to solving this type of
problem rests on the following key trick. Let the apples be represented by
1's since they are all identical. Now use two 0's to place between the 1's
to split them into 3 piles. Put the �rst pile into basket A, the second into
basket B and the remaining pile into basket C. The problem then reduces
to �nding the number of ways of placing the 0's among the 1's.
There are 10 + 1 = 11 positions to place the �rst 0, namely 9 spots
between the 10 1's and the front and back of the list of 1's. Now there are
11 numbers and 11 + 1 = 12 positions to place the second 0. However,
we are counting everything twice! If we place 0 in front and then a 0 after
the �fth 1, this is the same as if we place the �rst 0 after the �fth 1 and the
second 0 in the �rst spot. Hence, there are
11� 12
2
ways in total.
Solution to Problem 2. This is exactly the same as Problem 1 if we
consider the ice-cream scoops to be 1's and the type of avour it is by its
Copyright c 1998 Canadian Mathematical Society
297
position in one of the three piles of numbers. We use two 0's again to be
dividing points for the three piles. The number of ways of selecting scoops
from three avours becomes a problem of arranging the 10 1's and two 0's.
There are �12
2
�=
11� 12
2
ways of doing this.
Solution to Problem 3. Here, the problem and solution have become
completely algebraic. The standard way of solving this would be to use gen-
erating functions. Consider the following function:
(1 + x+ x2 + x
3 + � � � )3:Now consider the coe�cient of x10. How do you get this term? Take xa
in the �rst factor, xb in the second, and xc in the third so that a+b+c = 10.
Thus, we want all possible ways of doing this, so that a, b, and c are solutions
to the equation in the problem: x1 + x2 + x3 = 10. The coe�cient is the
number of nonnegative integer solutions to the equation!
Of course, one must check that the coe�cient of x10 is indeed�12
2
�.
Here are some problems to drive this idea home.
Standard Exercises
1. Reformulate the following problems to the other two types and solve them.
(a) How many ways are there to select 98 DNA codons if there are 4 di�erentcodons to choose from?
(b) Solve the diophantine equation x1 + x2 + � � � + x19 = 97, with xi � 3
for all i = 1, 2, : : : , 19.
(c) How many ways are there to put 98 pigeons into 97 pigeonholes?(Aside: Must one pigeonhole have more than one pigeon?)
2. How many ways are there to distribute 9 apples, 8 bananas and 7 oranges to 6
hungry school children? How would you reformulate this problem to the othertwo?
Other Exercises
1. Howmany ways are there to select 3 numbers from the integers 1 to 20 so thatno two are consecutive?
2. Howmany ways are there to distribute 10 distinct balls into three tubes labelledA, B, and C taking the order of the balls in each tube into account? (Careful!This is harder than meets the eye.)
Reference
[1] Tucker, Alan. Applied Combinatorics. John Wiley & Sons, Inc., Toronto. 1995pp. 205.
298
Swedish Mathematics Olympiad
1987 Qualifying Round
1. Kalle has a litre ask A full of orange juice and an empty litre ask B.
He pours part of the juice from the full askA to the empty askB. He
then adds water to B until it is full, and shakes it so that the mixture is
thoroughly blended. Lastly, he adds the blended mixture from ask B
to ask A until ask A is full. Show that ask A now contains at least
75% of the original juice.
2. Show that the number abc, which denotes a 3-digit number written in
the usual way, is divisible by 7 if and only if the number �a+ 2b� 3c
is divisible by 7.
3. Solve the system of equations
(x+ y)(x2� y2) = 1176
(x� y)(x2 + y2) = 696:
4. In the triangle ABC, arbitrarily choose a point P on the side BC.
Through the midpoint M of the same side, draw DM parallel to AP ,
where D is a point on one of the other two sides. Show that the line
segment DP divides the triangle into two regions of equal area.
5. Sixteen light bulbs are connected in a square network as shown in the
�gure. Each row and column has a switch which works as follows: one
half-turn of the switch extinguishes all the lighted bulbs in the row (or
column), and turns on all the bulbs that are o� in that row (or column).
{ { {
{ {
{ {
{ { {
In the initial state, six bulbs are on, as shown in the �gure. Prove that
it is impossible to turn on all 16 bulbs, using the switches. What is the
maximum number of bulbs that can be turned on?
299
6. A small school with 50 pupils has put the pupils into 8 groups. There are
4more pupils in the largest group than there are in the smallest group.
These groups are combined two by two into larger groups as follows.
The two smallest of the original groups are combined, numbers 3 and
4 form the next combined group, and so on. The largest of the four
groups so formed now has 5 more pupils than the smallest. The two
smallest of these groups form class A and the two largest form class B.Class B has 6 more pupils than class A.
Find the numbers of pupils in the original groups.
1987 Final Round
1. Sixteen real numbers are arranged in a \magic square" of side 4 in such
a way that the sum of the numbers in each row, in each column and in
each of the two main diagonals is always k. Prove that the sum of the
four numbers in the corners of the square is also k.
2. A circle of radius R is divided into two equal parts by the arc of another
circle. Show that the arc of this circle (which lies inside the given circle)
is longer than 2R.
3. Assume that 10 closed intervals all of length 1 are placed in the interval
[0; 4]. Show that there exists some point in the larger interval that
belongs to at least 4 of the smaller intervals.
4. f is a di�erentiable function de�ned on the interval 0 � x � 1 and
f(0) = f(1) = 1. Show that there exists at least one point y in the
interval such that
jf 0(y)j = 4
Z 1
0
jf(x)j dx:
5. The numbers a, b, c, and d are positive and abcd = 1. Show that there
exists a positive number t such that for all such a, b, c, and d,
1
1 + a+
1
1 + b+
1
1 + c+
1
1 + d> t:
Find the largest t with this property.
6. A baker with access to a number of di�erent spices bakes 10 cakes. He
uses more than half of the di�erent kinds of spices in each cake. No two
of the combinations of spices are exactly alike. Show that there exist
three spices a, b, c, such that every cake contains at least one of these.
300
J.I.R. McKnight Problems Contest 1983
1. Prove that in any triangleABC, whereAD is a median and jADj = m,
4m2 = b2 + c
2 + 2bc cosA.
2. Find the equations of the three normals to the parabola with the equa-
tion y2 = 4x, from (21;30).
Note: A normal ` to a curve C at a point P is a line ` which intersects
C at P and is perpendicular to the tangent of C at P .
3. Find the equations of the four common tangents of the circles having
equations x2 + y2 = 16 and (x� 25)2 + y
2 = 121.
4. A tank is �lled using two taps. It takes taps one and two t1 and t2
minutes respectively to �ll the tank each on its own. From the following
information, determine what the values t1 and t2 are:
(i) If the �rst tap is open for one-third of the time t2 and the second
tap is open for one-third of the time t1, then the fraction of the
volume of the tank that is �lled is 11=18.
(ii) If both taps are used simultaneously, it would take 3 hours and 36
minutes to �ll the tank.
5. Two bodies are moving with constant acceleration in a straight line go-
ing in the same direction. The �rst body starts 20 km ahead of the
other. It travels a distance of 25 and 50=3 km in 1 and 2 hours respec-
tively. The second body travels a distance of 30 and 59=2 km in 1 and 2
hours respectively. How long does it take for the second body to catch
up to the �rst?
Editor's comment. Problem4 of the above contest, as stated, has no solution.
The equation arising from condition (i) is not solvable for positive t1 and t2. However,
the problem is solvable with nice values if the fraction 11
18is replaced by 13
18.
301
PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, De-partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.
In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.
To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81
2"�11" or A4 sheets of paper.
These may be typewritten or neatly hand-written, and should be mailed tothe Editor-in-Chief, to arrive no later than 1 March 1999. They may alsobe sent by email to [email protected]. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication. Please note that we do not accept submissions sent by FAX.
2326? Correction. Proposed by Walther Janous, Ursulinengym-nasium, Innsbruck, Austria.
Prove or disprove that if A, B and C are the angles of a triangle, then
2
�<
Xcyclic
�1� sin A
2
� �1 + 2 sin A
2
�� � A
� 9
2�:
2329? Correction. Proposed by Walther Janous, Ursulinengym-nasium, Innsbruck, Austria.
Suppose that p and t > 0 are real numbers. De�ne
�p(t) := tp + t
�p + 2p and �p(t) :=�t+ t
�1�p + 2 :
(a) Show that �p(t) � �p(t) for p � 2.
(b) Determine the sets of p: A, B and C, such that
302
1. �p(t) � �p(t),
2. �p(t) = �p(t),
3. �p(t) � �p(t).
2351. Proposed by Paul Yiu, Florida Atlantic University, Boca Ra-ton, Florida, USA.
A triangle with integer sides is called Heronian if its area is an integer.
Does there exist a Heronian triangle whose sides are the arithmetic,
geometric and harmonic means of two positive integers?
2352. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.
Determine the shape of4ABC if
cosA cosB cos(A�B) + cosB cosC cos(B �C)
+ cosC cosA cos(C �A) + 2 cosA cosB cosC = 1 :
2353. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.
Determine the shape of4ABC if
sinA sinB sin(A�B) + sinB sinC sin(B � C)
+ sinC sinA sin(C �A) = 0 :
2354. Proposedby Herbert G �ulicher,WestfalischeWilhelms-Univer-sit�at, M�unster, Germany.
In triangle P1P2P3, the line joining Pi�1Pi+1 meets a line �j at the
point Si;j (i; j = 1; 2; 3, all indices taken modulo 3), such that all the points
Si;j , Pk are distinct, and di�erent from the vertices of the triangle.
1. Prove that if all the points Si;j are non-collinear, then any two of the
following conditions imply the third condition:
(a)P1S3;1
S3;2P2
� P2S1;2S1;3P3
� P3S2;3S2;1P1
= �1;
(b)S1;2S1;1
S1;1S1;3
� S2;3S2;2S2;2S2;1
� S3;1S3;3S3;3S3;2
= 1;
(c) �1, �2, �3 are either concurrent or parallel.
2. Prove further that (a) and (b) are equivalent if the Si;i are collinear.
Here, AB denotes the signed length of the directed line segment [AB].
303
2355. Proposed by G.P. Henderson, Campbellcroft, Ontario.For j = 1; 2; : : : ;m, let Aj be non-collinear points with Aj 6= Aj+1.
Translate every even-numbered point by an equal amount to get new points
A02, A
04, : : : , and consider the sequence Bj , where B2i = A
02i and
B2i�1 = A2i�1. The last member of the new sequence is either Am+1 or
A0m+1 according asm is even or odd.
Find a necessary and su�cient condition for the length of the path
B1B2B3 : : : Bm to be greater than the length of the path A1A2A3 : : : Am
for all such non-zero translations.
CRUX 1985 [1994: 250; 1995: 280] provides an example of such a con-
�guration. There, m = 2n, the Ai are the vertices of a regular 2n{gon and
A2n+1 = A1.
2356. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.Five points, A, B, C,K, L, with whole number coordinates, are given.
The points A, B, C do not lie on a line.
Prove that it is possible to �nd two points, M , N , with whole number
coordinates, such that M lies on the line KL and 4KMN is similar to
4ABC.
2357. Proposed by Gerry Leversha, St. Paul's School, London, Eng-land.
An unsteady man leaves a place to commence a one-dimensional ran-
dom walk. At each step he is equally likely to stagger one step to the east
or one step to the west. Let his expected absolute distance from the starting
point after 2n steps be a. Now consider 2n unsteady men each engaging in
independent random walks of this type. Let the expected number of men at
the starting point after 2n steps be b. Show that a = b.
2358. Proposed by Gerry Leversha, St. Paul's School, London, Eng-land.
In triangle ABC, let the mid-points of BC, CA, AB be L, M , N ,
respectively, and let the feet of the altitudes from A, B, C be D, E, F ,
respectively. Let X be the intersection of LE and MD, let Y be the inter-
section ofMF andNE, and let Z be the intersection ofND and LF . Show
that X, Y , Z are collinear.
2359. Proposed by Vedula N. Murty, Visakhapatnam, India.Let PQRS be a parallelogram. Let Z divide PQ internally in the ratio
k : l. The line through Z parallel to PS meets the diagonal SQ at X. The
line ZR meets SQ at Y .
Find the ratio XY : SQ.
304
2360. Proposed by K.R.S. Sastry, Dodballapur, India.In triangle ABC, let BE and CF be internal angle bisectors, and let
BQ and CR be altitudes, where F and R lie on AB, and Q and E lie on
AC. Assume that E, Q, F and R lie on a circle that is tangent to BC.
Prove that triangle ABC is equilateral.
2361. Proposed by K.R.S. Sastry, Dodballapur, India.The lengths of the sides of triangle ABC are given by relatively prime
natural numbers. Let F be the point of tangency of the incircle with side
AB. Suppose that \ABC = 60� and AC = CF . Determine the lengths of
the sides of triangle ABC.
2362. Proposed by Mohammed Aassila, Universit �e Louis Pasteur,Strasbourg, France.
Suppose that a, b, c > 0. Prove that
1
a(1 + b)+
1
b(1 + c)+
1
c(1 + a)� 3
1 + abc:
2363. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.
For natural numbers a; b; c > 0, let
q(a; b; c) := a+
a+
a+a+
a+:::b+:::
b+c+:::a+:::
b+c+
a+:::b+:::
a+c+:::a+:::
b+
c+a+
a+:::b+:::
b+c+:::a+:::
a+c+
a+:::b+:::
a+c+:::a+:::
(in the nth \column" above, from the third one onwards, we have, from top
to bottom, the sequence a; b; c; a repeated 2n�3 times), where it is assumed
that the right side (understood as an in�nite process) yields a well-de�ned
positive real number.
The original, a Talent Search Problem, asked to determine q(1;3; 5).
The value is3p2 (seeMathematics and Informatics Quarterly, 7 (1997), No. 1,
p. 53).
Determine whether or not there exist in�nitely many triples (a; b; c)
such that q(a; b; c) is the cube root of a natural number.
305
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.
R.P. Sealy was accidentally omitted from the list of solvers for problem 2338.
2015. [1995: 53, 129]Proposedby Shi-ChangShi and Ji Chen, NingboUniversity, China.
Prove that
(sinA+ sinB + sinC)
�1
A+
1
B+
1
C
�� 27
p3
2�;
where A, B, C are the angles (in radians) of a triangle.
Editor's comment. The �rst solution printed was incorrect [1996: 47,
solution I]. Its \correction" [1996: 125] was also incorrect. Finally, the sec-
ond solution printed, [1996: 48, solution II], was also incorrect. It is time
to correct these wrongs!!
In solution I, solver Grant claims that both terms
sinA+ sinB + sinC and 1=A+ 1=B + 1=C
are minimized when A = B = C(= 60�). This is true for the second term,
but NOT for the �rst! It is obvious that for the degenerate triangleA = 180�,B = C = 0�, the sine sum is zero while the sum is positive for A = B =
C. And of course this means that for `real' triangles su�ciently close to the
degenerate one, the sine sum will be less than the equilateral sine sum too.
Thus the proof falls apart.
In Solution II, the proposers �rst show that the function
y(x) = x�1=3 cosx is convex on (0; �=2]. However, in the second inequality
of their second displayed equation, they appear to apply Jensen's inequal-
ity to the function f(x) = log y(x), which is NOT convex everywhere in
this interval! That is, they want to prove that 6y(A=2)y(B=2)y(C=2) �6(y(�=6))3, which by taking logarithms is equivalent to f(A=2)+f(B=2)+
f(C=2)� 3f(�=6); that is,
1
3[f(A=2) + f(B=2)+ f(C=2)] � f [(1=3)(A=2+ B=2 + C=2)] :
For this we need that f is convex; it is not. In fact, the last inequality on
[1996: 48] is incorrect, as can be seen again for the degenerate triangle
A = 180, B = C = 0. However, this time we must use limits, so put
306
B=2 = C=2 = x and A=2 = �=2� 2x. Then you can �nd that the limit of
�cos(�=2� 2x) cos2(x)
(�=2� 2x)x2
�1=3
as x! 0 is ZERO, so the inequality fails.
Thanks to Waldemar Pompe and Bill Sands for pointing these out. So,
we must belatedly present a correct solution.
III Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Aus-tria.
SinceX
sinA =s
R, the given inequality is equivalent to
X 1
A� 27
p3
2�
R
s: (1)
Now, from item 6.60 on p. 188 in D.S. Mitrinovi�c et al., Recent Advancesin Geometric Inequalities, Kluwer Academic Publishers, 1989, the following
inequality (due to V. Mascioni) is known:
X 1
A� 9
�
rR
2r: (2)
We now show that (2) is stronger than (1). Indeed, we have:
9
�
rR
2r� 27
p3
2�
R
s; (3)
since this is equivalent to sp2 � 3
p3pRr; that is,
2s2 � 27Rr : (4)
Finally, (4) is simply item 5.12 on p. 52 in O. Bottema et al., GeometricInequalities, Groningen, 1969.
A very similar solution was submitted by Bob Prielipp, University of Wisconsin{
Oshkosh, Wisconsin, USA.
Remarks by Janous
(i) We can also show the inequality
3
qYsinA
X 1
A� 9
p3
2�: (5)
(Because of the AM{GM Inequality, (5) is stronger than (1).)
SinceY
sinA =sr
2R2
, we �nd that (5) can be re-written as
X 1
A� 9
p3
2�
3
r2R2
sr: (6)
307
We shall also show that
9
�
rR
2r� 9
p3
2�
3
r2R2
sr: (7)
A simple algebraic manipulation shows that this is the same as (4).
(ii) With the usual notation for the power mean
Mt(x; y; z) =
�xt + y
t + zt
3
� 1t
; t 6= 0 and M0(x; y; z) = (xyz)13 ;
we �nd that (5) is
Mt(sinA; sinB; sinC) � 3p3
2�M�1(A;B; C) with t = 0 : (8)
This raises the questions:
1. Is t = 0 the minimum value of t such that (8) holds?
2. What is the maximum value of t such that (8) holds with \�" instead of \�"?
3. What about analogous questions if we replace the right side of (8) by
3p3
2�Mp(A; B;C)?
(iii) Note thatY
sin(A=2) =r
4R: So, inequality (3) can be re-written as
1
3
XsinA �
p6
sYsin
�A
2
�:
This suggests a further question:
Let 0 < � � 1 be a given real number. Determine the optimum constantC = C(�) such that
1
3
XsinA � C
�Ysin(�A)
��
for all triangles. I conjecture that [Ed. using A = B = C = 60�]
C(�) =
p3
2
�sin
���
3
���3�
:
Similarly, inequality (7) is equivalent to
3
qYsinA �
p6
sYsin
�A
2
�;
leading to the analogous question of �nding an optimal constantD = D(�) such that
3
qYsinA � S
�Ysin(�A)
��
;
where 0 < � � 1. I conjecture that D(�) = C(�).
308
220A?.[1996: 363] Proposed by Ji Chen, Ningbo University, China.Let P be a point in the interior of the triangle ABC, and let
�1 = \PAB, �1 = \PBC, 1 = \PCA.
Prove or disprove that 3p�1�1 1 � �=6.
Solution by Kee-Wai Lau, Hong Kong (slightly modi�ed by the editor).
We prove the inequality. We write � = �1, � = �1, = 1. Applying
the Sine Rule to the triangles PAB, PBC, PCA we obtain
sin� sin� sin = sin(A� �) sin(B � �) sin(C � ) :
The function log sinx is concave for 0 < x < �; hence
log sin(A� �) + log sin(B � �) + log sin(C � )
� 3 log sin
�A+B+C�����
3
�:
Therefore, we get
log sin�+ log sin� + log sin � 3 log sin
�������
3
�: (1)
For �; �; > 0 and �+ � + < �, set
f(�;�; ) = log sin�+ log sin� + log sin � 3 log sin
�������
3
�;
and de�ne
D =n(�;�; ) 2 R3 j �; �; > 0; �+�+ <�; �� >
�3
63
o:
In view of (1), it is enough to prove the following
Proposition: If (�;�; ) 2 D, then f(�;�; ) > 0.
Proof: Fix " > 0, and let
D" =n(�; �; ) 2 R3 j �;�; > 0; �+�+ <��"; �� > �3
63
o:
It is easily seen that D"0 � D" if " < "0 and D =
S">0
D". Thus, it is enough
to prove that there exists � > 0 such that, if " < �, then the inequality
f(�;�; ) > 0 holds for all (�;�; ) 2 D".
[Editorial note: These "-� complications are caused by the fact that f is not de�ned
if �+�+ = �; that is, on the closure ofD inR3. Onemay avoid these complications
by putting f(�; �; ) = +1 if � + � + = �, and showing that such an f is
\continuous" on the closure of D. Nothing new, however, arises from this approach,
nor does the proof become simpler!]
Since f is continuous on D" (the closure of D" in R3), and since D" is a
compact subset of R3, the minimum value of f is attained on D". But if
309
(�; �; ) 2 D", then one of the numbers �, �, (say ) is less than �=2, so
that
@f
@ (�;�; ) = cot + cot
�������
3
�6= 0 : (2)
This means that f attains its minimum value on @D" | the boundary of D"
in R3. We have
@D" =n(�;�; ) 2 R3 j �; �; > 0; �+�+ = ��"; �� � �3
63
o[n
(�; �; ) 2 R3 j �;�; > 0; �+�+ <��"; �� = �3
63
o=: @D
0" [ @D00
" :
Assume that (�;�; ) 2 @D0". Then �; �; < �, so that ��2 > �� � �3
63,
which implies that � >�216
. Analogously: �; > �216
. Also:
� < � � � � < � � 2�216
= 214�216
:
Analogously: �; < 214�216
. Since there exists a real numberM such that
log sinx > M for x 2 [ �216
;214�216
];
we obtain that
f(�;�; ) > 3M � 3 log sin "3
for (�; �; ) 2 @D0" :
Since log sin "3! �1 as "! 0, there exists � > 0 such that M > log sin "
3
for all 0 < " < �; that is, if 0 < " < �, then
f(�;�; ) > 0 for (�;�; ) 2 @D0" : (3)
In order to complete the proof of the proposition, it is su�cient to show
that, if 0 < " < �, then f(�;�; ) � 0 on @D00" (because then (2) implies
that f(�;�; ) > 0 for (�;�; ) 2 D"). Set k = �3=63. De�ne
E" =n(�; �) 2 R2 j �; � > 0; �+�+ k
��<��"
oand
F (�;�) = log sin�+ log sin� + log sink
��� 3 log sin
������ k
��
3
!:
If (�; �; ) 2 @D00" , then (�;�) 2 E" and F (�;�) = f(�;�; ). Therefore
it is su�cient to prove that
F (�;�) � 0 for (�; �) 2 E":
Similarly, as before, if (�;�) 2 @E", where
@E" =n(�;�) 2 R2 j �; � > 0; �+�+ k
��=��"
o;
310
then (�;�; k��
) 2 @D0", and, by (3),
F (�;�) = f(�;�; k��
) > 0 for (�; �) 2 @E":
Therefore it su�ces to prove that F (�;�) � 0 for all stationary points
(�; �) 2 E". So, we assume that (�;�) 2 E". We have
@F
@�(�;�) = cot�+
�1� k
�2�
�cot
������ k
��
3
!�k cot k
��
�2�
and
@F
@�(�; �) = cot� +
�1� k
��2
�cot
������ k
��
3
!�k cot k
��
��2:
From the equations
@F
@�(�;�) =
@F
@�(�;�) = 0; (4)
we obtain that
� cot�� � cot� + (�� �) cot
������ k
��
3
!= 0 : (5)
We show that the above equality implies that � = �. Suppose that � 6= �
and without loss of generality assume that � > �. Set
G(�;�) =� cot�� � cot�
� � �� cot
������ k
��
3
!:
We have
2
qk�+ � < 2
qk�+ � � �+ k
��+ � < � ; (6)
which gives 0 < � < � � 2
qk�. The function g1(�) = � csc2 � � cot� is
non-decreasing in�0; ��2
qk�
�. This, together with (6), implies that
G(�;�) =1
���
Z �
�
(x csc2 x� cotx) dx� cot
������ k
��
3
!
< (� csc2 � � cot�)� cot
0@��2
qk���
3
1A < �0:75 :
[Ed: The last inequality can be veri�ed by computer algebra using, for exam-
ple, DERIVE.] Hence from (5) we deduce that � = �.
311
Now (4) becomes
H(�) := cot�+
�1� k
�3
�cot
��2�� k
�2
3
!� k
�3cot
k
�2= 0 :
(7)
We know that � � 2�� k�2
> 0. The equation � � 2�� k�2
= 0 has two
positive roots �0, �1 with 0:231 < �0 < 0:232 and 1:540 < �1 < 1:541, so
� 2 (�0; �1). We have
H0(�) = h1(�)� h2(�) + h3(�) + h4(�)� h5(�) ;
where
h1(�) = 3 cot
��2�� k
�2
3
!; h2(�) = � csc2 �; h3(�) = 3 cot� ;
h4(�) =2
3�5(�3�k)2 csc2
��2�� k
�2
3
!; h5(�) =
2k2
�5csc2
�k
�2
�:
We next verify [using again DERIVE] that
h1(�)� h2(�) > 2:97 and h3(�) + h4(�)� h5(�) > 0:995
for � 2 (�0; �1). Therefore, the equation H(�) = 0 has at most one so-
lution. Thus, (7) implies that � = �6. Hence (�; �) = (�
6;�6) is the only
stationary point of F in E" and we have F (�6;�6) = 0.
The proof is now complete.
2206. [1997: 46; 1998: 61, 62] Proposed by Heinz-J �urgen Sei�ert,Berlin, Germany.
Let a and b denote distinct positive real numbers.
(a) Show that if 0 < p < 1, p 6= 1
2, then
1
2
�apb1�p + a
1�pbp�< 4p(1� p)
pab+ (1� 4p(1� p))
a+ b
2:
(b) Use (a) to deduce P �olya's Inequality:
a� b
loga� log b<
1
3
�2pab+
a+ b
2
�:
Note: \log" is, of course, the natural logarithm.
III. Solution to part (a) by the proposer, slightly adapted by the editor.
312
With r =pab and x = log(a=r), we have x 6= 0, a = re
x, and
b = re�x. With q = 2p � 1 (so that jqj < 1), the desired inequality (after
dividing by r) becomes cosh(qx) < (1� q2) + q
2 cosh(x). Note that
(1� q2) + q
2 cosh(x)� cosh(qx) =
1Xk=1
x2k+1
2k+ 1!q2�1� q
2k�> 0
since�1� q
2k�> 0 for k � 1. Thus the desired inequality is proved.
2240. [1997: 243] Proposed by Victor Oxman, University of Haifa,Haifa, Israel.
Let ABC be an arbitrary triangle with the pointsD, E, F on the sides
BC, CA, AB respectively, so thatBD
DC� BF
FA� 1 and
AE
EC� AF
FB.
Prove that [DEF ] � [ABC]
4with equality if and only if two of the
three points D, E, F , (at least) are mid-points of the corresponding sides.
Note: [XY Z] denotes the area of triangle4XY Z.[Editor's note: Most solvers noted that the condition for equality should
actually be:
F plus at least one of D or E be the midpoints of the corresponding sides.]
Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.
There exist numbers u 2 [0; 1); v 2 (0; 1]; w 2 (0; 1) such that
D= (1� u)B+ uC ;
E = vA +(1� v)C ;
F =(1� w)A+ wB :
Then BD = uBC, DC = (1 � u)BC, AE = (1 � v)CA; EC = vCA,
AF = wAB and BF = (1�w)AB, so that the conditions
BD
DC� BF
FAand
AE
EC� AF
FB
give u=(1� u) � (1� w)=w and (1� v)=v � w(1�w) or
u+w � 1 � v + w: (1)
As is well-known, we have
[DEF ]
[ABC]=
������0 1� u u
v 0 1� v
1� w w 0
������ :
313
[Editor's note: A triangle, X = (x1; x2); Y = (y1; y2); Z = (z1; z2); has
area
[XY Z] =1
2
������x1 x2 1
y1 y2 1
z1 z2 1
������ ;see, for example, 13.45, H.S.M. Coxeter, Introduction to Geometry, JohnWiley and Sons, Inc., London (1961). Then
[DEF ] =1
2
������d1 d2 1
e1 e2 1
f1 f2 1
������ =1
2
������0 1� u u
v 0 1� v
1� w w 0
������������a1 a2 1
b1 b2 1
c1 c2 1
������ :]So
[DEF ]
[ABC]= w(1� w)� (1� u� w)(v+ w� 1): (2)
Since w(1� w) � 14; 0 < w < 1, with equality if and only if w = 1
2, from
(1) and (2) we obtain the desired inequality. There is equality if and only if
w = 12and u = 1
2or v = 1
2.
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid,
Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JIMMY CHUI, stu-
dent, Earl Haig Secondary School, North York, Ontario; WALTHER JANOUS, Ursuli-
nengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big
Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece;
GERRY LEVERSHA, St. Paul's School, London, England; GOTTFRIED PERZ, Pesta-
lozzigymnasium, Graz, Austria; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK,
Zaltbommel, the Netherlands; and the proposer. There was one incomplete solution.
2241. [1997: 243] Proposed by Toshio Seimiya, Kawasaki, Japan.
Triangle ABC (AB 6= AC) has incentre I and circumcentre O. The
incircle touches BC at D. Suppose that IO ? AD.
Prove that AD is a symmedian of triangle ABC. (The symmedian is
the re ection of the median in the internal angle bisector.)
Solution by D.J. Smeenk, Zaltbommel, the Netherlands.
Lemma. In a non-equilateral triangleABC with side lengths a; b; c, incentreI and circumcentre O, let P be the point on the half-line starting at B in thedirection of C for which BP = c, and Q be on the half-line from A to Cwith AQ = c; then PQ ? IO.
Proof. Let O0 and O00 be the projections of O on BC and AC respectively,
and let I0 and I00 be the projections of I on those sides. Let S be the point
where OO0 intersects II00. Consider triangles CPQ and SOI. Since CO0 =12a and CI0 = 1
2(a+ b� c); O0I0 = 1
2(b� c); similarly, O00I00 = 1
2(a� c).
314
Furthermore \OSI = \SOO00 = (because the sides of the �rst two angles
are perpendicular to the sides of the third). It follows that
OS =O00I00
sin =
a� c
2 sin ; and IS =
O0I0
sin =
b� c
2 sin :
This means that 4CPQ � 4SOI (by side-angle-side). Since SO ? CP
and SI ? CQ, it follows that OI ? PQ.
For the solution to our problem we must show that CD : BD = b2 : c2
(which, according to standard references, is a property that holds if and only
if AD is a symmedian). Because we take AD ? IO, the lemma implies that
AD k PQ, so that 4ACD � 4QPC. Thus CP : CQ = CD : AC, orc� a
c � b=a+ b� c
2b, so that
a =b2 + c
2
b+ c: (1)
From CD = s� c and BD = s� b we conclude
CD
BD=a+ b� c
a� b+ c: (2)
Plugging (1) into (2) gives the desired conclusion that CD : BD = b2 : c2.
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid,
Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MICHAEL
LAMBROU, University of Crete, Crete, Greece; and the proposer.
It is noteworthy that this result and that of 2246 (which appears later in this
issue) follow immediately from the same lemma, even though they seem to have
little else in common except for the same four solvers. Perhaps a peculiar kind of
sunspot activity would account for mathematicians as far apart as the Netherlands
and Japan conceiving such related problems at the same time. Smeenk believes that
his lemma is known, but he found it easier to prove than to �nd; in fact, he supplied
two proofs.
2242. [1997: 243] Proposed by K.R.S. Sastry, Dodballapur, India.
ABCD is a parallelogram. A point P lies in the plane such that
1. the line through P parallel to DA meets DC at K and AB at L,
2. the line through P parallel to AB meets AD at M and BC at N , and
3. the angle between KM and LN is equal to the non-obtuse angle of
the parallelogram.
Find the locus of P .
315
Editor's comment. Even though the number of solvers was relatively small,no two solutions were alike. This seemed to result, in part, from various interpre-tations of the proposal. For example, if one considers a parallelogram as the convexhull of its vertices (M is on the line segment AD, for example) and that the angle atvertex A is acute, then the locus of P is the empty set. Nonempty loci included var-ious conic arcs according to the aforementioned interpretations. It would not seeminstructive to try to report all of these; rather, we attempt to give a representativesummary.
The common approach was to use a standard analytic argument to derive theequation of a conic which passes through some, or all, of the vertices of the paral-lelogram, again, depending on interpretation. However, the actual derivation of theconic equations required the omission of the vertices so that they are not properly inthe locus of P . Smeenk was the only solver to point this out speci�cally.
Most of the solvers assumed that \BAD < �=2 which causes P to be outsidethe parallelogram. Smeenk, assuming further that \LSM = \BAD, where S =
MK \ LN; derives the locus equation
x(x� a) + y(y � b)� 2 cos\BAD(x � a)(y � b) = 0;
which is that of an ellipse through the vertices B(a; 0), C(a; b), D(0; b).
Con Amore pointed out that for P to be in the convex hull of the vertices\BAD > �=2. Also, assuming that MK \ LN lies in the half-plane from BD
containing A, they derive the equation of the composite quartic
[(a� x)x � (b� y)y] [(a� x)x+ (b� y)y � 2(a � x)(b� y) cos v] = 0 ;
where v = \ABC < �=2: The �rst component gives the locus as the intersection of ahyperbola with the interior of the parallelogram while the second (similar to Smeenk'sequation) gives the locus as an elliptic arc also in the interior of the parallelogram. IfMK\LN lies in the half-plane fromBD not containing A, then the resulting ellipticarc is symmetric to the derived one with respect to the centre of the parallelogram.
The proposer points out that, since the locus is the circumcircle of a rectangleif the angle at vertex A is a right angle, this could be the springboard for a moreelegant solution, presumably, via some a�ne transformation. Kone �cn �y remarks thatthe parallelogram can, in fact, be obtained by right-angled projection of the rectangleonto the plane which has in common the diagonal, say AC of the parallelogram.
Bradley derives the result that the lines KM , LN and AC are concurrent.
Solutions were received from CHRISTOPHER J. BRADLEY, Clifton College,
Bristol, UK; CON AMORE PROBLEM GROUP, Royal Danish School of Educational
Studies, Copenhagen, Denmark; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari,
Valladolid, Spain; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michi-
gan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; D.J. SMEENK,
Zaltbommel, the Netherlands; and the proposer.
316
2243. [1997: 243] Proposed by F.J. Flanigan, San Jose State Univer-sity, San Jose, California, USA.
Given f(x) = (x � r1)(x � r2) : : : (x � rn) and f0(x) = n(x � s1)
(x � s2) : : : (x � sn�1), (n � 2), consider the harmonic mean h of the
n(n� 1) di�erences ri � sj.
If f(x) has a multiple root, then h is unde�ned, because at least one
of the di�erences is zero.
Calculate h when f(x) has no multiple roots.
All solutions submitted were essentially the same.
From f(x) =
nYk=1
(x�rk), we get, by logarithmic di�erentiation, that
f0(x)
f(x)=
nXk=1
1
x� rk
;
giving
f0(sj)
f(sj)=
nXk=1
1
sj � rk
= 0 (1)
for k = 1, 2, : : : , n� 1.
Now, the harmonic mean h is de�ned by
1
h=
1
n(n� 1)
n�1Xj=1
nXk=1
1
rk � sj
:
In view of (1), we may say, according to temperament, that h is in�nite or
unde�ned.
Solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; CON
AMORE PROBLEM GROUP, Royal Danish School of Educational Studies, Copen-
hagen, Denmark; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;
MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA,
St. Paul's School, London, England; JOEL SCHLOSBERG, student, Robert Louis Steven-
son School, New York, NY, USA; DAVID STONE and VREJ ZARIKIAN, Georgia South-
ern University, Statesboro, Georgia, USA; and the proposer.
The proposer said that the problem was inspired by A.G. Clark's solution to
problem 3034 in the American Mathematical Monthly 1930, p. 317. See also the
American Mathematical Monthly 1923, p. 276 and 1930, p. 94.
317
2244. [1997: 243] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle and D is a point on AB produced beyond B such
that BD = AC, and E is a point on AC produced beyond C such that
CE = AB. The perpendicular bisector of BC meets DE at P .
Prove that \BPC = \BAC.
Solution by Stergiou Haralampos, Chalkis, Greece.Let jACj = b and jABj = c. We draw DF jjAC with jDF j = b and
F on the opposite side of DE from A. Then ACFD is a parallelogram; so
jCF j = b+c and\CAD = \CFD. LetH = CF^DE. SinceCHjjAD and
4ADE is isosceles, \CHE = \ADE = \CEH. It follows that jCHj = c,
so jHF j = b. If we draw BH, then ABHC is a parallelogram. Hence
\BHC = \BAC = \CFD (above). It is easy to see that BHFD is a
rhombus, so jPF j = jPBj = jPCj and \PBH = \PFH = \PCH.
But if \PBH = \PCH, it follows that BCHP is a cyclic quadrilateral, so
\BPC = \BHC = \BAC.
[Note that this proof works whether P is between D and E or not.]
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;
CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; CON AMORE PROBLEM
GROUP, Royal Danish School of Educational Studies, Copenhagen, Denmark;
CHARLES DIMINNIE and TREY SMITH, Angelo State University, San Angelo, Texas;
JORDI DOU, Barcelona, Spain; RICHARD I. HESS, Rancho Palos Verdes, Califor-
nia, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL
LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong;
VICTOR OXMAN, University of Haifa, Haifa, Israel; GOTTFRIED PERZ, Pestalozzi-
gymnasium, Graz, Austria; K.R.S. SASTRY and JEEVAN SANDHYA, Bangalore, India;
D.J. SMEENK, Zaltbommel, the Netherlands; KAREN YEATS, student, St. Patrick's
High School, Halifax, Nova Scotia; PAUL YIU, Florida Atlantic University, Boca
Raton, Florida, USA; and the proposer.
2245. [1997: 244] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,Alcorc �on, Madrid, Spain.
Prove that3n + (�1)(n2)
2� 2n is divisible by 5 for n � 2.
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.I will prove that the assertion is true for all non-negative integers n.
[Ed: with the usual convention that�0
2
�=�1
2
�= 0.]
Since 34 � 1 (mod 5), since 24 � 1 (mod 5) and since�n+ 4
2
�=
(n+ 4)(n+ 3)
2� n(n� 1)
2(mod 2) =
�n
2
�(mod 2) ;
we deduce that f(n+4)� f(n) (mod 5), where f(n) =3n + (�1)(n2)
2�2n.
318
Since f(0) = f(1) = f(2) = 0 and f(3) = 5, it follows that f(n) is
divisible by 5 for all non-negative integers n.
Also solved by CHARLES ASHBACHER, Cedar Rapids, Iowa, USA; SAM
BAETHGE, Nordheim, Texas, USA;MANSUR BOASE, student, St. Paul's School, Lon-
don, England; PAUL BRACKEN, CRM, Universit �e de Montr �eal, Montr �eal, Qu �ebec;
CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JAMES T. BRUENING,
Southeast Missouri State University, Cape Girardeau, MO, USA; MIGUEL ANGEL
CABEZ �ON OCHOA, Logro ~no, Spain; THEODORE CHRONIS, student, Aristotle Uni-
versity of Thessaloniki, Greece; GORAN CONAR, student, Gymnasium Vara�zdin,
Vara�zdin, Croatia; CHARLES R. DIMINNIE, Angelo State University, San Angelo,
TX, USA; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA;
RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ur-
sulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University,
Big Rapids, Michigan, USA;MICHAEL LAMBROU, University of Crete, Crete, Greece;
KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's School, London, England;
ALAN LING, student, University of Toronto, Toronto, Ontario; VEDULA N. MURTY,
Visakhapatnam, India; VICTOR OXMAN, University of Haifa, Haifa, Israel; BOB
PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; CHRISTOS
SARAGIOTIS, student, Aristotle University, Thessaloniki, Macedonia, Greece; JOEL
SCHLOSBERG, student, Robert Louis Stevenson School, New York, NY, USA; ROBERT
P. SEALY, Mount Allison University, Sackville, New Brunswick; HEINZ-J�URGEN
SEIFFERT, Berlin, Germany; REZA SHAHIDI, student, University of Waterloo, Wa-
terloo, Ontario; ZUN SHAN and EDWARD T.H. WANG, Wilfrid Laurier University,
Waterloo, Ontario; D.J. SMEENK, Zaltbommel, the Netherlands; DIGBY SMITH,
Mount Royal College, Calgary, Alberta; DAVID R. STONE, Georgia Southern Univer-
sity, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the
proposer.
2246. [1997: 244] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.
Suppose thatG, I and O are the centroid, the incentre and the circum-
centre of a non-equilateral triangle ABC.
The line throughB, perpendicular toOI intersects the bisector of\BAC
at P . The line through P , parallel to AC intersects BC at M .
Show that I, G andM are collinear.
Solution by the proposer.Lemma. In a non-equilateral triangleABC with side lengths a; b; c, incentreI and circumcentre O, let D be the point on the half-line starting at C inthe direction of A for which CD = a, and E be on the half-line from B toA with BE = a; thenDE ? IO.
This lemma was proved as part of the foregoing solution to 2241 (in
the notation of that problem). We assume that all points are well de�ned.
[See the remarks below, after the list of solvers.] From the lemma,
319
AD = a � b;AE = a � c, and ED ? IO so that (because we are given
BP ? IO) BP k ED. Let Q be the point where BP intersects AC. Then
4ADE � AQB. It follows that AB : AQ = AE : AD = (a� c) : (a� b).
Since AP bisects \BAQ;AB : AQ = BP : PQ, while PM k AC implies
BP : PQ = BM : MC. Thus, BM : MC = (a� c) : (a� b).
We introduce trilinear coordinates with respect to 4ABC. The coor-
dinates of I are (1;1; 1), of G are (bc; ca; ab), and ofM are
(0;MC sin ;MB sin�) = (0; c(a� b); b(a� c)):
Then I, G, andM are collinear if and only if
det
24 1 1 1
bc ca ab
0 c(a� b) b(a� c)
35 = 0;
which is easily con�rmed.
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid,
Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MICHAEL
LAMBROU, University of Crete, Crete, Greece; and TOSHIO SEIMIYA, Kawasaki,
Japan.
All solvers used trilinear coordinates. Orthogonality is generally awkward with
these coordinates, and each of the submitted solutions displayed clever insight to
overcome the di�culty that arises. The lemma in our featured solution simpli�es
matters considerably.
Note that onemust assume that4ABC is not equilateral | otherwiseO = I;
furthermore, as Bradley points out, IG must not be parallel to BC | otherwise
M is not de�ned. Smeenk's problem implies that the latter condition is equivalent
to forbidding AI ? IO. Bradley determined that the required assumption should
be that 2a 6= b + c. Putting these remarks together we deduce the unexpected
consequence,
For a triangle ABC with side lengths a; b; c, and with distinct centroid G, incentreI, and circumcentre O, IG k BC if and only if AI ? IO, if and only if 2a = b+ c.
This observation is closely related to problem 1506 in Mathematics Magazine 70 : 4
(October, 1997) 302-303: Prove that \AIO � 90� if and only if 2a � b + c, with
equality holding only simultaneously.
2248. [1997: 245] Proposed by Shawn Godin, St. Joseph ScollardHall, North Bay, Ontario.
Find the value of the sum1Xk=1
d(k)
k2;where d(k) is the number of positive
integer divisors of k.
Solution by David Doster, Choate Rosemary Hall, Wallingford, Con-necticut, USA.
320
The series1Xk=1
1
k2converges absolutely to �2=6. Hence the terms of the
series
1Xk=1
1
k2
!2
may be rearranged in any order without changing the value
of the sum �4=36. Thus
1Xk=1
1
k2
!2
=
1Xi=1
1
i2
!0@ 1Xj=1
1
j2
1A =
1Xk=1
Xij=k
1
i2j2=
1Xk=1
d(k)
k2;
since there are exactly d(k) pairs (i; j) such that ij = k. Thus
1Xk=1
d(k)
k2=�4
36:
Also solved by NIELS BEJLEGAARD, Stavanger, Norway; MANSUR BOASE,
student, St. Paul's School, London, England; THEODORE CHRONIS, student, Aris-
totle University of Thessaloniki, Greece; FLORIAN HERZIG, student, Perchtolds-
dorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER
JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, Univer-
sity of Crete, Crete, Greece; KEE-WAI LAU,Hong Kong; GERRYLEVERSHA, St. Paul's
School, London, England; ALAN LING, University of Toronto, Toronto, Ontario; BOB
PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; JOEL SCHLOSBERG,
student, Robert Louis Stevenson School, New York, NY, USA; HEINZ-J�URGEN
SEIFFERT, Berlin, Germany; DAVID R. STONE, Georgia Southern University, States-
boro, Georgia, USA; and the proposer.
Several solvers gave references to places where this problem or generalizations
were treated. Indeed, DOSTER gives a reference to the best generalization with his
comment:
The series is a Dirichlet series. Hardy and Wright (An Introduction to the The-
ory of Numbers, 5th ed., pp. 248-250) work out the general theory of this type of
sum and prove, more generally, that �(s)�(s � k) =
1Xn=1
�k(n)
nk, where s > 1, and
s > k+ 1, where �k(n) =Xdjn
dk and �(s) =
1Xn=1
1
ns.
Let s = 2 and k = 0 to get the required result.
Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell
Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer
Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil
Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,
J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia
321
Letter from the Editors
This edition is somewhat di�erent from usual. A little explanation is
necessary.
At the end of July 1998, before all the material for issue 5, 1998, was
complete, the Editor-in-Chief, Bruce Shawyer, was admitted to hospital to
await surgery. The Associate Editor, Clayton Halfyard, stepped into the
breach, and, with the assistance of Bill Sands and Memorial University sum-
mer students Paul Marshall and Trevor Rodgers, completed issue 5.
On 25 August 1998, the Editor-in-Chief underwent a quadruple coro-
nary artery bypass operation and is now at home for a two month recuper-
ation period. The Associate Editor has prepared this issue, but since the
Academy Corner is the sole responsibility of the Editor-in-Chief, it is missing
from this issue.
The Editor-in-Chief hopes to be able to respond to email messages in
the near future.
Bruce Shawyer Clayton Halfyard
Editor-in-Chief Associate Editor
Lettre de la r �edaction
Ce num�ero di� �ere quelque peu de l'ordinaire. Quelques explications
s'imposent.
La �n du mois de juillet 1998, avant que tous les textes du num�ero 5
soient prets, le r �edacteur en chef, Bruce Shawyer, a �et �e admis �a l'hopital
en vue d'une op �eration. Le r �edacteur en chef adjoint, Clayton Halfyard, a
donc pris la rel �eve. Avec l'aide de Bill Sands et des �etudiants de l'Universit �e
Memorial, PaulMarshall et Trevor Rodgers, il a r �eussi �a terminer le num�ero 5.
Le 25 aout, 1998, le r �edacteur en chef a subi un quadruple pontage
coronarien. Il est maintenant de retour �a la maison, pour une convalescence
de deux mois. Le r �edacteur en chef adjoint s'est occup �e du pr �esent num�ero,
mais, puisque la chronique Academy Corner rel �eve enti �erement du r �edacteur
en chef, elle ne paraitra pas dans ce num�ero.
Le r �edacteur en chef esp �ere etre en mesure de r �epondre �a son courriel
sous peu.
Bruce Shawyer Clayton Halfyard
R �edacteur en chef R �edacteur en chef adjoint
322
THE OLYMPIAD CORNERNo. 192
R.E. Woodrow
All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.
As a �rst Olympiad for this issue we give the problems of the 4th Math-
ematical Olympiad of the Republic of China (Taiwan) written April 13, 15,
1995. My thanks go to Bill Sands of the University of Calgary, who collected
these problems when he was assisting with the 1995 International Mathe-
matical Olympiad held in Canada.
4th MATHEMATICAL OLYMPIAD OF THE REPUBLIC
OF CHINA (TAIWAN)First Day | Taipei
April 13, 1995
1. Let P (x) = a0 + a1x+ � � � + an�1xn�1 + anx
n be a polynomial
with complex coe�cients. Suppose the roots ofP (x) are �1; �2; : : : ; �n withj�1j > 1, j�2j > 1; : : : ; j�j j > 1, and j�j+1j � 1; : : : ; j�nj � 1. Prove:
jYi=1
j�ij �pja0j2 + ja1j2 + � � �+ janj2 :
2. Given a sequence of integers: x1; x2; x3; x4; x5; x6; x7; x8, one con-structs a second sequence: jx2 � x1j; jx3 � x2j; jx4 � x3j; jx5 � x4j;jx6 � x5j; jx7 � x6j; jx8 � x7j; jx1 � x8j. Such a process is called a single
operation. Find all the 8-term integral sequences having the following prop-
erty: after �nitely many applications of the single operation the sequence
becomes an integral sequence with all terms equal.
3. Suppose n persons meet in a meeting, and every one among them
is familiar with exactly 8 other participants of that meeting. Furthermore
suppose that each pair of two participants who are familiar with each other
have 4 acquaintances in common at that meeting, and each pair of two par-
ticipants who are not familiar with each other have only 2 acquaintances in
common. What are the possible values of n?
323
Second Day | TaipeiApril 15, 1995
4. Given n distinct integers m1;m2; : : : ;mn, prove that there exists
a polynomial f(x) of degree n and with integral coe�cients which satis�es
the following conditions:
(1) f(mi) = �1, for all i, 1 � i � n.
(2) f(x) cannot be factorized into a product of two nonconstant poly-
nomials with integral coe�cients.
5. Let P be a point on the circumscribed circle of 4A1A2A3. Let
H be the orthocentre of 4A1A2A3. Let B1 (B2; B3 respectively) be the
point of intersection of the perpendicular from P to A2A3 (A3A1, A1A2
respectively). It is known that the three points B1, B2, B3 are collinear.
Prove that the lineB1B2B3 passes through the midpoint of the line segment
PH.
6. Let a, b, c, d be integers such that ad � bc = k > 0, (a; b) = 1,and (c; d) = 1. Prove that there are exactly k ordered pairs of real numbers
(x1; x2) satisfying0 � x1, x2 < 1 and for which both ax1+bx2 and cx1+dx2are integers.
As a second problem set this issue for your puzzling pleasure, we give
the XI Italian Mathematical Olympiad written May 5, 1995 at Cesenatico.
Thanks go to Bill Sands of the University of Calgary who collected them while
at the 1995 IMO in Canada.
XI ITALIANMATHEMATICAL OLYMPIADCesenatico, May 5, 1995
Time: 4.5 hours
1. Determine for which values of the integer n it is possible to cover
up, without overlapping, a square of side n with tiles of the type shown in
the picture
where each small square of the tile has side 1.
324
2. In a class of 20 students no two of them have the same ordered
pair (written and oral examinations) of scores in mathematics. We say that
student A is better than B if his two scores are greater than or equal to the
corresponding scores of B. The scores are integers between 1 and 10.
(a) Show that there exist three students A, B and C such that A is
better than B and B is better than C.
(b) Would the same be true for a class of 19 students?
3. In a town there are 4 pubs, A, B, C and D, connected as shown in
the picture.
B
D
C
A
A drunkard wanders about the pubs starting withA and, after having a drink,
goes to any of the pubs directly connected, with equal probability.
(a) What is the probability that the drunkard is at pub C at his �fth
drink?
(b) Where is the drunkard more likely to be after n drinks? (n > 5)
4. An acute-angled triangle ABC is inscribed in a circle with centre
O. Let D be the intersection of the bisector ofA with BC, and suppose that
the perpendicular to AO through D meets the line AC in a point P interior
to the segments AC. Show that AB = AP .
5. Two non-coplanar circles in Euclidean space are tangent at a point
and have the same tangents at this point. Show that both circles lie in some
spherical surface.
6. Find all pairs of positive integers x, y such that
x2 + 615 = 2y:
As a third set of problems for your attention we give the Third and
Fourth Grade and IMO Team selection rounds of the Yugoslav Federal Com-
petition for 1995. Thanks again go to Bill Sands, the University of Calgary,
for collecting them for me.
325
YUGOSLAV FEDERAL COMPETITION 1995Third and Fourth Grade
1. Let p be a prime number. Prove that the number
11 � � � 122 � � � 2 � � � 99 � � � 9� 123456789
is divisible by p, where dots indicate that the corresponding digit appears ptimes consecutively.
2. A polynomial P (x) with integer coe�cients is said to be divisible
by a positive integer m if and only if the number P (k) is divisible by m for
all k 2 Z. If the polynomial
P (x) = a0xn + a1x
n�1 + � � �+ an�1x+ an
is divisible bym, prove that ann! is divisible bym.
3. A chord AB and a diameter CD ? AB of a circle k intersect at
a point M . Let P lie on the arc ACB and let P 62 fA;B; Cg. Line PMintersects the circle k at P and Q 6= P , and line PD intersects chord ABat R. Prove that RD > MQ.
4. A tetrahedron ABCD is given. Let P and Q be midpoints of edges
AB and CD, and let O and S be the incentre and the circumcentre of the
tetrahedron, respectively. If points P , Q and S belong to the same line,
prove that the point O also belongs to that line.
Selection of the IMO Team
1. Find all the triples (x; y; z) of positive rational numbers such that
x � y � z and
x+ y + z;1
x+
1
y+
1
z; xyz 2 Z:
2. Let n be a positive integer having exactly 1995 1's in its binary
representation. Prove that 2n�1995 divides n!.
3. Let SABCD be a pyramid such that all of its edges are of the
same length. Let pointsM 2 BC andN 2 AS be such that the lineMN is
perpendicular to lineAD as well as to the lineBC. Find the ratiosBM=MCand SN=NA. [Editor's note: ABCD is the base of the pyramid.]
We now turn to readers' solutions to problems of the Swedish Mathe-
matics Contest, 1993 [1997: 196].
326
SWEDISH MATHEMATICS CONTEST 1993Final
November 20
1. The integer x is such that the sum of the digits of 3x is the same as
the sum of the digits of x. Prove that 9 is a factor of x.
Solutionsby Jamie Batuwantudawe, student, SirWinstonChurchillHighSchool, Calgary; by Michael Selby, University of Windsor, Windsor, Ontario;and by Enrico Valeriano Cuba, National University of Engineering, Lima,Peru. We give the solution of Valeriano.
Let S(n) be the sum of the digits of n. Working modulo 9
3x � S(3x) mod 9 :
Also
x � S(x) mod 9 :
Since S(3x) = S(x),2x � 0 mod 9 ;
and since (2; 9) = 1, we have x � 0 mod 9.
2. A railway line is divided into 10 sections by the stations A, B, C,
D, E, F , G, H, I, J and K. The distance between A and K is 56 km. A
trip along two successive sections never exceeds 12 km. A trip along three
successive sections is at least 17 km. What is the distance between B andG?z }| {A B C D E F G H I J K
Solutionsby Jamie Batuwantudawe, student, SirWinstonChurchillHighSchool, Calgary; by Michael Selby, University of Windsor, Windsor, Ontario;and by Enrico Valeriano Cuba, National University of Engineering, Lima,Peru. We give Batuwantudawe's solution.
A B C D E F G H I J K
Now AK = 56 and AK = AD + DG + GJ + JK. We know that
AD;DG;GJ � 17. Thus JK � 5 to satisfyAK = 56. We knowHK � 17,and since JK � 5,HJ � 12. But, we also knowHJ � 12. ThusHJ = 12.Since HK � 17 and HJ = 12, JK � 5. The only possibility is that
JK = 5.
Symmetrically we �nd that AB = 5 and BD = 12.
Now,
DH = AK � AB �BD �HJ � JK
= 56� 5� 12� 5� 12 = 22 :
327
Now GJ � 17 but HJ = 12. Hence GH � 5. Since DG � 17 and
DH = DG+GH = 22, we obtain DG = 17 and GH = 5.
Now
BG = BD +DG
= 12 + 17 = 29 :
3. Assume that a and b are integers. Prove that the equation a2+b2+x2 = y2 has an integer solution x, y if and only if the product ab is even.
Solutions by Bob Prielipp, University of Wisconsin{Oshkosh, Wiscon-sin, USA; Michael Selby, University of Windsor, Windsor, Ontario; by SreeSanyal, student, Western Canada High School, Calgary, Alberta; by EnriqueValeriano Cuba, National University of Engineering, Lima, Peru; and byMichael Lebedinsky, student, Henry Wise Wood High School, Calgary, Al-berta. We give Selby's solution.
First, we prove that this condition is necessary. Suppose ab is odd.
Then a, b are odd and a2 � b2 � 1 mod 4. Now x2 � 0 or 1 mod 4, andy2 � 0 or 1 mod 4. Therefore a2 + b2 + x2 = y2 is not possible, since if
we consider this modulo 4, 2 + x2 � y2 mod 4, which is impossible since
2 + x2 � 2 or 3 mod 4. Therefore ab must be even.
If ab is even, then, without loss of generality, a = 2k.
Consider 4k2 + b2 + x2 = y2.
If 4k2+b2 = 2t+1, t an integer, then set y�x = 1 and y+x = 2t+1,2y = (t+ 1)2, y = t+ 1 and x = t.
Then 2t+ 1+ t2 = (t+ 1)2. We are done.
If 4k2 + b2 is even, then b = 2s and 4k2 + b2 = 4(k2 + s2) = 4m.
Again, y2� x2 = 4m.
Set y�x = 2 and y+x = 2m. Then y = m+1 and x = y�2 = m�1.
Now 4m + (m � 1)2 = (m + 1)2, and again we are done. Hence
a2 + b2 + x2 = y2 always has a solution when ab is even.
4. To each pair of real numbers a and b, where a 6= 0 and b 6= 0, thereis a real number a � b such that
a � (b � c) = (a � b) � c ,a � a = 1 :
Solve the equation x � 36 = 216.
Solutions by Sree Sanyal and Aliya Walji, students, Western CanadaHigh School, Calgary, Alberta; and by Enrique Valeriano Cuba, National Uni-versity of Engineering, Lima, Peru.
328
Now a � (a � a) = (a � a) � a, so
a � 1 = 1 � a = a :
Also a � (b � b) = (a � b) � b and a = a � 1 = (a � b) � b so
a � b = a
b:
Finallyx
36= 216 =) x = 7776 :
5. A triangle with perimeter 2p has sides a, b and c. If possible, a newtriangle with the sides p� a, p� b and p� c is formed. The process is then
repeated with the new triangle. For which original triangles can the process
be repeated inde�nitely?
Solutions by Michael Selby, University of Windsor, Windsor, Ontario;by Enrique Valeriano, National University of Engineering, Lima, Peru; andby Sonny Chan, student, Western Canada High School, Calgary, Alberta. Wegive Valeriano's solution.
Let a � b � c and � be the di�erence between the longest and the
shortest side.
Original Triangle New Triangle
Perimeter = 2p Perimeter(1) = 3p� (a+ b+ c) = p
� = c� a �(1) = (p� a)� (p� c) = c� a
We can see that the perimeter of the new triangle is half the previous
perimeter, but � is the same. Then, if� > 0, repeating this process we can
obtain
Perimeter(k) =2p
2k< �(k) = c� a :
If c(k) is the longest side we obtain the absurd relation
c(k) < Perimeter(k) < �(k) < c(k). Finally, only with an equilateral tri-
angle as the original triangle (� = 0) can we repeat the process inde�nitely.
6. Let a and b be real numbers and let f(x) = (ax+b)�1. For which aand b are there three distinct real numbers x1, x2, x3 such that f(x1) = x2,f(x2) = x3 and f(x3) = x1?
Solutionsby FilipCrnogorac and SonnyChan, students,Western CanadaHigh School, Calgary, Alberta; and by Michael Selby, University of Windsor,Windsor, Ontario. We give Selby's write-up.
Consider the functions of the form
g(x) =�x+ �
x+ �:
329
Lemma. g(x) has at least 3 distinct �xed points if and only if
= � = 0, � = � 6= 0.
Proof. If = � = 0, � = � 6= 0, g(x) = x and it clearly has at least
3 distinct points x1; x2; x3 such that g(xi) = xi, i = 1; 2; 3. Conversely
consider the equation for a �xed point x, g(x) = x. This implies x2+�x =�x + � or x2 + (� � �)x� � = 0. Suppose this has three distinct roots.
Then the quadratic must be identically 0, or = � = 0 and � = �.
Now, if f(x) = 1
ax+b, then
f�f(x) = ax+ b
abx+ b2 + aand f�f�f(x) = abx+ a+ b2
a(a+ b2)x+ ab+ b(b2 + a):
The problem impliesf�f�f has three distinct real �xed points x1; x2; x3.By the above lemma, this is true if and only if
a+ b2 = a(a+ b2) = 0 and ab+ b(a+ b2) = ab 6= 0:
This is true if and only if a = �b2 and ab 6= 0.
To complete this number of the Olympiad Corner we turn to readers'
solutions to problems of the Dutch Mathematical Olympiad, second round,
September 1993 [1997: 197].
DUTCH MATHEMATICAL OLYMPIADSecond RoundSeptember, 1993
1. Suppose that V = f1; 2; 3; : : : ; 24; 25g. Prove that any subset of Vwith 17 or more elements contains at least two distinct numbers the product
of which is the square of an integer.
Solutionby Sonny Chan and Filip Crnogorac, students,Western CanadaHigh School, Calgary.
The set of numbers A = f1; 2; : : : ; 24; 25g contains a total of �ve per-
fect squares f1; 4; 9; 16; 25g. The product of any two of these will also be a
perfect square. There is one triplet, the product of any two of its elements
will result in a perfect square: f2; 8; 18g. The only other pairs of numbers
from A whose product is a perfect square are f3; 12g, f5; 20g, f6; 24g. Theother eleven elements of the setA are f7; 10; 11; 13; 14;15; 17; 19; 21; 22; 23gand they cannot form a perfect square when multiplied with any other ele-
ment of set A. Group the elements of set A as follows:
f1; 4; 9; 16; 25g; f2; 8; 18g; f3; 12g; f5; 20g; f6; 24g; f7g;
f10g; f11g; f13g; f14g; f15g; f17g; f19g; f21g; f22g; f23g:
330
If more than one number is chosen from a given group, a perfect square will
result. There is a total of 16 groups, so 16 numbers can be chosen without
creating a perfect square product. However, if any 17 numbers are chosen,
then two must be contained within the same group, and therefore will form
a perfect square product.
2. Given is a triangle ABC, \A = 90�. D is the midpoint of BC, Fis the midpoint of AB, E the midpoint of AF and G the midpoint of FB.
AD intersects CE, CF and CG respectively in P , Q and R. Determine the
ratio PQ
QR.
A E F G B
C
P QR
D
q
q
q
q
qqq
q q q
Solution by Filip Crnogorac, student, Western Canada High School,Calgary.
A E F G B
C
PQ
R
DJ
K
r
r
r
r
rr
r
r r r
r r
We know that two medians in a triangle divide each other in 2 : 1 ratio, or
in other words the point of intersection is 2
3the way from the vertex.
Since CF and AD are both medians in 4ABC, then AQ
QD= 2
1, where
Q is the point of intersection.
Also, since D is the midpoint of the hypotenuse in the right triangle
ABC, then it is the centre of the circumscribed circle with radius
DA = DC = DB.
Drop a perpendicular from D onto sides AB and CA. The feet of the
perpendiculars will be F and J , respectively, where J is the midpoint ofAC,
since DF and DJ are altitudes in isosceles triangles 4ADB and 4ADC,
respectively. Now consider4CFB. The segmentsCG and FD are medians
and therefore intersect at H say in the ratio 2 : 1 so, HD
FD= 1
3. From here
it can be seen that 4ARC and 4DRH are similar, since their angles are
331
the same. Also, since we know that FD = JA, and 2JA = AC then
HD = 1
6CA and 4ARC is 6 times bigger than 4DRH. Now we can see
that AR
RD= 6
1and since AR+RD = AD, then RD
AD= 1
7.
Similarly4APE � 4KPD, where medians DJ and CE meet at K.
We know that AE = 1
4AB, so then JK = 1
4JD, since JD is parallel to
AB. It now follows that AE
KD= 2
3, and from the similarity of the triangles
AP
PD= 2
3. Also, since AP + PD = AD, then AP
AD= 2
5. Combining these
results we have AP = 2
5AD, AQ = 2
3AD, QD = 1
3AD and RD = 1
7AD.
Thus
PQ = AQ� AP =2
3AD � 2
5AD =
4
15AD
and
QR = QD � RD =1
3AD � 1
7AD =
4
21AD:
From these PQ
QR= 7
5.
3. A series of numbers is de�ned as follows: u1 = a, u2 = b,un+1 = 1
2(un+un�1) for n � 2. Prove that limn!1 un exists. Express the
value of the limit in terms of a and b.
Solution by the Editors.
For the recurrence un+1 = 1
2(un + un�1) we obtain the associated
equation 2�2 � � � 1 = 0, which has roots � = �1
2; 1. Thus we seek a
solution of the form un = X��1
2
�n+ Y . From u1 = a and u2 = b we get
�1
2X + Y = a
1
4X + Y = b
so that X + 4
3(b� a) and Y = a+2b
3. It is now easy to check by induction
that un = 4
3(b� a)
��1
2
�n+ 1+2b
3and as n!1, un ! 1
3a+ 2
3b.
4. In a plane V a circle C is given with centre M . P is a point not on
the circle C.q
q
P
M
A
B
V
332
(a) Prove that for a �xed point P , AP2
+ BP2
is a constant for every
diameter AB of the circle C.
(b) Let AB be any diameter of C and P a point on a �xed sphere S
not intersecting V . Determine the point(s) P on S such that AP2
+BP2
is
minimal.
Solutionby Jamie Batuwantudawe, student, Sir WinstonChurchillHighSchool, Calgary.
(a) With 4PAB, we can join P and M to create two new triangles,
4PMA and 4PMB. Let \PMA = �. Then \PMB = 180� � �. Be-
causeM is the centre of circle C and A and B both lie on circle C, we have
MA =MB = r, the radius of the circle.
By the Law of Cosines,
BP2
= MP2
+ r2 � 2MPr cos(180� � �)
= MP2
+ r2 + 2MPr cos �
and
AP2
= MP2
+ r2 � 2MPr cos �
so AP2
+ BP2
= 2MP2
+ 2r2.
The right hand side is a constant depending only on the radius of the
circle and the distance of P from the centre.
(b) From (a), we know that AP2
+BP2
= 2MP2
+2r2. For any pointP on sphere S, the radius of the circle will remain constant. Therefore the
only variable a�ecting the sum AP2
+ BP2
is MP , the distance from the
point P to the centre of the circle. AP2
+BP2
will be a minimumwhenMPis minimum. Therefore we are looking for the point on the sphere closest
to M .
Let T be the centre of the sphere S, D be the point on the segment
MT that lies on the sphere, and D0 be any other point on S.
We know thatMD+DT < MD0+D0T because the shortest distance
between M and T is a straight line. We know that DT = D0T . Thus
MD <MD0. ThusD is the point on the sphere which minimizes the sum.
That completes the Corner for this issue. Send me your nice solutions
and generalizations as well as Olympiad contests.
333
BOOK REVIEWS
Edited by ANDY LIU
Juegos y acertijos para la ense~nanza de las Matem�aticas
(Games and riddles for the teaching of mathematics)
by Bernardo Recam�an Santos,
published by Grupo editorial Norma educativa, 1997, Bogot �a, Colombia.
ISBN#958-04-3731-9; softcover, 134 pages.
Reviewed by Francisco Bellot Rosado, Valladolid, Spain.
This book contains 75 games and riddles, uniformly distributed in �ve
chapters:
I mathematical games;
II arithmetical riddles;
III geometrical riddles;
IV logical riddles;
V algebraic riddles.
A last chapter with solutions (not very detailed), comments and alternatives
or variants, and a little Bibliography completes the work.
The di�culty level is indicated by one, two or three stars (?, ??, ? ? ?),meaning that the problem is for students of primary education, of the �rst
years of the secondary, either preuniversity level, respectively; although, as
the author said in the preface, this classi�cation is arbitrary.
The origins of the problems are very variable; there are some classical
examples (the game of Nim, the snake, a game by Paul Erd }os, the Kaprekar
algorithm, the Egyptian fractions, the age of the 3 daughters); others can be
found in popular books (Martin Gardner, Le petit Archim �ede) and some are
originals from the author. As he says, \los acertijos matem�aticos, como los
chistes callejeros, no tienen due ~no" (mathematical riddles, like street jokes,are not copyrightable).
The little book can be used with pro�t in mathematical clubs, extracur-
ricular activities and also in the classroom. I have used it myself inmy classes
of \Taller de Matem�aticas", a class of 2 hours each week for students of 12
years old in the new educational system of my country (Compulsory Sec-
ondary Education).
By the way, the book reveals the identity of one collaborator of CRUX:
Ignotus.
334
THE SKOLIAD CORNERNo. 32
R.E. Woodrow
As a contest this issue we give the Junior High School Mathematics
Contest, Preliminary Round 1998 of the British Columbia Colleges which was
written March 11, 1998. My thanks go to the contest organizer, Jim Totten,
the University College of the Cariboo, for forwarding the 1998 contest mate-
rials to me. Students are given 45 minutes to respond to the 15 questions.
BRITISH COLUMBIA COLLEGESJunior High School Mathematics Contest
Preliminary Round 1998Time: 45 minutes
1. A number is prime if it is greater than one and divisible only by one
and itself. The sum of the prime divisors of 1998 is:
(a) 5 (b) 14 (c) 42 (d) 66 (e) 122
2. Successive discounts of 10% and 20% are equivalent to a single
discount of:
(a) 15% (b) 25% (c) 28% (d) 30% (e) 32%
3. Suppose that kA = A2 and A 2 B = A� 2B. Then the value of
k7 2 k3 is:
(a) 1 (b) 16 (c) 961 (d) 43 (e) 31
4. The expression that is not equal to the value of the four other ex-
pressions listed is:
(a) 1p9 + 9� 8 (b) (1 + 9)� (�
p9 + 8) (c) �1� 9 +
p9 + 8
(d) (1�p9)� (9� 8) (e) 19� 9� 8
5. The sum of all of the digits of the number 1075 � 75 is:
(a) 8 (b) 655 (c) 664 (d) 673 (e) 675
335
6. A circle is divided into three equal parts and one part is shaded as in
the accompanying diagram. The ratio of the perimeter of the shaded region,
including the two radii, to the circumference of the circle is:
(a) 1 (b) 2
3(c) 1
3(d) 3+�
3�(e) �
3
7. The value of1
2� 1
2� 1
2�1
2
is:
(a) 3
4(b) 4
5(c) 5
6(d) 6
7(e) 6
5
8. If each small square in the accompanying grid is one square cen-
timetre, then the area in square centimetres of the polygon ABCDE is:
A
B
C D
E
(a) 38 (b) 39 (c) 42 (d) 44 (e) 46
9. A point P is inside a square ABCD whose side length is 16. P is
equidistant from two adjacent vertices, A and B, and the side CD opposite
these vertices. The distance PA equals:
(a) 8:5 (b) 6p3 (c) 12 (d) 8 (e) 10
10. A group of 20 students has an average mass of 86 kg per person.
It is known that 9 people from this group have an average mass of 75 kg
336
per person. The average mass in kilograms per person of the remaining 11people is:
(a) 94 (b) 95 (c) 96 (d) 97 (e) none of these
11. In the following display each letter represents a digit:
3 B C D E 8 G H I
The sum of any three successive digits is 18. The value of H is:
(a) 3 (b) 4 (c) 5 (d) 7 (e) 8
12. In the accompanying diagram \ADE = 140�. The sides are
congruent as indicated. The measure of \EAD is:
A E
D
(a) 30� (b) 25� (c) 20� (d) 15� (e) 10�
13. The area (in square units) of the triangle bounded by the x-axisand the lines with equations y = 2x+ 4 and y = �2
3x+ 4 is:
(a) 8 (b) 12 (c) 15 (d) 16 (e) 32
14. Two diagonals of a regular octagon are shown in the accompanying
diagram. The total number of diagonals possible in a regular octagon is:
(a) 8 (b) 12 (c) 16 (d) 20 (e) 28
337
15. A local baseball league is running a contest to raise money to send
a team to the provincial championship. To win the contest it is necessary
to determine the number of baseballs stacked in the form of a rectangular
pyramid. The �fth and sixth levels from the base of the stack of baseballs are
shown. If the stack contains a total of seven levels, the number of baseballs
in the stack is:
(a) 100 (b) 112 (c) 166 (d) 168 (e) 240
In the May number of the Corner we gave the problems of the 15th
W.J. Blundon Contest written by students in Newfoundland and Labrador.
Next we give the \o�cial" solutions. My thanks go to Bruce Shawyer for
forwarding the contest and solutions to me.
15th W.J. BLUNDON CONTESTFebruary 18, 1998
1. (a) Find the exact value of
1
log2 36+
1
log3 36:
Solution. 1
log236
+ 1
log336
= log36 2 + log36 3 = log36 6 = 1
2.
(b) If log15 5 = a, �nd log15 9 in terms of a.
Solution. 1 = log15 15 = log15(5 �3) = log15 5+log15 3 = a+log15 3.
log15 3 = 1� a =) log15 9 = log15 32 = 2 log15 3 = 2(1� a).
2. (a) If the radius of a right circular cylinder is increased by 50% and
the height is decreased by 20%, what is the change in the volume?
Solution. V2 = �(1:5r)2(:8h) = 1:8(�r2h) = 1:8V1. So the volume is
increased by 80%.
(b) How many digits are there in the number 21998 � 51988 ?Solution. 21998 � 51988 = 210 � 21988 � 51988 = 1024 � 101988, which has
1988 + 4 = 1992 digits.
3. Solve: 32+x + 32�x = 82.
338
Solution.
32+x + 32�x = 82
9 � 3x + 9
3x= 82
9(3x)2 � 82(3x) + 9 = 0
(9 � 3x � 1)(3x � 9) = 0
3x =1
9; 3x = 9
x = �2 x = 2
4. Find all ordered pairs of integers such that x6 = y2 + 53.
Solution.
x6 = y2 + 53
x6 � y2 = 53
(x3 � y)(x3 + y) = 53
x3 � y = 53 x3 � y = 1 x3 � y = �53 x3 � y = �1x3 + y = 1 x3 + y = 53 x3 + y = �1 x3 + y = �53
x = 3 x = 3 x = �3 x = �3y = �26 y = 26 y = 26 y = �26
The pairs are (3;�26); (3; 26); (�3; 26); (�3;�26).5. When one-�fth of the adults left a neighbourhood picnic, the ratio
of adults to children was 2 : 3. Later, when 44 children left, the ratio of
children to adults was 2 : 5. How many people remained at the picnic?
Solution. Let A be the number of adults and C be the number of chil-
dren initially at the picnic. After one-�fth of the adults left, four-�fths re-
main. So4
5A
C=
2
3=) 6A = 5C:
After 44 children left
C � 444
5A
=2
5=) 8A = 25C � 1100:
Solving the two equations gives A = 50, C = 60. The number remaining is
then4
5(50) + (60� 44) = 40 + 16 = 56:
339
6. Find the area of a rhombus for which one side has length 10 and the
diagonals di�er by 4.
Solution.
10
10
10
10
b
b
b+ 2
b+ 2
(b+ 2)2 + b2 = 100
2b2 + 4b� 96 = 0
b2 + 2b� 48 = 0
(b� 6)(b+ 8) = 0
b = 6; b 6= �8
Since the area of a rhombus is one half the product of the diagonals we get
A =1
2(2b)(2b+ 4) =
1
2(12)(16) = 96:
7. In how many ways can 10 dollars be changed into dimes and quar-
ters, with at least one of each coin being used?
Solution. Let q be the number of quarters and d be the number of
dimes. Then
25q + 10d = 1000
d = 100� 5
2q:
Since d must be an integer, q must be even. Also d must be positive. So
100� 2
5q > 0
q < 40:
So q must be an even positive integer less than 40, of which there are 19.
340
8. Solve:px+ 10 + 4
px+ 10 = 12.
Solution. Let y = 4px+ 10. Then y2 =
px+ 10, and the equation
becomesy2 + y = 12 Then: 4
px+ 10 = 3
y2 + y� 12 = 0 x+ 10 = 81(y+ 4)(y� 3) = 0 x = 71y 6= �4; y = 3
9. Find the remainder when the polynomial x135+x125�x115+x5+1is divided by the polynomial x3 � x.
Solution.
x135 + x125 � x115 + x5 + 1 = (x3 � x)Q(x) + ax2 + bx+ c
= x(x� 1)(x+ 1)Q(x) + ax2 + bx+ c
This must be valid for all values of x. Substituting in x = 0, x = 1, andx = �1 gives:
x = 0 : 1 = 0 + c =) c = 1x = 1 : 3 = 0 + a+ b+ c =) a+ b = 2x = �1 : �1 = 0 + a� b+ c =) a� b = �2
Solving the system
a+ b = 2a� b = �2
gives a = 0, b = 2. So the remainder is 2x+ 1.
10. Quadrilateral ABCD below has the following properties: (1) The
midpoint O of side AB is the centre of a semicircle; (2) sides AD, DC and
CB are tangent to this semicircle. Prove that AB2 = 4AD � BC.
Solution. First join the obvious lines in the given �gure:
r
r r
r
r r
r
r
A O B
D
C
E
F
G
By the properties of tangents, DE = DF and CF = CG. Therefore
\EDO = \FDO = � and \FCO = \GCO = . Since OA = OB, we
have \EAO = \GBO = �.
341
Summing the angles of quadrilateral ABCD, we get �+2�+2 +� =360�. Hence � + �+ = 180�; that is, they are the angles of a triangle.
Considering triangles AOD, DOC and COB, we get \AOD = ,\DOC = � and \COB = �. Thus the three triangles are similar.
Considering the triangles ADO and BOC, we have AD
AO= OB
BC, or
AD � BC = AO � OB.
Since AO = OB = 1
2AB, we get the result.
Last issuewe gave the problems of theU.K. Intermediate Mathematical
Challenge. Here are the solutions.
1. C 2. B 3. A 4. A 5. E
6. C 7. C 8. A 9. C 10. B
11. C 12. E 13. D 14. D 15. D
16. D 17. D 18. C 19. A 20. E
21. E 22. D 23. A 24. A 25. B
That completes the Skoliad Corner for this number. Sendme your com-
ments, suggestions, and most importantly, suitable contest materials for use
in future issues.
Advance Announcement
The 1999 Summer Meeting of the Canadian Mathematical Society will
take place at Memorial University in St. John's, Newfoundland, from Satur-
day, 29 May 1999 to Tuesday, 1 June 1999.
The Special Session on Mathematics Education will feature the topic
What Mathematics Competitions do for Mathematics.The invited speakers are
Ed Barbeau (University of Toronto),
Ron Dunkley (University of Waterloo),
Tony Gardiner (University of Birmingham, UK), and
Rita Janes (Newfoundland and Labrador Senior Mathematics League).
Requests for further information, or to speak in this session, as well as sug-
gestions for further speakers, should be sent to the session organizers:
Bruce Shawyer and Ed Williams
CMS Summer 1999 Meeting, Education Session
Department of Mathematics and Statistics, Memorial University
St. John's, Newfoundland, Canada A1C 5S7
342
MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by
High School and University Students. It continues, with the same emphasis,
as an integral part of Crux Mathematicorum with Mathematical Mayhem.
All material intended for inclusion in this section should be sent to the
Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,
PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic
address is still
The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).
The rest of the sta� consists of Adrian Chan (Upper Canada College), Jimmy
Chui (Earl Haig Secondary School), Richard Hoshino (University ofWaterloo),
David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).
Mayhem Problems
The Mayhem Problems editors are:
Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.
Note that all correspondence should be sent to the appropriate editor |
see the relevant section. In this issue, you will �nd only solutions | the
next issue will feature only problems.
We warmly welcome proposals for problems and solutions.
High School Solutions
Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,
Canada. L3S 3E8 <[email protected]>
First, we present a very elegant solution to a past problem:
H224. Consider square ABCD with side length 1. Select a point Mexterior to the square so that \AMB is 90�. Let a = AM and b = BM .
Now, determine the point N exterior to the square so that CN = a and
DN = b. Find, as a function of a and b, the length of the line segmentMN .
343
Solution by Hoe Teck Wee, Singapore.Locate point P exterior to the square so that BP = a and CP = b.
Locate point Q exterior to the square so that DQ = a and AQ = b. Then
MPNQ is a square of side length a+b, and it follows thatMN is a diagonal
of this square. Thus, the desired length ofMN isp2(a+ b).
H225. In Cruxmayhemland, stamps can be bought only in two de-
nominations, p and q cents, both of which are at least 31 cents. It is known
that if p and q are relatively prime, the largest value that cannot be created
by these two stamps is pq � p � q. For example, when p = 5 and q = 3,one can a�x any postage that is higher than 15� 5� 3, or 7 cents, but not
7 cents itself. The governor of Cruxmayhemland tells you that 1997 is the
largest value that cannot be created by these stamps. Find all possible pairs
of positive integers (p; q) with p > q.
SolutionbyMiguel Carri �on �Alvarez, Universidad Complutense deMad-rid, Spain.
If p and q are not relatively prime, then they have a common divisor d.Then any value that does not divide d cannot be created by these two stamps.
Hence there are in�nitely many values that cannot be created. Hence, we
must have p and q being relatively prime. We are seeking all possible pairs
(p; q) with p > q and p and q relatively prime such that 1997 = pq� p� q.Adding 1 to both sides of this equation, we have 1998 = pq � p� q + 1 =(p� 1)(q� 1). Since 1998 = 21 � 33 � 371 has 2 � 4 � 2 = 16 positive divisors,
there are exactly eight ways to write down 1998 as the product of two positiveintegers, namely 1998 = 1 � 1998 = 2 � 999 = 3 � 666 = 6 � 333 = 9 � 222 =18 � 111 = 27 � 74 = 37 � 54.
Now p and q are at least 31, so we must have p � 1 and q � 1 both
being at least 30. The only pair for which this is true is p � 1 = 54 and
q�1 = 37. And we see that 55 and 38 are relatively prime integers. Hence,
the only solution is (p; q) = (55;38).
H226. In right-angled triangle ABC, with BC as hypotenuse,
AB = x and AC = y, where x and y are positive integers. SquaresAPQB,
BRSC, and CTUA are drawn externally on sides AB, BC, and CA, re-spectively. When QR, ST , and UP are joined, a hexagon is formed. Let Kbe the area of the hexagon PQRSTU .
(a) Prove that K cannot equal 1997. (Hint: Try to �nd a general formula
for K.)
(b) Prove that there is only one solution (x; y) with x > y so that K =1998.
SolutionbyMiguel Carri �on �Alvarez, Universidad Complutense deMad-rid, Spain.
(a) The hexagon PQRSTU is the union of the right-angled triangle
ABC, the three squares, and trianglesQRB, STC, and UPA. We will now
344
show that all four triangles have equal area. Consider triangle QRB. We
have
\ABC + \QBA+ \RBQ+ \CBR = 360�
=) \RBQ = 180� � \ABC:
Let [P ] denote the area of triangle P .
Now, [QRB] = 1
2�QB �RB � sin\RBQ = 1
2�AB �CB � sin\ABC =
[ABC]. The argument does not rely on ABC being right-angled, so simi-
larly, we have [STC] = [UPA] = [ABC] = xy=2. It follows that K =4 � xy
2+ x2+ y2+BC2, and BC2 = x2 + y2 impliesK = 2(x2+xy+ y2).
Therefore, K is necessarily an even integer, so it cannot be 1997.
(b) Using our formula from the previous part, we seek an integer so-
lution to 999 = x2 + xy + y2. Taking congruences modulo 3, we have
x2 + xy + y2 � 0 (mod 3).
Suppose x � 0 (mod 3). Then y2 � 0 (mod 3), which implies that
y � 0 (mod 3). Suppose x � 1 (mod 3). Then y2 + y + 1 � 0 (mod3), which implies that y � 1 (mod 3). Suppose x � 2 (mod 3). Then
y2 + 2y + 4 � y2 + 2y + 1 � (y + 1)2 � 0 (mod 3), which implies that
y � 2 (mod 3).
Hence, it follows that we must have x � y (mod 3).
Then we make the substitution x = 3u+ a, y = 3v + a where a is 0,1 or �1, and this yields 999 = 9u2 + 9au + 3a2 + 9v2 + 9av + 9uv. All
terms are divisible by 9 except possibly 3a2. If 3a2 does not divide 9, thenthe left side, 999, is divisible by 9, but the right side is not. Hence, we must
have 3 dividing a2, which implies that a = 0, since neither 12 nor (�1)2 isdivisible by 3. Hence, a = 0, and so x = 3u and y = 3v, which implies that
u2 + uv + v2 = 111. Multiplying by 4, we get 444 = 4u2 + 4uv + 4v2 =(3u2 + 6uv + 3v2) + (u2 � 2uv + v2) = 3(u + v)2 + (u � v)2. Since
(u� v)2 < (u+ v)2, it follows that
3(u+ v)2 < 444 < 4(u+ v)2
=) 111 < (u+ v)2 < 148
=) 11 � u+ v � 12:
Now if u + v = 12, then (u � v)2 = 444 � 3(u + v)2 = 12, andbecause u and v are integers, this has no solution. However, if u+ v = 11,then (u � v)2 = 81, and so u � v = 9, because u > v (which follows
directly from x > y). Solving for u and v, we get u = 10 and v = 1, whichimplies that x = 30 and y = 3. Checking, (x; y) = (30; 3) is a solution to
2x2 + 2xy + 2y2 = 1998.
Therefore, there is a unique solution to K = 1998, and it is
(x; y) = (30;3).
345
H227. The numbers 2, 4, 8, 16, : : : , 2n are written on a chalkboard. A
student selects any two numbers a and b, erases them, and replaces them by
their average, namely (a+b)=2. She performs this operation n�1 times until
only one number is left. Let Sn and Tn denote the maximum and minimum
possible value of this �nal number, respectively. Determine a formula for
Sn and Tn in terms of n.
Solution.
We shall prove by induction that if we perform the operation on the
numbers a1, a2, : : : , an�1, an, where a1 < a2 < � � � < an, then
Tn =a1
2+a2
4+a3
8+ � � �+ an�2
2n�2+an�1
2n�1+
an
2n�1:
For n = 2, we have two numbers, a1 and a2, and thus T2 must be
(a1 + a2)=2. Thus the claim holds for n = 2.
Assume that for n = 2, 3, : : : , k�2, k�1, the identity holds. Then forthe case n = k, that is, when we start o� with the numbers a1, a2, a3, : : : ,ak, we will perform the operation k�2 times and be left with two numbers,
x and y. Say x comes from performing the operation on a set of p numbers
from the set. Then y comes from performing the operation on the remaining
set of k� p numbers.
To illustrate this, say we start o� with 2, 4, 8, 16, and 32. Then we can
replace 2 and 8 by 5, 16 and 32 by 24, then 4 and 24 by 14. Then we are left
with two numbers, 5 and 14. We got 5 from performing the operation on the
numbers 2 and 8, and we got 14 from performing the operation on the other
three numbers, namely 4, 16, and 32.
Let b1, b2, b3, : : : , bp be the set of numbers that we used to get the
number x, where the bi's are in increasing order, and let c1, c2, c3, : : : , ck�pbe the other numbers from the set that we used to get y, where the ci's arein increasing order. Then, by our induction hypothesis, we have
x � b1
2+b2
4+ � � �+ bp�2
2p�2+bp�1
2p�1+
bp
2p�1;
and similarly,
y � c1
2+c2
4+ � � �+ ck�p�2
2k�p�2+ck�p�1
2k�p�1+
ck�p
2k�p�1:
Note that the bi's and the ci's are just some permutation of the set a1,a2, a3, : : : , ak. Hence, to minimize the value of (x + y)=2, the average of
the two numbers will be minimized when the denominators are as large as
possible. Hence, we want either p or k � p to be 1, since we will then have
a denominator of 2k�1 in one of the terms.
346
Without loss of generality, assume that k � p = 1. Thus, we want to
now show that
x+ y
2� c1
2+b1
4+b2
8+ � � �+ bk�3
2k�2+bk�2
2k�1+bk�1
2k�1
� a1
2+a2
4+a3
8+ � � �+ ak�2
2k�2+ak�1
2k�1+
ak
2k�1;
because this will prove the case for n = k.
Note that the bi's are in increasing order, so if c1 = ar for some r,then b1 = a1, b2 = a2, : : : , br�1 = ar�1, br = ar+1, br+1 = ar+2, : : : ,bk�1 = ak.
Noting that the ai's are in increasing order, we have
ar
2=
ar
4+ar
8+ar
16+ � � �
� ar
4+ar
8+ar
16+ � � �+ ar
2r
� a1
4+a2
8+a3
16+ � � �+ ar�1
2r:
Adding
a1
4+a2
8+a3
16+ � � �+ ar�1
2r+ar+1
2r+1+ar+2
2r+2+ � � �+ ak�1
2k�1+
ak
2k�1
to both sides, we get the desired inequality.
Thus, the induction is proved and hence, for our question, we take
ai = 2i for i = 1, 2, : : : , k, and we have
Tn =2
2+
4
4+
8
8+ � � �+ 2n�2
2n�2+
2n�1
2n�1+
2n
2n�1
= (n� 1) � 1 + 2 = n+ 1:
Thus, our formula for Tn is obtained by taking the ai's in increasing
order. In contrast, to get Sn, we want to take the ai's in decreasing order,
namely a1 = 2n, a2 = 2n�1, : : : , an�1 = 4, an = 2. This is so that we can
get the maximum possible value at the end. Thus, we have
Sn =2n
2+
2n�1
4+
2n�2
8+ � � �+ 23
2n�2+
22
2n�1+
21
2n�1
= 2n�1 + 2n�3 + 2n�5 + � � �+ 25�n + 23�n + 22�n
= 23�n � (1 + 22 + 24 + � � �+ 22n�4) + 22�n
= 23�n4n�1 � 1
3+ 22�n
=8(4n�1 � 1)
3 � 2n +4
2n
=22n+1 + 4
3 � 2n :
347
H228. Verify that the following three inequalities hold for positive
reals x, y and z:
(i) x(x�y)(x�z)+y(y�z)(y�x)+z(z�x)(z�y) � 0 (this is known
as Schur's Inequality),
(ii) x4 + y4 + z4 + xyz(x+ y + z) � 2(x2y2 + y2z2 + z2x2),
(iii) 9xyz + 1 � 4(xy+ yz + xz), where x+ y + z = 1.
Solution by Vedula N. Murty, Visakhapatnam, India.We shall show that all three inequalities are equivalent to showing that
(x+ y � z)(x� y)2 + (y+ z � x)(y� z)2 + (z + x� y)(z� x)2 � 0:
And we shall then show that this inequality holds for all positive reals x, y,and z. Let
p = (x+ y � z)(x� y)2 + (y+ z � x)(y� z)2 + (z + x� y)(z� x)2:
Expanding and rearranging terms, we �nd that
2x(x� y)(x� z) + 2y(y� z)(y� x) + 2z(z� x)(z � y)
= 2x3 + 2y3 + 2z3 + 6xyz� 2(x2y+ xy2 + x2z + xz2 + y2z + yz2)
= (x3 + y3 + 2xyz� x2y � x2z � y2x� y2z)
+(y3 + z3 + 2xyz� y2x� y2z � z2x� z2y)
+(z3 + x3 + 2xyz � z2x� z2y� x2y � x2z)
= (x+ y� z)(x� y)2 + (y + z � x)(y � z)2 + (z + x� y)(z� x)2
= p :
Therefore, we have shown that x(x� y)(x� z) + y(y� z)(y� x) +z(z � x)(z � y) = p=2. Thus (i) holds if and only if p � 0.
Note that x4 + y4+ z4�x2y2� y2z2� z2x2 = 1
2[(x2� y2)2+ (y2�
z2)2 + (z2� x2)2]. Furthermore, x2y2+ y2z2 + z2x2 � xyz(x+ y+ z) =1
2[x2(y� z)2 + y2(x� z)2 + z2(x� y)2]. Subtracting the second equation
from the �rst, we have
x4+ y
4+ z
4+ xyz(x + y + z)� 2(x
2y2+ y
2z2+ z
2x2)
=1
2[(x
2 � y2)2+ (y
2 � z2)2+ (z
2 � x2)2 � x
2(y � z)
2 � y2(x� z)
2
�z2(x � y)2]
=1
2f(x� y)
2[(x + y)
2 � z2] + (y � z)
2[(y + z)
2 � x2]
+(z � x)2[(z + x)
2 � y2]g
=x + y + z
2� [(x� y)
2(x + y � z) + (y � z)
2(y + z � x)
+(z � x)2(z + x � y)]
=p(x + y + z)
2:
348
Since x, y, and z are positive, x+ y+ z is also positive, and it follows that
(ii) holds if and only if p � 0.
Suppose that x + y + z = a. Then if we replace x by x=a, y by y=aand z by z=a, then the inequality will remain the same. Thus, there is no
loss in generality in assuming that x + y + z = 1, for we can always divide
each term by some constant to ensure that the sum of the terms becomes 1.
Thus, letting x+ y + z = 1, we have:
p
2=
1
2[(x + y � z)(x� y)
2+ (y + z � x)(y � z)
2+ (z + x� y)(z � x)
2]
= (x3+ y
3+ z
3) + 3xyz � (x
2y + x
2z + y
2x + y
2z + z
2x + z
2y)
= (x3 + y3 + z3) + 3xyz � (x2 + y2 + z2)(x+ y + z) + (x3 + y3 + z3)
= 2(x3+ y
3+ z
3) + 3xyz � (x
2+ y
2+ z
2)
= 2(x+ y+ z)(x2+ y
2+ z
2 � xy � yz � zx) + 6xyz + 3xyz
�(x2 + y2+ z
2)
= x2+ y
2+ z
2 � 2(xy + yz + zx) + 9xyz
= (x+ y + z)2 � 4(xy + yz + zx) + 9xyz
= 1� 4(xy + yz + zx) + 9xyz :
Therefore, (iii) holds if and only if p � 0. Hence, we have shown that all
three inequalities are equivalent to p � 0. Thus, if we can show that this
inequality holds, then (i), (ii), and (iii) all hold.
To show that p � 0, we separate the problem into two cases. If x, y,and z are the sides of a triangle, then x+ y � z, x + z � y, and y + z � xare all positive, and thus it immediately follows that p � 0. Otherwise, thethree sides do not form the sides of a triangle, and so if we assume without
loss of generality that x � y � z, then x+ y � z � 0.
And we have
p = (x+ y � z)(x� y)2 + (y+ z � x)(y� z)2 + (z + x� y)(z� x)2
= (x+ y � z)[(x� z) + (z � y)]2 + (y+ z � x)(y� z)2
+(z + x� y)(z� x)2
= (x� z)2(x+ y� z) + (z � y)2(x+ y � z)
+2(x� z)(z � y)(x+ y � z) + (y + z � x)(y � z)2
+(z + x� y)(z� x)2
= (x� z)2(2x) + (y� z)2(2y) + 2(z � x)(z � y)(z� x� y) :
And this last expression is non-negative, since x � 0, y � 0,z � x � 0, z � y � 0 and z � x � y � 0. Therefore, p � 0, and we
have proven that all three inequalities hold.
349
Advanced Solutions
Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada. M1G
1C3 <[email protected]>
The following problems �rst appeared in Volume 23, Issue 5.
A201. Consider an in�nite sequence of integers a1, a2, : : : , ak, : : :with the property that every m consecutive numbers sum to x and every nconsecutive numbers sum to y. If x and y are relatively prime, then show
that all numbers are equal.
Erratum. As noted by some of our solvers, the problem as stated is
incorrect . The correct problem should read as follows:
Consider an in�nite sequence of integers a1, a2, : : : , ak, : : : with the
property that every m consecutive numbers sum to x and every n consecu-
tive numbers sum to y. If m and n are relatively prime, then show that all
numbers are equal.
Solution.
Since every m consecutive numbers sum to the same value, we must
have a periodic sequence with ak = am+k, for all natural numbers k. Like-wise, ak = an+k, for all natural numbers k. What this means is that the
sequence is periodic with periods m and n. It may be intuitively clear now
why the sequence must be a constant sequence. Here is a rigorous proof.
It is enough to show that the n numbers a0, a1, : : : , an�1 are equal since
every other number is equal to one of these by periodicity. Now a0, am,
a2m, : : : are equal, so it is enough to show that ai, for i = 0, 1, : : : , n� 1,is equal to akm for some integer k.
We show that each of the n numbers a0, am, : : : , a(n�1)m, leaves a
di�erent remainder upon division by n. In other words, the set of numbers
f0; am; : : : ; a(n�1)mg is a complete residue system modulo n. To see this,
im � jm (mod n) holds if and only if i � j (mod n) since m and n are
relatively prime. Further, for 0 � i; j < n, we must then have i = j.In other words, the n values a0, am, : : : , a(n�1)m must leave a di�erent
remainder upon division by n.
Now, putting it all together, suppose lm = nq+ r, where 0 � r < n.Then alm = anq+r = ar. Now we know that all possible remainders r, from0 to n�1, are achieved for some integer l, so we can conclude that all ar areequal.
A202. Let ABC be an equilateral triangle and � its incircle. IfD and
E are points on AB and AC, respectively, such that DE is tangent to �,show that
AD
DB+AE
EC= 1:
(8th Iberoamerican Mathematical Olympiad, Mexico '93)
350
Solution I by Miguel Carri �on �Alvarez, Universidad Complutense deMadrid, Spain.
B C
A
I
DE
F G
H
�
Let I be the incentre and r the inradius. Let F , G, and H be the
points where � is tangent to AB, AC, and DE. Then, DF = DH and
EG = EH (this is a well-known property of the two tangents from a point
to a circle). This implies that \FID = \DIH and \GIE = \EIH. Since
\FIG = 120�, \FID+\GIE = 60�. We can write \FID = 30�+�, and\GIE = 30� � �. Now, FD = r tan\FID and AF = r tan 60�, so that
AD
DB=
AF � FD
AF + FD=
tan60� � tan\FID
tan60� + tan\FID
=
p3� tan 30� + tan�
1� tan30� tan�p3 +
tan 30� + tan�
1� tan30� tan�
=
p3� 1 +
p3 tan�p
3� tan�
p3 +
1 +p3 tan�p
3� tan�
=1�p3 tan�
2;
and similarly
AE
EC=
1 +p3 tan�
2:
We then have
AD
DB+AE
EC=
1�p3 tan�
2+
1 +p3 tan�
2= 1
as required.
351
Solution II. Using the same diagram as above, assume, without loss of
generality, that the sides of the equilateral triangle ABC have length 1. Letx = AD and y = AE, where 0 � x; y � 1=2. Now using the Cosine Law in
triangle ADE, we have
(1� x� y)2 = x2 + y2 � 2xy cos 60�:
This is equivalent to each of the following:
1 + x2 + y2 � 2x� 2y + 2xy = x2 + y2 � xy ;
2x+ 2y � 1 = 3xy ;
x� xy + y� xy = 1� x� y+ xy ;
and �nally,x
1� x+
y
1� y= 1 :
The last equation is valid since x and y cannot be equal to 1. This is
our desired result.
Also solved by Alexandre Trichtchenko, student, Brook�eldHigh School,Ottawa, Ontario.
A203. Let Sn = 1 + a + aa + � � � + aa:::a
, where the last term is a
tower of (n� 1) a's. Find all positive integers a such that Sn = naSn�1=n.
Solution.
Claim: a = 1 is the only possibility.
Since a is positive, we have by the AM-GM inequality
1 + a+ aa + � � �+ aa:::a
n��a0+1+a+���+aa
:::a� 1
n
;
with equality if and only if 1 = a = aa = � � � = aa:::a
. Using the de�nition
of Sn, we have
Sn
n� �
aSn�1
�1=n:
Since equality occurs, we must have 1 = a = aa = � � � = aa:::a
, Sn = n, forall natural numbers n and it follows that Sn = n(1Sn�1=n).
A204. Given a quadrilateral ABCD as shown, with AD =p3,
AB + CD = 2AD, \A = 60� and \D = 120�, �nd the length of the
line segment from D to the mid-point of BC.
Solution I by Miguel Carri �on �Alvarez, Universidad Complutense deMadrid, Spain and Alexandre Trichtchenko, student, Brook�eld High School,Ottawa, Ontario.
352
:
:
:
A
B
C
D
E
F
60�
60�p3
Observe that AB and CD are parallel. Then by Thales' Theorem, a
line through the mid-points E and F of AD and BC respectively will also
be parallel to AB, and the length ofEF will be (AB+CD)=2 = AD. Now,
applying the Cosine Law to triangle DEF , we have
DF 2 = DE2 + EF 2� 2 �DE � EF cos\DEF
=
�AD
2
�2+ AD2 � AD2 cos 60�
=3AD2
4;
so DF = 3=2.
Solution II.
:
:
x
2x
?
U
A
B
C
D
E
60�
D0
Let the mid-point of BC be E. Note that linesAB and CD are paral-
lel. Rotate triangle DEC about point E so that C coincides with B, and Dcoincides withD0 as shown in the �gure. This is possible since E was chosen
to be the midpoint of line segment BC.
Now A, B, and D0 are collinear since lines CD and AB are parallel,
so that \D0BE + \ABE = \DCE + \ABE = 180�.Let x = AD, thenAD0 = AB+BD0 = AB+CD = 2x. Thus triangle
DAD0 is similar to the ubiquitous 1: 2:p3 triangle. Hence,
DE = DD0=2 =p3=2AD = 3=2.
353
Challenge Board Solutions
Editor: David Savitt, Department of Mathematics, Harvard University,
1 Oxford Street, Cambridge, MA, USA 02138 <[email protected]>
C75.
(a) Let n be an integer, and suppose a1, a2, a3, and a4 are integers such
that a1a4 � a2a3 � 1 (mod n). Show that there exist integers Ai,
1 � i � 4, such that each Ai � ai (mod n) and A1A4 � A2A3 = 1.
(b) Let SL(2;Z) denote the group of 2 � 2 matrices with integer entries
and determinant 1, and let �(n) denote the subgroup of SL(2;Z) of
matrices which are congruent to the identity matrix modulo n. (Two
matrices are congruent modulo n if each pair of corresponding entries
is congruent modulo n.) What is the index of �(n) in SL(2;Z)?
Solution.
(a) Without loss of generality, we suppose that the ai are non-zero.
(They may, of course, nevertheless be congruent to 0 (mod n).) Choose
A1 = a1. Our �rst order of business is to �nd A2 � a2 (mod n) withA1 and A2 relatively prime. Let d = gcd(a2; n). Then a2=d and n=d are
relatively prime, and so by Dirichlet's Theorem on the in�nitude of primes
in arithmetic progressions, we can choose a prime p distinct from the prime
divisors of A1 and satisfying
p � a2
d
�mod
n
d
�:
Set A2 = dp. It is immediate that A2 � a2 (mod n). Since a1 and a2 must
not share any common divisors with n, we see that d and A1 are relatively
prime, and therefore so are A1 and A2.
Select any x, y such that A1x � A2y = 1. We must �nd k such that
x + A2k � a4 (mod n) and y + A1k � a3 (mod n). Rewriting these
congruences as
A2k � a4 � x (mod n)
and
A1k � a3 � y (mod n);
if we multiply the �rst congruence by�a3, the second by a4, and we add the
equations, then we �nd
A1a4k� A2a3k � a4(a3 � y)� a3(a4 � x) (mod n);
and since A1a4 �A2a3 � a1a4 � a2a3 � 1 (mod n), we obtain
k � a3x� a4y (mod n):
354
One readily checks that with such a choice of k, setting A4 = x + A2k and
A3 = y +A1k gives the Ai the desired properties.
(b) Let SL(2;Z=nZ) denote the group of 2 � 2 matrices with entries
in Z=nZ (the ring of integers modulo n) and determinant 1. The reduction-
mod-n homomorphism 'n : SL(2;Z) ! SL(2;Z=nZ), via which the en-
tries of 'n(A) are the congruence classes of the corresponding entries of A,evidently has kernel equal to �(n). Additionally, part (a) of this problem
proves that 'n is surjective. So, to compute the index of �(n) in SL(2;Z),we need only compute the order of SL(2;Z=nZ). Call this order C(n).
Let n = pa11� � � pak
kbe the prime factorization of n. Given an ele-
ment of SL(2;Z=nZ), we obtain an element in each SL(2;Z=paiiZ) by re-
ducing modulo paii. Conversely, given one element in each SL(2;Z=pai
iZ),
we can use the Chinese Remainder Theorem to obtain a uniquematrix whose
modulo paii
reductions are our given matrices. It follows, therefore, that
C(n) = C(pa11) � � �C(pak
k).
Let �a bc d
�
be a representative of an element of SL(2;Z=pkZ) with p prime and k > 0,and suppose �
A BC D
�
is an element of SL(2;Z=pk+1Z) whose reduction modulo pk is�
a bc d
�:
Choose representatives a0, b0, c0, and d0 in Z=pk+1Z of a, b, c, and d, and
write
A = a0 + a0pk;
B = b0 + b0pk;
C = c0 + c0pk;
D = d0 + d0pk;
with a0, b0, c0, d0 in Z=pZ.Then
det
�a0 b0
c0 d0
�= 1 +mpk
in Z=pk+1Z, and the condition that
det
�a0 + a0p
k b0 + b0pk
c0 + c0pk d0 + d0p
k
�= 1
in Z=pk+1Z is transformed into the condition that
a0d0 + a0d0 � b0c
0 � b0c0 � �m (mod p) :
355
One of a0, b0, c0, d0 must be invertible modulo p, say d0, without loss of
generality. Then for any arbitrary choice of b0, c0, and d0, there exists exactlyone a0 which satis�es this condition. Consequently, we have shown that eachelement of SL(2;Z=pkZ) lifts to precisely p3 elements in SL(2;Z=pk+1
Z),and so C(pk+1) = p3C(pk). Inductively, C(pk+1) = p3kC(p).
To calculate the number of elements�a bc d
�
in Z=pZ, note that to have determinant 1, the �rst row of�a bc d
�
must be non-zero. However, the reader may verify directly that to each of
the remaining p2 � 1 possible choices for the pair (a; b), there are exactly pchoices for the pair (c; d) which make
det
�a bc d
�= 1 :
Thus,
C(p) = (p2 � 1)p = p3�1� 1
p2
�;
and putting together all our work, we conclude that
C(n) = n3Ypjn
�1� 1
p2
�;
where the product is taken over the primes p dividing n.
C76. Let X be any topological space. The nth symmetric power of X,
denoted X(n), is de�ned to be the quotient of the ordinary n-fold product
Xn by the action of the symmetric group on n letters|that is, it is the space
of unordered n-tuples of points of X. Show that the symmetric power C(n)
is actually homeomorphic to the ordinary product Cn.
Solution.
Every unordered n-tuple hr1; : : : ; rni of points in C is the solution set
to a unique monic polynomial, speci�cally, the polynomial
f(z) = (z � r1) � � � (z � rn) whose coe�cients are given by the elementary
symmetric polynomials in the ri. Furthermore, by the Fundamental Theo-
rem of Algebra, every polynomial of degree n over C has exactly n solutions
in C, counting multiplicity. Therefore, the map ' : C(n) ! Cn is a bijection,
where ' is de�ned as
'(hr1; : : : ; rni) = (s1; : : : ; sn)
356
with the si given by
f(z) = (z � r1) � � � (z � rn) = zn � s1zn�1 + � � �+ (�1)nsn:
It is evident that ' is continuous, and so to prove that ' is a homeomor-
phism, it remains to show that '�1 is continuous as well. This amounts to
proving that the roots of a polynomial vary continuously with the coe�cients
of the polynomial. The reader may attempt to prove this using elementary
methods; instead, we employ a useful theorem from complex analysis:
Theorem. (Rouch �e) Let D be a closed disk in C. Suppose that f(z)and g(z) are complex-valued functions which are complex-di�erentiable inthe interior of D, and that at any point z on the boundary of D we havejf(z)� g(z)j < jf(z)j. Then f(z) and g(z) have the same number of zerosin D, counting multiplicity.
Given � > 0, let Di be a disk centred at ri and of radius at most �,chosen so that no other roots of f lie on the boundary of Di. Since jf(z)jachieves its minimum on the boundary ofDi, this minimum is positive. Call-
ing this minimummi, we can choose �i > 0 to be su�ciently small such that
�i(zn�1 + � � �+ z + 1) < mi � jf(z)j
on the boundary of Di. Let � be the minimum of the �i, and let g(z) be anymonic polynomial whose coe�cients each di�er from the coe�cients of f(z)by at most �. By construction, the polynomials f(z) and g(z) satisfy the
conditions of Rouch �e's Theorem on each disk Di, and consequently f and ghave the same number of zeros in each Di. Thus, the roots of g(z) all aredistant from corresponding roots of f(z) by no more than �, and so indeed
'�1 is continuous and ' is a homeomorphism.
Note for advanced readers: Without too much di�culty, one may use the above
result to prove, more generally, that if X is any 2-dimensional manifold, then the
symmetric product X(n) is always a manifold again. However, if X is a manifold of
any positive dimension k other than 2 and if n > 1, then X(n) is never a manifold.
One argues this roughly as follows. Let U be a neighbourhood of an unordered n-
tuple of points of X with two of the points the same and the rest di�erent. Then U
looks like (Rk)n�2 � (Rk � Rk)=f(x; y) � (y; x)g, where, in the last term of the
product, the action of the symmetric group causes (x; y) and (y; x) to be identi�ed.
Changing variables in the last term of the product by putting u = x+y and v = x�y,
we �nd thatU is in fact homeomorphic to (Rk)n�2�Rk�V , where V is the quotient
ofRk under the identi�cation v = �v. Then V with the origin removed is homotopic
to the projective space RP k�1, whereas Rk with any point removed is homotopic to
the sphere Sk�1. Since RP k�1 and Sk�1 are homotopic only for k = 2, we conclude
that U is not Euclidean space if k 6= 2, and so X(n) can only be a manifold if X is
2-dimensional.
357
Pushing the Envelope
Naoki Satostudent, Yale University
If C is a di�erentiable curve in the plane, then we can form a family
of lines F by taking the set of tangents at all points of C. This family F is
called the envelope of C (see Figure 1). Obtaining F from C is easy enough,
but consider the converse problem; that is, given a family of lines F , does
there exist a curve C which is tangent to every line in F , so making F the
envelope of C.
F
C
Figure 1
To solve the problem, we assume that F is a one-parameter family
of lines; that is, the lines are parameterized by one parameter, say t, andfurthermore, that the lines vary smoothly in t. It should be noted that with
respect to the point-line duality in projective geometry, the concept of the
envelope is dual to the concept of the locus.
Before tackling the problem, we �rst give a few examples.
Example 1. The parabola y = x2. The derivative is given by
dy
dx= 2x :
Hence, the tangent at a particular point (x0; y0) = (t0; t2
0) is given by
y � y0 = 2t0(x� x0) ;
Copyright c 1998 Canadian Mathematical Society
358
or
y = 2t0x� t20:
Example 2. The general ellipse is given by the following parameteriza-
tion: x = a cos t, y = b sin t. The derivative is given by
dy
dx=
dy
dt
dx
dt
=b cos t
�a sin t = � bacot t :
Hence, the tangent at a particular point (x0; y0) = (a cos t0; b sin t0) is givenby
y � y0 = � ba(cot t0)(x� x0) ;
or
y = � ba(cot t0)x+ b csc t0 :
Now, we tackle the problem. Assume F is of the form fm(t)x+ b(t)g,so in Example 1, m(t) = 2t and b(t) = �t2. A diagram makes the next step
visually clear. If we �x one particular line ` in F , and consider other lines in
F that approach `, then the intersections converge to a point on the desired
curve C (see Figure 2).
`
C
Figure 2
Mathematically, let ` be the line corresponding to a �xed t, namely
y = m(t)x + b(t). We consider the lines y = m(t + h)x + b(t + h) as happroaches 0. Solving for their intersection, we obtain
m(t)x+ b(t) = m(t+ h)x+ b(t+ h) ;
359
which implies that
x = � b(t+ h)� b(t)
m(t+ h)�m(t)= �
b(t+h)�b(t)h
m(t+h)�m(t)
h
:
By de�nition, this approaches � b0(t)
m0(t)as h approaches 0; hence, this is the
x-coordinate of the point on C that is in fact tangent to `. The y-coordinateis
� b0(t)
m0(t)�m(t) + b(t) =
m0(t)b(t)�m(t)b0(t)
m0(t):
Therefore,
C =
��� b0(t)
m0(t);m0(t)b(t)�m(t)b0(t)
m0(t)
��:
Checking, for Example 1, we have
C =
��2t
2;(2)(�t2)� (2t)(�2t)
2
��= f(t; t2)g ;
and for Example 2,
C =
( a csc t cot t
1 + cot2 t;b2(1 + cot2 t) csc t� b2 cot2 t csc t
b(1 + cot2 t)
!)
= f(a cos t; b sin t)g :
There is an alternative method for �nding curves of envelopes. If the
envelope is given in the form F (x; y; t) = 0, then the curve is found by
solving the equations F = 0 and @F
@t= 0. (There are also some regularity
conditions which we will ignore here.) In our case, we can take F (x; y; t) =y �m(t)x� b(t), so the curve is found by solving the system of equations
y �m(t)x� b(t) = 0 ;
m0(t)x+ b0(t) = 0 :
Solving for x and y, we �nd the same solution as above. Why does this more
general method work?
Orthogonal Curves
A related topic, orthogonal curves, can be interesting in their own right.
IfF and G are each a family of curves in the plane, then they are orthogonal ifevery curve inF is orthogonal to every curve inG. Two curves are orthogonal
if their tangents are perpendicular at any point they intersect. For example,
the family of concentric circles centred at the origin and the family of lines
passing through the origin are orthogonal (see Figure 3).
360
Figure 3
The natural question that arises is: For a family of curves F , does there
exist a family G which is orthogonal to F. Most smooth one-parameter fam-
ilies will have a solution.
Example 3. We take the family of parabolas fy = tx2g; in this case, tis again a parameter. For each parabola, di�erentiating we �nd
dy
dx= 2tx =
2y
x:
We wish to eliminate t from the expression, because we want to �nd curves
which are orthogonal to every such parabola, and so the expression must be
independent of t. As every high school student knows, two lines are perpen-
dicular if and only if the product of their slopes is �1, so we want to solve
the di�erential equation
dy
dx= � x
2y
2y dy = �x dxZ2y dy = �
Zx dx
y2 = �x2
2+C
x2
2+ y2 = C or
x2
2C+y2
C= 1 :
The orthogonal family is a family of ellipses. In other words, each parabola
is orthogonal to each ellipse.
Example 4. We take the family of circles f(x � t)2 + y2 = t2 � a2g,where a is a �xed non-negative integer, and jtj � a. This is a general family
361
of coaxial circles. For each circle, di�erentiating we �nd
dy
dx= �x� t
y;
and by re-arranging the equation of the circle, we obtain
t =x2 + y2 + a2
2x;
sody
dx=
y2� x2 + a2
2xy:
The orthogonal family must then satisfy
dy
dx=
2xy
x2 � y2� a2
(x2 � y2 � a2) dy = 2xy dx
2xy dx+ (�x2 + y2 + a2) dy = 0
By some elementary theory of di�erential equations, we �nd y�2 is an inte-
grating factor of the above equation, so the equation becomes�2x
y
�dx+
��x
2
y2+ 1 +
a2
y2
�dy = d
�x2
y+ y � a2
y2
�= 0
x2
y+ y � a2
y= 2C
We put in a constant of 2C, because on simpli�cation, the above equa-
tion becomes x2 + (y � C)2 = a2 + C2. This is the family of circles, with
centre on the y-axis, passing through the points (�a; 0).In general, families of the same type of curve can have di�erent orthog-
onal families; it depends on how the curves are \stacked".
Problems
1. For each of the following families of lines, �nd the curve which it is the
envelope of:
(a) fy = � a
t2x+ 2a
tg. This is the family of lines for which the product
of the x-intercept and y-intercept is a constant, 4a
(b) fy = (cos t)x� t cos t+ sin tg(c) fy = �(tan t)x+a sin tg. This is the family of lines for which the
axes are cut o� a chord of constant length a.
362
2. For each of the following families of lines, �nd the orthogonal family of
lines:
(a) fy = tx3g(b) fy = texg.
3. Let c be a positive real. Consider the graphs of the equation
x2
t2+
y2
t2 � c2= 1 ;
as t varies. For jtj < c, the graph is an ellipse; for jtj > c, it is a
hyperbola, all with foci (�c; 0). Show that the family of such ellipses
is orthogonal to the family of such hyperbolae.
363
PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, De-partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.
In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.
To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81
2"�11" or A4 sheets of paper.
These may be typewritten or neatly hand-written, and should be mailedto the Editor-in-Chief, to arrive no later than 1 April 1999. They may alsobe sent by email to [email protected]. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication. Please note that we do not accept submissions sent by FAX.
2364. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.
A sequence fxng is given by the recursion: x0 = p, xn+1 = qxn+q�1(n � 0), where p is a prime and q � 2 is an integer.
(1) Suppose that p and q are relatively prime. Prove that the sequence
fxng does not consist of only primes.
(2)? Suppose that pjq. Prove that the sequence fxng does not consist of
only primes.
2365. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.Triangle DAC is equilateral. B is on the line DC so that \BAC =
70�. E is on the line AB so that \ECA = 55�. K is the mid-point of
ED. Without the use of a computer, calculator or protractor, show that
60� > \AKC > 57:5�.
364
2366. Proposed by Catherine Shevlin, Wallsend-upon-Tyne, Eng-land.
Triangle ABC has area p, where p 2 N. Let
� = min�AB2 + BC2 + CA2
�;
where the minimum is taken over all possible triangles ABC with area p,and where � 2 N.
Find the least value of p such that � = p2
2367. Proposed by K.R.S. Sastry, Dodballapur, India.In triangle ABC, the Cevians AD, BE intersect at P . Prove that
[ABC]� [DPE] = [APB]� [CDE] :
(Here, [ABC] denote the area of4ABC, etc.)
2368. Proposed by Iliya Bluskov, Simon Fraser University, Burnaby,BC.
Let (a1; a2; :::; an) be a permutation of the integers from 1 to nwith the
property that ak+ak+1+ :::+ak+s is not divisible by (n+1) for any choiceof k and s where k � 1 and 0 � s � n� k� 1. Find such a permutation
(a) for n = 12;
(b) for n = 22.
2369?. Proposed by Federico Arboleda, student, Bogot �a, Colombia(age 11).
Prove or disprove that for every n 2 N, there exists a 2n-omino such
that every n-omino can be placed entirely on top of it.
(Ann-omino is de�ned as a collection ofn squares of equal size arranged
with coincident sides.)
2370. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.
Determine the exact values of the roots of the polynomial equation
x5 � 55x4 + 330x3 � 462x2 + 165x� 11 = 0 :
2371. Proposed by Bill Sands, University of Calgary, Calgary, Al-berta.
For n an integer greater than 4, let f(n) be the number of �ve-element
subsets, S, of f1; 2; : : : ; ng which have no isolated points, that is, such that
if s 2 S, then either s� 1 or s+ 1 (not taken modulo n) is in S.
Find a \nice" formula for f(n).
365
2372. Proposed by Bill Sands, University of Calgary, Calgary, Al-berta.
For n and k positive integers, let f(n; k) be the number of k{element
subsets S of f1; 2; : : : ; ng satisfying:(i) 1 2 S and n 2 S; and(ii) whenever s 2 S with s < n� 1, then either s+ 2 2 S or s+ 3 2 S.
Prove that f(n; k) = f(4k� 2� n; k) for all n and k; that is, the sequence
f(k; k); f(k+ 1; k); f(k+ 2; k); : : : ; f(3k� 2; k)
of non-zero values of ff(n;k)g1n=1
is a palindrome for every k.
2373. Proposed by Toshio Seimiya, Kawasaki, Japan.Given triangle ABC with AB > AC. LetM be the mid-point of BC.
Suppose thatD is the re ection ofM across the bisector of \BAC, and that
A, B, C and D are concyclic.
Determine the value ofAB � AC
BC.
2374. Proposed by Toshio Seimiya, Kawasaki, Japan.Given triangle ABC with \BAC < 60�. Let M be the mid-point of
BC. Let P be any point in the plane of4ABC.
Prove that AP + BP + CP � 2AM .
2375. Proposed by Toshio Seimiya, Kawasaki, Japan.Let D be a point on side AC of triangle ABC. Let E and F be points
on the segments BD and BC respectively, such that \BAE = \CAF . LetP and Q be points on BC and BD respectively, such that EPkDC and
FQkCD.
Prove that \BAP = \CAQ.
366
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.
2247. [1997: 244] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.
(a) Suppose that n � 3 is an odd natural number.
Show that the only polynomial P 2 R[x] satisfying the functional equa-tion:
(P (x+ 1))n= (P (x))
n+
n�1Xk=0
�n
k
�xk; for all x 2 R;
is given by P (x) = x.
(b)? Suppose that n � 1 is a natural number.
Show that the only polynomialP 2 R[x] satisfying the functional equa-tion:
(P (x+ 1))n = (P (x))n +
n�1Xk=0
�n
k
�xk; for all x 2 R;
is given by P (x) = x.
(c)? Suppose that n � 1 is a natural number.
Show that the only polynomialP 2 R[x] satisfying the functional equa-tion:
P ((x+ 1)n) = (P (x))n+
n�1Xk=0
�n
k
�xk; for all x 2 R;
is given by P (x) = x.
I. Solutionby David Stone and Vrej Zarikian, Georgia SouthernUniver-sity, Statesboro, Georgia, USA.
(a) First observe that
n�1Xk=0
�n
k
�xk = (x+ 1)n � xn:
Thus the equation satis�ed by P can be written as
Q(x+ 1) = Q(x) for all x 2 R;
367
where Q(x) = (P (x))n � xn. But Q 2 R[x] and is periodic implies that
Q(x) = c for some constant c, so
(P (x))n = xn + c: (1)
By degree considerations, P (x) = ax + b for constants a and b; if P were
of degree 2, for instance, the left side of (1) would be of degree 2n, whichexceeds the degree of the right side. Thus (ax+ b)n = xn + c. Since n � 3and the right side of this equation has no \middle terms", it must be that
b = 0. Hence (ax)n = xn + c, so c = 0 and an = 1. Since n is odd, a = 1.In other words, P (x) = x.
(b) The proposition is not true as stated. Arguing as in part (a), we
reach the same conclusion | (P (x))n = xn + c implies P (x) = ax + b. Ifn = 1, then P (x) = x + c so a = 1 and b = c. If n > 1, it must be that
b = 0, so c = 0 and an = 1. If n is odd, a = 1. If n is even, a = 1 or �1.In summary:
n P (x)
1 x+ c (c arbitrary)> 1, odd x> 1, even x or �x
(c) This proposition is not true as stated, either. For example, take n = 1and P (x) = x+ c (c arbitrary).
II. Correction by the editor.
Unfortunately, part (b) of this problem was incorrectly transcribed into
print: the functional equation should have been
(P (x+ 1))n = P (xn) +
n�1Xk=0
�n
k
�xk:
That is, the �rst term on the right side should be P (xn) instead of (P (x))n.With this correction, P (x) = �x is never a solution, although P (x) = x+ cis still a counterexample when n = 1. We apologize to the proposer for
stating his problem incorrectly.
It appears from the proposer's (partial) solution that he had intended
n � 2 to be assumed in part (b) and (c) as well as (a), though his problem
statement did not make this clear. In any case, readers are invited to prove
or disprove the revised part (b) for n � 2. For part (c), see below!
III. Solution to part (c) byMichael Lambrou, University of Crete, Crete,Greece.
[Lambrou �rst solved (a) and (b), giving the same answer for (b) as in
Solution I. | Ed.]
In this case we show that for n > 1 the only polynomial satisfying the
368
modi�ed identity
P ((x+ 1)n) = (P (x))n+
n�1Xk=0
�n
k
�xk ;
that is, equivalently
P ((x+ 1)n) = (P (x))n+ (x+ 1)n � xn; (2)
isP (x) = x. But for n = 1 the polynomial could be of the formP (x) = x+cfor any constant c.
For a start it is easy to verify identity (2) for the stated polynomials,
so let us concentrate on the more interesting converse. The case n = 1 is
easy: (2) becomes P (x + 1) = P (x) + 1 which is the same equation as in
part (b) for the case n = 1. So this case has already been dealt with, giving
the stated conclusion for the form of P .
Assume then that n � 2.
We �rst observe that (2), valid for the real variable x, can be lifted to
an identity valid also for complex numbers z:
P ((z+ 1)n) = (P (z))n+ (z + 1)n � zn: (3)
For a quick proof, set
R(z) = P ((z+ 1)n)� (P (z))n� (z + 1)n + zn;
which is a polynomial in z. By (2) this polynomial vanishes for an in�nity of
values of z (all real z!). Hence it is identically zero, proving (3).
We now show that any polynomial P satisfying (3) must be of degree 1.
For this purpose, suppose on the contrary that it is of degree m � 2.Write
P (z) = amzm + am�1z
m�1 + � � �+ a0 (am 6= 0):
Given any real x, set
z = e2�i=n(1 + x)� 1
so that
(z + 1)n =�e2�i=n(1 + x)
�n= (x+ 1)n
and so P ((z + 1)n) = P ((x+ 1)n). Using (3) we �nd
(P (z))n+ (z + 1)n � zn = (P (x))n+ (x+ 1)n � xn;
that is,
(P (z))n� (P (x))n = zn � xn: (4)
369
Hence
n�1Yk=0
�P (z)� e2k�i=nP (x)
�= zn � xn: (5)
We have used the fact that the equationwn�an = 0 has roots w = e2k�i=na(0 � k � n� 1) so wn � an factors as
Qn�1k=0
(w� e2k�i=na).
The right hand side of (5) as a polynomial in x is
zn � xn =�e2�i=nx+ (e2�i=n� 1)
�n� xn
which is of degree n or less. In fact it is of degree n � 1 or less, as the
coe�cient of xn is e2�i�1 = 0, leaving powers of x up to n�1 at most. We
shall derive a contradiction to (5) by showing its left hand side is of degree
larger than n� 1.
For 0 � k � n� 1 we have
P (z)� e2k�i=nP (x) = (amzm + � � �+ a0)� e2k�i=n(amx
m + � � �+ a0)
= am
�e2�i=nx+ (e2�i=n� 1)
�m+ � � �+ a0
�e2k�i=n(amxm + � � �+ a0);
and the coe�cient of xm is clearly
am
�e2�mi=n � e2k�i=n
�= ame
2k�i=n
�e2�(m�k)i=n � 1
�: (6)
Observe that n divides exactly one of the n consecutive numbers m � 0,m� 1, : : : ,m� (n� 1). Say n j (m� k0) for some k0 2 f0; 1; : : : ; n� 1g,but n does not divide any of the remaining n� 1 terms. Thus the coe�cient
in (6) is non-zero for k 6= k0, 0 � k � n � 1. In other words for k 6= k0,0 � k � n � 1, the polynomial P (z)� e2k�i=nP (x) in x is of degree m.
Moreover the polynomial P (z)� e2k0�i=nP (x) is not identically zero, since
if P (z) = e2k0�i=nP (x) for all real x then (P (z))n = (P (x))n for all real
x, and so by (4) we would get zn = xn for all real x, which is clearly false;
otherwise z = e2k0�i=nx for some 0 � k0 � n � 1, which is incompatible
with z = e2�i=n(1 + x)� 1, as the constant term of this last is non-zero.
To summarize, we have shown:
� the degree of P (z)� e2k�i=nP (x) for k 6= k0, 0 � k � n� 1, ism;
� P (z)� e2k0�i=nP (x) is not identically zero.
Hence the left hand side of (5) is of degree at least m(n� 1) > n� 1. Thisis a contradiction. Hence we conclude that P is a �rst degree polynomial,
P (x) = ax + b for some a; b. Recall that we are seeking a polynomial
P 2 R[x], so a; b are restricted to be real.
370
By (2) we have
a(x+ 1)n + b = (ax+ b)n+ (x+ 1)n � xn
for all real x. Comparing coe�cients of xn and xn�1 and the constant term
(recall n � 2) we �nd in turn a = an, na = nan�1b + n (that is,
a = an�1b + 1), and a + b = bn + 1. The second equation gives a 6= 0so our equations become
an�1 = 1; (7)
a = 1 � b+ 1; (8)
a+ b = bn + 1: (9)
By (7), we have a = �1 (in fact for even n it only gives a = 1). But if
a = �1 then by (8), we would get b = �2 and in (9), we would have
�1 � 2 = (�2)n + 1. This last is clearly impossible as (�2)n is never
equal to �4. Hence a = 1 and (8) gives b = 0. To summarize, we have
shown P (x) = 1 �x+0, showing that P (x) = x is the only (real) polynomial
satisfying our condition, and the proof is complete.
Remarks. (1) For even n = 2t we do not have to go through the con-
sideration of complex numbers. By setting z = �(1 + x) � 1 = �x � 2and making the obvious adaptations and simpli�cations of the above we can
arrive at the proof. For example the analysis into products in (5) can be
replaced by
(P (z))2t� (P (x))2t =�(P (z))t� (P (x))t
� �(P (z))t+ (P (x))t
�and then estimating the degree of the right hand side as at leastmt.
The di�culty of the above approach for odd n arises from the fact that
(1 + x)n is, for the real variable x, a 1-1 function, so cannot be put in the
form (1 + z)n for a di�erent (real) z. Thus we had to go through complex
numbers.
(2) If one could somehow show P (0) = 0 (as expected), then a quick
way to complete the proof would be as follows. Set x0 = 0 and
xk+1 = (1 + xk)n for k = 0; 1; : : : . It is easy to see that each xk is a
�xed point of P ; that is, P (xk) = xk, inductively based on
P (xk+1) = P ((1+xk)n) = (P (xk))
n+(1+xk)n�xn
k= (1+xk)
n = xk+1:
But the values of xk are all distinct (the xk's are strictly increasing), so
P (x) = x identically as required. Unfortunately, however, I do not have
an easy proof of P (0) = 0, other than a proof resembling the proof given
above. Can the readers �nd such an easy proof of P (0) = 0 (or, just as well,
P (1) = 1) ?
371
Parts (a) and the incorrectly stated (b) were also solved by CHRISTOPHER
J. BRADLEY, Clifton College, Bristol, UK; FLORIAN HERZIG, student, Perchtolds-
dorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; and DIGBY
SMITH, Mount Royal College, Calgary, Alberta.
Hess and Smith found all functions satisfying the given equation, as did our
featured solvers Stone and Zarikian (Solution I) and Lambrou; Bradley and Herzig
just showed that P (x) = x is not the only such function. Part (a) only was solved
by the proposer. There was also one incorrect solution sent in.
Stone and Zarikian also consider the similar functional equation
P ((x+ 1)n
) = P (xn
) +
n�1Xk=0
n
k
!xk
;
which, in view of the correction to part (b), is the obvious fourth variation on the
proposer's original three equations. They prove easily that the only polynomials
satisfying this equation for all real x are of the form P (x) = x + c for c constant.
Readers may like to check this out for themselves, and then try �nishing o� part (b)!
2249. [1997: 245] Proposed by K.R.S. Sastry, Dodballapur, India.
How many distinct acute angles � are there such that
cos� cos 2� cos 4� = 1
8?
Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,Ontario.
There are exactly three such values: � = �
9, 2�
7, and �
3.
Let P = cos� cos 2� cos 4�. Then, for all � such that sin� 6= 0, wehave P = 1
8if and only if
sin� = 8P sin� = sin(8�) :
This is equivalent to having 8� = 2k�+� or 8� = (2k+1)��� for some
integer k. That is, either
(i) � = 2k�
7, or
(ii) � = (2k+1)�
9.
Note that � = 0 is not a solution. Since �
2< 5�
9< 4�
7, we see that the
only possible values for k, to ensure that 0 < � � �
2, are k = 1 in (i) and
k = 0; 1 in (ii). These yield � = 2�
7, �9, and �
3, respectively. Since sin� 6= 0
for these values, the result P = 1
8follows.
Also solved by (NOTE: a dagger y before a name indicates that the solver's
solution is very similar to the one above,) HAYO AHLBURG, Benidorm, Spain;
GERALD ALLEN, CHARLES DIMINNIE, TREY SMITH and ROGER ZARNOWSKI,
Angelo State University, San Angelo, TX, USA; MIGUEL AMENGUAL COVAS, Cala
Figuera, Mallorca, Spain; �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo,
372
Bosnia and Herzegovina; SAM BAETHGE, Nordheim, Texas, USA; yMICHAEL
BATAILLE, Rouen, France; y FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-
ladolid, Spain; ADRIAN BIRKA, student, Lakeshore Catholic High School, Port Col-
bourne, Ontario; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN
CHAN, student, Upper Canada College, Toronto, Ontario; JIMMY CHUI, student,
Earl Haig Secondary School, North York, Ontario; y THEODORE CHRONIS, student,
Aristotle University of Thessaloniki, Greece; y GORAN CONAR, student, Gymnasium
Vara�zdin, Vara�zdin, Croatia; DAVID DOSTER, Choate Rosemary Hall, Wallingford,
Connecticut, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;
V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; y KEE-
WAI LAU, Hong Kong; y ALAN LING, student, University of Toronto, Toronto, On-
tario; VEDULA N. MURTY, Visakhapatnam, India; VICTOR OXMAN, University of
Haifa, Haifa, Israel; y BOB PRIELIPP, University of Wisconsin{Oshkosh, Wiscon-
sin, USA; DIGBY SMITH, Mount Royal College, Calgary, Alberta; y CHRISTOPHER
SO, student, Francis Libermann Catholic High School, Scarborough, Ontario; DAVID
R. STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E.
TSAOUSSOGLOU, Athens, Greece; DAVIDVELLA, Skidmore College, Saratoga Springs,
NY, USA; and the proposer. There were also two incorrect and four incomplete sub-
missions, two of which gave the correct answer without proof.
2250. [1997: 245] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a scalene triangle with incentre I. Let D, E, F be the points
where BC, CA, AB are tangent to the incircle respectively, and let L, M ,
N be the mid-points of BC, CA, AB respectively.
Let l,m, n be the lines throughD,E, F parallel to IL, IM , IN respectively.
Prove that l, m, n are concurrent.
Solution by Michael Lambrou, University of Crete, Greece.
Introduce position vectors, ~c =�!BC, ~a =
�!BA, ~d =
�!BD,
~e =�!BE, ~f =
�!BF , all originating fromB. If we denote, as usual, the lengths
jBCj = a, jCAj = b, and jABj = c, then as jBDj = jBF j = a+c�b2
, we
have ~d =�a+c�b2a
�~c and ~f =
�a+c�b
2c
�~a. Also ~e =
�(a+b�c)~a+(b+c�a)~c
2b
�.
If the bisector, AI, cutsBC atP , then jBP j = ac
b+c, so
�!BP =
�c
b+c
�~c.
So, asjAIjjIP j =
jABjjBP j =
b+c
a, we have,
�!BI =
�!BP +
�!PI =
�a
a+b+c
�~a+
�c
a+b+c
�~c:
Consider X, where��!BX = ~x =
�3a�b�ca+b+c
�~a+
�3c�b�aa+b+c
�~c. We show
that X is on lines l; m and n, so these lines are concurrent, as is required.
All we need to do is to verify that: ~x� ~d, ~x�~e and ~x� ~f are parallel to�!IL,�!
IM and�!IN , respectively.
373
Then
~x� ~d =
�3a� b� c
a+ b+ c
�~a+
�3c� b� a
a+ b+ c� a+ c� b
a
�~c
and
�!IL =
�!BL��!
BI =~c
2��
a
a+ b+ c~a+
c
a+ b+ c~c
�
=
�a+ b� c
2(a+ b+ c)
�~c�
�a
a+ b+ c
�~a
=
� �a3a� b� c
�(~x� ~d):
That ~x � ~f is parallel to�!IN follows immediately from the above by
interchanging the roles of c and a. Similar routine calculation shows ~x�~e isparallel to
�!IM .
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid,
Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WALTHERJANOUS,
Ursulinengymnasium, Innsbruck, Austria; D.J. SMEENK, Zaltbommel, the Nether-
lands; and the proposer. There was one incomplete solution.
2251. [1997: 300] Proposed by Victor Oxman, University of Haifa,Haifa, Israel.
In the plane, you are given a circle (but not its centre), and points A,K, B, D, C on it, so that arc AK = arc KB and arc BD = arc DC.
Construct, using only an unmarked straightedge, the mid-point of arc AC.
Preliminary comment. There are two arcs AC; the midpoint of one of
them will always be easily constructed; the other is more challenging. Most
solvers (including the proposer) constructed only the easier one, which does
not require the lemma. Note that it is essentially the same as the lemma in
Oxman's solutions to his earlier problem: 2234 [1998: 247].
Solution by Toshio Seimiya, Kawasaki, Japan.Lemma. If we are given points A, M , B, and C on a line with
AM = MB, then using only an unmarked ruler we can construct D onthe line such that BC = CD.
Construction. Let P be a point not on the lineAB, and letQ be a point
on the segment AP .
Let R = BQ \MP; S = AR \BP;N = PM \QS; T = NC \ SB;D = QT \AB: Then BC = CD:
374
Proof. By Ceva's Theorem we havePQ
QA� AMMB
� BSSP
= 1. Since
AM = MB we getPQ
QA=
PS
SB, so that QS k AB. It follows that since
QD, NC, SB are concurrent at T ,
BC
CD=SN
NQ;
and since SA, NM , QB are concurrent at R,
SN
NQ=AM
MB:
Using these two equalities together with AM =MB, we conclude that
BC
CD=AM
MB= 1;
so that BC = CD.
We now turn to the main construction.
Construction. We denote the given circle by �. Draw AD meeting CKat I. LetM be the second intersection ofBI with �; thenM is the midpoint
of the arc AC not containing B. Let N be the intersection ofKD with BI;then BN = NI. By the lemma we can construct the point IB such that
IM = MIB. Let IC be the intersection of AIB with KC, and let L be the
second intersection of ICB with �. Then L is the midpoint of the arc ABC.
Proof. Since arc AK = arc KB, we have that CK bisects \ACB.
Similarly AD bisects \BAC, so that I is the incentre of4ABC.
[Editor's comment. The statement of the problem might be construed as
allowing C and K to lie on the same side of the line AB, in which case Iwould be an excentre. Seimiya's argument is based on the interpretation
that the points lie on the circle in the prescribed order, namely AKBDC; it
easily can be modi�ed to accommodate the alternative interpretation.]
Hence BM bisects \ABC, so that M is the midpoint of the arc AC not
containing B.
Since I is the incentre and K is the midpoint of arc AB, we have
KI = KB. Similarly we have DI = DB. Hence KD is the perpen-
dicular bisector of BI, and N is the midpoint of BI. By the lemma we
can therefore construct point IB such that IM = MIB using only an un-
marked ruler. SinceM is the midpoint of arc AC while I is the incentre andIM = MIB ; IB is an excentre. Thus IC is an excentre, ICB is the exterior
bisector of \ABC, and L is the midpoint of arc ABC.
Also solved by NIELS BEJLEGAARD, Stavanger, Norway; CHRISTOPHER J.
BRADLEY, Clifton College, Bristol, UK; HANS ENGELHAUPT, Franz{Ludwig{Gym-
nasium, Bamberg, Germany; SHAWN GODIN, St. Joseph Scollard Hall, North Bay,
375
Ontario; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; V �ACLAV KONE �CN �Y,
Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, Univer-
sity of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England;
D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens,
Greece; and the proposer.
2252. [1997: 300] Proposed by K.R.S. Sastry, Dodballapur, India.
Prove that the nine-point circle of a triangle trisects a median if and
only if the side lengths of the triangle are proportional to its median lengths
in some order.
The �rst part of the solution is by Christopher J. Bradley, Clifton Col-lege, Bristol, UK. The rest is a compilation of ideas from other solutionsliberally sprinkled with comments by editor, Cathy Baker.
If l; m and n are the lengths of the medians from A;B and C respec-
tively, then Apollonius' Theorem gives the formulae
4l2 = 2b2 + 2c2 � a2
4m2 = 2c2 + 2a2 � b2
4n2 = 2a2 + 2b2 � c2:
Suppose a � b � c. Then clearly from these formulae, n � m � l. If
the medians are proportional to the sides, these orderings must be preserved
and so a constant k exists such that
2a2 + 2b2 � c2 = ka2
2c2 + 2a2 � b2 = kb2
2b2 + 2c2 � a2 = kc2:
Adding, one �nds k = 3 and each of the three equations reduces to
a2 + c2 = 2b2:
Note that if the triangle has two equal sides, then it must be equilateral.
If, for example, a = b � c, then 2b2 = a2+ c2 and 2a2 = b2+ c2, so adding
gives 2c2 = a2 + b2, and subtraction, a = b = c.
Conversely, if a2 + c2 = 2b2, then substitution in the formulae gives
4l2 = 3c2; 4m2 = 3b2 and 4n2 = 3a2, so l
c= m
b= n
a=
p3
2; that is,
the medians are proportional to the sides. Note that in this case, b must
lie between a and c. If, for example, b < c � a [c � a < b], then
2b2 = a2 + c2 � 2c2 [2b2 = a2 + c2 � 2a2]; a contradiction.
We have shown that a triangle with sides of length a; b; c is self median
if and only if 2b2 = a2 + c2, where b lies between a and c.
Let D and E be the midpoints of AC and BC, respectively; F the foot
of the altitude from A; P the other point where the nine point circle meets
376
BD. Then BD = kBP , for some k > 0. Then using the power of B with
respect to the nine-point circle, we get:
BP � BD = BF � BE:
But BF = c cosB and, by the Cosine Law, cosB = a2+c
2�b22ac
. Since
BD2 = m2 = 2c2+2a
2�b24
and BE = a
2; we have
2c2 + 2a2 � b2
4k=ac(a2 + c2 � b2)
4ac
(k� 1)b2 = (k� 2)(a2 + c2):
If the nine-point circle trisects BD, then k = 3, so 2b2 = a2 + c2,b lies between a and c, and the medians are proportional to the sides.
Conversely, if 2b2 = a2 + c2, then (k� 3)b2 = 0 and k = 3. Hence if
2b2 = a2 + c2, then the nine-point circle trisects the median from B.
Solved by FRANCISCO BELLOTROSADO, I.B.Emilio Ferrari, Valladolid, Spain;
CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; CON AMORE PROBLEM
GROUP, Royal Danish School of Educational Studies, Copenhagen, Denmark;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; GERRY LEVERSHA,
St. Paul's School, London, England; ISTV �AN REIMAN, Budapest, Hungary; TOSHIO
SEIMIYA, Kawasaki, Japan; and the proposer.
2253. [1997: 300] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle and Ib, Ic are the excentres of 4ABC relative to
sides CA, AB respectively.
Suppose that
IbA2 + IbC
2 = BA2 + BC2 and that IcA2 + IcB
2 = CA2 + CB2.
Prove that 4ABC is equilateral.
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.Let us turn the problem around and start from 4ABC (with sides
a; b; c) and the triangle formed by the feet of the altitudes D, E, F . Then
this4ABC plays the role of the given4IaIbIc, while4DEF corresponds
to the original 4ABC (since IaA, etc. are the altitudes in 4IaIbIc). In
this notation our given condition is that BD2 + BF 2 = ED2 + EF 2 and
CD2+CE2 = FD2+FE2, and we are to show that4DEF is equilateral.
SinceAE : AF = c cosA : b cosA = AB : AC, we have4AEF � 4ABCand, hence, EF = a cosA. Similarly FD = b cosB and DE = c cosC.
Hence we get
(a2 + c2) cos2B = a2 cos2A+ c2 cos2C; (1)
(a2 + b2) cos2C = a2 cos2A+ b2 cos2B: (2)
377
Adding b2 cos2B to the �rst and c2 cos2C to the second equation yields
(a2 + b2 + c2) cos2B + (a2 + b2 + c2) cos2C;
and, hence, B = C(since B + C 6= �). Thus (2) becomes
(a2 + b2) cos2B = a2 cos2A+ b2 cos2B;
whence A = B. Therefore 4ABC and also4DEF are equilateral.
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;
FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MAR �IA ASCENSI �ON L �OPEZ
CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHER J. BRADLEY,
Clifton College, Bristol, UK; GORANCONAR, student, Gymnasium Vara�zdin, Vara�zdin,
Croatia; THEODORECHRONIS, student, Aristotle University of Thessaloniki, Greece;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y,
Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, Univer-
sity of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England;
VEDULA N. MURTY, Visakhapatnam, India; ISTV �AN REIMAN, Budapest, Hungary;
MARAGOUDAKIS PAVLOC, Pireas Greece; D.J. SMEENK, Zaltbommel, the Nether-
lands; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. There was one
incorrect solution.
Chronis notes that the result can be generalized by replacing the exponent 2 by
any positive real: 4ABC is equilateral if for some t > 0, we have IbAt + IbC
t =
BAt+BCt and IcAt+ IcB
t = CAt+CBt. Indeed, our featured solution extends
to the proof of Chronis's more general statement.
2254. [1997: 300] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is an isosceles triangle with AB = AC. Let D be the point on
sideAC such that CD = 2AD. Let P be the point on the segment BD such
that \APC = 90�.
Prove that \ABP = \PCB.
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.Let Q be the midpoint of BC and complete the rectangle AQCS: It
follows that B;D and S are collinear, as ASjjBC and AS : BC =AD : DC. Now observe that A;P; Q;C and S are concyclic by Thales'
Theorem [\APC = \AQC = \ASC = 90�, so P;Q and S are all on the
circle with diameter AC] and ABjjSQ. Hence
\ABP = \PSQ = \PCQ = \PCB:
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;
FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; MANSUR
BOASE, student, St. Paul's School, London, England; CHRISTOPHER J. BRADLEY,
Clifton College, Bristol, UK; THEODORE CHRONIS, student, Aristotle University of
Thessaloniki, Greece; CON AMORE PROBLEM GROUP, Royal Danish School of Edu-
cational Studies, Copenhagen, Denmark; RUSSELL EULER and JAWAD SADEK, NW
378
Missouri State University, Maryville, Missouri, USA;WALTHER JANOUS, Ursulinen-
gymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA; DAG
JONSSON, Uppsala, Sweden; GEOFFREY A. KANDALL, Hamden, Connecticut, USA;
V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; MICHAEL
LAMBROU, University of Crete, Crete, Greece (2 solutions); KEE-WAI LAU, Hong
Kong; GERRY LEVERSHA, St. Paul's School, London, England; VICTOR OXMAN,
University of Haifa, Haifa, Israel; PAVLOS MARAGOUDAKIS, Hatzikeriakio, Pireas,
Greece; LUIS A. PONCE, Santos, Brazil; D.J. SMEENK, Zaltbommel, the Netherlands
(2 solutions); PANOS E. TSAOUSSOGLOU, Athens, Greece; ENRIQUE VALERIANO,
National University of Engineering, Lima, Peru; JOHN VLACHAKIS, Athens, Greece;
PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; and the proposer.
2255. [1997: 300] Proposed by Toshio Seimiya, Kawasaki, Japan.
Let P be an arbitrary interior point of an equilateral triangle ABC.
Prove that j\PAB � \PACj � j\PBC � \PCBj.Solution by Theodore Chronis, student, Aristotle University of Thessa-
loniki, Greece.Without loss of generality we suppose that \PAB � \PAC.
A
B CC0 D
P
In the �gure PC = PC0; we deduce that \BPD � \CPD [since
\BPD � \C0PD � \CPD]. Therefore
\PAB + \ABP � \PAC + \ACP [opposite interior angles];
which implies
\PAB + (60� � \PBC) � \PAC + (60� � \PCB);
which in turn implies
\PAB � \PAC � \PBC � \PCB:
379
Editor's comment. It is clear from Chronis's argument that the result
continues to hold for any isosceles triangle ABC with \B = \C. Just re-
place the 60� angle by the base angle.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; CON AMORE PROBLEM GROUP, Royal Danish School of EducationalStudies, Copenhagen, Denmark; JORDI DOU, Barcelona, Spain; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON,Beaverton, Oregon, USA; MICHAEL LAMBROU, University of Crete, Crete,Greece; KEE-WAI LAU, Hong Kong; VICTOR OXMAN, University of Haifa,Haifa, Israel; PANOS E. TSAOUSSOGLOU, Athens, Greece; ENRIQUEVALERIANO, Lima, Peru; and the proposer.
2256. [1997: 300] Proposed by Russell Euler and Jawad Sadek, De-partment of Mathematics and Statistics, Northwest Missouri State Univer-sity, Maryville, Missouri, USA.
If 0 < y < x � 1, prove thatln(x)� ln(y)
x� y> ln
�1
y
�.
I. Solution by David Doster, Choate Rosemary Hall, Wallingford, Con-necticut, USA.
First we show that lnu > 1� 1
ufor u > 1. Let g(u) = lnu� 1 + 1
u.
Then g0(u) = u�1u2
> 0, which implies that g is strictly increasing on [1;1).Hence g(u) > g(1) = 0 for u > 1.
Now, hold y �xed at y = a, where 0 < a < x � 1, and let
f(x) =lnx� lna
x� a+ lna. Then f 0(x) =
1� a
x� ln
�x
a
�(x� a)2
:
Since x
a> 1, we have f 0(x) < 0 by the inequality above. Hence f is
strictly decreasing on (a; 1] and so f(x) > f(1) = �a lna1�a > 0.
II. Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.
For a �xed y 2 (0; 1), consider the function f(x) = x lny+lnx, wherey < x � 1.
Since 0 < y < 1, we have y lny+ lny < lny; that is, f(y)< f(1).
Since f 00(x) = � 1
x2< 0, we have that f is strictly concave on [y;1],
and since y < x � 1, we get f(x) > f(y), which is readily seen to be
equivalent to the proposed inequality.
Also solved by PAUL BRACKEN, CRM, Universit �e de Montr �eal, Montr �eal,
Qu �ebec; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORE
CHRONIS, student, Aristotle University of Thessaloniki, Greece; CON AMORE
PROBLEM GROUP, Royal Danish School of Educational Studies, Copenhagen, Den-
mark; GORAN CONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croatia; LUZ M.
DeALBA, Drake University, Des Moines, IA, USA; HANS ENGELHAUPT, Franz{Lud-
380
wig{Gymnasium, Bamberg, Germany; NOEL EVANS, GERALD ALLEN, CHARLES
R. DIMINNIE, TREY SMITH and ROGER ZARNOWSKI (jointly), Angelo State Uni-
versity, San Angelo, TX, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;
RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, NewMex-
ico Highlands University, Las Vegas, NM, USA; WALTHER JANOUS, Ursulinengym-
nasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,
Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI
LAU, Hong Kong; NICK LORD, Tonbridge School, Tonbridge, Kent, England; PHIL
McCARTNEY, Northern Kentucky University, Highland Heights, KY, USA; VICTOR
OXMAN, University of Haifa, Haifa, Israel; M.PERISASTRY, Maharaja's College,
Vizianagaram, Andhra Pradesh, India; RENATO ALBERTO RODRIGUES, Pra�ca Os-
waldo Cruz, Paraiso, S~ao Paulo, Spain; DIGBY SMITH, Mount Royal College, Calgary,
Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; UNIVERSITY OF ARIZONA
PROBLEM SOLVING LAB, University of Arizona, Tucson, AZ, USA; VREJ ZARIKIAN
and DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA; and
the proposers.
Most of the submitted solutions are similar to either I or II given above. Herzig
proved the stronger result thatlnx� lny
x� y> ln
�1 +
1
y
�for all distinct x, y with
0 < x; y � 1 by using the following inequality known as Bernoulli's Inequality:
(1 + x)� > 1 + �x, where x > �1, x 6= 0 and � > 1 or � < 0.
[Ed. See, for example, page 34 of D.S. Mitrinovi �c et al., Recent Advances in Geomet-
ric Inequalities, Kluwer Academic Publishers, 1989.]
2258. [1997: 301] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.
In a right-angled triangle ABC (with \C = 90�), D lies on the seg-
ment BC so that BD = ACp3. E lies on the segment AC and satis�es
AE = CDp3. Find the angle between AD and BE.
I. Solutionby D. Kipp Johnson,Valley CatholicHigh School,Beaverton,Oregon.
Without loss of generality we put A at (0;1), C at (0;0), and D at
(x; 0). Then BD = ACp3 =
p3, so B is at (x +
p3; 0). Since
AE = CDp3 = x
p3, E is at (0;1 � x
p3). The slope of the line AD
is �1=x and the slope of the line BE is (xp3 � 1)=(x +
p3). If � is the
angle between AD and BE, then:
tan � =
xp3� 1
x+p3
+1
x
1 +
��1
x
� xp3� 1
x+p3
! =
p3(x2 + 1)
x2 + 1=p3;
implying � = 60�.
381
II. Solution by Niels Bejlegaard, Stavanger, Norway.
Let�!CA = a~| and
�!EA = y~|. Then we have that
�!CD =
yp3~{ and
�!CB =
�ap3 +
yp3
�~{. Thus
�!EB =
�!EC +
�!CB =
�!EA+
�!AC +
�!CB
=�!EA��!CA+
�!CB = (y� a)~|+
�ap3 +
yp3
�~{ ;
�!AD =
�!AC +
�!CD = �a~|+ yp
3~{ :
Then the angle � between the vectors�!EB and
�!AD satis�es:
cos � =
�!EB � �!ADj�!EBj � j�!ADj
=
(y� a)(�a) +�ap3 +
yp3
�� yp
3q4a2 + 4
3y2 �
qa2 + 1
3y2
=a2 + 1
3y2
2hqa2 + 1
3y2i2 =
1
2:
Therefore, � = 60�.
III. Solution by Toshio Seimiya, Kawasaki, Japan.Let P be a point such that AP k CB and DP k CA. Since CAPD is
a rectangle, we get PD = AC, AP = CD, and \CDP = 90� = \CAP .
As BD = ACp3 we get BD = PD
p3, so that \DPB = 60�. Since
AE = CDp3 we have AE = AP
p3, so that \APE = 60�. Since
4PAE � 4PDB, we have PA : PD = PE : PB, and
\APD = 60� + \EPD = \EPB. Therefore 4PAD � 4PEB. Thus we
have
\PAD = \PEB: (1)
Let O be the intersection of AD and BE. Then we know from (1) that
\EOA = \EPA = 60�. Therefore the angle between AD and BE is 60�.
A
E
C
D
B
O
60�
60�
P
382
IV. Solution by Luiz A. Ponce, Santos, Brazil.This is a generalization to the case where BD = AC � m and
AE = CD � m for some m > 0. Let AC = b, CD = a, and AD = c.De�ne F to be the point of intersection of AD and BE. Let � = \DFB,
= \CDA, and � = \CAD.
6
?
am
b
6
?
-�
a bm
-�
C
E
A
D B
H
K
F
N�
�
�
Since AE = CD � m and BD = AC � m, we have AE = am and
BD = bm. Since ACD is a right-angled triangle, we have
+ � = 90�: (1)
Through the point B construct a line parallel to AD intersecting the line
AC (extended) at H, and through the point A construct a line parallel to
CB intersecting the line BH at K. It is now clear that \EAK = 90�,\EBK = �, AK = BD = bm, and
BK = AD = c : (2)
Note thatAE
CD=am
a= m =
bm
b=AK
AC:
Since we also have \DCA = \EAK = 90�, it follows that
4AEK � 4CDA (SAS). According to similarity we can write
EK = AD �m = cm ; (3)
and also \KEA = \ADC = = \NEA, where N is the point of inter-
section of AD with EK. Moreover, in triangle ANE we have
\N + \NEA+ \EAN = 180�;
that is, \N + + � = 180�;
or \N = 90�
383
by considering (1). Thus AD ? EK, which implies that EK ? BK since
AD k BK. Consequently we conclude that EKB is a right-angled triangle.
This together with (2) and (3) yield
tan� = tan(\EBK) =EK
BK=cm
c= m:
Therefore, the acute angle between AD and BE is �, such that tan� = m,
where m > 0 is the given ratio.
V. Solutionby PaulYiu, Florida AtlanticUniversity, Boca Raton, Florida,USA.
Let r = BC
AC. It is possible to locate the points D and E if and only
ifp3 < r < 4
3
p3. Then the angle between AD and BE is 60�. This is
a special case of the following situation. Let ACB be a right triangle with
r = BC
AC> 1. For any positive t < r satisfying t2 � rt+ 1 > 0, let D and E
be points on the segments BC and AC respectively, so that BD = t � AC,
and AE = t � CD. Then, the angle between AD and BE is arctan t.
q
q
K
B D C
HA
P 0
E
Complete the right triangle into a rectangle ACBK. For t < r, wechoose pointsD on BC andH onKA such that DBKH is a rectangle with
DB = t �KB. Since t2 � rt + 1 > 0, AC > t(AK � t � BK) = t � AH,
there is a point E on AC such that AE = t � AH. Note that HE and HBare perpendicular to each other, since the right triangles HAE and BKHare similar. Consider the circumcircles of the rectangleDBKH and the right
triangle AEH. The centres of these circles lie onHB andHE respectively.
Let P 0 be the intersection of these circles other than H. Note that HP 0EandHP 0B are both right angles. This means thatB, P 0, andE are collinear.
Note also that
\BP 0D = \BKD = \AHE = \AP 0E:
384
From this we conclude that A, P 0, andD are collinear. This means that P 0 isthe intersection of the segments AD and BE, and the angle between them
is the same as \BKD, which is clearly arctan t.
Remark. If r � 2, the condition t2 � rt + 1 > 0 is always satis�ed for t < r.
For r > 2, this is satis�ed by values of t in either of the ranges
�0;
r�
pr2�4
2
�
and
�r+p
r2�4
2; r
�. For example, for r = 5
2, t must satisfy either 0 < t < 1
2or
2 < t < 52.
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;
SAM BAETHGE, Nordheim, Texas, USA; FRANCISCO BELLOT ROSADO, I.B.
Emilio Ferrari, Valladolid, Spain; PAUL BRACKEN, Universit �e de Montr �eal,
Qu �ebec; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGEL
CABEZ �ON OCHOA, Logro ~no, Spain; THEODORE CHRONIS, student, Aristotle Uni-
versity of Thessaloniki, Greece; CON AMORE PROBLEM GROUP, Royal Dan-
ish School of Educational Studies, Copenhagen, Denmark; HANS ENGELHAUPT,
Franz{Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Per-
chtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y,
Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, Univer-
sity of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St.
Paul's School, London, England; VICTOR OXMAN, University of Haifa, Haifa, Israel;
ISTV �AN REIMAN, Budapest, Hungary; �ANGEL JOVAL ROQUET, LaSeud'Urgell,
Spain; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; D.J. SMEENK, Zaltbommel, the
Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; ENRIQUE VALERIANO,
National University of Engineering, Lima, Per �u; JOHN VLACHAKIS, Athens, Greece;
PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; and the proposer.
Both Hess and Lambrou consider the case when the pointsD andE are external
to the segmentsBC andAC. Neither of them consider all the possible combinations,
however. Perhaps the interested reader would like to pursue this investigation.
Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell
Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer
Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil
Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,
J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia
385
THE OLYMPIAD CORNERNo. 193
R.E. Woodrow
All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.
As a �rst Olympiad to give you puzzling pleasure, we give the 18th
Austrian-Polish Mathematics Competition written in Austria June 28{30,
1995. My thanks go to Bill Sands, University of Calgary, who collected this
contest while assisting at the International Olympiad in Toronto in 1995, as
well as to Walther Janous, Ursulinengymnasium, Innsbruck, Austria.
18thAUSTRIAN-POLISH MATHEMATICS
COMPETITION
Problems of the Individual ContestJune 28-29, 1995 (Time: 4.5 hours)
1. For a given integer n � 3 �nd all solutions (a1; : : : ; an) of the
system of equations
a3 = a2 + a1; a4 = a3 + a2; : : : ; an = an�1 + an�2
a1 = an + an�1; a2 = a1 + an
in real numbers.
2. Let A1; A2; A3; A4 be four distinct points in the plane and let
X = fA1; A2; A3; A4g. Show that there exists a subset Y of the set X
with the following property: there is no disc K such that K \ X = Y .
Note: All points of the circle limiting a disc are considered to belong to the
disc.
3. Let P (x) = x4 + x3 + x2 + x + 1. Show that there exist polyno-
mials Q(y) and R(y) of positive degrees, with integer coe�cients, such that
Q(y) � R(y) = P (5y2) for all y.
4. Determine all polynomials P (x) with real coe�cients, such that
(P (x))2 + (P (1=x))2 = P (x2)P (1=x2) for all x 6= 0:
5. An equilateral triangle ABC is given. Denote the mid-points of
sides BC, CA, AB respectively by A1, B1, C1. Three distinct parallel lines
p; q; r are drawn through A1, B1, C1, respectively. Line p cuts B1C1 at A2;
386
line q cuts C1A1 at B2; line r cuts A1B1 at C2. Prove that the lines AA2,
BB2, CC2 concur at a point D lying on the circumcircle of triangle ABC.
6. The Alpine Club consisting of n members organizes four high-
mountain expeditions for its members. Let E1, E2, E3, E4 be the four teams
participating in these expeditions. How many ways are there to compose
those teams, given the condition thatE1\E2 6= ;,E2\E3 6= ;,E3\E4 6= ;?
Problems of the Team ContestJune 30, 1995 (Time: 4 hours)
7. For every integer c consider the equation 3y4 + 4cy3 + 2xy +
48 = 0, with integer unknowns x and y. Determine all integers c for which
the number of solutions (x; y) in pairs of integers satisfying the additional
conditions (A) and (B) is a maximum:
(A) the number jxj is the square of an integer;
(B) the number y is square-free (that is, there is no prime p with p2
dividing y).
8. Consider the cube with vertices f�1;�1;�1); that is, the set
f(x; y; z) : jxj � 1; jyj � 1; jzj � 1g. Let V1; : : : ; V95 be points of that
cube. Denote by vi the vector from (0; 0; 0) to Vi. Consider the 295 vectors
of the form s1v1 + s2v2 + � � �+ s95v95, where si = 1 or si = �1.(a) Let d = 48. Show that among all such vectors one can �nd a vector
w = (a; b; c) with a2 + b2 + c2 � d.
(b) Find a number d < 48 with the same property.
Note: The smaller d, the better mark will be attracted by the solution.
9. Prove that the following inequality holds for all integers n;m � 1
and all positive real numbers x; y:
(n� 1)(m� 1)(xn+m + yn+m) + (n+m� 1)(xnym + xmyn)
� nm(xn+m�1y + xyn+m�1):
The next contest we give was also collected by Bill Sands while he
was assisting at the IMO in Toronto. These are the problems of the 9th
Iberoamerican Mathematical Olympiad held September 20, 21 in Fortaleza,
Brazil. Students were given 41
2hours each day.
9thIBEROAMERICAN MATHEMATICAL OLYMPIAD
Fortaleza, Brazil, September 20{21, 1994First Day | Time: 4.5 hours
1. (Mexico): A natural number n is called brazilian if there exists an
integer r, with 1 < r < n� 1, such that the representation of the number
387
n in base r has all the digits equal. For example, 62 and 15 are brazilian,
because 62 is written 222 in base 5 and 15 is 33 in base 4. Prove that 1993
is not brazilian, but 1994 is brazilian.
2. (Brazil): Let ABCD be a cyclic quadrilateral. We suppose that
there exists a circle with centre in AB, tangent to the other sides of the
quadrilateral.
(i) Show that AB = AD + BC.
(ii) Calculate, in terms of x = AB and y = CD, the maximal area that
such a quadrilateral can reach.
3. (Brazil): In each cell of an n�n chessboard is a lamp. When a lamp
is touched, the state of this lamp, and also the state of all the lamps in its
row and in its column, is changed (switched from OFF to ON and vice versa).
At the beginning, all the lamps are OFF. Show that it is always possible, with
suitable sequence of touches, to turn ON all the lamps of the chessboard,
and �nd, in terms of n, the minimal number of touches in order that all the
lamps of the chessboard are ON.
Second Day | Time: 4.5 hours
4. (Brazil): The triangle ABC is acute, with circumcircle k. Let P be
an internal point to k. The lines AP , BP , CP meet k again at X, Y , Z.
Determine the point P for which triangle XY Z is equilateral.
5. (Brazil): Let n and r be two positive integers. We wish to con-
struct r subsets of f0; 1; : : : ; n� 1g, called A1; : : : ; Ar, with card(Ai) = k
and such that, for each integer x, 0 � x � n � 1, there exist x1 2 A1;
x2 2 A2; : : : ; xr 2 Ar (an element in each subset), with
x = x1 + x2 + � � �+ xr:
Find, in terms of n and r, the minimal value of k.
6. (Brazil): Show that all natural numbers n � 21000000 can be
obtained beginning at 1 with less than 1100000 sums; that is, there exists
a �nite sequence of natural numbers x0; x1; : : : ; xk, with k < 1100000,
x0 = 1, xk = n, such that for each i = 1; 2; : : : ; k, there exists r, s, with
0 � r < i, 0 � s < i, and xi = xr + xs.
As a �nal problem set to challenge you, we present the problems of the
IX, X and XI Grade of the Georgian Mathematical Olympiad, Final Round
for 1995. It is interesting that 60% of the Grade XI problems come from the
Grade IX paper. My thanks again go to Bill Sands, University of Calgary, for
collecting these problems while he assisted with the IMO in Toronto in 1995.
388
GEORGIAN MATHEMATICAL OLYMPIAD 1995
Final RoundGRADE IX
1. A three-digit number was decreased by the sum of its digits. Then
the result was decreased by the sum of its digits and so on. Show that on
the 100th step of this procedure the result will be zero, whatever the initial
three-digit number is chosen. How many repetitions are enough to get zero?
2. Two circles of the same size are given. Seven arcs, each of them
of 3� measure, are taken on the �rst circle and 10 arcs, each of them of 2�
measure, are taken on the second one. Prove that it is possible to place one
circle on the other so that these arcs do not intersect. Is it or is it not possible
to prove the same if the number of arcs with measure 2� is 11?
3. Prove that if the product of three positive numbers is 1 and their
sum ismore than the sumof their reciprocals, then only one of these numbers
can be more than 1.
4. Prove that in any convex hexagon there exists a diagonal which cuts
from the hexagon a triangle with area less than 16of the area of the hexagon.
5. The set M of integers has the following property: if the numbers
a and b are inM , then a + 2b also belongs to M . It is known that the set
contains positive as well as negative numbers. Prove that if the numbers a,
b and c are inM , then a+ b� c is also inM .
GRADE X
1. (a) Five di�erent numbers are written in one line. Is it always pos-
sible to choose three of them placed in increasing or decreasing order?
(b) Is it always possible to do the same, if we have to choose four num-
bers from nine?
2. (Same as IX.2)
3. Prove that for any natural number n, the average of all its factors
lies between the numberspn and n+1
2.
4. The incircle of a triangle divides one of its medians into three equal
parts. Find the ratio of the sides of the triangle.
5. The function f is given on the segment [0; 1]. It is known that
f(x) � 0 and f(1) = 1. Besides that, for any two numbers x1 and x2, if
x1 � 0, x2 � 0 and x1 + x2 � 1, then f(x1 + x2) � f(x1) + f(x2).
(a) Prove that f(x) � 2x for any x.
(b) Does the inequality f(x) � 1:9x hold for every x?
389
GRADE XI
1. (Same as IX.3)
2. (Same as IX.1)
3. How many solutions has the equation x = 1995 sinx+ 199?
4. (Same as IX.4)
5. A natural number is written in each square of anm�n rectangular
table. By onemove, it is allowed to double all numbers of any row or subtract
1 from all numbers of any column. Prove that by repeating these moves
several times, all numbers in the table become zeros.
Next a bit of housekeeping. After the columns were set, and before they
appeared in print form, I received solutions from PavlosMaragoudakis to the
six problems of the Swedish contest for which we published the solutions last
issue [1997: 196; 1998: 328{329]. He also submitted solutions to problems
1, 2, 3 and 5 of the Dutch Mathematical Olympiad, Second Round, 1993
[1997: 197; 1998: 329{332]. Last issue we gave solutions to all but the last
problem. It was rather unfortunate timing in terms of acknowledging his
contribution, but we are able to close out the �le by having a complete set
of solutions from the readers.
5. P1; P2; : : : ; P11 are eleven distinct points on a line. PiPj � 1 for
every pair Pi, Pj. Prove that the sum of all (55) distances PiPj , 1 � i <
j � 11 is smaller than 30.
Solution by Pavlos Maragoudakis, Pireas, Greece.
Without loss of generality we suppose that P1; P2; : : : ; P11 are adja-
cent.
P1 P2 P11qq q
Now if 1 � i < j � 11, then PiPj = PjP1 � PiP1. So
X1�i<j�11
PiPj =X
1�i<j�11(PjP1 � PiP1)
= 10P11P1 + 9P10P1 � P10P1 + 8P9P1 � 2P9P1
+7P8P1 � 3P8P1 + 6P7P1 � 4P7P1 + 5P6P1
�5P6P1 + 4P5P1 � 6P5P1 + 3P4P1 � 7P4P1
+2P3P1 � 8P3P1 + P2P1 � 9P2P1
390
= 10P11P1 + 8P10P1 + 6P9P1 + 4P8P1 + 2P7P1
�2P5P1 � 4P4P1 � 6P3P1 � 8P2P1
= 10P11P1 + 8(P10P1 � P2P1) + 6(P9P1 � P3P1)
+4(P8P1 � P4P1) + 2(P7P1 � P5P1)
= 10P11P1 + 8P10P2 + 6P9P3 + 4P8P4 + 2P7P5
< 10 � 1 + 8 � 1 + 6 � 1 + 4 � 1 + 2 � 1 = 30
Also setting the record straight, I found amongst the solutions for an-
other contest, the solution by Miguel Amengual Covas, Cala Figuera, Mal-
lorca, Spain to problem 2 of the Dutch Mathematical Olympiad, Second
Round, for which we published a solution last issue [1997: 197, 1998: 330{
331]. My apologies.
While we do not normally give solutions to problems of the USAMO,
I am giving two comments/solutions from our readers to problems of the
USAMO 1997 [1997: 261, 262].
2. Let ABC be a triangle, and draw isosceles triangles BCD, CAE,
ABF externally toABC, withBC;CA;AB as their respective bases. Prove
that the lines through A;B;C perpendicular to the lines !EF;
!FD;
!DE,
respectively, are concurrent.
Comment by Mansur Boase, student, St. Paul's School, London, Eng-land.
The result is immediate from Steiner's Theorem:
If the perpendiculars from the verticesA,B,C of a triangleABC to the
sides B1C1, C1A1, and A1B1, respectively, of a second triangle A1; B1; C1
are concurrent, then the perpendiculars from the vertices A1; B1; C1 of the
triangle A1B1C1 to the sides BC, CA, AB are also concurrent.
5. Prove that for positive real numbers a, b, c,
(a3 + b3 + abc)�1 + (b3 + c3 + abc)�1 + (c3 + a3 + abc)�1 � (abc)�1:
Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land and by Murray S. Klamkin, University of Alberta, Edmonton, Alberta.We give Klamkin's presentation.
Since the inequality is homogeneous, we can assume abc = 1. Then if
we let x = a3, y = b3, z = c3, the inequality becomes
1
1 + x+ y+
1
1 + y+ z+
1
1 + z + x� 1 (1)
where xyz = 1 and x, y, z are positive. On expanding, (1) is equivalent to
(x+ y + z)(xy+ yz+ zx� 2) � 3 :
391
This follows from the known elementary inequalities
x+ y+ z
3��yz + zx+ xy
3
�1=2� (xyz)1=3 :
There is equality if and only if x = y = z = 1.
Comment: The inequality in the form (1) was also given in the Spring
1997, Senior A-Level Tournament Of The Towns competition. A generaliza-
tion to
1
1 + S � x1+
1
1 + S � x2+ � � �+
1
1 + S � xn� 1 ;
where S = x1+x2+ � � �+xn, x1x2 : : : xn = 1, and xi > 0 is due to Dragos
Hrimiuc, University of Alberta, and will probably appear as a problem in
Math. Magazine.
Next we give two solutions by our readers to two problems of the 3rd
Ukrainian Mathematical Olympiad, March 26{27, 1994 given in [1997: 262].
2. (9{10) A convex polygon and point O inside it are given. Prove that
for any n > 1 there exist points A1; A2; : : : ; An on the sides of the polygon
such that��!OA1 +
��!OA2 + : : :+
��!OAn =
�!0 .
Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta.
It follows by continuity that there always exists a chord A1OA01 such
that A1O = A01O and hence��!OA1 +
��!OA01 =
�!0 . Similarly, there exists
a chord A2A03 which is bisected by the midpoint O1 of OA01. It follows by
the parallelogram law that��!OA2 +
��!OA03 =
��!OA01 and hence
��!OA1 +
��!OA2 +��!
OA03 =�!0 . Again similarly there exists a chord A3A
04 which is bisected by
the midpoint of OA03 so that��!OA1 +
��!OA2 +
��!OA3 +
��!OA04 =
�!0 , and so on
for any number of vectors n > 1.
3. (10) A sequence of natural numbers ak, k � 1, such that for each k,
ak < ak+1 < ak + 1993 is given. Let all prime divisors of ak be written for
every k. Prove that we receive an in�nite number of di�erent prime numbers.
Solution by Pavlos Maragoudakis, Pireas, Greece.We suppose that there is a sequence of natural numbers such that
ak < ak+1 < ak + 1993, k � 1, and the set of all prime divisors of all
ak is �nite. Let p1; p2; : : : ; pr list all the prime divisors of all ak. Now every
ak has the form pa11 : : : parr , ai = 0; 1; 2; : : : , i = 1; : : : ; r.
Let S = fpa11 pa22 : : : parr j ai = 0; 1; 2; : : : ; i = 1; 2; : : : ; rg:De�ne (xn) with x1 < x2 < : : : such that S = fxn = n 2 N�g. We
have that ak < ak+1 and ak 2 S, k � 1. Thus (ak) is a subsequence of
392
(xk). But ak < ak�1+1993 < ak�2+2 �1993 < � � � < a1+(k�1)1993 <
k(a1 + 1993); k � 1.
Hence ak < k(a1 + 1993), k � 1. Therefore
1Xn=1
1
an>
1Xn=1
1
n(a1 + 1993)=
1
a1 + 1993
1Xn=1
1
n= +1
and
1Xn=1
1
an�
1Xn=1
1
xn=
Xa1;::: ;ar�0
1
p�11 pa22 : : : pa4r
=
1Xa1=0
1
p�11
! 1Xa2=0
1
pa22
!: : :
1X�r=0
1
parr
!
=p1
p1 � 1�
p2
p2 � 1� � �
p4
p4 � 1< +1 ;
which is a contradiction.
We now turn our attention to the solutions by readers to problems of
the Mock Test of the Hong Kong Committee for the IMO 1994 [1997: 322{
323].
INTERNATIONALMATHEMATICAL OLYMPIAD1994
Hong Kong Committee | Mock Test, Part ITime: 4.5 hours
1. In a triangle 4ABC, \C = 2\B. P is a point in the interior of
4ABC satisfying that AP = AC and PB = PC. Show that AP trisects
the angle \A.
Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;
and by D.J. Smeenk, Zaltbommel, the Netherlands. We give the solution byAmengual Covas.
Let \PAC and \BAP be 2� and � respectively. Then, since
\C = 2\B, we deduce from A+ B +C = 180� that
2�+ � + 3B = 180�: (1)
A
B C
P
� � �
393
The angles at the base of the isosceles triangle PAC are each 90�� �. Also
4BPC is isosceles, having base angles
C � (90� � �) = 2B + �� 90�;
and so
\BPA = 180� � (\PBA+ \BAP )
= 180� � [B � (2B + �� 90�) + 180� � 2�� 3B]
= 4B + 3�� 90�:
As usual, let a, b and c denote the lengths of the sidesBC,AC andAB.
By the Law of Cosines, applied to 4BPA, where PA = b and
PB = PC = 2b sin�,
c2 = b2 + (2b sin�)2 � 2 � b � 2b sin� � cos(4B + 3�� 90�) ;
so that
c2 = b2[1 + 4 sin2 �� 4 sin� sin(4B + 3�)] : (2)
We now use the fact that \C = 2\B is equivalent to the condition
c2 = b(b + a), which has appeared before in CRUX [1976: 74], [1984:
278] and [1996: 265{267]. Since a = 2 � PC � cos(2B + � � 90�) =
4b sin� sin(2B + �), we have
c2 = b2[1 + 4 sin� sin(2B + �)] : (3)
Therefore, from (2) and (3), we get
b2[1 + 4 sin2 �� 4 sin� sin(4B + 3�)] = b2[1 + 4 sin� sin(2B + �)] ;
which simpli�es to
sin�� sin(4B + 3�) = sin(2B + �) :
Since sin�� sin(4B+3�) = �2 cos(2B+ 2�) sin(2B+ �), this equation
may be rewritten as
sin(2B + �) � [1 + 2 cos(2B + 2�)] = 0:
Since, from (1), 2B+� < 180�, we must have 1+2 cos(2B+2�) = 0,
giving cos(2B + 2�) = �1=2 ; that is,
2B + 2� = 120� (4)
since, again from (1), 2B + 2� < 180�.Finally, we may eliminate B between (1) and (4) to obtain � = �. The
result follows.
394
Mock Test, Part II
Time: 4.5 hours
1. Suppose that yz+ zx+ xy = 1 and x, y, and z � 0. Prove that
x(1� y2)(1� z2) + y(1� z2)(1� x2) + z(1� x2)(1� y2) �4p3
9:
Solutions by Murray S. Klamkin, University of Alberta, Edmonton,Alberta; and Pavlos Maragoudakis, Pireas, Greece. We give the solutionby Klamkin.
We �rst convert the inequality to the following equivalent homoge-
neous one:
x(T2 � y2)(T2 � z2) + y(T2� z2)(T2 � x2) + z(T2 � x2)(T2 � y2)
� (4p3=9)(T2)
5=2
where T2 = yz+zx+xy, and for subsequent use T1 = x+y+z, T3 = xyz.
Expanding out, we get
T1T22 � T2
Xx(y2 + z2) + T2T3 � (4
p3=9)(T2)
5=2 ;
or
T1T22 � T2(T1T2 � 3T3) + T2T3 = 4T2T3 � (4
p3=9)(T2)
5=2 :
Squaring, we get one of the known Maclaurin inequalities for symmetric
functions:3pT3 � 2
pT2=3 :
There is equality if and only if x = y = z.
To �nish this number of the Corner we give two solutions to problems
of the 45th Mathematical Olympiad in Poland, Final Round [1997: 323{324].
1. Determine all triples of positive rational numbers (x; y; z) such that
x+ y + z, x�1 + y�1 + z�1 and xyz are integers.
Solution by Murray S. Klamkin, University of Alberta, Edmonton,Alberta.
Let x+ y+ z = n1,1x+ 1
y+ 1
z= n2, and xyz = n3, where n1, n2, n3
are integers. Then yz+ zx+ xy = n2n3 and x, y, z are roots of the cubic
t3 � n1t2 + n2n3t� n3 = 0 :
395
As known, the only rational roots of the latter are factors of n3, and
consequently x, y, z are integers.
The only triples of integers (x; y; z), aside from permutations, which
satisfy 1x+ 1
y+ 1
z= n2 are
(1; 1; 1); (1; 2; 2); (2;3; 6); (2; 4; 4); and (3; 3; 3):
5. Let A1; A2; : : : ; A8 be the vertices of a parallelepiped and let O be
its centre. Show that
4(OA21 +OA2
2 + � � �+ OA28) � (OA1 +OA2 + � � �+ OA8)
2 :
Solution by Murray S. Klamkin, University of Alberta, Edmonton,Alberta.
Let one of the vertices be the origin and let the vectors B + C,
C+A, A+B denote the three coterminal edges emanating from this origin.
Then the vectors to the remaining four vertices are S + A, S + B, S + C,
and 2S where S = A + B + C and which is also the vector to the centre.
The inequality now becomes
2(S2 +A2 + B2 + C2) � (jSj+ jAj+ jBj+ jCj)2 ;
or
S2+A2+B2+C2 � 2jSjfjAj+jBj+jCjg+2fjBj jCj+jCj jAj+jAj jBjg:
Since
S2 = A2 +B2 +C2 + 2B � C + 2C � A+ 2A � B;the inequality now becomes
S2�B�C�C�A�A�B � jSjfjAj+jBj+jCjg+fjBj jCj+jCj jAj+jAj jBjg:
Clearly,
S2 � jSjfjAj + jBj+ jCjgand
�B � C �C � A� A � B � jBj jCj jAj+ jAj jBj :There is equality if and only if the parellelepiped is degenerate,
for example, B = C = O.
That completes this number of the Olympiad Corner. Send me your
nice solutions and Olympiad contests.
396
BOOK REVIEWS
Edited by ANDY LIU
Dissections : Plane & Fancy
by Greg N. Frederickson,
published by Cambridge University Press, 1997,
ISBN# 0-521-57197-9, hardcover, 310+ pages, $34.95.
Reviewed by Andy Liu, University of Alberta.
This is a much awaited sequel to Harry Lindgren's 1964 classic work,
Geometric Dissections, which the author (G.N.F.) revised and augmented in
the 1972 Dover edition. Actually, the current volume is much more than just
a sequel. It is the most comprehensive treatise on the subject of geometric
dissections. It may be enjoyed on at least three levels.
First and foremost, this book is a collection of interesting dissection
puzzles, old and new. Only some background in high school geometry is
needed to fully enjoy these problems. Can you cut an octagon into �ve piecesand rearrange them into a square? How about turning a star into a pentagon?The solutions, which are both appealing but for somewhat opposing reasons,
are shown below.
Octagon to square
Star to pentagon
This book is also an instructive manual on the art and science of geo-
metric dissections. While one may admire the ingenuity which produced
the spectacular solutions, the author probes into the underlying fabric which
might have led to such incisive insight. Many techniques are discussed, too
many to enumerate here. A favourite is that of tessellation. Below are two
tilings which might have suggested the dissections above.
397
Tilings for octagon, square
Cut the star for tiling
Tilings for star, pentagon
Finally, this book is an important historical document, detailing the
inter-cultural development of the subject. Travel from the palace school oftenth-century Baghdad to the mathematical puzzle columns in turn-of-the-century newspapers, from the 1900 Paris Congress of Mathematicians to thenight sky of Canberra. Readers puzzled by this quote need look no further
than the illustrious names of Ab �u'l Waf�a, Henry Dudeney/Sam Loyd, David
Hilbert/Max Dehn and Harry Lindgren. Biographical sketches of Waf�a, Du-
deney, Loyd and Lindgren are provided, along with those of over forty other
people who have made signi�cant contributions to geometric dissections.
The writing style is very engaging, and the book is good reading even if one
skips over some of the more complicated technical details.
In conclusion, the reviewer echoes Martin Gardner that this book will
be a classic. It comes with the highest recommendation.
398
THE SKOLIAD CORNERNo. 33
R.E. Woodrow
As a contest this issue we give the Senior High School Mathematics
Contest, Preliminary Round 1998 of the British ColumbiaColleges. My thanks
go to the organizer, Jim Totten, The University College of the Cariboo, for
forwarding me the contest materials. Time allowed is 45 minutes!
BRITISH COLUMBIA COLLEGES
Senior High School Mathematics Contest
Preliminary Round 1998
Time: 45 minutes
1. The integer 1998 = (n� 1)nn(10n+ c) where n and c are positive
integers. It follows that c equals:
(a) 2 (b) 5 (c) 6 (d) 7 (e) 8
2. The value of the sum log 12+ log 2
3+ log 3
4+ � � �+ log 9
10is:
(a) �1 (b) 0 (c) 1 (d) 2 (e) 3
3. Four basketballs are placed on the gym oor in the form of a square
with each basketball touching two others. A �fth basketball is placed on top
of the other four so that it touches all four of the other balls, as shown. If the
diameter of a basketball is 25 cm, the height, in centimetres, of the centre of
the �fth basketball above the gym oor is:
(a) 25p2 (b) 25
2
p2 (c) 20 (d) 25
2(1 +
p2) (e) 25(1 +
p2)
4. Last summer, I planted two trees in my yard. The �rst tree came in
a fairly small pot, and the hole that I dug to plant it in �lled one wheelbarrow
load of dirt. The second tree came in a pot, the same shape as that of the
�rst tree, that was one-and-a-third times as deep as the �rst pot and one-
and-a-half times as big around. Let us make the following assumptions:
i) The hole for the second tree was the same shape as for the �rst tree.
399
ii) The ratios of the dimensions of the second hole to those of the �rst
hole are the same as the ratios of the dimensions of the pots.
Based on these assumptions, the number of wheelbarrows of dirt that I �lled
when I dug the hole for the second tree was:
(a) 2 (b) 2:5 (c) 3 (d) 3:5 (e) none of these
5. You have an unlimited supply of 5-gram and 8-gram weights that
may be used in a pan balance. If you use only these weights and place them
only in one pan, the largest number of grams that you cannot weigh is:
(a) 22 (b) 27 (c) 36 (d) 41 (e) there is no largest number of grams
6. If all the whole numbers from 1 to 1;000;000 are printed, the num-
ber of times that the digit 5 appears is:
(a) 100;000 (b) 500;000 (c) 600;000 (d) 1;000;000 (e) 2;000;000
7. The perimeter of a rectangle is x centimetres. If the ratio of two
adjacent sides is a : b, with a > b, then the length of the shorter side, in
centimetres, is:
(a) bxa+b
(b) x2� b (c) 2bx
a+b(d) ax
2(a+b)(e) bx
2(a+b)
8. The sum of the positive solutions to the equation xxpx = (x
px)x
is:
(a) 1 (b) 112
(c) 214
(d) 212
(e) 314
9. Two circles, each with a radius of one unit, touch as shown. AB
and CD are tangent to each circle. The area, in square units, of the shaded
region is:
A B
C D
(a) � (b) �4
(c) 2� �2
(d) 4� � (e) none of these
10. A parabola with a vertical axis of symmetry has its vertex at (0;8)
and an x{intercept of 2. If the parabola goes through (1; a), then a is:
(a) 5 (b) 5:5 (c) 6 (d) 6:5 (e) 7
11. A �ve litre container is �lled with pure orange juice. Two litres of
juice are removed and the container is �lled up with pure water and mixed
400
thoroughly. Then two litres of the mixture are removed and again the con-
tainer is �lled up with pure water. The percentage of the �nal mixture that
is orange juice is:
(a) 27 (b) 25 (c) 30 (d) 36 (e) 24
12. The lengths of the sides of a triangle are b+ 1, 7� b and 4b� 2.
The number of values of b for which the triangle is isosceles is:
(a) 0 (b) 1 (c) 2 (d) 3 (e) none of these
13. The number of times in one day when the hands of a clock form a
right angle is:
(a) 46 (b) 22 (c) 24 (d) 44 (e) 48
14. In my town some of the animals are really strange. Ten percent of
the dogs think they are cats and ten percent of the cats think they are dogs.
All the other animals are perfectly normal. One day, I tested all the cats and
dogs in the town and found that 20% of them thought that they were cats.
The percentage of the dogs and cats in the town that really are cats is:
(a) 12:5 (b) 18 (c) 20 (d) 22 (e) 22:5
15. A short hallway in a junior high school contains a bank of lockers
numbered one to ten. On the last day of school the lockers are emptied and
the doors are left open. The next day, a malicious math student walks down
the hallway and closes the door of every locker that has an even number.
The following day, the same student again walks down the hallway and, for
every locker whose number is a multiple of three, closes the door if it is open
and opens it if it is closed. On the next day, the student does the same thing
with every locker whose number is divisible by four. If the student continues
this procedure for a total of nine days, the number of lockers that are closed
after the ninth day is:
(a) 4 (b) 5 (c) 6 (d) 7 (e) 8
Last issue we gave the Junior High School Mathematics Contest, Pre-
liminary Round 1998 of the British Columbia Colleges. Here are the \o�cial
solutions", which come our way from the organizer, Jim Totten, The Univer-
sity College of the Cariboo.
BRITISH COLUMBIA COLLEGES
Junior High School Mathematics Contest
Preliminary Round 1998
Time: 45 minutes
1. A number is prime if it is greater than one and divisible only by one
and itself. The sum of the prime divisors of 1998 is: (c)
401
Solution. We can factor the number 1998 as follows: 1998 = 2 �999 = 2� 32 � 111 = 2� 33 � 37. Hence, the sum of its prime divisors is
2 + 3 + 37 = 42.
2. Successive discounts of 10% and 20% are equivalent to a single
discount of: (c)
Solution. If P denotes the initial price then the new price after deduct-
ing the two consecutive discounts is P (1�0:1)(1�0:2) = 0:72P . This gives
the total discount of (1� 0:72)� 100% = 28%.
3. Suppose that kA = A2 and A 2 B = A� 2B. Then the value of
k7 2k3 is: (e)
Solution. According to our de�nitions, k7 2 k3 = 72 2 32 = 492 9 =
49� 2� 9 = 31.
4. The expression that is not equal to the value of the four other ex-
pressions listed is: (d)
Solution. The values of the �ve expressions are:
(a) 1p9 + 9� 8 = 2,
(b) (1 + 9)� (�p9 + 8) = 2,
(c) �1� 9 +p9 + 8 = 2,
(d) (1�p9)� (9� 8) = �2,
(e) 19� 9� 8 = 2.
Thus,the expression (d) has a di�erent value than the other four expres-
sions.
5. The sum of all of the digits of the number 1075 � 75 is: (c)
Solution. Consider the procedure for subtracting 75 from 1075 \by
hand":100 : : : 000
�7599 : : : 925
Hence, the decimal digits of 1075 � 75 are: 5, 2 and seventy-three copies
of 9. The sum of the digits is 5 + 2 + 73� 9 = 664.
6. A circle is divided into three equal parts and one part is shaded as in
the accompanying diagram. The ratio of the perimeter of the shaded region,
including the two radii, to the circumference of the circle is: (d)
402
Solution. The ratio is given by
2�r3
+ r + r
2�r=
2�r+6r3
2�r=
2�r + 6r
6�r=� + 3
3�:
7. The value of1
2� 1
2� 1
2�12
is: (b)
Solution. We can simplify this compound fraction by working succes-
sively from the bottom to the top of the expression:
1
2� 1
2� 1
2� 12
=1
2� 1
2� 1
( 32)
=1
2� 1
2� 23
=1
2� 1
(43)
=1
2� 34
=154
=4
5:
8. If each small square in the accompanying grid is one square cen-
timetre, then the area in square centimetres of the polygon ABCDE is: (a)
Solution. We can �nd the area by decomposing the polygon ABCDE
into simpler �gures, for example, into three triangles: ABE, BCE, and
CDE.
A E
DC
B
If we chooseAE,CE, andCD as bases of the triangles then the lengths
of the corresponding perpendicular heights are 5, 3, and 8 cm. Hence, the
area of the polygon is 1
2� 4� 5 + 1
2� 8� 3 + 1
2� 4� 8 = 38.
9. A point P is inside a square ABCD whose side length is 16. P is
equidistant from two adjacent vertices, A and B, and the side CD opposite
these vertices. The distance PA equals: (e)
403
Solution. The situation is illustrated by the following diagram, where
x denotes the distance PA.
A B
CD
x
x x
E
P16
The Pythagorean Theorem applied to trianglePEB gives (16�x)2+82 = x2,
so that 162 � 32x+ x2 + 82 = x2, and x = 162+82
32= 10.
10. A group of 20 students has an average mass of 86 kg per person.
It is known that 9 people from this group have an average mass of 75 kg
per person. The average mass in kilograms per person of the remaining 11
people is: (b)
Solution. If m1; m2; : : : ;m20 denote the masses of the students thenm1+m2+���+m20
20= 86. Hence, m1+m2+ � � �+m20 = 20� 86 = 1720. We
can assume, without loss of generality, that the average mass of the �rst nine
students is 75; that is, m1+m2+���+m9
9= 75. Hence, m1+m2+ � � �+m9 =
9 � 75 = 675. The total mass of the remaining 11 people is 1720� 675 =
1045. This gives the average of 104511
= 95.
11. In the following display each letter represents a digit:
3 B C D E 8 G H I
The sum of any three successive digits is 18. The value of H is: (a)
Solution. We have 3 + B + C = 18. Consequently, B + C = 15.
By subtracting this equation from B + C + D = 18 we get D = 3. Now,
D+E+8 = 18 givesE = 10�D = 10�3 = 7. Finally,E+8+G = 18 gives
G = 10�E = 10� 7 = 3, and 8+G+H = 18 givesH = 18� 8� 3 = 7.
12. In the accompanying diagram \ADE = 140�. The sides are
congruent as indicated. The measure of \EAD is: (e)
A E
D
C
B
404
Solution. If \EAD = �, then also \ACB = �, since triangle ABC
is isosceles. Hence, \CBD = 2� as an external angle of triangle ABC.
Consequently, \ADC = 2�, since triangle BCD is isosceles. Further,
\ECD = \ADC + \CAD as an external angle of triangle ADC. Hence,
\ECD = 2� + � = 3�. Now, \AED = \ECD = 3�, because tri-
angle CDE is isosceles. This implies that \CDE = 180� � 6�. Finally,
\ADE = \ADC + \CDE = 2� + 180� � 6� = 180� � 4�. Thus,
180� � 4� = 140�. This yields � = 10�.
13. The area (in square units) of the triangle bounded by the x{axis
and the lines with equations y = 2x+ 4 and y = �23x+ 4 is: (e)
Solution. Two vertices of the triangle lie on the x{axis, so they are
the x{intercepts of the lines. The x{intercept of the �rst line, determined
by the equation 2x + 4 = 0, is �2. Similarly, the second x{intercept, de-
termined by �23x2 + 4 = 0, is 6. Consequently, the length of the base of
the triangle is 6 � (�2) = 8. Since both lines have the same y{intercept
4, they intersect each other and the y{axis at level 4 and, consequently, the
length of the height of the triangle is 4. Therefore, the area of the triangle is
A = 1
2(4� 8) = 16.
14. Two diagonals of a regular octagon are shown in the accompanying
diagram. The total number of diagonals possible in a regular octagon is: (d)
Solution. Let di, for i = 1; 2; : : : ; 8, denote the number of diagonals
connected to the ith vertex. Then d1 = d2 = : : : = d8 = 5, since each
vertex of the octagon is connected to �ve diagonals. On the other hand, each
diagonal joins two vertices. Therefore, in the sum d1 + d2 + � � � + d8 =
8� 5 = 40, each diagonal is counted twice. Hence, the number of diagonals
in the octagon is 20.
15. A local baseball league is running a contest to raise money to send
a team to the provincial championship. To win the contest it is necessary
to determine the number of baseballs stacked in the form of a rectangular
pyramid. The �fth and sixth levels from the base of the stack of baseballs are
shown. If the stack contains a total of seven levels, the number of baseballs
in the stack is: (d)
405
Solution. The �fth level has 3 � 4 = 12 balls, the sixth 2 � 3 = 6
balls, and the seventh 1� 2 = 2 balls. We notice that the number of balls
in both sides of the rectangle they form increases by one each time we move
one level down. Thus, the total number of balls is 1� 2 + 2� 3 + 3� 4 +
4� 5 + 5� 6 + 6� 7 + 7� 8 = 168.
That completes the Skoliad Corner for this issue. We need suitable
contests and solutions. I welcome any comments, criticisms, or suggestions
for the future direction of this feature.
Advance Announcement
The 1999 Summer Meeting of the Canadian Mathematical Society will
take place at Memorial University in St. John's, Newfoundland, from Satur-
day, 29 May 1999 to Tuesday, 1 June 1999.
The Special Session on Mathematics Education will feature the topic
What Mathematics Competitions do for Mathematics.
The invited speakers are
Ed Barbeau (University of Toronto),
Ron Dunkley (University of Waterloo),
Tony Gardiner (University of Birmingham, UK),
Rita Janes (Newfoundland and Labrador Senior Mathematics League), and
Shannon Sullivan (student, Memorial University).
Requests for further information, or to speak in this session, as well as sug-
gestions for further speakers, should be sent to the session organizers:
Bruce Shawyer and Ed Williams
CMS Summer 1999 Meeting, Education Session
Department of Mathematics and Statistics, Memorial University
St. John's, Newfoundland, Canada A1C 5S7
406
MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by
High School and University Students. It continues, with the same emphasis,
as an integral part of Crux Mathematicorum with Mathematical Mayhem.
All material intended for inclusion in this section should be sent to the
Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,
PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic
address is still
The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).
The rest of the sta� consists of Adrian Chan (Upper Canada College), Jimmy
Chui (Earl Haig Secondary School), Richard Hoshino (University ofWaterloo),
David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).
Shreds and Slices
An Algebraic Relation with a Geometric Twist
Cyrus Hsia
Consider the following algebraic relationship between the positive real num-
bers a, b, and c:1
c=
1
a+
1
b:
If we consider line segments with lengths of a, b, and c, then they are related
to each other as shown in the following diagram.
��
�
a b
c
The diagram shows the three line segments parallel to each other and
emanating from a common line. This �gure and relationship between the
line segments appear a lot. The reader is encouraged to prove this.
407
As a corollary of this fact, here is another geometric example. Let P ,
A, B, and C be points in the plane such that \APC = \CPB = 60� andA, C, and B are collinear. Show that 1=PC = 1=PA+ 1=PB.
A
P
BC
60�60�
Another algebraic relation between three positive numbers with an interest-
ing geometric interpretation is the following:
1pc
=1pa+
1pb:
The reader is encouraged to �nd a geometric interpretation for the
above relation before looking at the diagram below. Use a, b, and c as the
length of three line segments, and determine a geometric �gure that relates
the three.
q
a
qb
cq
It turns out that if a, b, and c are considered to be the lengths of the
radii of three circles, then the circles may all be tangent to a common line
and to each other as shown. Again readers are encouraged to prove this
themselves.
Now what about a generalization? Consider the following algebraic
relation between positive reals a, b, c, and a real number x:
cx = ax + bx:
The �rst case then corresponds to the value x = �1 and the second case to
x = �12.
408
The case x = 1 is trivial, as we could interpret it geometrically as a line
segment of length c is made up of the sum of its parts of lengths a and b.
c
a
b
If we wanted to get fancy, we could give the following geometric exam-
ple instead. Consider an equilateral triangle ABC inscribed in a circle, as
shown. P is a point on arc BC. Prove that PA = PB + PC.
A
B
P
C
The reader is probably already familiar with the famous case x = 2
known as the Pythagorean Theorem: A triangle with sides a, b, and c is a
right-angled triangle if and only if the lengths satisfy a2 + b2 = c2.
a
bc
Of course, no discussion about algebraic relations in the above form is
complete without mentioning the notorious Fermat's Last Theorem and the
recent announcement that it has �nally been laid to rest. If x is an integer
with x > 2, then the claim is that no solution in the natural numbers exists
for a, b, and c. However, in our general case, the values are real, so we are
not limited by the above result to �nding wild and wacky geometric or other
interpretations for it.
409
If the reader is curious, as are we, then try the following exercises to
�nd geometric interpretations for special cases of the above relation. The
exercises are explorational and may not have nice solutions, if any. Readers
are welcome to submit any interesting results they �nd.
Exercises
1. Let a, b, and c be the lengths of three line segments. Determine how
these three line segments are related geometrically if they satisfy the
relation
(a) c3 = a3 + b3,
(b)pc =
pa+
pb,
(c) 1c2
= 1a2
+ 1b2.
2. It is clear that algebraically, all the relations are similar. However,
the geometric interpretations do not appear to be related. Is there a
general geometric description where each of the above geometric �gures
is a special case?
3. The algebraic relation clearly does not work for the case x = 0. Is
there a way to de�ne the relation so that it would be consistent with
everything else mentioned so far?
We extend congratulations to Ravi Vakil and Alice Staveley, who were
married at St. John's, NF on Monday, 12 October 1998.
410
Mayhem Problems
The Mayhem Problems editors are:
Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.
Note that all correspondence should be sent to the appropriate editor |
see the relevant section. In this issue, you will �nd only problems | the
next issue will feature only solutions.
We warmly welcome proposals for problems and solutions. We request
that solutions from this issue be submitted by 1 September 1999, for publi-
cation in issue 8 of 1999.
High School Problems
Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,
Canada. L3S 3E8 <[email protected]>
H245. Determine how many distinct integers there are in the set��12
1998
�;
�22
1998
�;
�32
1998
�; : : : ;
�19982
1998
��:
H246. Let S(n) denote the sum of the �rst n positive integers. We
say that an integer n is fantastic if both n and S(n) are perfect squares. For
example, 49 is fantastic, because 49 = 72 and S(49) = 1+2+3+ � � �+49 =
1225 = 352 are both perfect squares. Find another integer n > 49 that is
fantastic.
H247. Say that the integers a, b, c, d, p, and r form a cyclic set
(a; b; c; d; p; r) if there exists a cyclic quadrilateral with circumradius r, sides
a, b, c, and d, and diagonals p and 2r.
(a) Show that if r < 25, no cyclic set exists.
(b) Find a cyclic set (a; b; c; d; p; r) for r = 25.
H248. Consider a tetrahedral die that has the four integers 1, 2, 3,
and 4 written on its faces. Roll the die 2000 times. For each i, 1 � i � 4,
let f(i) represent the number of times that i turned up. (So, f(1) + f(2)+
f(3) + f(4) = 2000.) Also, let S denote the total sum of the 2000 rolls.
If S4 = 6144 � f(1)f(2)f(3)f(4), determine the values of f(1), f(2),
f(3), and f(4).
411
Advanced Problems
Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.
M1G 1C3 <[email protected]>
A221. Construct, using straightedge and compass only, the common
tangents of two non-intersecting circles.
A222. Does there exist a set of n consecutive positive integers such
that for every positive integer k < n, it is possible to pick k of these numbers
whose mean is still in the set?
A223. Proposed by Mohammed Aassila, Universit �e Louis Pasteur,Strasbourg, France.
Suppose p is a prime with p � 3 (mod 4). Show that for any set of
p � 1 consecutive integers, the set cannot be divided into two subsets so
that the product of the members of the one set is equal to the product of the
members of the other set.
(Generalization of Question 4, IMO 1970)
A224. Proposed by Waldemar Pompe, student, University of War-saw, Poland.
Let P be an interior point of triangle ABC such that \PBA =
\PCA = (\ABC + \ACB)=3. Prove that
AC
AB + PC=
AB
AC + PB:
Challenge Board Problems
Editor: David Savitt, Department of Mathematics, Harvard University,
1 Oxford Street, Cambridge, MA, USA 02138 <[email protected]>
C81. Let fang be the sequence de�ned as follows: a0 = 0, a1 = 1,
and an+1 = 4an � an�1 for n = 1, 2, 3, : : : .
(a) Prove that a2n � an�1an+1 = 1 for all n � 1.
(b) Evaluate
1Xk=1
arctan
�1
4a2k
�:
C82. Find the smallest multiple of 1998 which appears as a partial
sum of the increasing sequence
1; 1; 2; 2; 2; 4; 4; 4; 4; 8; : : : ;
in which the number 2k appears k+ 2 times (for k a non-negative integer).
412
IMO Report
Adrian Chan, student, UCC, TorontoThis year's Canadian IMO team began with a week of training at the Univer-
sity of Calgary with lavish meal tickets. Then they were o� to beautiful and
rocky Kananaskis where we all had an \adventurous" time. Once the team
stepped o� the air-conditioned plane and into hot andmuggy Taipei, Taiwan,
it marked the team's o�cial arrival to the 39th International Mathematical
Olympiad.
The team consisted of the following members: Adrian \Oops I dropped
my..." Birka, Adrian \If You Will" Chan, Jimmy \Nuclear Aerial Strike" Chui,
Mihaela \Baia" Enachescu, Jessie \So Cute" Lei, and Adrian \Nailing Radar"
Tang. Team leader Dr. Christopher \Focus" Small was driven to the edge,
while deputy leader J.P. \It's So Easy" Grossman calmly polished o� old
competitions one by one. Special thanks to leader observer Arthur \Rubik's
Cube" Baragar and deputy observer Dorette \Dutch" Pronk for their coaching
and experience. Also, thanks must go to Dr. Bill Sands of the University of
Calgary for organizing such a fun training session.
The contest itself seemed to continue the trend of di�cult IMO's and
low medal cuto�s. With 76 countries competing, Canada fared extremely
well, bringing back 1 gold, 1 silver, 2 bronze and an honourable mention.
The scores were as follows:
CAN 1 Adrian Birka 10
CAN 2 Adrian Chan 31 Gold Medal
CAN 3 Jimmy Chui 14 Bronze Medal
CAN 4 Mihaela Enachescu 30 Silver Medal
CAN 5 Jessie Lei 13 Honourable Mention
CAN 6 Adrian Tang 15 Bronze Medal
In this year's contest, Canada placed 20th out of 76 countries, up from
last year's 29th ranking. Best of luck to CAN 1, 4, and 6 as they continue
university studies at MIT, Harvard, and Waterloo respectively. CAN 2, 3,
and 5 are all eligible for next year's team. Hopefully there won't be as much
moaning of \Where did I go wrong?" next year around!
Special thanks must also go to Dr. Graham Wright of the Canadian
Mathematical Society for again taking care of the tab, and Professor Ed
Barbeau for his hard work and dedication to the training of potential IMO
candidates through his year-long correspondence program.
Although sometimes things didn't make sense, and the IMO ag some-
how disappeared, the 39th International Mathematical Olympiad ran
smoothly and was de�nitely a success. The new experience of a place half-
way around the world with a stimulating culture was new to most of us. Best
of luck to all IMO hopefuls for the 1999 team, as yet another Canadian IMO
journey begins next July in Bucharest, Romania.
413
Bogus Arguments and Arcane Identities
Ravi VakilPrinceton University
In Euler's time, mathematics was faster and looser than today, and
niceties such as limits were blatantly ignored. Here is an argument of Euler's
that seems to have no right to work, but does nonetheless. We conclude with
some avenues for exploration and an open question.
If r1, : : : , rn are the zeros of a polynomial a0+ a1x+ � � �+ anxn, and
none of the ri are zero (or a0 6= 0), then the negative of the linear term over
the constant term is the sum of the reciprocals of the roots:
�a1
a0=
1
r1+ � � �+
1
rn:
What about power series? For example, cosx = 1�x2=2+x4=24�� � � ,so that
cospx = 1�
x
2+x2
24� � � � :
The zeroes of this function are the squares of the odd multiples of �=2:
((2n + 1)�=2)2, n = 0, 1, 2, : : : . One might hope that the principle for
polynomials given above still holds:
1
2=
1Xn=0
1
((2n+ 1)�=2)2
which can be rewritten as
1Xn=0
1
(2n+ 1)2=
�2
8: (1)
This is actually true!
Another possibility is to use sinx, which has power series expansion
sinx = x�x3
6+
x5
120� � � � :
Can you use the power series for (sinpx)=x to \prove" the identity
1Xn=1
1
n2=
�2
6? (2)
Copyright c 1998 Canadian Mathematical Society
414
Can you relate (2) to (1) by arguing that (2) minus a quarter of (2) is (1)?
If you write a short computer program to computeP1000
n=11n2, and com-
pare it to �2=6, you'll see that they are indeed very close. In fact, they di�er
by almost exactly 1=1000. Can you explain why this might be? Can you
guesstimate the di�erence betweenP1000n=1
1(2n+1)2
and �2=8?
And �nally, can you conjure up other examples of this sort of argument,
to \prove" other arcane identities? If so, please let us know!
Acknowledgements. This note was inspired by the example of cospx
given in [A].
References
[A] S. Abhyankar, Historical ramblings in algebraic geometry and re-lated algebra, Amer. Math. Monthly 83 (1976), no. 6, 409{448.
[E] L. Euler, Introductio inAnalysin In�nitorum, Berlin Academy, 1748.
415
The Fibonacci Sequence
Wai Ling Yeestudent, University of Waterloo
The sequence de�ned by F0 = 0, F1 = 1, and Fn = Fn�1 + Fn�2 forn � 2 is called the Fibonacci sequence. Named after Leonardo of Pisa, who
is also known as Fibonacci (unsurprisingly), it is one of the most widely stud-
ied sequences of all time. The Fibonacci sequence is an excellent topic with
which to begin learning some basic number theory and various techniques
for working with recurrence relations.
Basic Results
Theorem 1. For all n � 1,
F 2n � Fn�1Fn+1 = (�1)n�1 :
Proof by Induction. When n = 1, F 21 � F0F2 = 12 � 0 � 1 = 1 =
(�1)1�1, so the formula holds for n = 1. Assume that the formula holds for
some n = k, k � 1. For n = k + 1,
F 2k+1 � FkFk+2 = F 2
k+1 � Fk(Fk+1 + Fk)
= (Fk+1 � Fk)Fk+1 � F 2k
= Fk�1Fk+1 � F 2k
= �(�1)k�1 by the induction hypothesis
= (�1)k;
so the formula holds for n = k + 1. Therefore, by mathematical induction,
the formula holds for all n � 1.
Theorem 2. For all n � 1,
F1 + F3 + � � �+ F2n�1 = F2n :
Proof. Using the recurrence relation n times, we have
F2n = F2n�1 + F2n�2= F2n�1 + F2n�3 + F2n�4= � � �= F2n�1 + F2n�3 + � � �+ F3 + F1 + F0
= F2n�1 + F2n�3 + � � �+ F3 + F1 :
Copyright c 1998 Canadian Mathematical Society
416
Exercise 1. ProvenXj=1
F 2j = FnFn+1 :
Divisibility
Theorem 3. For allm, n � 1,
Fn+m = FnFm+1 + Fn�1Fm :
Proof. We will prove this by induction onm. Whenm = 1,
Fn+1 = Fn � 1 + Fn�1 � 1 = FnF2 + Fn�1F1;
so the formula holds for all n whenm = 1. Assume that the formula holds
for all n whenm = M . For m = M + 1,
Fn+M+1 = Fn+M + F(n�1)+M= FnFM+1 + Fn�1FM + Fn�1FM+1 + Fn�2FM
by the induction hypothesis
= FnFM+1 + (Fn�1 + Fn�2)FM + Fn�1FM+1
= Fn(FM + FM+1) + Fn�1FM+1
= FnFM+2 + Fn�1FM+1;
so the formula holds for all n for m = M + 1. By mathematical induction,
the formula holds for allm, n � 1.
Corollary 4. For allm, n � 1, FnjFnm.
Proof. We will prove this by induction onm. Whenm = 1, Fn certainly
divides itself for every positive integer n. Suppose the statement holds for
all n whenm = M . For m = M + 1,
Fn(M+1) = FnM+n = FnMFn+1 + FnM�1Fn
by Theorem 3. Since FnM is divisible by Fn by the induction hypothe-
sis, FnMFn+1 + FnM�1Fn is also divisible by Fn. This is equivalent to
FnjFn(M+1), so the result holds for m = M + 1. By mathematical induc-
tion, the formula holds for allm, n � 1.
Exercise 2. Prove that for every positive integer n, there exist n con-
secutive, composite Fibonacci numbers.
Number Theory 101
We will now de�ne a few terms in the interests of formality. For in-
tegers a and b, we say that a divides b if there exists an integer q such that
b = aq, and a is called a divisor of b. Given two non-zero integers a and b,
the largest number which divides both of them, denoted gcd(a; b), is called
417
their greatest common divisor. If gcd(a; b) = 1, then a and b are said to be
relatively prime.
Theorem 5. (The Division Algorithm) Given a positive integer a and
an integer b, there exist unique integers q and r such that b = aq + r and
0 � r < a. Then q is called the quotient, and r is called the remainder upondivision of b by a.
Proof. Consider the set
S = fs : s = b� aq � 0; q 2 Zg:
S cannot be empty. If b � 0, then select q = 0 to give b 2 S. Otherwise,
if b < 0, select q = b so that b � ab = b(1 � a) � 0, which means that
b� ab 2 S. Since S is non-empty and contains only non-negative integers,
we can �nd the smallest element in S. Call it r.
Suppose r � a. Then 0 � r� a = b� aq� a = b� a(q+1) for some
q, so r � a 2 S and it is smaller than r, contradiction. Thus 0 � r < a.
Suppose that we can �nd 0 � r1 < r2 < a in S and corresponding q1 and
q2. Then b = aq1+ r1 = aq2+ r2, which implies that a(q1� q2) = r2� r1.
Thus a divides r2 � r1. However, we also know that 0 < r2 � r1 < a; that
is, r2 � r1 lies between two consecutive multiples of a and thus cannot be
divisible by a, contradiction. We have shown the existence and uniqueness
of q and r.
Exercise 3. Prove that gcd(a; b) = gcd(a; b � aq) for any non-zero
integers a and b and any integer q.
Exercise 4. Prove that if a and q are relatively prime, then gcd(a; qb) =
gcd(a; b), where a; b; q are non-zero integers.
Exercise 5. Prove that Fn and Fn+1 are relatively prime.
Number Theory 102
The Euclidean Algorithm. The Euclidean Algorithm is an algorithm used
to determine the greatest common divisor of two numbers. Suppose we
have two distinct positive integers a and b where, without loss of generality,
a < b. By the Division Algorithm, b = aq1+r1 where 0 � r1 < a for unique
integers q1 and r1. If r1 = 0, then a divides b so our greatest common divisor
is a. Otherwise, by Exercise 3, gcd(a; b) = gcd(a; b� aq1) = gcd(a; r1) =
gcd(r1; a). In this case, we then repeat the same argument using r1 and a
where we used a and b before, respectively. We have a = r1q2 + r2 where
0 � r2 < r1 for unique integers q2 and r2 by the Division Algorithm, and
gcd(r1; a) = gcd(r1; a� q2r1) = gcd(r2; r1). Continue applying this argu-
ment. Since the ris are strictly decreasing and non-negative, there must be
a last remainder, say rn, that is bigger than 0. So we have
418
b = aq1 + r1; 0 � r1 < a; gcd(a; b) = gcd(r1; a);
a = r1q2 + r2; 0 � r2 < r1; gcd(r1; a) = gcd(r2; r1);
r1 = r2q3 + r3; 0 � r3 < r2; gcd(r2; r1) = gcd(r3; r2);
� � � � � �rn�2 = rn�1qn + rn; 0 � rn < rn�1; gcd(rn�1; rn�2) = gcd(rn; rn�1);
rn�1 = rnqn+1; gcd(rn; rn�1) = rn:
We have found that gcd(a; b) = rn.
Theorem 6. For all a, b � 1,
gcd(Fa; Fb) = Fgcd(a;b):
Proof. If a and b are equal, the result is immediate, so assume that
a < b. Apply the Euclidean Algorithm to obtain
b = aq1 + r1; 0 � r1 < a;
a = r1q2 + r2; 0 � r2 < r1;
r1 = r2q3 + r3; 0 � r3 < r2;
� � �rn�2 = rn�1qn + rn; 0 � rn < rn�1;rn�1 = rnqn+1:
We have
gcd(Fa; Fb) = gcd(Fa; Faq1+r1) = gcd(Fa; Faq1�1Fr1 + Faq1Fr1+1)
by Theorem 3. Since Faq1Fr1+1 is a multiple of Fa by Corollary 4,
gcd(Fa; Fb) = gcd(Fa; Faq1�1Fr1 + Faq1Fr1+1) = gcd(Fa; Faq1�1Fr1)
by Exercise 3. By Exercise 5, gcd(Faq1; Faq1�1) = 1. Since Fa divides Faq1,
gcd(Fa; Faq1�1) = 1 also. By Exercise 4, since Fa and Faq1�1 are relatively
prime,
gcd(Fa; Faq1�1Fr1) = gcd(Fa; Fr1) :
We conclude that gcd(Fa; Fb) = gcd(Fr1; Fa). Repeating this argument, we
obtain
gcd(Fr1; Fa) = gcd(Fr2; Fr1) = � � � = gcd(Frn; Frn�1) :
Since rn divides rn�1, Frn dividesFrn�1, which implies that gcd(Frn; Frn�1)
= Frn. Thus,
gcd(Fa; Fb) = Frn = Fgcd(a;b) :
Finding Fn Explicitly
The monic quadratic with roots � and � is
(x� �)(x� �) = x2 � (�+ �)x+ �� :
419
Let � and � be the roots of x2 � x� 1 in particular. Then, comparing coef-
�cients, �+ � = 1 and �� = �1. Using this, we can rewrite the recurrence
relation Fn = Fn�1 + Fn�2 as Fn = (�+ �)Fn�1 � ��Fn�2. From this
equation, we obtain
Fn � �Fn�1 = �(Fn�1 � �Fn�2) :
Let sn�1 = Fn��Fn�1 for alln � 2. Rewriting the above equation in terms
of the si, we obtain sn�1 = �sn�2. In other words, the sequence fsng is ageometric sequence with common ration �. We conclude that sn = �n�1s1.
Similarly, Fn � �Fn�1 = �(Fn�1 � �Fn�2), and if we let
tn�1 = Fn � �Fn�1 for n � 2, then tn = �n�1t1 for n � 1. Hence,
Fn =�� �
�� �Fn +
��
�� �Fn�1 �
��
�� �Fn�1
=�(Fn� �Fn�1)� �(Fn � �Fn�1)
�� �
=�tn�1 � �sn�1
�� �
=�n�1t1 � �n�1s1
�� �
=�n�1(F2 � �F1)� �n�1(F2� �F1)
�� �
=�n�1(1� �)� �n�1(1� �)
�� �:
Recall that �+ � = 1, so the equation above is, in fact,
�n � �n
�� �:
Solving x2 � x � 1 = 0 for the values of � and �, we have, without loss of
generality,
� =1+
p5
2and � =
1�p5
2:
Substituting these values, we conclude that
Fn =
�1+p5
2
�n��1�p5
2
�np5
:
This is called Binet's Formula for the Fibonacci sequence.
Exercise 5. Let � = (1 +p5)=2. Prove that Fn is the integer closest
to�np5:
420
Problems
1. Prove that the product of every four consecutive Fibonacci numbers is
the area of a Pythagorean triangle.
2. Prove that every positive integer can be written as a sum of distinct
Fibonacci numbers.
3. Prove
Fn =
bn+12 cXk=1
�n� k
k � 1
�:
4. Prove that if Fn is prime and n � 5, then n is prime.
5. Prove that Fn + 1 is always composite for n � 4.
6. Show that for any positive integer n, among the �rst n2 Fibonacci num-
bers, there exists at least one that is divisible by n.
7. De�ne a Fibonacci prime to be a Fibonacci number that is prime. Prove
or disprove: There are in�nitelymany Fibonacci primes. (Note that this
is an open problem.)
421
J.I.R. McKnight Problems Contest 1984
1. Find the real roots of the equation:qx+ 3� 4
px� 1 +
qx+ 8� 6
px� 1 = 1 :
(Note: All square roots are to be taken as positive.)
2. Consider a reservoir in the shape of an inverted cone as shown in the
diagram below. Water runs into the reservoir at the constant rate of
2 m3 per minute. How fast is the water level rising when it is 6metres
deep?
r
r6
?
6m
6
?
10m
-� 5m
3. Three forces of magnitudes 10 N, 15 N, and 10 N act at angles of 30�,70�, and 120� respectively, to the real axis Ox. Using the complex
numbers and the imaginary axisOy �nd the magnitude and direction of
the resultant force.
4. Normals are drawn from the point�154;�3
4
�to the parabola whose
equation is y2 = 4x. Find the coordinates of the points where the
normals meet the parabola.
5. The horizontal base of a triangular pyramid is an equilateral triangle
QRS, each of whose sides is 20 cm long. The sloping edges of the
pyramid PQRS are respectively 20 cm, 20 cm, and 12 cm long.
(a) Calculate the perpendicular height of the pyramid to the nearest
millimetre.
(b) Calculate the angle of inclination of each of the three edges with
the base to the nearest tenth of a degree.
6. Prove that if tanA = tan3B and tan2B = 2 tanC, thenA+B�C =
n� for some n 2 Z.
422
Swedish Mathematics Olympiad
1988 Qualifying Round
1. Show that the function
f(x) =
qx� 4
px� 1 + 3 +
qx� 6
px� 1 + 8
is constant on the closed interval 5 � x � 10.
2. Find the rational root of the equation
(2x)log2 = (3x)log3; x > 0
in the form p
q, where p and q are integers.
3. We will call two squares on a chessboard \neighbours" if they have a
side or corner in common. The numbers 1 to 64 are arbitrarily placed
on the 64 squares of a chessboard. Show that there are always two
\neighbours" whose numbers have positive di�erence at least 9.
4. A car's tires wear proportionally with the distance driven. Furthermore,
front tires last a km and back tires b km, where a < b. If, after an
appropriate distance is driven, the tires are rotated (that is, back tires
placed on front wheels and front tires on back wheels), the distance
which can be driven without needing to replace any of the tires can be
increased. What is the longest distance which can be driven with a set
of tires, before any new tires must be bought?
5. P , Q, and R are points on the circumference of a circle such that PQR
is an equilateral triangle. S is an arbitrary point on the circumference
of the circle. Consider the lengths of the line segments PS, QS, and
RS. Show that one of them is the sum of the other two.
6. Show that for every positive integer n, there exist positive integers x
and y such that px2 + nxy + y2
is an integer.
423
1988 Final Round
1. The sides of a triangle have lengths a > b > c, and the corresponding
perpendiculars have lengths ha, hb, and hc. Show that
a+ ha > b+ hb > c+ hc :
2. Six ducklings swim on the surface of a pond, which is in the shape of a
circle with radius 5m. Show that, at every instant, two of the ducklings
swim at a distance of at most 5 m from each other.
3. Show that for aribtrary real numbers x1, x2, and x3,
if x1 + x2 + x3 = 0; then x1x2 + x2x3 + x3x1 � 0 :
Find all n � 4 for which the statement
if x1+x2+ � � �+xn = 0; then x1x2+x2x3+ � � �+xn�1xn+xnx1 � 0
is true. (Both sums have n terms.)
4. Let P (x) be a polynomial of degree 3 with exactly three distinct real
roots. Find the number of real roots of the equation
(P 0(x))2� 2P (x)P 00(x) = 0 :
5. Let m and n be positive integers. Show that there exists a constant
� > 1, independent ofm and n, such that
m
n<p7 implies that 7�
m2
n2�
�
n2:
6. The sequence a1, a2, : : : , is de�ned by the recursion formula
an+1 =
sa2n +
1
ann � 1 ;
and a1 = 1. Show that one can choose � such that
1
2�
an
n�� 2 for all n � 1:
424
PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, De-partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.
In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.
To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81
2"�11" or A4 sheets of paper.
These may be typewritten or neatly hand-written, and should be mailedto the Editor-in-Chief, to arrive no later than 1 April 1999. They may alsobe sent by email to [email protected]. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication. Please note that we do not accept submissions sent by FAX.
2374. [1998: 365] (Correction)Proposed by Toshio Seimiya, Kawasaki,Japan.
Given triangle ABC with \BAC > 60�. Let M be the mid-point of
BC. Let P be any point in the plane of4ABC.
Prove that AP + BP + CP � 2AM .
2376. Proposed by Albert White, St. Bonaventure University, St.Bonaventure, NY, USA.
Suppose that ABC is a right-angled triangle with the right angle at C.
Let D be a point on hypotenuse AB, and let M be the mid-point of CD.
Suppose that \AMD = \BMD. Prove that
1. AC2MC
2+ 4[ABC] [BCD] = AC
2MB
2;
2. 4AC2MC
2 � AC2BD
2= 4[ACD]2 � 4[BCD]2,
where [XY Z] denotes the area of 4XY Z.(This is a continuation of problem 1812, [1993: 48].)
425
2377. Proposed by Nikolaos Dergiades, Thessaloniki, Greece.
Let ABC be a triangle and P a point inside it. Let BC = a, CA = b,
AB = c, PA = x, PB = y, PC = z, \BPC = �, \CPA = � and
\APB = .
Prove that ax = by = cz if and only if ��A = � �B = �C = �3.
2378. Proposed by David Doster, Choate Rosemary Hall, Wall-ingford, Connecticut, USA.
Find the exact value of: cot
��
22
�� 4 cos
�3�
22
�:
2379. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.
Suppose thatM1,M2 andM3 are the mid-points of the altitudes from
A to BC, from B to CA and from C to AB in4ABC. Suppose that T1, T2and T3 are the points where the excircles to 4ABC opposite A, B and C,
touch BC, CA and AB.
Prove thatM1T1,M2T2 andM3T3 are concurrent.
Determine the point of concurrency.
2380. Proposed by Bill Sands, University of Calgary, Calgary, Al-berta.
When the price of a certain book in a store is reduced by 1=3 and
rounded to the nearest cent, the cents and dollars are switched. For ex-
ample, if the original price was $43.21, the new price would be $21.43 (this
does not satisfy the \reduced by 1=3" condition, of course). What was the
original price of the book? [For the bene�t of readers unfamiliar with North
American currency, there are 100 cents in one dollar.]
2381. Proposed by Angel Dorito, Geld, Ontario.
Solve the equation log2 x = log4(x+ 1).
2382. Proposed by Mohammed Aassila, Universit �e Louis Pasteur,Strasbourg, France.
If4ABC has inradius r and circumradius R, show that
cos2�B �C
2
��
2r
R:
2383. Proposed by Mohammed Aassila, Universit �e Louis Pasteur,Strasbourg, France.
Suppose that three circles, each of radius 1, pass through the same point
in the plane. Let A be the set of points which lie inside at least two of the
circles. What is the least area that A can have?
2384. Proposed by Paul Bracken, CRM, Universit �e de Montr �eal,Qu �ebec.
Prove that 2(3n� 1)n � (3n+ 1)n for all n 2 N.
426
2385. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Madrid, Spain.
A die is thrown n � 3 consecutive times. Find the probability that the
sum of its n outcomes is greater than or equal to n+6 and less than or equal
to 6n� 6.
2386?. Proposed by Clark Kimberling, University of Evansville,Evansville, IN, USA.
Write
1 ! 1
1! 3
1! 4 1
1 3! 6 2 1
1 3 4! 8 1 3 2 1
1 2 3 4 6!
(The last ten numbers shown indicate that up to this point, eight 1's,
one 2, three 3's, two 4's and one 6 have been written.)
(a) If this is continued inde�nitely, will 5 eventually appear?
(b) Will every positive integer eventually be written?
Note: 11 is a number and not two 1's.
2387. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.
For �xed p 2 N, consider the power sums
Sp(n) :=
nXk=1
(2k� 1)p; where n � 1 ;
so that Sp(n) is a polynomial in n of degree p+ 1 with rational coe�cients.
Prove that
(a) If all coe�cients of Sp(n) are integers, then p = 2m � 1 for some
m 2 N.
(b)? The only values of p yielding such polynomials are p = 1 and p = 3
(with S1(n) = n2 and S3(n) = 2n4 � n2).
427
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.
1637. [1991: 114; 1992: 125; 1994: 165] Proposed by GeorgeTsintsifas, Thessaloniki, Greece.
Prove that X sinB + sinC
A>
12
�
where the sum is cyclic over the angles A;B;C (measured in radians) of a
nonobtuse triangle.
Comment byWaldemar Pompe, student,University ofWarsaw, Poland.
We can use Crux 2015 [1998: 305] to derive the strengthening of Crux
1637 given on [1994: 165], namely that
X sinB + sinC
A�
9p3
�
for all triangles ABC.
Without loss of generality, we can assume that A � B � C. Then
1
A� 1
B� 1
C
and
sinB + sinC � sinA+ sinC � sinA+ sinB :
By Chebyshev's Inequality and Crux 2015, we have
3X sinB + sinC
A��2X
sinA���X 1
A
��
27p3
�;
and the result follows.
2257. [1997: 300] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.
The diagonals AC and BD of a convex quadrilateral ABCD intersect
at the pointO. LetOK, OL,OM , ON , be the altitudes of triangles4ABO,4BCO, 4CDO, 4DAO, respectively.
Prove that if OK = OM and OL = ON , then ABCD is a parallelo-
gram.
428
I. Solution by Con Amore Problem Group, Royal Danish School of Ed-ucational Studies, Copenhagen, Denmark.
Let OK = OM = h, OA = a, OB = b, OC = c, OD = d, and
\AOB = v. Expressing in two ways the area of 4AOB, we get
12AB � h = 1
2ab sinv;
and so
1
h=
pa2 + b2 � 2ab cos v
ab sinv=
1
sinv
r1
b2+
1
a2� 2 �
1
b�1
a� cos v :
Similarly,
1
h=
1
sinv
r1
d2+
1
c2� 2 �
1
d�1
c� cos v ;
so thatr1
b2+
1
a2� 2 �
1
b�1
a� cos v =
r1
d2+
1
c2� 2 �
1
d�1
c� cos v ; (1)
and similarly,r1
a2+
1
d2+ 2 �
1
a�1
d� cos v =
r1
b2+
1
c2+ 2 �
1
b�1
c� cos v : (2)
Now, consider another convex quadrilateral A1B1C1D1, with diagonals in-
tersecting in O1, and such that \A1O1B1 = v, O1A1 = 1a, O1B1 = 1
b,
O1C1 = 1c, and O1D1 = 1
d. The equalities (1) and (2) imply that the oppo-
site sides of A1B1C1D1 are equal in length, which means that A1B1C1D1
is a parallelogram. So 1a= 1
c, and 1
b= 1
d, implying a = c and b = d. This
proves that ABCD is a parallelogram.
II. Solution by the proposer (slightly edited).If ABCD is a trapezoid with AB k CD, then
AB
CD=
OK
OM= 1 ;
which means that ABCD is a parallelogram. Hence, assume that ABCD is
not a trapezoid and set P = AB\CD, Q = AD\BC. Indeed, P 6= Q. The
assumption on the given quadrilateral says exactly that PO bisects \BPCand QO bisects \AQB. Thus
AP
PC=
AO
OC=
AQ
QC;
implying that P , O, and Q lie on the Apollonius circle with centre on the
line AC. Similarly, since
BP
PD=
BO
OD=
BQ
QD;
429
P; O, and Q lie on the Apollonius circle with centre on the lineBD. This im-
plies that O is the circumcentre of4POQ; that is, points P and Q coincide,
a contradiction.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
MICHAEL LAMBROU, University of Crete, Crete, Greece; TOSHIOSEIMIYA, Kawasaki, Japan; and D.J. SMEENK, Zaltbommel, the Netherlands. There
were also �ve incorrect solutions submitted.
Most of the submitted solutions are similar to the proposer's solution.
2260. [1997: 301] Proposed by Vedula N. Murty, Visakhapatnam,India.
Let n be a positive integer and x > 0. Prove that
(1 + x)n+1 �(n+ 1)n+1
nnx :
Solution by Florian Herzig, student, Cambridge, UK; Gerry Leversha,St. Paul's School, London, England; Nick Lord, Tonbridge School, Tonbridge,Kent, England; and Panos E. Tsaoussoglou, Athens, Greece.
By the AM{GM Inequality applied to the n + 1 positive numbers
x;1
n;1
n; : : : ;
1
n, we have
�x+ 1
n+ 1
�n+1
�x
nn, with equality if and only
if x = 1
n. This is clearly equivalent to the given inequality.
Also solved by PAUL BRACKEN, CRM, Universit �e de Montr �eal, Montr �eal,
Qu �ebec; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORE
CHRONIS, student, Aristotle University of Thessaloniki, Greece; KEITHEKBLAW, Walla Walla, Washington, USA; RUSSELL EULER and JAWAD
SADEK, NW Missouri State University, Maryville, Missouri, USA; RICHARD
I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, New Mex-ico Highlands University, Las Vegas, NM, USA; WALTHER JANOUS, Ur-
sulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University
of Crete, Crete, Greece; PAVLOS MARAGOUDAKIS, Hatzikiriakio, Pireas,Greece; PHIL McCARTNEY, Northern Kentucky University, Highland Heights,
KY, USA; VICTOR OXMAN, University of Haifa, Haifa, Israel; MICHAEL
PARMENTER, Memorial University of Newfoundland, St. John's, Newfound-land; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DAVID R. STONE, Georgia
Southern University, Statesboro, Georgia, USA; JOHN VLACHAKIS, Athens, Greece;
EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; ROGERZARNOWSKI, TREY SMITH, CHARLES DIMINNIE and GERALD ALLEN (jointly),
Angelo State University, San Angelo, TX, USA; and the proposer. There were also
three incomplete solutions.
The majority of solvers used a standard calculus approach to establish the given
inequality. The only exceptions are the �ve listed in the solutions above plus Lambrou
andMaragoudakis, who used Bernoulli's Inequality. Both Janous and Lambrou noted
that the given inequality holds for all positive real n.
430
Janous also generalized the problem by showing that if a > 0, and � > � > 0
are given real numbers, then the largest constant C = C(a; �; �) such that
(a + x)� � Cx� holds for all x > 0 is given by C =
�a
�� �
������
��.
The given problem is the special case when a = � = 1 and � = n+ 1.
By applying the AM{GM Inequality to the n + 1 positive numbers:
1;x
n;x
n; : : : ;
x
n, Lord and the proposer obtained (1 + x)
n+1 � (n+ 1)n+1
nnxn,
which is stronger than the proposed inequality for x > 1.
Zarnowski et al. commented that when n is odd, the inequality is true for all
real x, while if n is even, there is a number xn � �2 such that the inequality holds
for all x � xn.
2261. [1997: 301] Proposed by Angel Dorito, Geld, Ontario.
Assuming that the limit exists, �nd
limN!1
1 +
2 + N+:::1+:::
N + 1+:::2+:::
!;
where every fraction in this expression has the form
a+ b+:::
c+:::
b+ c+:::
a+:::
for some cyclic permutation a, b, c of 1, 2, N .
[Proposer's comment: this problem was suggested by Problem 4 of
Round 21 of the International Mathematical Talent Search,Mathematics andInformatics Quarterly, Vol. 6, No. 2, p. 113.]
Solution by Keith Ekblaw, Walla Walla, Washington, USA.It will be shown that
1 +2 + N+���
1+���N + 1+���
2+����!
1 +p5
2as N !1:
First, consider
JN = N +1 + 2+���
N+���2 + N+���
1+���:
Note that1 + 2+���
N+���2 + N+���
1+���> 0:
Thus JN > N and hence JN !1 as N !1. Now let
KN = 1 +2 + N+���
1+���N + 1+���
2+���= 1+
2 + JNKN
JN= 1 +
2
JN+
1
KN
:
431
Thus as N ! 1 (and hence JN ! 1), KN ! 1 + 1=KN. Letting K =
limN!1KN , we haveK = 1+1=K orK2�K�1 = 0 and so the required
limit is
K =1 +
p5
2:
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
CON AMORE PROBLEM GROUP, Royal Danish School of Educational Studies,Copenhagen, Denmark; FLORIAN HERZIG, student, Cambridge, UK; RICHARD
I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengym-
nasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY
LEVERSHA, St. Paul's School, London, England; and the proposer.
Hess calculates that ifN is replaced by 3:5, then the expression inside the limit
(which is itself a limit, actually) is equal to 2. Readers may like to �nd other \nice"
triples of numbers a; b; c so that the expression
a+b+
c+���
a+���
c+a+���
b+���
is rational, say.
The proposer notes (as can be seen from the above proof) that the answer is
still the golden ratio if the 2's in the given expression are replaced by any constant
real number.
2262. [1997: 301] Proposed by Juan-Bosco Romero M�arquez, Uni-versidad de Valladolid, Valladolid, Spain.
Consider two triangles4ABC and4A0B0C0 such that \A � 90� and\A0 � 90� and whose sides satisfy a > b � c and a0 > b0 � c0. Denote the
altitudes to sides a and a0 by ha and h0a.
Prove that (a)1
hah0a�
1
bb0+
1
cc0, (b)
1
hah0a�
1
bc0+
1
b0c.
Solution by Christopher J. Bradley, Clifton College, Bristol, UK.
(a) By the Cauchy-Schwarz inequality on the vectors
�1
b;1
c
�and
�1
b0;1
c0
�,
we have
1
bb0+
1
cc0��1
b2+
1
c2
�12�
1
b02
+1
c02
� 12
:
Now1
b2+
1
c2=b2 + c2
b2c2�
a2
b2c2, since \A � 90�.
Alsoa2
b2c2�
a2
b2c2 sin2A=
1
h2a, since 1
2aha and 1
2bc sinA are both
formulae for the area of 4ABC. Similarly for 4A0B0C0.Hence
1
bb0+
1
cc0�
1
hah0a, as required.
432
The proof of part (b) is the same, except that it starts with the vectors�1
b;1
c
�and
�1
c0;1
b0
�.
Also solved by CLAUDIO ARCONCHER, Jundia��, Brazil; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete,
Crete, Greece; FLORIAN HERZIG, student, Cambridge, UK; RICHARD I. HESS, Ran-
cho Palos Verdes, California, USA; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA,St. Paul's School, London, England; VICTOR OXMAN, University of Haifa, Haifa,
Israel; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki,Japan; GEORGE TSAPAKIDIS, Agrinio, Greece; JOHN VLACHAKIS, Athens, Greece;
and the proposer.
Most of the submitted solutions are similar to the one given above. Severalsolvers pointed out that the restrictions on the sides are unnecessary, and that equal-
ity in (a) occurs if and only if \A = \A0 = 90� and b=c = b
0=c
0 and in (b) if and
only if \A = \A0 = 90� and b=c = c
0=b
0.
Janous proved more generally that for any real number p � 1
1
(hah0a)p
� 1
(bb0)p+
1
(cc0)p; and
1
(hah0a)p
� 1
(bc0)p+
1
(b0c)p.
2263. [1997: 364] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle, and the internal bisectors of\B, \C, meetAC, AB
at D, E, respectively. Suppose that \BDE = 30�. Characterize 4ABC.
Solution by the proposer.Let F be the re ection of E across BD. Since \EBD = \CBD, it
follows that F lies on BC. So \BDF = \BDE = 30�, and DF = DE.
Then 4DEF is equilateral, so EF = ED and \FED = 60�. By the Law
of Sines for 4EFC and 4EDC we obtain that
EC
sin\EFC=
EF
sin\ECF=
ED
sin\ECD=
EC
sin\EDC;
which gives sin\EFC = sin\EDC. It follows that we have either that
\EFC = \EDC, or that \EFC + \EDC = 180�.
Case 1. \EFC = \EDC. Then
\FEC = \DEC = 12\FED = 30�:
Let I be the intersection of BD and CE. Then
\DIC = \IED + \IDE = 30� + 30� = 60�:
Since \DIC = 90� � 12\A, we obtain 90� � 1
2\A = 60�, so \A = 60�.
Case 2. \EFC + \EDC = 180�. Then\FED+ \FCD = 180�, so that 60� + \FCD = 180�.Thus \FCD = 120�; that is, \ACB = 120�.
433
Therefore, ABC is a triangle with either \A = 60� or \C = 120�.
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;
FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; FLORIAN HERZIG, stu-
dent, Cambridge, UK; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria;
MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAILAU, Hong Kong;GERRY LEVERSHA, St. Paul's School, London, England; VICTOR OXMAN, Univer-
sity of Haifa, Haifa, Israel; D.J. SMEENK, Zaltbommel, the Netherlands. There were
also two incomplete solutions submitted.
Herzig and Lambrou have also shown that the characteristic condition is su�-
cient, that is, \A = 60� or \C = 120� implies \BDE = 30�.
2265. [1997: 364] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.
Given triangle ABC, let ABX and ACY be two variable triangles
constructed outwardly on sidesAB and AC of4ABC, such that the angles
\XAB and \Y AC are �xed and \XBA + \Y CA = 180�. Prove that all
the linesXY pass through a common point.Solution by Toshio Seimiya, Kawasaki, Japan.
�
DrA r
B
rC
r
Xr
Y
r
Z
r
ErrP
We denote the circumcircle of 4ABC by �. Let BX and CY meet
at Z. Since \XBA + \Y CA = 180�, we get \XBA = 180� � \Y CA= \ACZ, so that A;B;Z;C are concyclic, that is, Z lies on �. Let D;E
be the second intersections of AX;AY respectively with �. Since AX and
AY are �xed lines, D and E are �xed points. Let P be the intersection
of BE and CD. Since hexagon ADCZBE is inscribed in �, by Pascal's
Theorem the intersections of AD and BZ, of DC and BE, and of CZ and
EA are collinear. Therefore variable line XY always passes through the
�xed point P . [Editorial note: if the diagram di�ers from the one shown,
for example if Z lies between X and B, the proof still works with minor
changes.]
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
FLORIAN HERZIG, student, Cambridge, UK; MICHAEL LAMBROU, University of
Crete, Crete, Greece; MAR �IA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano,
434
Valladolid, Spain; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.
One incorrect solution and one comment were sent in.
Seimiya and the proposer had similar solutions.
Herzig notes that, if the �xed angles are chosen to be AXB and AY C instead,
then the lines XY still pass through a �xed point. Readers may like to show this
themselves.
2266. [1997: 364] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.
BCLK is the square constructed outwardly on side BC of an acute
triangle ABC. Let CD be the altitude of 4ABC (with D on AB), and let
H be the orthocentre of4ABC. If the lines AK and CD meet at P , show
thatHP
PD=
AB
CD:
I. Solution by Florian Herzig, student, Cambridge, UK.
K B
L C
AHP
X
D
Y
M
N
Construct a square ABMN outwardly on 4ABC. Let X and Y be
the points of intersection of AK;BC and AB;CM respectively. The ro-
tation through a right angle about B maps 4MBC onto 4ABK. Hence
AK and CM are perpendicular and it follows that AP and CD are alti-
tudes in 4AY C. Therefore P is the orthocentre in that triangle, and as a
consequence Y P ? AC or Y P k BH. Thus
HP
PD=
BY
Y D=
BM
CD=
AB
CD:
435
II. Solution by Toshio Seimiya, Kawasaki, Japan.
K B
L C
AP
H
S
D
Let S be a point on AK such that DS ? BC and soDS k AH k BK.
Since AH k DS we get
HP
PD=
AH
DS: (1)
Since DS k BK and BK = BC, we have
AD
AB=
DS
BK=
DS
BC: (2)
Since AH ? BC and CD ? AB we get \HAD = \BCD. Moreover we
have \HDA = \BDC(= 90�), so that 4HAD � 4BCD. Thus
AH
BC=
AD
CD: (3)
From (2) and (3) we have
AH
DS=
AH
BC�BC
DS=
AD
CD�AB
AD=
AB
CD;
so that we obtain from (1) thatHP
PD=
AB
CD:
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid,
Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WALTHERJANOUS,
Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State Univer-sity, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete, Crete,
Greece; GERRY LEVERSHA, St. Paul's School, London, England; D.J. SMEENK, Zalt-
bommel, the Netherlands; JOHN VLACHAKIS, Athens, Greece; and the proposer.
The proposer's solution was the same as Herzig's. Most other solvers used
either similar triangles (as in II), trigonometry, or coordinates.
436
2268. [1997: 364] Proposed by Juan-Bosco Romero M�arquez, Uni-versidad de Valladolid, Valladolid, Spain.
Let x, y be real. Find all solutions of the equation
2xy
x+ y+
sx2 + y2
2=pxy +
x+ y
2:
Solution by Nikolaos Dergiades, Thessaloniki, Greece.
Let A =
sx2 + y2
2and B =
pxy. Then
2(A2 + B2) = (x+ y)2 and 2(A2 � B2) = (x� y)2
and the given equation yields
A� B =x+ y
2�
2xy
x+ y
or2(A2 � B2)
A+ B=
(x� y)2
x+ y
(x� y)2
A+ B=
(x� y)2
x+ y:
Therefore, we have either (x � y)2 = 0 (which implies that x = y) or
A+ B = x+ y. Let us consider A+ B = x+ y:
A+ B = x+ y () (A+ B)2 = 2(A2 + B2)
() (A� B)2 = 0 () A = B
()
sx2 + y2
2=pxy
() (x� y)2 = 0 () x = y
In conclusion, all solutions have x = y 6= 0, because x+ y 6= 0.
Also solved by HAYO AHLBURG, Benidorm, Spain; PAUL BRACKEN,CRM, Universit �e de Montr �eal; CHRISTOPHER J. BRADLEY, Clifton Col-
lege, Bristol, UK; CON AMORE PROBLEM GROUP, Royal Danish School
of Educational Studies, Copenhagen, Denmark (2 solutions); CHARLES R.
DIMINNIE, Angelo State University, San Angelo, TX, USA; DAVID DOSTER,
Choate Rosemary Hall, Wallingford, Connecticut, USA; RUSSELL EULER and
JAWAD SADEK, NW Missouri State University, Maryville, Missouri, USA;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL
LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong
Kong; VICTOR OXMAN, University of Haifa, Haifa, Israel; HEINZ-J�URGENSEIFFERT, Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands; DAVID
R. STONE, Georgia Southern University, Statesboro, Georgia, USA; ROGER
437
ZARNOWSKI, Angelo State University, San Angelo, Texas, USA; and the proposer.
There were 14 incorrect solutions submitted, 11 of which simply did NOT exclude
the origin from the solution set.
2270. [1997: 365] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.
Given 4ABC with sides a, b, c, a circle, centre P and radius � inter-
sects sidesBC, CA, AB in A1 and A2, B1 and B2, C1 and C2 respectively,
so thatA1A2
a=B1B2
b=C1C2
c= � � 0:
Determine the locus of P .
Solution by the proposer.[Assume that no two sides of the triangle are equal.] The distance from
P to BC is x =
s�2 �
�2a2
4;
P to CA is y =
s�2 �
�2b2
4;
P to AB is z =
s�2 �
�2c2
4:
It follows thatx2 � y2
y2 � z2=
b2 � a2
c2 � b2, or
x2(c2 � b2) + y2(a2 � c2) + z2(b2� a2) = 0 : (1)
Considering x; y; z to be the triangular coordinates of P with respect to
4ABC, we conclude that (1) represents a conic K. Note that K passes
through the incentre I(1; 1; 1) and the excentres Ia(�1; 1; 1); Ib(1;�1; 1),and Ic(1;1;�1), [and also the circumcentre O(cosA; cosB; cosC)]. So K
is the conic through O of the pencil determined by I; Ia; Ib; Ic. Since the
degenerate conics of the pencil are degenerate orthogonal hyperbolas (that
is, pairs of perpendicular lines), K must be an orthogonal hyperbola.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK.
Neither solver mentioned the obvious special cases:
The locus is a pair of perpendicular lines when 4ABC is isosceles, and just
the four points I; Ia; Ib; Ic when equilateral.
Bradley points out that it is clear from the statement of the problem (with no
need for coordinates) that the locus includes I; Ia; Ib; Ic (when � = 0) and O (when
� = 1).
438
2271. [1997: 365] Proposed by F.R. Baudert, Waterkloof Ridge,South Africa.
A municipality charges householders per month for electricity used ac-
cording to the following scale:
�rst 400 units | 4.5cj per unit;
next 1100 units | 6.1cj per unit;
thereafter | 5.9cj per unit.
IfE is the total amount owing (in dollars) for n units of electricity used,
�nd a closed form expression, E(n).
Solution by Michael Lambrou, University of Crete, Greece.We may view the charges as consisting of
(i) 4.5cj per unit and, additionally,
(ii) a surcharge of 6:1� 4:5 = 1:6cj per unit, but with
(iii) a refund of 6:1 � 5:9 = 0:2cj per unit for units consumed in excess of
400 + 1100 = 1500.
So (in dollars) the amount owing for n units is:
45
1000n+
16
1000maxf0; n� 400g �
2
1000maxf0; n� 1500g:
Now writing maxfa; bg = 12(ja� bj+ a+ b), this simpli�es to:
1
1000f52n+ 8jn� 400j � jn� 1500j � 1700g :
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; HANS
ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG,
student, Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-
tria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; KEE-
WAI LAU, Hong Kong; and the proposer. There were two incorrect solutions submit-
ted. Some solvers used unit step functions instead of absolute value to express the
answer.
2272?. [1997: 365] Proposer unknown (please identify yourself!)
Write r - s if there is an integer k satisfying r < k < s. Find, as a
function of n (n � 2), the least positive integer satisfying
k
n-
k
n� 1-
k
n� 2- � � � -
k
2- k:
439
Solution by Florian Herzig, student, Cambridge, UK (modi�ed by theeditor).
Let kn denote the least positive integer k satisfying
k
n-
k
n� 1-
k
n� 2- � � � �
k
2- k (1)
We claim that kn =
$�n+ xn
2
�2+ 1
%, where xn = bn + 1 � 2
pn� 1c.
We �rst show that kn satis�es (1). To this end, let u =�p
kn�. Then for all
integers a = 1; 2; � � � ; u � 1 we have, from a �pkn � 1 that a(a + 1) �
pkn(
pkn � 1) < kn. Hence
kn
a�
kn
a+ 1> 1, which implies
kn
a+ 1-kn
a.
Therefore we have
kn
u-
kn
u� 1- � � � -
kn
2� kn: (2)
Next we show that for all integers a = u+ 1; u+ 2; � � � ; n
n+ xn � a <kn
a< n+ xn � a+ 1: (3)
In fact, the left inequality in (3) holds for all a = 1; 2; � � � ; n. To see this, note
that from kn >
�n+ xn
2
�2we get a2�axn�an+kn =
�a�
n+ xn
2
�2+
kn ��n+ xn
2
�2> 0, and thus n+ xn � a <
kn
a:
On the other hand, note that the right inequality in (3) is equivalent to�a�
n+ xn + 1
2
�2<
�n+ xn + 1
2
�2� kn: (4)
Since kn >
�n+ xn
2
�2, we have
a � u+ 1 =jp
kn
k+ 1 �
�n+ xn
2
�+ 1 �
n+ xn + 1
2:
[Ed: The last inequality holds since n+ xn is an integer.]
Hence it su�ces to establish (4) for a = n.
Substituting a = n into the right inequality of (3), we need to show thatkn < nxn+n. From n� 2
pn� 1 < xn < n we get (n� xn)
2 < 4(n� 1)
or (n+ xn)2 + 4 < 4(nxn + n). Hence kn �
�n+ xn
2
�2+ 1 < nxn + n.
Therefore (3) holds, and by setting a = u+ 1; u+ 2; � � � ; n we get
kn
n< xn + 1 <
kn
n� 1< � � � < xn + n� u� 1 <
kn
u+ 1< xn + n� u <
kn
u:
440
Hence
kn
n-
kn
n� 1� � � -
kn
u+ 1-kn
u: (5)
From (2) and (5) we conclude that kn satis�es (1).
Now we show that if k is any integer satisfying (1), then k � kn. To
this end, let x =
�k
n
�.We �rst show that x � xn. Since there exists an
integer z such thatk
n< z <
k
n� 1and x =
�k
n
�< z, we have z � x � 1
and hencek
n� 1> x+ 1. Similarly,
k
n� 2> x+ 2,
k
n� 3> x+ 3, � � � ,
k
1> x+ n� 1. That is, for all a = 1; 2; � � � ; n we have k > (n� a)(x+ a)
or k � (n� a)(x+ a) + 1 = �a2 + a(n� x) + nx+ 1: (6)
Hence
k ��n� x
2
�2+ nx+ 1�
�a�
n� x
2
�2: (7)
Note that x � 1. [Ed: If k � n�1, then 0 <k
n<
k
n� 1� 1, contradicting
k
n-
k
n� 1. Hence k � n].
On the other hand, it is clear that xn � n � 1. Suppose, contrary to
what we claim, that x < xn. Then we have 1 � x < xn � n � 1 and so
2 � n � x < n or 1 �n� x
2<
n
2. (Here we must assume that n � 3.
The case when n = 2 can be treated separately, since it is easy to verify that
k2 = 3.) Hence we may let a =
�n� x
2
�in (6) and (7) and obtain
k ��n� x
2
�2+ nx+ 1�
��n� x
2
���n� x
2
��2: (8)
Since the right side of (7) is an integer and since the last squared term in (8)
is either 0 or1
4we get
k �$�
n� x
2
�2+ nx+ 1
%=
$�n+ x
2
�2+ 1
%: (9)
Thus x =
�k
n
��$1
n
$�n+ x
2
�2+ 1
%%=
$1
n
�n+ x
2
�2+
1
n
%.
441
[Ed: It is known and easy to show that
�bzcn
�=
�z
n
�for all real
numbers z and positive integers n.]
Hence 0 �$1
n
�n+ x
2
�2+
1
n
%� x =
$1
n
�n� x
2
�2+
1
n
%, which
implies that1
n
�n� x
2
�2+ 1
!� 1 or
�n� x
2
�2� n� 1.
Thus n� x < 2pn� 1 or x > n� 2
pn� 1, from which we get x �
bn+1�2pn� 1c = xn, a contradiction. Hence x � xn. Therefore we may
replace x by xn in (6), (7), and (9) and conclude that
k �$�
n+ xn
2
�2+ 1
%= kn. This completes the proof.
Also solved by PETER TINGLEY, student, University of Waterloo, Waterloo,
Ontario. There was one incorrect solution. Tingley gave the answer
kn = nbn� 2pn� 1 + 1c+
$�n� bn� 2
pn� 1 + 1c
2
�2+ 1
%
which is readily seen to be the same as the one obtained by Herzig. The proposer
had conjectured that
kn =
�1 + (n�m)2 if m2 � n� 2
1 + (n�m)2 + (n�m) otherwise
where m =
�1 +
p4n� 7
2
�and had veri�ed it for 2 � n � 600 using a computer.
In a private communication Tingley has actually proved that this conjectured formula
is equivalent to the answer given by Herzig and himself. Interested readers may �nd
the proof of this fact quite challenging.
2273. [1997: 366] Proposed by Tim Cross, King Edward's School,Birmingham, England.
Consider the sequence of positive integers: f1, 12, 123, 1 234, 12 345,: : : g, where the next term is constructed by lengthening the previous term at
its right-hand end by appending the next positive integer. Note that this next
integer occupies only one place, with \carrying" occurring as in addition: thus
the ninth and tenth terms of the sequence are 123 456789 and 1 234 567900
respectively.
Determine which terms of the sequence are divisible by 7.
Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.The sequence fang under consideration satis�es the recurrence
a1 = 1 and an = 10an�1 + n for n � 2:
442
A simple induction argument shows that
81an = 10n+1 � 9n� 10; n 2 N:
Let n 2 N. Applying the Euclidean Algorithm twice, we see that there ex-
ist non-negative integers j; k; r such that 0 � k � 6, 0 � r � 5, and
n = 42j + 6k + r. Since 342j � 36k � 1 (mod 7) by Fermat's Little
Theorem, it follows that
4an � 3r+1 + 2k� 2r � 3 (mod 7):
The following table gives the remainder when the expression on the right
hand side of the above congruence is divided by 7:
rnk 0 1 2 3 4 5 6
0 0 2 4 6 1 3 5
1 4 6 1 3 5 0 2
2 6 1 3 5 0 2 4
3 2 4 6 1 3 5 0
4 1 3 5 0 2 4 6
5 2 4 6 1 3 5 0
Inspecting this table and using the above congruence, we see that an is
divisible by 7 if and only if n � 0; 22; 26; 31; 39; or 41 (mod 42).
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;
MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; HANS ENGELHAUPT, Franz{
Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Cambridge,
UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS,
Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete,
Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's
School, London, England; J.A. MCCALLUM, Medicine Hat, Alberta; JOEL
SCHLOSBERG, student, Robert Louis Stevenson School, New York, NY, USA; TREY
SMITH, GERALD ALLEN, NOEL EVANS, CHARLES DIMINNIE, AND ROGER
ZARNOWSKI (jointly), Angelo State University, San Angelo, Texas; DAVIDR. STONE,
Georgia Southern University, Statesboro, Georgia, USA; PAUL YIU, Florida Atlantic
University, Boca Raton, Florida, USA; and the proposer. There was one incorrect
solution submitted.
2274. [1997: 366] Proposed by V�aclav Kone �cn �y, Ferris State Uni-versity, Big Rapids, Michigan, USA.
A. Letm be a non-negative integer. Find a closed form for
nXk=1
mYj=0
(k+j).
B. Let m 2 f1; 2; 3; 4g. Find a closed form for
nXk=1
mYj=0
(k+ j)2.
443
C?. Let m and �j (j = 0; 1; : : : ;m) be non-negative integers. Prove or
disprove that
nXk=1
mYj=0
(k+ j)�j is divisible by
m+1Yj=0
(n+ j).
I. Solution by Florian Herzig, student, Cambridge, UK.
A. Note that
nXk=1
mYj=0
(k+ j) =
nXk=1
(m+ k)!
(k� 1)!= (m+ 1)!
nXk=1
�m+ k
k � 1
�
= (m+ 1)!
��m+ 1
0
�+
�m+ 2
1
�+
�m+ 3
2
�+ � � �+
�m+ n
n� 1
��:
This combinatorial sum [inside the square brackets] is well-known and can
be evaluated as follows. Note that the �rst two terms add to�m+3
1
�which in
turn adds with the third term to�m+4
2
�, and so on until we obtain the desired
closed form�m+n+1
n�1�in the end. Thus
nXk=1
mYj=0
(k+ j) = (m+ 1)!
�m+ n+ 1
n� 1
�
=(m+ n+ 1)!
(m+ 2)(n� 1)!=
n(n+ 1) : : : (n+m+ 1)
m+ 2:
B. The expressions reduce to sums of the formPn
k=1 km. With the help
of a calculator I got n+ 2
3
!2(3n2 + 6n+ 1)
5form = 1;
n+ 3
4
!12(2n+ 3)(5n2 + 15n+ 1)
35form = 2;
n+ 4
5
!8(35n4 + 280n3 + 685n2 + 500n+ 12)
21form = 3;
n+ 5
6
!40(126n5 + 1575n4 + 6860n3 + 12075n2 + 7024n+ 60)
77form = 4:
C. I assume that \is divisible by" means polynomial division.
Clearly the �j should be positive integers, and in this case I will prove
that the claim is true. De�ne
P (n) =
nXk=1
mYj=0
(k+ j)�j ;
444
a polynomial of degree (Pn
j=1 �j) + 1 (since eachPn
k=1 kl is a polynomial
of degree l+ 1). Then
P (n) =
nXk=�m
mYj=0
(k+ j)�j =
m+n+1Xk=1
mYj=0
(k�m� 1 + j)�j ;
as in the �rst of these sums all the terms for k � 0 vanish. Hence, for all
integers a such that �m � a � 0 we get
P (a) =
m+a+1Xk=1
(k�m� 1)�0(k�m)�1 : : : (k� 1)�m = 0 ;
since each term is zero. [Thus n� a must be a factor of P (n) for each such
a, so P (n) must be divisible by each of n; n + 1; : : : ; n +m. | Ed.] For
n = �m � 1 the above sum is empty and so P (�m � 1) = 0 as well (to
avoid empty sums one can instead write P (n) asPm+n+1
k=0 p(k)�p(0)wherep(k) is the above product). Therefore P (n) is divisible by
Qm+1j=0 (n+ j) as
claimed.
II. Solution to part A byMichael Lambrou, University of Crete, Greece.A. From the identity
mYj=0
(k+ j) =1
m+ 2
24m+1Yj=0
(k+ j)�m+1Yj=0
(k� 1 + j)
35
(easily veri�ed by considering the common factorQm
j=0(k + j) of the two
products on the right), we obtain telescopically
nXk=1
mYj=0
(k+ j) =1
m+ 2
24m+1Yj=0
(n+ j)�m+1Yj=0
j
35 =
1
m+ 2
m+1Yj=0
(n+ j):
[Editorial note: Lambrou also solved parts B and C.]
All three parts also solved by G. P. HENDERSON, Garden Hill, Ontario. PartsA
and B only solved by THEODORE N. CHRONIS, Athens, Greece; WALTHER JANOUS,
Ursulinengymnasium, Innsbruck, Austria; and the proposer.
As Herzig mentions above, part A at least is a fairly familiar result. (For ex-
ample, see formula 2.50, page 50 of Concrete Mathematics by Graham, Knuth andPatashnik.) In fact it was proposed for publication in CRUX in 1991 by an Edmonton
high school student, Jason Colwell, but was not accepted by the then editor.
Chronis notes that, in the solution for part B, when m = 4, the �fth degree
polynomial has 2n+ 5 as a factor.
445
2275. [1997: 366] Proposed byM. Perisastry, Vizianagaram, AndhraPradesh, India.
Let b > 0 and ba � ba for all a > 0. Prove that b = e.
I. Solution by Gerald Allen, Charles Diminnie, Trey Smith and RogerZarnowski (jointly), Angelo State University, San Angelo, TX, USA; RussellEuler and Jawad Sadek (jointly), NW Missouri State University, Maryville,Missouri, USA; Michael Parmenter, Memorial University of Newfoundland,St. John's, Newfoundland; Reza Shahidi, student, University of Waterloo,Waterloo, Ontario; George Tsapakidis, Agrinio, Greece; and John Vlachakis,Athens, Greece.
Let f(x) = bx � bx for x > 0. Since f(x) � 0 for all x > 0 and
f(1) = 0, it follows that f has a relative (as well as an absolute) minimum
at x = 1.
Since f 0(1) exists, we have f 0(1) = 0; that is, b lnb � b = 0. Since
b > 0, we get ln b = 1 or b = e.
II. Solution by Theodore Chronis, student, Aristotle University ofThessaloniki, Greece; Florian Herzig, student, Cambridge, UK; MichaelLambrou, University of Crete, Crete, Greece; Gerry Leversha, St. Paul'sSchool, London, England; Vedula N. Murty, Visakhapatnam, India;Heinz-J �urgen Sei�ert, Berlin, Germany; and David R. Stone, Georgia South-ern University, Statesboro, Georgia, USA.
The given inequality is equivalent to ba�1 � a for all a > 0. Letting
a = 1 +1
nwhere n 2 N, we get b
1n � 1 +
1
n, or b �
�1 +
1
n
�n. Hence
b � limn!1
�1 +
1
n
�n= e.
On the other hand, letting a =
�1 +
1
n
��1=
n
n+ 1, where n 2 N,
we get from b1�a �1
a, that b
1n+1 � 1 +
1
n, or b �
�1 +
1
n
�n+1
. Hence
b � limn!1
�1 +
1
n
�n= e. Therefore, b = e.
Also solved by FRANK P. BATTLES, Massachusetts Maritime Academy, Buz-zards Bay, MA, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; CON
AMORE PROBLEM GROUP, Royal Danish School of Educational Studies, Copen-
hagen, Denmark; LUZ M. DeALBA, Drake University, Des Moines, IA, USA; DAVIDDOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; KEITH EKBLAW,
Walla Walla, Washington, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium,
Bamberg, Germany; F.J. FLANIGAN, San Jose State University, San Jose, Califor-
nia, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD,
New Mexico Highlands University, Las Vegas, NM, USA; WALTHER JANOUS, Ur-
sulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University,
Big Rapids, Michigan, USA; KEE-WAI LAU, Hong Kong; D.J. SMEENK, Zaltbommel,
the Netherlands; DIGBY SMITH, Mount Royal College, Calgary, Alberta; and the
446
proposer. There was one incorrect solution.
Although the problem did not ask to show that the condition b = e is both
necessary and su�cient, a few solvers did provide a proof of the simple fact that
ex � ex for all x > 0.
2277. [1997: 431] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,Alcorc �on, Madrid, Spain.
For n � 1, de�ne
un =
�1
(1; n);
2
(2; n); : : : ;
n� 1
(n� 1; n);
n
(n; n)
�;
where the square brackets [ ] and the parentheses ( ) denote the least com-
mon multiple and greatest common divisor respectively.
For what values of n does the identity un = (n� 1)un�1 hold?
Solution by Florian Herzig, student, Cambridge, UK.
We �rst introduce some notation:
for any prime p, let (n)p = maxf�jp� � ng, [n]p = maxf�jp�divides ng.Thus p[n]p jjn. For each k = 1; 2; � � � ; n we determine a = a(k) such that
pajjk. If a � [n]p, then since pajn, we have pajj(k; n) and so p 6 jk
(k; n).
If a > [n]p, then since p[n]pjk we have p[n]pjj(k; n) and so pa�[n]p jjk
(k; n).
Thus the highest power of p in anyk
(k;n)arises when a = (n)p. This shows
that un =Yp
p(n)p�[n]p, where the product is over all primes. Note that
n� 1 =Yp
p[n�1]p and hence un = (n� 1)un�1 is equivalent to
(n)p� [n]p = (n� 1)p for all primes p (1)
We distinguish two cases:
Case (i) Suppose n is a prime power, say n = qb where q is a prime and
b > 0. For p 6= q, (1) is satis�ed since (n)p = (n� 1)p and [n]p = 0. For
p = q we have n = pb and so (n)p = [n]p = b and (n� 1)p = b� 1. Hence
(1) holds if and only if b� 1 = 0; that is, b = 1.
Case (ii) If n is not a prime power, then (n)p = (n� 1)p for all primes
p. Hence (i) holds if and only if [n]p = 0 for all primes p, and so n = 1.
[Ed: Clearly n = 1 is not a solution, since u0 is unde�ned.]
Therefore un = (n� 1)un�1 if and only if n is a prime.
447
Also solved by ED BARBEAU, University of Toronto, Toronto; NIKOLAOS
DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinengymnasium, Inns-
bruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,USA; JOEL SCHLOSBERG, student, Robert Louis Stevenson School, New York, NY,
USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA,
St. Paul's School, London, England; DAVID R. STONE, Georgia Southern University,Statesboro, Georgia, USA; and the proposer.
From the proof given above, it is not di�cult to see that, in fact, we have
un = [1; 2; � � � ; n]=n. This was explicitly pointed out by Kone �cn �y, Lambrou, and
the proposer, but only Lambrou and the proposer actually gave a proof.
2278. [1997: 431] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,Alcorc �on, Madrid, Spain.
Determine the value of an, which is the number of ordered n{tuples
(k2; k3; : : : ; kn; kn+1) of non-negative integers such that
2k2 + 3k3 + : : :+ nkn + (n+ 1)kn+1 = n+ 1:
I. Solution by Michael Lambrou, University of Crete, Crete, Greece.We show that an = p(n+ 1)� p(n) for n � 1 where p(m) denotes
the number of partitions of m into positive integral parts. Our argument is
based on the well-known observation that to a partition of m where lk k's
appear (k = 1; 2; : : : ; m), so that
1l1 + 2l2 + � � �+mlm = m; (1)
corresponds the ordered m-tuple (l1; l2; : : : ; lm). Conversely, to any given
ordered m-tuple (l1; l2; : : : ; lm) of positive integers satisfying (1), there cor-
responds a partition of m.
For �xed n � 1 consider the partitions of n+ 1 as above. They are of
two types:
(a) those for which the number 1 is absent in the decomposition; or
(b) those for which the number 1 appears at least once.
The number of partitions of type (a) is clearly an. Moreover, for each
partition in case (b), if we delete one 1, we get a partition of n. Conversely,
every partition of n with an extra 1 added on gives a partition of n + 1 of
type (b). Clearly then p(n+ 1) = an + p(n), as required.
II. Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Aus-tria.
We deal more generally with the equation:
k1 + 2k2 + � � �+ (j � 1)kj�1 + (j+ 1)kj+1 + � � �+ (n+ 1)kn+1 = n+ 1
where j 2 f1; 2; : : : ; n+ 1g is �xed and determine the number an(j) of its
non-negative solutions in Zn.
448
For this we recall that (since Euler's days) such problems are dealt with
best by generating functions, namely:
E(x) = (1 + x1 + x2�1 + : : : ) � (1 + x2 + x2�2 + : : : ) � : : :
=1
1� x�
1
1� x2�
1
1� x3: : : =
1Xk=0
p(k)xk
where p(k) denotes the number of partitions of k; that is, the number of
unordered representations of k as k = s1 + s2 + � � �+ se with sj a positive
integer for j = 1; 2; : : : ; e, or equivalently k = 1n1+2n2+ � � �+knk, where
nj � 0 is the number of appearances of summand j. Therefore, all partitions
with summand j forbidden are obtained via
(1�xj)E(x) = (1�xj)1Xk=0
p(k)xk =
j�1Xk=0
p(k)xk+
1Xk=j
�p(k)�p(k�j)
�xk :
Hence the desired amount an(j) equals:
an(j) =
�p(n+ 1); if n+ 1 � j � 1
p(n+ 1)� p(n+ 1� j); if n+ 1 � j
Also solved by HEINZ-J�URGEN SEIFFERT, Berlin, Germany. There was one
incorrect solution submitted.
Janous remarks how his ideas above can be extended to include the case where
the excluded summand can be a subset of the values from 1 to n+ 1.
Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell
Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer
Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil
Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,
J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia
449
THE ACADEMY CORNERNo. 21
Bruce Shawyer
All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7
THE BERNOULLI TRIALS 1998
Hints and Answers
Ian Goulden and Christopher Small
University of Waterloo
The questions were printed earlier this year [1998: 257].
1. False. The diameter of the larger circle is 50.
Let CD = x.
Then AC = 9 + x
and CE = 4+ x.
Consider similar triangles in
AED.
A B
F
E
C D
2. True. LetX be the number of rounds until his �rst error, Y the number
of subsequent rounds until his second error. Then
E(X + Y ) = E(X) + E(Y ) = 2E(X)
But E(X) = 2, which can be found by summing the series.
3. True. Use 1, 2, 4, 8, : : : , 210 and a binary expansion of any number
from 8 to 1998.
4. True. Write the points in coordinates and use the Pythagorean formula.
Everything will cancel!
450
5. True. The volume of the tetrahedron isp2=12. To prove this, imbed
the tetrahedron into a cube so that the vertices of the tetrahedron are
four of the eight vertices of the cube. Show that the tetrahedron has a
volume which is one third the volume of the cube. On the other hand,
the volume of the sphere is 4�r3=3 where r = 1=�.
6. False. Actually 15032 = 2259009 is the smallest.
7. True. Draw a graph of the function in the unit square.
��������
8. True. The choice k = 9 works:9n(n+ 1)
2+ 1 =
(3n+ 1)(3n+ 2)
2.
9. True. The maximum area is achieved with a cyclic quadrilateral. The
area of such a cyclic quadrilateral can be determined by Brahmagupta's
formula
A =p(s� a)(s� b)(s� c)(s� d)
=p4! = 2
p6 .
10. False. Write �(x) = x=(1 + x). The equation can be written as
f [�(x)] = �[f(x)] .
This is clearly satis�ed by f(x) = �n(x). In particular, the choice of
f(x) = x=(1 + x) works.
11. True. 323
= 38 > 2� 29 = 2� 232
.
So 2323
> 22�232
= 4232
> 3232
.
So 32323
> 33232
> 2� 23232
.
Finally, we have 232323
> 22�23232
= 423232
> 323232
.
Obviously, this can be continued.
12. True. WriteZ 10
x dx
ex � 1=
Z 10
x e�x
1� e�xdx
=
Z 10
x(e�x + e�2x + e�3x + � � � ) dx
= 1 +1
22+
1
32+ � � � =
�2
6
451
13. False.
The answer is one square metre
exactly.
To prove this consider the �gure
to the right:
The thin triangle has the re-
quired dimensions and its area is 2
3
2 2
2
2
1
2� 4� 5�
�2� 2 +
1
2� 2� 3 +
1
2� 2� 2
�= 1 .
14. False. Suppose there were such a function.
The function ex � eex
can be seen to be a one-to-one decreasing func-
tion. So f must be one-to-one:
f(x) = f(y) =) f [f(x)] = f [f(y)] =) x = y .
As f is continuous and one-to-one, it must be strictly increasing or
strictly decreasing.
But either way, f [f(x)] must then be a strictly increasing function.
Contradiction.
15. False.
Consider a third random point
X in the circle.
The region R1 corresponds to
the points for X where the an-
gle at X is obtuse. The region
R2 corresponds to the points
for X where the angle at B is
obtuse.
BX
A
C
As the probabilities are the same, the average areas of R1 and R2 must
be the same.
16. False. The second player wins by forcing bilateral symmetry on remain-
ing petals. For example, if the �rst player starts by taking petal 1, thesecond player takes petals 7 and 8 together. If the �rst player chooses
1 and 2 the second player takes petal 8, etc. This ensures that the �rst
player can never take the last petal.
452
THE OLYMPIAD CORNERNo. 194
R.E. Woodrow
All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.
We begin this number with the problems from the Bi-National Israel
Hungary Competition from 1995. My thanks go to Bill Sands, University of
Calgary, who collected these problems for me while assisting at the IMO in
Canada in 1995.
BI-NATIONAL ISRAEL-HUNGARY COMPETITION1995
1. Denote the sum of the �rst n prime numbers by Sn. Prove that
there exists a whole square between Sn and Sn+1.
2. Let P , P1, P2, P3, P4 be �ve points on a circle. Denote the distance
of P from the line PiPk by dik. Prove that d12d34 = d13d24.
3. Consider the polynomials f(x) = ax2 + bx + c which satisfy
jf(x)j � 1 for all x 2 [0; 1]. Find the maximal value of jaj+ jbj+ jcj.
4. Consider a convex polyhedron, whose faces are triangles. Prove thatit is possible to colour the edges by red and blue in a way that one can travel
from any vertex to any other vertex, passing only through red edges, and also
one can travel only through blue edges.
Next we give two rounds of the 31st Spanish Mathematical Olympiad.
Both these contests were collected for me by Bill Sands, University of Calgary,
while he was assisting at the IMO in Canada in 1995. I also received them
from Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid, Spain, one of
the contest organizers. Many thanks.
31st SPANISH MATHEMATICAL OLYMPIADFirst Round: December 2{3, 1994
(Proposed by the Royal Spanish Mathematical Society)
First Day | Time: 4 hours
1. Let a, b, c be distinct real numbers and P (x) a polynomial with real
coe�cients. If:
� the remainder on division of P (x) by x� a equals a,
453
� the remainder on division of P (x) by x� b equals b,
� and the remainder on division of P (x) by x� c equals c;determine the remainder on division of P (x) by (x� a)(x� b)(x� c).
2. Show that, if (x+px2 + 1)(y+
py2 + 1) = 1, then x+ y = 0.
3. The squares of the sides of a triangle ABC are proportional to the
numbers 1, 2, 3.
(a) Show that the angles formed by the medians of ABC are equal to
the angles of ABC.
(b) Show that the triangle whose sides are the medians of ABC is sim-
ilar to ABC.
4. Find the smallest natural numberm such that, for all natural num-
bers n � m, we have n = 5a+ 11b, with a, b integers � 0.
Second Day | Time: 4 hours
5. A subset A �M = f1; 2; 3; : : : ; 11g is good if it has the following
property:
\If 2k 2 A, then 2k� 1 2 A and 2k+ 1 2 A" .
(The empty set andM are good). How many good subsets hasM?
6. Consider the parabolas y = cx2 + d, x = ay2 + b, with c > 0,d < 0, a > 0, b < 0. These parabolas have four common points. Show that
these four points are concyclic.
7. Show that there exists a polynomial P (x), with integer coe�cients,
such that sin 1� is a root of P (x) = 0.
8. An aircraft of the airline \Air Disaster" must y between two cities
with m + n stops. At each stop, the aircraft must load or unload 1 ton of
goods. In m of the stops, the aircraft loads; in n of the stops, the aircraft
unloads. Nobody from the sta� has observed that the aircraft cannot handle
a load of more than k tons (n < k < m+n), and the stops where the plane
loads and unloads are randomly distributed. If the aircraft takes o� with n
tons of goods, �nd the probability of the aircraft arriving at its destination.
31st SPANISH MATHEMATICAL OLYMPIADSecond Round: February 24{25, 1995
First Day | Time: 4 hours
1. Consider sets A of 100 distinct natural numbers, such that the fol-
lowing property holds: \If a, b, c are elements of A (distinct or not), there
exists a non-obtuse triangle of sides a, b, c." Let S(A) be the sum of the
454
perimeters of the triangles considered in the de�nition of A. Find the mini-
mal value of S(A).
2. We have several circles of paper on a plane, such that some of them
are overlapped, but no one circle is contained in another. Show that it is
impossible to form disjoint circles with the pieces which result from cutting
o� the non-overlapped parts and re-assembling them.
3. A line through the barycentre G of the triangle ABC intersects the
side AB at P and the side AC at Q. Show that
PB
PA�QC
QA�
1
4:
Second Day | Time: 4 hours
4. Find all the integer solutions of the equation
p(x+ y) = xy
in which p is a prime number.
5. Show that, if the equations
x3 +mx� n = 0 ;
nx3 � 2m2x2 � 5mnx� 2m3 � n2 = 0 ; (m 6= 0; n 6= 0)
have a common root, then the �rst equation would have two equal roots,
and determine in this case the roots of both equations in terms of n.
6. AB is a �xed segment and C a variable point, internal to AB.
Equilateral triangles ACB0 and CBA0 are constructed, in the same half-
plane de�ned by AB, and another equilateral triangle ABC0 is constructedin the opposite half-plane. Show that:
(a) The lines AA0, BB0 and CC0 are concurrent.
(b) If P is the common point of the lines of (a), �nd the locus of P when
C varies on AB.
(c) The centres A00, B00, C00 of the three equilateral triangles also form
an equilateral triangle.
(d) The points A00, B00, C00 and P are concyclic.
As a �nal problem set for your puzzling pleasure over the Christmas
break we include the Final Round problems of the 46th Polish Mathematical
Olympiad. Again these problems come to us from more than one source.
Bill Sands collected them while assisting at the IMO in Canada. Marcin
E. Kuczma, Warszawa, Poland, one of Poland's premiere contest people, also
sends us copies of contest materials regularly. Again, many thanks.
455
46th POLISH MATHEMATICAL OLYMPIAD 1994{5Problems of the Final Round (March 31-April 1, 1995)
First Day | Time: 5 hours
1. Find the number of those subsets of f1, 2, : : : , 2ng in which the
equation x+ y = 2n+ 1 has no solutions.
2. A convex pentagon is partitioned by its diagonals into eleven re-
gions: one pentagon and ten triangles. What is the maximum number of
those triangles that can have equal areas?
3. Let p � 3 be a given prime number. De�ne a sequence (an) by
an = n for n = 0; 1; 2; : : : ; p� 1 ;
an = an�1 + an�p for n � p :
Determine the remainder left by ap3 on division by p.
Second Day | Time: 5 hours
4. For a �xed integer n � 1 compute the minimum value of the sum
x1 +x22
2+x33
3+ � � �+
xnn
n;
given that x1; x2; : : : ; xn are positive numbers satisfying the condition
1
x1+
1
x2+ � � �+
1
xn= n :
5. Let n and k be positive integers. From an urn containing n tokens
numbered 1 through n, the tokens are drawn one by one without replace-
ment, until a number divisible by k appears. For a given n determine all
those k � n for which the expected number of draws equals exactly k.
6. Let k, l, m be three non-coplanar rays emanating from a common
origin P and let A be a given point on k (other than P ). Show that there
exists exactly one pair of points B, C, with B lying on l and C on m, such
that
PA+AB = PC +CB and PB + BC = PA+ AC :
Next an apology. Somehow when preparing the list of solvers of prob-
lems for the October issue of CRUX withMAYHEM, I omitted listingMiguel
Amengual Covas, Cala Figuera, Mallorca, Spain as a solver of problem 3 of
the Swedish Mathematical Olympiad [1998: 327] and for problem 2 of the
Dutch Mathematical Olympiad [1998: 330]. Sorry!
456
Now we turn to solutions by the readers to problems of the IrishMath-
ematical Olympiad 1994 [1997: 388{389].
1. Let x, y be positive integers with y > 3 and
x2 + y4 = 2[(x� 6)2 + (y+ 1)2] :
Prove that x2 + y4 = 1994.
Solution by Pavlos Maragoudakis, Pireas, Greece.Rewriting we get
x2 � 24x� y4 + 2y2 + 4y + 74 = 0 : (1)
Now (1) has integer solutions only if the discriminant 4(y4� 2y2� 4y+70)is a perfect square. It is easy to prove that for y � 4,
(y2� 2)2 < y4 � 2y2� 4y+ 70 < (y2 + 1)2: (�)
(Indeed (�)() y2� 2y+33 > 0 and 4y(y+1) > 69. The �rst inequalityis true. Since y � 4, 4y(y + 1) � 4 � 4 � 5 = 80 > 69.) The only perfect
squares between (y2� 2)2 and (y2 + 1)2 are (y2� 1)2 and (y2)2. Now
(y2� 1)2 = y4� 2y2� 4y + 70() y =69
462 Z ;
and y4� 2y2� 4y+70 = y4() y2+2y� 35 = 0() y = 5 or y = �7.Thus, y = 5. Now (1) gives x = 37 and
x2 + y4 = 372 + 54 = 1369 + 625 = 1994 :
Comment by Jim Totten, The University College of the Cariboo, Kam-loops, BC.
The result also works for y = 1 and y = 2 as well, but fails for y = 3with x = 1.
2. Let A, B, C be three collinear points with B between A and C.
Equilateral triangles ABD, BCE, CAF are constructed with D, E on one
side of the line AC and F on the opposite side. Prove that the centroids
of the triangles are the vertices of an equilateral triangle. Prove that the
centroid of this triangle lies on the line AC.
Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;and by Pavlos Maragoudakis, Pireas, Greece. We give the solution ofAmengual Covas.
We introduce a rectangular Cartesian system with origin at B and
x{axis along AC. Let AB = a and BC = b. The centroids O1, O2, O3
of equilateral triangles ABD, BCE and CAF are
G1 =
��a2;a
2p3
�; G2 =
�b
2;
b
2p3
�and
457
G3 =
��a+ b
2;�
a+ b
2p3
�:
Hence the centroid G of triangle4G1G2G3 is
G =
��a+ b
3; 0
�:
It is straightforward to verify that
G1G2 = G2G3 = G3G1 =
sa2 + ab+ b2
3:
Therefore 4G1G2G3 is equilateral.
Finally, since the y{coordinate ofG is 0, clearlyG is on the x{axis; that
is to say, G lies on the line AC.
-
6
x{axisp
B G C(b; 0)
A(�a; 0)
D
G2
G1
G3
3. Determine with proof all real polynomials f(x) satisfying the equa-tion
f(x2) = f(x)f(x� 1) :
Solution by Pavlos Maragoudakis, Pireas, Greece.We will prove that f(x) = 0 or f(x) = (x2 + x+1)k, k = 0; 1; 2; : : : .
Consider any (possibly complex) root p of f(x). Then
f(p2) = f(p) � f(p� 1) = 0 � f(p� 1) = 0
and
f((p+ 1)2) = f(p+ 1) � f(p) = f(p+ 1) � 0 = 0 :
458
So p2, (p+1)2 are also roots of f(x). Thus p2n
and (p+1)2n
are roots of f(x),n = 0; 1; 2; : : : . If jpj 6= 1 or jpj 6= jp+ 1j then we get an in�nite number of
roots, so f(x) is a constant polynomial, and having a root p, f(x)� 0.
If jpj 6= 1 or jp+ 1+ 6= 1; that is, if jpj = 1 = jp + 1j then p � p = 1and p � p = (p + 1)(p + 1), so p + p = �1 and p = �(p + 1), and now
p(�p�1) = 1. Therefore p2+p+1 = 0. It follows that f(x) = (x2+x+1)k,for some k � 1.
On the other hand if f(x) has no roots, then f(x) = c 6= 0, is a non-
zero constant. Then f(x2) = f(x) � f(x � 1) gives c = c � c, and c 6= 0gives c = 1. Thus f(x) = (x2 + x + 1)0. In any case f(x) = 0 or
f(x) = (x2 + x+ 1)k for some k = 0; 1; 2; : : : .
6. A sequence fxng is de�ned by the rules
x1 = 2
and
nxn = 2(2n� 1)xn�1 ; n = 2; 3; : : : :
Prove that xn is an integer for every positive integer n.
Solutions by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta; and by Pavlos Maragoudakis, Pireas, Greece. We give Maragoudakis'solution.
Nowx2 = 2 � 3
2x1
x3 = 2 � 53x2
� � �
xn�1 = 2 � 2n�3n�1 xn�2
xn = 2 � 2n�1n
xn�1
It follows that
xn = 2n�1(2n� 1)(2n� 3) � � � 5 � 3
n � (n� 1) � � � 3 � 2� 2
=(2n)!
(n!)2=
�2n
n
�2 Z
since�2n
n
�is a binomial coe�cient.
7. Let p, q, r be distinct real numbers which satisfy the equations
q = p(4� p) ;
r = q(4� q) ;
p = r(4� r) :Find all possible values of p+ q+ r.
459
Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta.
From the discriminant of each quadratic equation, it follows that
p, q, r are all less than 4. If one of p, q, r is zero, all are zero, so we now
assume none are zero. Also, it follows that p, q, r all have the same sign.
From the product of the three equations we get 1 = (4� p)(4� q)(4� r)so that p, q, r are all positive.
We now let p = 4sin2 �, so then successively, q = 4 sin2 2�,r = 4 sin2 4�, p = 4sin2 8�. Hence, sin � = � sin 8� so that we have
(sin7�=2)(cos 9�=2) = 0 or (sin9�=2)(cos 7�=2) = 0 :
Solving for � leads to only the following possible non-zero values of p+q+r :4(sin2 �=7 + sin2 2�=7 + sin2 3�=7);4(sin2 �=9 + sin2 2�=9 + sin2 4�=9)and 3 � 4 sin2 �=3; that is, 9.
9. Let w, a, b, c be distinct real numbers with the property that there
exist real numbers x, y, z for which the following equations hold:
x + y + z = 1xa2 + yb2+ zc2 = w2
xa3 + yb3+ zc3 = w3
xa4 + yb4+ zc4 = w4:
Express w in terms of a, b, c.
Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta.
As known, the following determinant must vanish:��������1 1 1 1a2 b2 c2 w2
a3 b3 c3 w3
a4 b4 c4 w4
��������Also, from a known expansion theorem [1] for \alternant" determinants, we
have
(w� a)(w� b)(w� c)(a� b)(a� c)(b� c)(w(bc+ ca+ ab)+ abc) = 0 :
Hence, w = a, or b, or c, or �abc=[bc+ ca + ab].Reference
[1] T. Muir, A Treatise on the Theory of Determinants, Dover, N.Y., p. 333,#337.
Now we turn to readers' solutions of problems from the December 1997
number of the Corner. We give some solutions to problems proposed to
the Jury but not used at the 37th International Olympiad at Mumbai, India
[1997; 450{453].
460
1. Let a, b and c be positive real numbers such that abc = 1. Prove
thatab
a5 + b5 + ab+
bc
b5 + c5 + bc+
ca
c5 + a5 + ca� 1 :
When does equality hold?
Solutions by Pierre Bornsztein, Courdimanche, France; and by PanosE. Tsaoussoglou, Athens, Greece. We give the solution by Tsaoussoglou.
We note �rst that
a5 + b5
2��a3 + b3
2
��a2 + b2
2
�;
since a5 � a3b2 � a2b3 + b5 = (a � b)2(a + b)(a2 + ab + b2) � 0 with
equality if and only if a = b.
Similarlya3 + b3
2��a+ b
2
��a2 + b2
2
�
because a3 � a2b � ab2 + b3 = (a � b)2(a + b) � 0, with equality if and
only if a = b. Thus
a5 + b5
2�
�a3 + b3
2
��a2 + b2
2
�� ab
�a3 + b3
2
�
� ab
�a+ b
2
��a2 + b2
2
��
a2b2(a+ b)
2:
It is enough, therefore, to prove
ab
ab(a+ b)ab+ ab+
bc
bc(b+ c)bc+ bc+
ca
ca(c+ a)ca+ ca� 1 ;
or
1
ab(a+ b) + abc+
1
bc(b+ c) + abc+
1
ca(c+ a) + abc� 1 :
Equivalently,
1
ab(a+ b+ c)+
1
bc(a+ b+ c)+
1
ca(a+ b+ c)� 1 ;
orc
abc(a+ b+ c)+
a
abc(a+ b+ c)+
b
abc(a+ b+ c)� 1 :
Again, because abc = 1 we get
a+ b+ c
a+ b+ c� 1 ;
461
which is true.
The equality requires a = b = c = 1.
2. Let a1 � a2 � � � � � an be real numbers such that for all integers
k > 0,
ak1 + ak2 + � � �+ akn � 0 :
Let p = maxfja1j; : : : ; janjg. Prove that p = a1 and that
(x� a1)(x� a2) � � � (x� an) � xn � an1
for all x > a1.
Solution by Pierre Bornsztein, Courdimanche, France.
Pour k 2 N�, on note Sk = ak1 + � � �+ akn.
On a S1 � 0, donc na1 � s1 � 0 et alors a1 � 0.
Si an � 0, a1 � � � � � an � 0, donc p = a1.
Par contre, si an < 0, par l'absurde, supposons que p 6= a1.
Alors p = janj. Pour i 2 f1; : : : ; n�1g, si jaij < janj, limk!1�aian
�k=
0. Cependant, si jaij = janj, limk!1�aian
�k= 1.
Par suite pour
Tk �nXi=1
�ai
an
�k= 1 +
n�1Xi=1
�ai
an
�k;
on a limk!1 Tk = l avec l 2 [1;+1). En particulier, il existe k 2 N� tel queT2k+1>0 et comme an<0 on a a2k+1
n <0, d'o �u s2k+1=a2k+1n � T2k+1<0,
qui est une contradiction. Par cons �equent, p = a1.
Soit x > a1. Alors x� ai > 0 pour i = 1; 2; : : : ; n.
D'apr �es AM/GM
nYi=2
(x� ai) ��Pn
i=2(x� ai)
n� 1
�n�1
=
x�
1
n� 1
nXi=2
ai
!n�1:
Or s1 � 0, donc a1 � �Pn
i=2 ai, et alors
nYi=2
(x� ai) ��x+
a1
n� 1
�n�1=
n�1Xi=0
�n� 1
i
��a1
n� 1
�ixn�1�i :
462
Mais pour i 2 f1;2; : : : ; n � 1g,�n�1i
�1
(n�1)i �(n�1)(n�2):::(n�i)
(n�1)i � 1:
DoncnYi=2
(x� ai) �n�1Xi=0
ai1xn�1�i
et alors
nYi=1
(x� ai) � (x� a1)n�1Xi=0
ai1xn�1�i
= xn � an1 :
3. Let a > 2 be given, and de�ne recursively:
a0 = 1; a1 = a; an+1 =
a2n
a2n�1� 2
!an :
Show that for all integers k > 0, we have
1
a0+
1
a1+
1
a2+ � � �+
1
ak<
1
2
�2 + a�
pa2 � 4
�:
Solution by Pierre Bornsztein, Courdimanche, France.Comme a > 2, il existe b > 1 tel que a = b+ 1
bet on a
b2 � ab+ 1 = 0 ;
c. �a.d. b =a+pa2 � 4
2(b > 1) ;
et comme le produit des racines est 1,
1
b=
a�pa2 � 4
2
d'o �u1
2(2 + a�
pa2 � 4) = 1 +
1
b:
On pose Sn =Pn
i=01ai
pour n � 1.
On veut donc prouver que Sn < 1 + 1b.
Or a1 = b+ 1b, donc a2 = (b+ 1
b)(b2+ 1
b2) et par une r �ecurrence rapide
pour n � 1 :
an = (b+1
b)
�b2 +
1
b2
�: : :
�b2
n�1
+1
b2n�1
�:
463
On en d �eduit que
an =b2 + 1
b�b4 + 1
b2: : :
b2
n
+ 1
b2n�1
!
=(b2 + 1)(b4 + 1) � � � (b2n + 1)
b2n�1
et donc pour n � 2
Sn = 1 +b
b2 + 1+
nXi=2
b2i�1
(b2 + 1) � � � (b2i + 1):
Or1
(b2 + 1) � � � (b2i�1 + 1)=
b2i
+ 1
(b2 + 1) � � � (b2i + 1)
=b2
i
(b2 + 1) � � � (b2i + 1)+
1
(b2 + 1) � � � (b2i + 1);
d'o �u pour i � 2
b2i�1
(b2 + 1) � � � (b2i + 1)
=1
b
�1
(b2 + 1) � � � (b2i�1 + 1)�
1
(b2 + 1) � � � (b2i + 1)
�
et ainsi
Sn = 1 +b
b2 + 1+
1
b
�1
b2 + 1�
1
(b2 + 1) � � � (b2n + 1)
�
< 1 +b
b2 + 1+
1
b(b2 + 1)
= 1 +b
b2 + 1+
1
b�
b
b2 + 1
d'o �u Sn < 1 +1
bpour n � 2 ;
et comme S1 = 1 +1
a= 1+
b
b2 + 1< 1 +
1
b;
le r �esultat est vrai pour tout n � 1.
Remarque : Puisque b > 1, limn!+1 b2n
= +1, et donc limn!1 Sn =1 + 1
b.
4. Let a1; a2; : : : ; an be non-negative real numbers, not all zero.
(a) Prove that xn� a1xn�1 � � � � � an�1x� an = 0 has precisely one
positive real root.
464
(b) Let A =Pn
j=1 aj and B =Pn
j=1 jaj , and let R be the positive real
root of the equation in (a). Prove that AA � RB.
Solution by Pierre Bornsztein, Courdimanche, France.(a) Pour x > 0, xn � a1x
n�1 � : : :� an = 0 est �equivalent �a
1� a1x� � � � � an
xn= 0.
Si l'on pose f(x) = 1�Pn
i=1aixi, alors :
(i) f est d �erivable sur R+ et pour x > 0, f 0(x) =Pn
i=1iaixi+1 > 0
puisqueai � 0, non tous nuls.Donc, f est continue, et strictement croissante
sur R+�.
(ii) De plus limx!0+
f(x) = �1; limx!+1
f(x) = 1 . If existe un
unique r �eel strictement positif, not �e R, tel que f(R) = 0.
(b) Pour j 2 f1; : : : ; ng, on pose xj =ajA� 0 et alors
Pn
j=1 xj = 1.
Comme ln est une fonction concave sur R+�, on a
nXj=1
xj ln
�A
Rj
�� ln
0@ nXj=1
xjA
Rj
1A
= ln
0@ nXj=1
aj
Rj
1A
= ln(1) (cf. (a))
= 0 :
DoncnXj=1
[xj ln(A)� jxj ln(R)] � 0 ;
c. �a.d.
ln(A) � A � ln(R) � Bou encore
ln(AA) � ln(RB)
d'o �u le r �esultat suit.
5. Let P (x) be the real polynomial, P (x) = ax3+bx2+cx+d. Provethat if jP (x)j � 1 for all x such that jxj � 1, then
jaj+ jbj+ jcj+ jdj � 7 :
Solution by Pierre Bornsztein, Courdimanche, France.On pose s(P ) = jaj+ jbj+ jcj + jdj � 0.
Quitte �a changerP en (�P ) et/ou x en (�x) on peut toujours supposerque a � 0 et b � 0.
465
On a P (0) = d, P (1) = a + b + c + d, P (�1) = �a + b � c + d,
8P (12) = a+ 2b+ 4c+ 8d, 8P (�1
2) = �a+ 2b� 4c+ 8d, d'o �u
a = 2
3P (1)� 2
3P (�1) + 4
3P (1
2)� 4
3P (�1
2)
b = 2
3P (1) + 2
3P (�1)� 2
3P (1
2)� 2
3P (�1
2)
c = �1
6P (1) + 1
6P (�1) + 4
3P (1
2)� 4
3P (�1
2)
d = �1
6P (1)� 1
6P (�1) + 2
3P (1
2) + 2
3P (�1
2) = P (0) :
Maintenant si c � 0 et d � 0, S(P ) = a+ b+ c+ d = P (1) � 1.
Cependant si c � 0 et d � 0, on obtient
s(P ) = a+ b+ c� d
= a+ b+ c+ d� 2d
= P (1)� 2P (0)
d'o �u S(P ) � jP (1)j+ 2jP (0)j = 3.
On proc �ede au cas c � 0 et d � 0, o �u
S(P ) = a+ b� c+ d
= a+ b+ c+ d� 2c
et d'apr �es les relations pour a, b, c, d ci-haut,
S(P ) =4
3P (1)�
1
3P (�1)�
8
3P
�1
2
�+
8
3P
��1
2
�
�4
3+
1
3+
8
3+
8
3;
c. �a.d. S(P ) � 7 :
En�n, si c � 0 et d � 0, S(P ) = a+b�c�d, et suivant la meme proc �edure,
on obtient
S(P ) =5
3P (1)� 4P
�1
2
�+
4
3P
��1
2
�
�5
3+ 4 +
4
3;
c. �a.d: S(P ) � 7 :
Finalement, dans tous les cas S(P ) � 7.
Remarque : On peut montrer (voir Olympiades su �edoises 1965 { �nale) que
pour un tel polynome, on a toujours jaj � 4 (ce qui est imm�ediat avec
l'expression de a ci-haut).
Ces deux in �equalit �es sont simultan �ement v �eri� �ees par
P (x) = 4x3 � 3x, qui respecte jP (x)j � 1 pour jxj � 1.
466
7. Let f be a function from the set of real numbers R into itself such
that for all x 2 R, we have jf(x)j � 1 and
f
�x+
13
42
�+ f(x) = f
�x+
1
6
�+ f
�x+
1
7
�:
Prove that f is a periodic function (that is, there exists a non-zero real number
c such that f(x+ c) = f(x) for all x 2 R).Solutions by Mohammed Aassila, Strasbourg, France; and by Pierre
Bornsztein, Courdimanche, France. We give Aassila's solution.We will prove that f is 1-periodic. We have
f
�x+
k
6+l + 1
7
�+ f
�x+
k+ 1
6+
l
7
�
= f
�x+
k
6+
l
7
�+ f
�x+
k + 1
6+l+ 1
7
�:
If k runs through 1; 2; : : : ;m�1, wherem 2 N, and adding these equations
we obtain
f
�x+
l + 1
7
�+ f
�x+
m
6+
l
7
�= f
�x+
m
6
�+ f
�x+
m
6+l + 1
7
�:
Similarly when l runs 1; 2; : : : ; n� 1, (n 2 N) and adding these equa-
tions, we obtain
f
�x+
n
7
�+ f
�x+
m
6
�= f(x) + f
�x+
m
6+n
7
�:
We choose n = 7 andm = 6, and �nd
2f(x+ 1) = f(x) + f(x+ 2) :
This means that the sequence f(x+n) is an arithmetic sequence with com-
mon di�erence f(x + 1) � f(x). But, since f is bounded, we must have
f(x + 1) � f(x) = 0. Hence 1 is a period of f . Finally, 1 is the \best"
period because the function
f(x) =f6xg+ f7xg
2
satis�es all the hypotheses of the problem. (Here, fxg denotes the fractionalpart of x.)
8. Let the sequence a(n), n = 1; 2; 3; : : : ; be generated as follows:
a(1) = 0, and for n > 1,
a(n) = a([n=2]) + (�1)n(n+1)=2 :
(Here [t] means the greatest integer less than or equal to t.)
467
(a) Determine the maximum and minimum value of a(n) over n � 1996 and
�nd all n � 1996 for which these extreme values are attained.
(b) How many terms a(n), n � 1996, are equal to 0?
Solution by Pierre Bornsztein, Courdimanche, France.Parti (a) On v �eri�e facilement que a1 = 0, a2 = �1, a3 = 1 et, pour
k � 1
a4k = a2k + 1a4k+1 = a2k � 1a4k+2 = a2k+1 � 1a4k+3 = a2k+1 + 1
9>>=>>; (1)
Pour p 2 N�, on pose Ep = f2p; 2p + 1; : : : ; 2p+1 � 1g.Pour p = 1, on pose '1(2) = (�1), '1(3) = (1), et pour p = 2, on pose
'2(4) = (�1; 1), '2(5) = (�1;�1), '2(6) = (1;�1), '2(7) = (1; 1).
On proc �ede �a construire 'p : Ep ! Fp pour p � 2 comme suite o �u
Fp = f("1; "2; : : : ; "p) j 8i 2 f1; : : : ; pg, "i 2 f�1; 1gg. Alors pour
n 2 Ep+1�n
2
�2 Ep; et on pose 'p+1(n) = ("1; : : : ; "p; "p+1)
o �u
("1; : : : ; "p) = 'p
��n
2
��et "p+1 = (�1)(n(n+1))=2 :
On construit ainsi une application 'p+1 : Ep+1 ! Fp+1.
Propri �et �e 1. Pour p � 1, 'p est une bijection de Ep sur Fp.
Preuve. Par r �ecurrence sur p, '1 est clairement une bijection de E1 sur F1.
Supposons que 'p soit une bijection de Ep sur Fp pour p � 1 �x �e.
Soient n, k dans Ep+1 tels que 'p+1(n) = 'p+1(k). On peut toujours sup-
poser que n � k. Par l'absurde, si n > k, 'p+1(n) = 'p+1(k) entraine'p([
n2]) = 'p([
k2]) et comme 'p est bijective, [
n2] = [k
2], d'o �u n = k+1, car
n, k sont entiers et si n � k+ 2 alors [n2] > [k
2].
Or si k est impair, k = 2q + 1 alors n = 2q + 2 et [n2] = q + 1 6= q = [k
2],
qui est contradictoire.
Si, cependant, k = 4l, alors n = 4l + 1 d'o �uk(k+1)
2= 2q(4q+ 1), qui est
pair etn(n+1)
2= (2q+ 1)(4q+ 1) qui est impair. Donc "p+1(k) 6= "p+1(n)
qui est aussi contradictoire.
Quand k = rq + 2, alors n = 4q + 3 d'o �uk(k+1)
2= (2q + 1)(4q + 3) qui
est impair etn(n+1)
2= (4q+3)(2q+2) qui est pair, et on obtient la meme
contradiction.
468
Finalement n = k, et 'p+1 est injective. Car Ep+1 et Fp+1 ont la meme
cardinalit �e, 2p, 'p+1 est bijective.
Propri �et �e 2. 8p 2 N, p � 1, 8n 2 Ep, 'p(n) = ("1; : : : "p). Alors
an =Pp
i=1 "i.
Preuve. R �ecurrence imm�ediate sur p en utilisant la construction de 'p et
an = a[n=2] + (�1)(n(n+1))=2.
Propri �et �e 3. 8p 2 N�, 8n 2 Ep, an � p mod 2.
Preuve. Par r �ecurrence sur p. C'est imm�ediat pour p = 1. Supposons le
r �esultat vrai pour p � 1 �x �e. Alors pour n 2 Ep+1 on a [n2] 2 Ep et
an = a[n=2] + (�1)(n(n+1))=2 � a[n=2] + 1 � p + 1 mod 2 d'o �u le r �esultat
au rang p+ 1.
Propri �et �e 4. Pour n � 1, a2n�1 = n� 1.
Preuve. Par r �ecurrence sur n. C'est imm�ediat pour n = 1 et n = 2. Suppo-sons le r �esultat vrai pour n � 2 �x �e. Alors, d'apr �es (1)
a2n+1�1 = a4(2n�1�1)+3 = a2(2n�1�1)+1 + 1
= a2n�1 + 1 = n ;
d'o �u la propriet �e au rang n+ 1.
On a 210 = 1024 < 1996 < 2047 = 211 � 1.
Or si n < 29 � 1, n = 1 ou n 2 Ep avec p � 8, donc an = 0 ou
an = 'p(n) � 8 (cf P2).
Cependant si n 2 E9, an = '9(n) � 9 (d'apr �es P2) avec �egalit �e ssi n =210 � 1 (d'apr �es P4 et car '9 est bijective).
Si n 2 E10, an = '10(n) � 10 et an est pair (cf. P2 et P3) avec an = 10 ssi
n = 211 � 1 (P4 et '10 bijective).
d'o �u an � 9 pour n � 1996.
Finalement la valeur maximale de an, pour n � 1996 est 9, avec an = 9 ssi
n = 1023.
On passe des valeurs de an pour n 2 Ep �a celles de an pour n 2 Ep+1
en ajoutant ou en retranchant 1. On peut ainsi pr �evoir que l'on obtiendra
les valeurs minimales en utilisant une suite (Un) d'indices avec U1 = 2et pour n � 1
Un+1 =
(2Un + 1 si Un est pair,
2Un si Un est impair.
Propri �et �e 5. Pour p � 1,
(a) Up et p sont des entiers de parit �es contraires.
469
(b) Up = 2p+2�13
si p est pair ; Up = sp+2�23
si p est impair.
Preuve. R �ecurrence sur p.
Propri �et �e 6. Pour n � 1, aUn = �n.
Preuve. Il est clair que pour tout n � 1,hUn+1
2
i= Un (par d �e�nition de
Un+1).
Pour n � 1, quand n est pair
Un(Un + 1)
2=
2n+2 � 1
3�2n+1 + 1
3est impair ;
et de meme quand n est impair
Un(Un + 1)
2=
2n+1 � 1
3�2n+2 + 1
3qui est aussi impair :
Donc pour tout n � 1
(�1)Un(Un+1)
2 = �1 :Alors pour tout n � 1,
aUn+1= aUn � 1
et comme aU1 = a2 = �1, une r �ecurrence imm�ediate sur n permet de
conclure que
� si n � 210 � 1, alors n = 1 o �u n 2 Ep avec p � 9 donc an = 0 ou
an = 'p(n) � �9 (cf. P2)
� si n 2 E10, alors an = '10(n) � �10 et an est pair (cf. P2 et P3). De plus
U10 = 1365 (cf. P5) donc U10 2 E10, et a1365 = aU10 = �10 (cf. P6)
an = �10 ssi n = 1365 :
Finalement la valeur minimale de an, pour n � 1996, est �10, avec
an = �10 ssi n = 1365.
Parti (b) Remarquons d'abord que a1 = 0.
Si n 2 Ep, 'p(n) = ("1; : : : "p) o �u "i 2 f�1; 1g et an = "1 + � � � + "p(cf. P2). Donc an = 0 ssi il y a autant de "i = 1 que de "j = �1. Comme
'p est une bijection de Ep sur Fp, il y a autant de n 2 Ep (pour p � �x �e)
tels que an = 0 que de fa�cons de choisir p
2places pour les \+1" parmi les p
places possibles.
Ainsi, d'un part si p est impair 8n 2 Ep, an 6= 0 (on retrouve P3).
D'autre part,
p = 2 donne C12 = 2 possibilit �es,
p = 4 donne C24 = 6 possibilit �es,
p = 6 donne C36 = 20 possibilit �es,
470
p = 8 donne C48 = 70 possibilit �es,
p = 10 donne C510 = 252 possibilit �es.
Mais 211 � 1 = 2047 > 1996. Il reste �a d �eterminer combien de
n 2 f1997; : : : ; 2047g v �eri�ent an = 0.
Or
a62 = a30 + 1 = a15 = a7 + 1 = 3 (on utilise (1)), et
a63 = a26�1 = 5 (cf. P4):
En utilisant (1) on en d �eduit
a124 = 4 = a126; a125 = 2; a127 = 6 :
Puisa249 = a251 = a253 = 3; a250 = 1;a252 = a254 = 5; a255 = 7:
Et encore
a499 = a503 = a505 = a507 = a509 = 4;a504 = a508 = a510 = 6;a511 = 8;a500 = a502 = a506 = 2;a501 = 0:
Mais si n 2 f1997; : : : ; 2047g alors [n2] 2 f998; : : : ; 1023g ;
c. �a.d.
�1
2
�n
2
��2 f499; : : : ; 511g :
Donc an = 0 ssi
an = a501 + "9 + "10 avec "9 + "10 = 0 (deux possibilit �es) ,
ou
an = ak + "9 + "10 avec k 2 f500;502; 506g et "9 = "10 = �1(trois possibiliti �es) :
Il ya a donc 5 indices n 2 f1997; : : : ; 2047g tels que an = 0 et comme
1 + 2 + 6 + 20 + 70 + 252� 5 = 346 ;
le nombre de n � 1996 tels que an = 0 est N = 346.
9. Let triangle ABC have orthocentre H, and let P be a point on its
circumcircle, distinct from A, B, C. Let E be the foot of the altitudeBH, let
PAQB and PARC be parallelograms, and let AQ meet HR in X. Prove
that EX is parallel to AP .
471
Solution by Toshio Seimiya, Kawasaki, Japan.
A
B C
E
P
Q R
X
H
Since AXkBP and A, B, P , C are concyclic we have
\XAP = \APB = \ACB :
As AH ? BC and BH ? AC, we get \ACB = \AHE, so that
\XAP = \AHE : (1)
Since APCR is a parallelogram and A, B, P , C are concyclic we have
\ARC = \APC = \ABC. As AH ? BC and CH ? AB, we get
\AHC + \ABC = 180� ;
so that
\AHC + \ARC = 180� :
Hence A, H, C, R are concyclic.
It follows that
\AHR = \ACR = \CAP : (2)
From (1) and (2) we have
\XAE = \XAP � \CAP = \AHE � \AHR = \XHE :
Hence X, A, H, E are concyclic, and
\AEX = \AHX = \AHR :
Hence we have, from (2), that \AEX = \CAP . Thus EXkAP .
472
10. Let ABC be an acute-angled triangle with jBCj > jCAj, andlet O be the circumcentre, H its orthocentre, and F the foot of its altitude
CH. Let the perpendicular to OF at F meet the side CA at P . Prove that
\FHP = \BAC.
Solutions by Toshio Seimiya, Kawasaki, Japan; and by D.J. Smeenk,Zaltbommel, the Netherlands. We give Smeenk's solution.
C
A BMF
P
Q
H
��
�
O
�2��! �
2��
�2��!
�2�(���) !
We denote \CFP = \OFB = �. M is the mid-point of AB. Now
OM = R cos ; FM = R sin(�� �) ;
so that
tan' =cos
sin(�� �): (1)
From the Law of Sines in4CPF ,
CF = 2R sin� sin�
\FCP =�
2� � (2)
\FPC =�
2+ �� '
so CP : CF = sin' : cos(�� ').
473
With (2)
CP =2R sin� sin� sin'
cos(�� ')=
2R sin� sin�
cos� cot'+ sin(�): (3)
From (1) and (3),
CP =2R sin� cos
sin�(sin2�� cos2 �)=�2R sin� cos
sin� cos 2�(4)
OQ ? OB; \OBQ =�
2� � =) \OQB = � :
Now
CQ = a� QB = R
�2 sin��
1
sin�
�= �R
cos 2�
sin�:
Furthermore, CH = 2R cos , CO = R.
It is easy to verify that CP : CH = CO : CQ; and
�2Rsin� cos
sin� cos 2�: 2R cos = R :
�R cos 2�
sin�: (5)
We also have
\PCH = \OCQ =�
2� � : (6)
From (5) and (6), we have that 4PCH and 4OCQ are similar. Thus
\PHC = \OQC = � � �. Thus \FHP = �.
11. Let ABC be equilateral, and let P be a point in its interior. Let
the lines AP , BP , CP meet the sides BC, CA, AB in the points A1, B1,
C1 respectively. Prove that
A1B1 � B1C1 � C1A1 � A1B � B1C � C1A :
Solutions by Toshio Seimiya, Kawasaki, Japan; and by PierreBornsztein, Courdimanche, France. We give Seimiya's solution.
A
B CA1
B1
C1
P
a a0
c0
c
b0
b
474
We put A1B = a, A1C = a0, B1C = b, B1A = b0, C1A = c and
C1B = c0. Then we have by Ceva's Theorem
abc = a0b0c0 : (1)
Since \B1AC1 = 60� we have
B1C21 = (b0)2 + c2 � 2b0c cos 60�
= (b0)2 + c2 � b0c � b0c:
Similarly we have
C1A21 � c0a and A1B
21 � a0b :
Multiplying these three inequalities, we get
B1C21 � C1A
21 � A1B
21 � b0c � c0a � a0b : (2)
From (1) and (2) we have
B1C21 �C1A
21 � A1B
21 � a2b2c2 :
Thus we have
B1C1 � C1A1 � A1B1 � abc :
That is
B1C1 � C1A1 � A1B1 � A1B � B1C �C1A :
That completes the Corner for this issue. Send me your nice solutions
as well as Olympiad contests and materials.
Professor Dan Pedoe
The Editors of CRUX with MAYHEM are saddened to learn that
Dan Pedoe died a few weeks ago, on 27 October 1998. He made major
contributions to the success of CRUX during its �rst ten years, and always
maintained his interest in the journal.
Readers may be interested to read his autobiographical notes in the
recent COLLEGE MATH. JOURNAL 29:3 (May 1998) 170-188.
475
BOOK REVIEWS
Edited by ANDY LIU
This is my last column as editor of the Book Reviews column. Looking
back over the past �ve years, I am happy that many books have been brought
to the attention of the readers, obscure titles as well as famed tomes. While
some are of marginal quality, and it is not our duty to just sing praises here,
the great majority are outstanding publications. They are tremendous re-
sources in our hands. Let us make good use of them.
Grateful acknowledgement is due to the publishers who provided the
volumes for review in the �rst place. The Mathematical Association of Amer-
ica should be singled out for supplying the column with every title it has
published in popular mathematics, and it is an ample and steady source.
Gratitude is owed to all reviewers, solicited from as wide a spectrum as pos-
sible, particularly geographically. My friend and mentor Murray Klamkin
regularly graced these pages with his much-esteemed opinion, and deserves
a special thank you.
Book Reviews columns as a rule do not generate much fan mail. How-
ever, I have been pleasantly surprised how often I hear remarks from satis-
�ed readers, o�ering corrections, suggesting titles, and providing construc-
tive criticism. Recently, a reader from France called and pointed out that
the Mathematical Association of America has not acquired Martin Gardner's
Sixth Book of Mathematical Diversions from Scienti�c American, as claimed
in the Mini-reviews in the February issue, but the right belongs to the Uni-
versity of Chicago Press.
The Last Recreations by Martin Gardner,
published by Springer-Verlag, New York, 1997,
ISBN# 0-387-94929-1, hardcover, 392+ pages, $25.00.
Reviewed by Andy Liu.
This is the �fteenth and last of Martin's Scienti�c American antholo-
gies, but the �rst with this publishing giant. This may have something to do
with the move from W. H. Freeman to Springer-Verlag of Jerry Lyons, the
outstanding editor of mathematics books. The subtitle is Hydras, Eggs, and
Other Mathematical Mysti�cations.
The material is from the Scienti�c American columns from December
1979 to his retirement in December 1981, though that particular column,
titled \The La�er Curve", was in an earlier volume, Knotted Doughnuts.
Included also are three chapters Martin later wrote as guest columnist, on
August and September 1983 as well as June, 1986. The last chapter, titled
\Trivalent Graphs, Snarks, and Boojums", is a drastically revised version of
476
his April 1976 column, which he had originally intended to leave out, in view
of rapid advancements on the status of the Four Colour Problem. The re-
viewer is glad to see its inclusion. Thus only the October 1975 column, titled
\Extrasensory Perception by Machines", is missing from the �fteen volumes.
It is anthologized in a science �ction puzzle collection.
By now, everything that can be said about Martin's writing has been
said many times over. Buy this book at once and treasure it! Although Mar-
tin continues to contribute to various journals, this is The Last Recreations
under his own banner!
The Editor-in-Chief, Bruce Shawyer, and his predecessor, Bill Sands,
would like to take this opportunity of thanking Andy Liu for his long and
sterling work in this column.
\Thank you very much, Andy. All the best!"
The 1998 Crossword PuzzlePeter Hurthig, Columbia College, Burnaby, BC.
Four Mathematicians, a Detective, a King and a French Lady
Across:
1. A word on a towel.4. An abbreviated diophantine solu-
tion.5. A Russian who solved number
problems.9. King Zog's country.10. A graduate student who impressed
Gauss!11. Ruth Rendell's inspector (to his
friends).12. One third of a communist or one
half of a y.
Down:
1. A German whose problems were
numbered.
2. Brands with dishonour.
3. British weight.
5. A bony �sh.
6. A high priest.
7. She wrote Delta of Venus.
8. Professor Pedoe.
4
5
3
6
2
9
1
10
7 8
11
12
477
THE SKOLIAD CORNERNo. 34
R.E. Woodrow
We begin this issue with the problems of the Old Mutual Mathemati-
cal Olympiad 1991 Final Paper 1 and Final Paper 2. The Preliminary Round
of this contest is in a multiple choice format and I plan to give an exam-
ple with the 1992 paper in the new year. My thanks go to John Grant
McLoughlin, of the Faculty of Education, Memorial University of Newfound-
land, for collecting this contest and forwarding it for use in the Corner.
OLD MUTUAL MATHEMATICAL OLYMPIAD 1991Final Paper 1
Time: 2 hours
1. In the �gure shown ABC and AEB are semi-circles and F is the
mid-point of AC and AF = 1 cm. Find the area of the shaded region.
CA F
BE
2. What is the value ofp17� 12
p2 +
p17 + 12
p2 in its simplest
form?
3. In a certain mathematics examination, the average grade of the
students passing was x%, while the average of those failing was y%. The
average of all students taking the examination was z%. Find the percentage
who failed in terms of x, y and z.
4. In the �gure shownAB = AD =p130 cm andBEDC is a square.
A
D
B
CE
478
Also the area of 4AEB = area of square BEDC.
Find the area of BEDC.
Final Paper 2Time: 2 hours
1. If the pattern below of dot-�gures is continued, how many dots will
there be in the 100th �gure?
w w
w
w
w
w
w
w w
w
w
w
w
w
w w
w
w
w
w
w
w
w
w
2. It is required to place a small circle in the space left by a large circle
as shown. If the radius of the large one is a and that of the small one is b,
�nd the ratio a=b.
3. Find all solutions to the simultaneous equations
x+ y = 2 ;
xy � z2 = 1 ;
and prove that there are no other solutions.
4. If a, b, c and d are numbers such that
a+ b < c+ d ;
b+ c < d+ e ;
c+ d < e+ a ;
and d+ e < a+ b ;
prove that the largest number is a and the smallest is b.
479
5. The diagram below [rotated through 90�] shows a container whoselower part is a hemisphere and whose upper part is a cylinder.
6
?20cm
-�220cm
The cylindrical part has internal diameter of 20 cm and is 220 cm long. Water
is poured into it and rises to a height of 20 cm in the cylindrical part. The
top is then sealed with a at cover and the container is turned upside down.
The water is now 200 cm high in the cylindrical part.
(i) Calculate the volume of the hemisphere in terms of �.
(ii) Find the total height of the container.
[Note: The volume of a sphere of radius R is 43�R3.]
Last number we gave the problems of the British Columbia Colleges,
Senior High School Mathematics Contest Preliminary Round, 1998. Next we
give the \o�cial" solutions, courtesy of Jim Totten, The University College
of the Cariboo, and an organizer of the contest.
BRITISH COLUMBIA COLLEGESSenior High School Mathematics Contest
Preliminary Round 1998Time: 45 minutes
1. The integer 1998 = (n� 1)nn(10n+ c) where n and c are positive
integers. It follows that c equals: (d).
Solution. By comparing the expression 1998 = (n � 1)nn(10n + c)with 1998 = 2� 33 � 37, we get n = 3, since 33 is the only power dividing
1998. Hence, 10n+ c = 37, which gives c = 7.
2. The value of the sum log 12+ log 2
3+ log 3
4+ � � �+ log 9
10is: (a).
Solution. We have:
log 12+ log 2
3+ log 3
4+ � � �+ log 9
10= log(1
22334� � � 9
10) = log 1
10= �1.
3. Four basketballs are placed on the gym oor in the form of a square
with each basketball touching two others. A �fth basketball is placed on top
of the other four so that it touches all four of the other balls, as shown.
480
If the diameter of a basketball is 25 cm, the height, in centimetres, of the
centre of the �fth basketball above the gym oor is: (d).
Solution. Let us �rst take a look at the four basketballs lying on the
oor:
A
C
B
We �nd AC =pAB2 + BC2 =
p252 + 252 = 25
p2. Now, in order to
�nd the height of the centre of the �fth ball, we analyze the cross-section of
the pyramid by a vertical plane passing through A and C:
CA
D
E
F
The heightDF = DE+EF , where EF = 252andDE =
qAD2 � (1
2AC)2
=q252 � (1
225p2)2 = 25
2
p2. Hence, DF = 25
2(1 +
p2).
4. Last summer I planted two trees in my yard. The �rst tree came in
a fairly small pot and the hole that I dug to plant it in �lled one wheelbarrow
load of dirt. The second tree came in a pot, the same shape as that of the
�rst tree, that was one-and-a-third times as deep as the �rst pot and one-
and-a-half times as big around. Let us make the following assumptions:
(i) The hole for the second tree was the same shape as for the �rst tree.
(ii) The ratios of the dimensions of the second hole to those of the �rst
hole are the same as the ratios of the dimensions of the pots.
481
Based on these assumptions, the number of wheelbarrows of dirt that I �lled
when I dug the hole for the second tree was: (c).
Solution. To solve the problem we need to �nd the ratio of the volume
of the second pot to the volume of the �rst. We can assume that both pots
are generalized cylinders with bases of the same shape. It is convenient to
think that the second pot was obtained from the �rst by stretching it hori-
zontally and vertically. The change of the perimeter is proportional to the
change of horizontal dimensions, so that both horizontal dimensions have
been stretched one-and-a-half times. Therefore, the area of the base has
increased 11
2� 11
2= 9
4times. Since the height has been stretched 11
3times,
the volume increased 9
4� 11
3= 3 times.
5. You have an unlimited supply of 5-gram and 8-gram weights that
may be used in a pan balance. If you use only these weights and place them
only in one pan, the largest number of grams that you cannot weigh is: (b).
Solution. IfW is the weight in grams then W can be weighed ifW =5x + 8y, where x and y are non-negative integers. Furthermore, y can be
written in the form 5z+ r, where z is a non-negative integer and r is either
0; 1; 2; 3, or 4. Thus W = 5(x + 8) + 8(0), W = 5(x + 8) + 8(1), W =5(x+8)+8(2),W = 5(x+8)+8(3), orW = 5(x+8)+8(5). These can be
simpli�ed toW = 5k,W = 5(k+1)+3,W = 5(k+3)+1,W = 5(k+4)+4and W = 5(k + 6) + 2. By putting k = x + 8z, we conclude that W can
be weighed if W = 5k + 8r, with k � 0 and r � 4. Thus, according to the
value of r,W = 5k+8(0),W = 5k,W = 5(k+1)+3,W = 5(k+3)+1,W = 5(k + 4) + 4, W = 5(k + 6) + 2. On the other hand, W (as well
as every integer) can take exactly one of the forms: 5K, 5K + 1, 5K + 2,5K + 3, or 5K + 4. By comparing these forms with the expressions for W
that can be weighed, we conclude:
(i) ifW is of the form 5K, then it can be weighed;
(ii) if W is of the form 5K + 1, then it can be weighed if K = k + 3;that is, when K � 3;
(iii) if W is of the form 5K + 2, then it can be weighed if K = k + 6;that is, when K � 6;
(iv) ifW is of the form 5K + 3; then it can be weighed if K = k + 1;that is, when K � 1;
(v) if W is of the form 5K + 4, then it can be weighed if K = k + 4;that is, when K � 4.
Consequently, the largest values ofW that cannot be weighed in categories:
(ii), (iii), (iv), and (v), are 5(2) + 1 = 11, 5(5) + 2 = 27, 5(0) + 3 = 3,5(3) + 4 = 19, respectively. The largest of them is 27.
6. If all the whole numbers from 1 to 1,000,000 are printed, the num-
ber of times that the digit 5 appears is: (c).
482
Solution. Method I. The number of times the numeral 5 appears will
not change if instead of printing the numbers from 1 to 1,000,000 we will
print the numbers from 0 to 999,999. Furthermore, the number of appear-
ances of 5 will not change if we complete the decimal representation of each
number to a six-digit sequence, by writing some zeros in front of the number,
if necessary. For example, 1998will be represented by 001,998. The printoutof all of the six-digit sequences will have a total of 6 � 106 digits and each
digit will appear the same number of times. This gives 6�10610
= 600,000appearances of 5.
Method II. (Standard, but more tedious.) Again, as above, we repre-
sent the numbers by six-digit sequences. At �rst, we count the number of
sequences that have precisely k copies of 5, for 1 � k � 6. The numeral 5can appear in
�6
k
�di�erent positions in such a sequence, while in each of the
remaining 6 � k positions we can have one of the nine digits other than 5.This gives
�6
k
�96�k di�erent sequences. They provide k
�6
k
�96�k copies of 5.
Hence, the number of appearances of 5 in all of the sequences is given by the
sum:
1
�6
1
�95 + 2
�6
2
�94 + 3
�6
3
�93 + 4
�6
4
�92 + 5
�6
5
�91 + 6
�6
6
�90 = 600,000 :
7. The perimeter of a rectangle is x centimetres. If the ratio of two
adjacent sides is a : b, with a > b, then the length of the shorter side, in
centimetres, is: (e).
Solution. If a1 and b1 are the lengths of the longer and the shorter
sides of the rectangle, then b1 : x = b : [2(a+ b)]. This gives b1 = bx2(a+b)
.
8. The sum of the positive solutions to the equation xxpx = (x
px)x
is: (e)
Solution. From xxpx = (x
px)x we get that that xxx
1=2
= (xx1=2)x,
which simpli�es to xx3=2
= x(3=2)x. This has the trivial solution x = 1. If
x 6= 1 then the exponents must be equal; that is x3=2 = 3
2x. By squaring
both sides, x3 = 9
4x2. Therefore, x = 9
4, since x is positive. The sum of the
solutions is 1 + 9
4= 31
4.
9. Two circles, each with a radius of one unit, touch as shown. (See
diagram on next page.) AB and CD are tangent to each circle. The area, in
square units, of the shaded region is: (d).
Solution. The square ABCD has area of 2 � 2 = 4 square units. To
�nd the shaded area we need to remove the area of two semicircles from the
area of ACBD. Thus, the area of the shaded region is 4 � 2(12)� = 4� �
square units.
483
A B
C D
10. A parabola with a vertical axis of symmetry has its vertex at (0;8)and an x{intercept of 2. If the parabola goes through (1; a), then a is: (c)
Solution. The parabola has another x{intercept of �2 symmetric to
the x{intercept of 2 with respect to the vertical axis of symmetry passing
through (0; 8). This implies that the equation of the parabola is of the form
y = A(x � 2)(x + 2). The parabola has the vertex at (0;8): therefore
8 = A(0 � 2)(0 + 2). This gives A = �2, so that the equation of the
parabola is y = �2(x � 2)(x + 2). If a is the value of y corresponding to
x = 1, then a = y = �2(1� 2)(1 + 2) = 6.
11. A �ve litre container is �lled with pure orange juice. Two litres of
juice are removed and the container is �lled up with pure water and mixed
thoroughly. Then two litres of the mixture are removed and again the con-
tainer is �lled up with pure water. The percentage of the �nal mixture that
is orange juice is: (d).
Solution. Initially, there were 5 litres of juice, and after the removal of
the �rst two litres, 3 litres of juice are left. After mixing this amount with 2litres of pure water, each litre of the mixture will contain 3
5litres of pure juice.
Thus, when we remove 2 litres of the mixture, 3 � 2(35) = 9
5litres of pure
juice will remain. This gives the concentration of [(95)� 5]� 100% = 36%.
12. The lengths of the sides of a triangle are b+ 1, 7� b and 4b� 2.The number of values of b for which the triangle is isosceles is: (b)
Solution. Potentially, we have three possibilities for the triangle to be
isosceles:
(1) b+ 1 = 7� b,
(2) b+ 1 = 4b� 2, and
(3) 7� b = 4b� 2.
The �rst possibility gives b = 3 and, consequently, 4, 4, and 10 as possible
lengths of the sides of the triangle. The second possibility gives b = 1, and2, 2, and 6 as possible lengths. Finally, the third possibility gives b = 9
5, and
145, 26
5, and 26
5as the corresponding lengths. However, the sum of lengths of
any two sides in a triangle is greater than or equal to the length of the third
side. This leaves b = 95as the only possibility.
484
13. The number of times in one day when the hands of a clock form a
right angle is: (d).
Solution. Suppose that we measure the time from midnight to the next
midnight. If �1 is the angle made by the minute hand and �2 the angle
made by the hour hand turned clockwise from 12 then �1 = 2�t and �2 =2�12t = 1
6�t, where t represents time measured in hours. The angle between
both hands is 90� when �1 � �2 = 12� + 2k� or �1 = �2 = 3
2�t + 2k�.
By solving these equations for t, we get t = 611
+ 1211k or t = 18
11+ 12
11k.
Since 0 � t � 24, each of the equations yields 22 solutions in non-negative
integers k. This corresponds to the total of 44 solutions.
14. In my town some of the animals are really strange. Ten percent of
the dogs think they are cats and ten percent of the cats think they are dogs.
All the other animals are perfectly normal. One day I tested all the cats and
dogs in the town and found that 20% of them thought that they were cats.
The percentage of the dogs and cats in the town that really are cats is: (a)
Solution. Let d and c denote the number of dogs and cats in the town,
respectively. Then the number of animals who responded that they are cats
is 0:2(d+ c). This corresponds to 0:1d+ 0:9c, according to the truthfulness
of our animals. By equating both expressions we get 0:1d = 0:7c, or d = 7c.Thus, the percentage of cats is c
c+d� 100 = c
c+7c� 100 = 100
8% = 12:5%.
15. A short hallway in a junior high school contains a bank of lockers
numbered one to ten. On the last day of school the lockers are emptied and
the doors are left open. The next day a malicious math student walks down
the hallway and closes the door of every locker that has an even number.
The following day the same student again walks down the hallway and for
every locker whose number is a multiple of three closes the door if it is open
and opens it if it is closed. On the next day the student does the same thing
with every locker whose number is divisible by four. If the student continues
this procedure for a total of nine days, the number of lockers that are closed
after the ninth day is: (d).
Solution. A locker number k changes its status from open to closed or
vice-versa n(k) times, where n(k) denotes the number of distinct divisors of
k that are greater than 1 and less or equal to 10. Thus, n(1) = 0, n(2) = 1,n(3) = 1, n(4) = 2, n(5) = 1, n(6) = 3, n(7) = 1, n(8) = 3, n(9) = 2,
n(10) = 3. Since all lockers are initially open, the kth locker is closed after
the period of nine days ifn(k) is odd. Therefore, seven lockers will be closed.
That completes the Skoliad Corner for this issue. Please send me your
comments and suggests for the evolution of the Corner. Also, I need suitable
contest material at the pre-Olympiad level for use in future Corners.
485
MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by
High School and University Students. It continues, with the same emphasis,
as an integral part of Crux Mathematicorum with Mathematical Mayhem.
All material intended for inclusion in this section should be sent to the
Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,
PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic
address is still
The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).
The rest of the sta� consists of Adrian Chan (Upper Canada College), Jimmy
Chui (Earl Haig Secondary School), Richard Hoshino (University ofWaterloo),
David Savitt (Harvard University) and Wai Ling Yee (University of Waterloo).
Mayhem Problems
The Mayhem Problems editors are:
Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,David Savitt Mayhem Challenge Board Problems Editor.
Note that all correspondence should be sent to the appropriate editor |
see the relevant section. In this issue, you will �nd only solutions | the
next issue will feature only problems.
We warmly welcome proposals for problems and solutions. We request
that solutions from the previous issue be submitted in time for publication
in issue 8 of 1999.
High School Solutions
Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,
Canada. L3S 3E8 <[email protected]>
H229. Here is a simple way to remember how many books there are
in the Bible. Remember that there are x books in the Old Testament, where
x is a two-digit integer. Then multiply the digits of x to get a new integer y,
which is the number of books in the New Testament. Adding x and y, you
end up with 66, the number of books in the Bible. What are x and y?
486
Solutionby Katya Permiakova, student, Lisgar Collegiate Institute, Ot-tawa, Ontario.
Let x = 10a + b, where a and b are integers satisfying the conditions
1 � a � 9 and 0 � b � 9. Then y = ab. Now we are given that 10a+ b+ab = 66. Rearranging terms and solving for b, we get b(a+ 1) = 66� 10a,so b = 66�10a
a+1= �10 + 76
a+1. Now in order for b to be an integer, a + 1
must divide 76. The only positive divisors of 76 are 1, 2, 4, 19, 38, and 76.Since our choice for a is limited to the integers between 1 and 9, the only
possibilities for a are 1 and 3 (since that gives us a+ 1 = 2 and a + 1 = 4,respectively).
If a = 1, then we have b = �10 + 76
2= 28, but this does not satisfy
b � 9. However, if a = 3, then b = �10 + 76
4= 9, and this is legitimate.
Hence x = 39 and y = 27.
Also solved by KEON CHOI, student, A.Y. Jackson Secondary School, NorthYork, Ontario; LINO DEMASI, student, St. Ignatius High School, Thunder Bay, On-
tario; and WENDY YU, student, Woburn Collegiate Institute, Scarborough, Ontario.
H230. Dick and Cy stand on opposite corners (on the squares) of a
4� 4 chessboard. Dick is telling too many bad jokes, so Cy decides to chase
after him. They take turns moving one square at a time, either vertically or
horizontally on the board. To catch Dick, Cy must land on the square Dick is
on. Prove that:
(i) If Dick moves �rst, Cy can eventually catch Dick.
(ii) If Cy moves �rst, Cy can never catch Dick.
(Can you generalize this to a 2m� 2n chessboard?)
Solution.
(i) Place coordinates on the board so that Cy is standing on (0; 0) andDick is standing on (3;3). We shall show that after a few moves, Cy can catch
Dick on a turn. Regardless of what Dick does on his �rst two turns, Cy can
move to (1;1) after two moves. Now it is Dick's turn. At that time, Dick
must be on (1;3), (2; 2), (3; 1), or (3; 3). So if on his next move, Dick goes
to either (1; 2) or (2; 1), Cy is standing one square away and so Cy moves
into Dick's square on his next move, and catches Dick. So Dick must move
to one of (0; 3), (2; 3), (3; 2), or (3; 0).
If Dick goes to (0; 3) or (2;3), then Cy can go to (1;2), and from here
it is easy to see that Dick can last at most two moves before he gets caught
(since Cy can trap him into a corner). If Dick goes to (3; 2) or (3; 0), then Cy
can go to (2;1), by the same argument, Cy can catch Dick. Thus no matter
what, if Dick moves �rst, then Cy can eventually catch Dick.
487
(i)
(0; 0)
(1; 1) (1; 3)
(2; 2)
(3; 1) (3; 3)�
?�
?
6
(ii)
C
D
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.................................................................................
Dick
Cy!
(ii) Colour the 4�4 board in black and white, as in a regular chessboard
(so adjacent squares are of di�erent colours). Thus, if (0; 0) is a black square,then (3; 3) must be a black square as well. So for each move, if a person is
on a square of a certain colour, then he will move to a square of the other
colour.
Hence, Dick and Cy both start o� on a black square. If Cy moves �rst,
then Cy moves onto a white square (while Dick remains on a black square).
Then Dick moves to some white square. Now Dick and Cy are both on white
squares, and so on Cy's next move he must move onto a black square (while
Dick remains on a white square). Thus whenever Dick is on a square of a
certain colour, Cy is moving to a square of the other colour. And so, on any
given move, Cy can never move to a square that Dick is currently on, and so
Cy will not be able to catch Dick.
Also solved by KEON CHOI, student, A.Y. Jackson Secondary School, NorthYork, Ontario; LINO DEMASI, student, St. Ignatius High School, Thunder Bay, On-
tario; and WENDY YU, student, Woburn Collegiate Institute, Scarborough, Ontario.
H231. Let O be the centre of the unit square ABCD. Pick any
point P inside the square other than O. The circumcircle of PAB meets the
circumcircle of PCD at points P andQ. The circumcircle of PADmeets the
circumcircle of PBC at points P and R. Show that QR = 2 � OP .
Solution by Wendy Yu, student, Woburn Collegiate Institute, Scarbor-ough, Ontario and Keon Choi, student, A.Y. Jackson Secondary School,NorthYork, Ontario.
q
D A
BC
Q
P
O L
BA
D C
QP
R
O
Construct line L through O parallel to AD and BC. All points on this
line are the same distance fromA as fromB, and the same distance fromC as
488
from D. Thus this line contains the centres of the circumcircles of PAB and
PCD. Hence, the line L bisects segment PQ. So the point Q must be the
re ection of P about the line L, and it follows that OP = OQ. Similarly,
if we construct line M through O parallel to AB and CD, then R is the
re ection of P about the lineM . Hence, OP = OR. Because PQ and PR
are perpendicular (since the lines L and M are perpendicular), PQR is a
right-angled triangle. Furthermore, OP = OQ = OR, which implies that O
is the midpoint of the hypotenuse QR. Hence, we have QR = OQ+OR =OP +OP = 2OP , and so QR = 2OP , as desired.
Also solved by LINO DEMASI, student, St. Ignatius High School, Thunder Bay,
Ontario.
H232. Lucy and Anna play a game where they try to form a ten-digit
number. Lucy begins by writing any digit other than zero in the �rst place,
then Anna selects a di�erent digit and writes it down in the second place,
and they take turns, adding one digit at a time to the number. In each turn,
the digit selected must be di�erent from all previous digits chosen, and the
number formed by the �rst n digits must be divisible by n. For example,
3, 2, 1 can be the �rst three moves of a game, since 3 is divisible by 1, 32 is
divisible by 2 and 321 is divisible by 3. If a player cannot make a legitimate
move, she loses. If the game lasts ten moves, a draw is declared.
(i) Show that the game can end up in a draw.
(ii) Show that Lucy has a winning strategy and describe it.
I. Solutionby Lino Demasi, student, St. IgnatiusHigh School, ThunderBay, Ontario.
The number 3,816,547,290 has the property that the number formed
by the �rst n digits is divisible by n, for n = 1, 2, 3, : : : , 10. Thus, if the
moves are carried out in this order, then the game can end up in a draw.
Here is Lucy's winning strategy. First note that Anna must play an even
digit on each of her moves. So Lucy's goal is to play as many even numbers
as possible. So Lucy plays a 6 to start. There are three cases to be considered
for Anna's second move:
(1) If Anna plays a 4 or a 2, then Lucy plays the other on the third
move. Anna must now play an even number because her number now has to
be divisible by 4, so if Anna plays an 8, then Lucy plays a 0, and Anna loses
because on the sixth move, she would have to play an even number and there
are none left. If Anna plays a 0, then Lucy plays a 5, and then Anna also loses
because she must now play an even number on the sixth move, the only one
of which is an 8, but neither 642,058 or 624,058 is divisible by 6.
(2) If Anna plays a 0, then Lucy plays a 9. Then Anna must play a 2 to
make the four-digit number divisible by 4. Lucy then plays a 5. Anna must
now play an 8 to make her number divisible by 6. Then Lucy can counter
with a 3, since 6,092,583 is divisible by 7. The only even number Anna can
now play is a 4, but 60,925,834 is not divisible by 8, so she loses.
489
(3) If Anna plays an 8, then Lucy plays a 4. Anna's only choice now is
a 0. Then Lucy plays a 5. Now Anna's only choice is a 2, but 684052 is not
divisible by 6, so she loses.
Note that this covers all the cases because Anna must play an even digit
on the second move. Thus, Lucy can always force a win.
II. Solution by Wendy Yu, student, Woburn Collegiate Institute, Scar-borough, Ontario.
As before, if the game is played in the following order: 3, 8, 1, 6, 5, 4,7, 2, 9, 0, then the game will end up in a draw.
For Lucy's winning strategy, she can start o� with a 4. Then, Anna must
counter with an even number. So if she responds with a 2 or an 8, then Lucy'snext move is a 0. If Anna's response is a 0 or a 6, then Lucy's next move is
a 2. Now, in the case where the number 480 has been written, Anna cannot
�nd a digit to make a four-digit number divisible by 4, so she immediately
loses. In the other three cases, there is at least one digit that Anna can pick
to remain in the game.
Thus, after four moves, if the game lasts that long, one of the following
numbers will be on the board: 4028, 4208, 4620, or 4628. Then Lucy picks
a 5, and in each of those four cases, it will be impossible for Anna to then
make a move so that the new six-digit number is divisible by 6, since the
digits she needs are all taken. Thus, if Lucy follows this strategy, she can
always force a win.
Also solved by KEON CHOI, student, A.Y. Jackson Secondary School, NorthYork, Ontario.
Advanced Solutions
Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.
M1G 1C3 <[email protected]>
A205. Find all functions f : R ! R such that f(x) � x and
f(x+ y) � f(x) + f(y) for all reals x and y.
Solution.We have
f(x) � x ; (1)
f(x+ y) � f(x) + f(y) : (2)
Let x = y = 0 in (2). We have f(0) � 2f(0) which implies 0 � f(0).But for x = 0 in (1), we have f(0) � 0 so f(0) = 0.
Now take y = �x in (2) to get f(0) � f(x) + f(�x) for all real x.In other words, �f(x) � f(�x). But f(�x) � �x by (1) so �f(x) � �x
490
implies that f(x) � x for all real x. Thus combining this with (1) we have
f(x) = x for all real x. Now it can be easily checked that this function
satis�es the two conditions (1) and (2) in the problem.
A206. Let n be a power of 2. Prove that from any set of 2n � 1positive integers, one can choose a subset of n integers such that their sum
is divisible by n.
Solution.
Let n = 2m, for some integer m. We will prove this result by mathe-
matical induction onm. For m = 0, n = 1 and the result clearly holds.
Now assume the result is true for some arbitrary k � 0. In other words,for n = 2k, any set of 2k+1 � 1 positive integers has a subset of n integers
whose sum is divisible by n.
Now consider any set, A, of 2k+2 � 1 positive integers. Let A1 and
A2 be the two subsets of A consisting of the �rst 2k+1 � 1 integers and the
last 2k+1 � 1 integers respectively. By the induction hypothesis, each of
these sets has 2k integers whose sum is divisible by 2k. Call the subset of 2k
numbers, from A1, B1 and the subset of 2k numbers, from A2, B2. Now call
the set of integers remaining from A when the 2k+1 integers from B1 and
B2 are removed from A3. Now A3 also has (2k+2 � 1)� 2(2k) = 2k+1 � 1elements, so again by the induction hypothesis it has a subset, call it B3, of
size 2k whose sum is divisible by 2k.
It remains to show that two of the three sets B1, B2, and B3 can be
combined to form a set of 2k+1 integers whose sum is divisible by 2k+1. Let
the sum of the three sets be s1, s2, and s3 respectively. Each is divisible
by 2k, so let ti = si=2k, for i = 1, 2, and 3. Now by the Pigeonhole
Principle, at least two of these numbers must have the same parity, even or
odd. Without loss of generality, let the two sets be A1 and A2. The sum of
two numbers with the same parity is even, so t1 + t2 is even. Multiplying
by 2k, we have s1 + s2 is divisible by 2k+1. Thus there are 2k+1 integers
from the original set A whose sum is divisible by 2k+1. Our induction onm
is complete.
A207. Given triangle ABC, let A0, B0, and C0 be on the sides BC,
AC, and AB respectively such that 4A0B0C0 � 4ABC. Find the locus of
the orthocentre of all such triangles A0B0C0.
Solution by Alexandre Trichtchenko, student, Brook�eld High School,Ottawa, Ontario.
Here we give the solution when the triangle ABC is acute. A similar
argument can be given for an obtuse triangle.
Let \BAC = �, \ABC = �, and \BCA = . Let A00, B00, andC00 be the feet of the altitudes from vertices A0, B0, and C0 respectively of
triangle A0B0C0. Further, let H0 be the point of the orthocentre of triangle
A0B0C0, the point of intersection of its altitudes.
491
A
B CA0
B0
C0
A00
B00
C00
H0
Since trianglesB0H0C00 andB0A0B00 are similar \B0H0C00 = �. Like-
wise, \A0H0C00 = �. Thus \B0H0A0 = \B0H0C00 + \A0H0C00 = �+ �.
Also, \A0H0B0 + \B0CA0 = �+ � + = 180�. Hence, the quadri-
lateral CB0H0A0 is cyclic. Since \B0A0H0 and \B0CH0 are inscribed in
the circumcircle of CB0H0A0 and subtended by the same arc H0B0, we have\H0CB0 = \H0A0B0 = 90� � �. Similarly, \H0AB0 = 90� � �. So
\H0CB0 = \H0AB0, and so AH0 = CH0. By similar reasoning, we can
show that AH0 = BH0 = CH0. Thus H0 is the circumcentre of triangle
ABC and is independent of the choice of triangle A0B0C0. Thus the locus ofthe orthocentre of all triangles A0B0C0 is just the single point H0 = O, the
circumcentre of triangle ABC.
Also solved by D.J. SMEENK, Zaltbommel, the Netherlands.
A208. Let p be an odd prime, and let Sk be the sum of the products
of the elements f1; 2; : : : ; p � 1g taken k at a time. For example, if p = 5,then S3 = 1� 2� 3 + 1� 2� 4 + 1� 3� 4 + 2� 3� 4 = 50. Show that
pjSk for all 2 � k � p� 2.
Solution.Consider the monic polynomial of degree p�1, xp�1�1 � 0 (mod p).
There are precisely p� 1 incongruent solutions modulo p of this polynomial
equation, namely, x = 1, 2, : : : , p� 1. Each of these follows from Fermat's
Little Theorem, which states that ap�1 � 1 (mod p), where p is a prime and
a and p are relatively prime. Thus, modulo p, we have
xp�1 � 1 � (x� 1)(x� 2) � � � [x� (p� 1)]
� xp�1 � S1xp�2 + S2x
p�3
� � � �+ (�1)p�2Sp�2 + (�1)p�1Sp�1 (mod p) :
Equating coe�cients, we have pjSk for all 2 � k � p� 2.
Note: We also get Wilson's Theorem for free. (Why?)
Also solved by ALEXANDRETRICHTCHENKO, student, Brook�eld High School,
Ottawa, Ontario.
492
Challenge Board Solutions
Editor: David Savitt, Department of Mathematics, Harvard University,
1 Oxford Street, Cambridge, MA, USA 02138 <[email protected]>
C77. | Corrected problem (not a solution!)
Let Fi denote the ithFibonacci number, with F0 = 1 and F1 = 1. (Then
F2 = 2, F3 = 3, F4 = 5, etc.)
(a) Prove that each positive integer is uniquely expressible in the form
Fa1 + � � �+ Fak , where the subscripts form a strictly increasing sequence of
positive integers no pair of which are consecutive.
(b) Let � = 12(1 +
p5), and for any positive integer n, let f(n) equal
the integer nearest to n� . If n = Fa1 + � � � + Fak is the expression for n
from part (a) and if a2 6= 3, prove that f(n) = Fa1+1 + � � �+ Fak+1.
(c) Keeping the notation from part (b), if a2 = 3 (so that a1 = 1), it isnot always true that the formula f(n) = Fa1+1 + � � � + Fak+1 holds. For
example, if n = 4 = F3 + F1 = 3 + 1, then the closest integer to n� =6:47 : : : is 6, not F4 + F2 = 5 + 2 = 7. Fortunately, in the cases where the
formula fails, we can correct the problem by setting a1 = 0 instead of a1 = 1:for example, 4 = F3+ F0 = 3+ 1 as well, and indeed 6 = F4+ F1 = 5+ 1.Determine for which sequences of ai this correction is necessary.
C78. Let n be a positive integer. An n�nmatrix A is amagic matrixof order m if each entry is a non-negative integer and each row and column
sum is m. (That is, for all i and j,P
kAik =Pk Akj = m.) Let A be a
magic matrix of order m. Show that A can be expressed as the sum of m
magic matrices of order 1.
I. Solution by Christopher Long, graduate student, Rutgers University.
Consider the magic matrix A as the adjacency matrix of a weighted
bipartite graph G between two sets (\left" and \right") of n vertices: If the
(i; j)th entry ofA is greater than 0, place an edge inG between the ith vertex
on the left and the jth vertex on the right and give the edge a weight equal
to the (i; j)th entry of A. If the (i; j)th entry of the A is 0, do not place an
edge at all. The condition that the matrix A is a magic matrix implies that
total weight of all the edges emanating from any single vertex of G, left or
right, is equal to m.
Given a subsetS of the left-hand vertices ofG, let us compute the size of
its neighbourhood (the collection of all vertices on the right which are joined
by an edge to a vertex in S). Remove fromG all of the edges whose left-hand
vertices are not in S. Then the total weight of the remaining edges is exactly
mjSj, and the neighbourhood of S is exactly the set of right-hand vertices
whose weight is still non-zero. (By the weight of a vertex, we mean the
weight of all the edges touching that vertex.) But each right-hand vertex has
493
weight at most m, so by the Pigeonhole Principle the number of right-hand
vertices with non-zero weight must be at leastmjSj=m = jSj. That is, everysubset on the left has a neighbourhood on the right which is at least as big.
Thus, the conditions of the following famous theorem (phrased traditionally,
and thus thoroughly objectionable) are satis�ed, with the left-hand vertices
as boys, the right-hand vertices as girls, and edges as acceptable marriages:
Theorem (Hall's Marriage Theorem). Suppose there are n boys and n
girls, and that each boy knows precisely which (possibly more than one) of
the girls he is willing to marry. Suppose further that given any set S of boys,
the total number of di�erent girls that boys in S are willing to marry is at
least S. Then there exists a way of pairing all the boys with the girls in such
a way that each boy is willing to marry the girl to whom he is paired.
The proof of the Marriage Theorem is an excellent exercise, and can
also be found in almost any graph theory book, so we omit it here. In our
case, if the pairing obtained from the Marriage Theorem pairs the vertex i
on the left with the vertex �i on the right, then we know that the (i; �i)th
entry of A is positive. Let A0 be the matrix whose (i; �i)th entry is 1 for all
i and whose other entries are all 0. Then A0 is a magic matrix of order 1 and
A�A0 is a magic matrix of orderm�1, and the result follows by induction.
II. Solution.
As in the previous solution, we prove the result by induction by showing
that there exists a permutation � of f1; 2; : : : ; ng such that the (i; �i)-entryof A is positive{that is, by constructing a \magic submatrix" of order 1 in
A. We do this for all magic matrices A by another induction, this time on
NZ(A), the number of non-zero entries of A. If the order of the magic
matrix is m > 0, then there is a non-zero entry in every row, so the total
number of non-zero entries is at least n. However, if the number of non-zero
entries is exactly n, then certainly A is m times a magic matrix of order 1,and this completes our base case.
Now assume the result holds for NZ(A) < k, where k > n. Consider
a magic matrix A of order m with exactly k non-zero entries. As k > n, by
the Pigeonhole Principle, there is a row with at least two non-zero entries,
and each is less than m. Let (i1; j1) be the position of one of them, and let
(i1; j2) be the position of the other. As the (i1; j2)th entry of A is less than
m, there is a non-zero entry at some position (i2; j2) in the same column
(and it is also less thanm). By the same argument, there is a non-zero entry
in the same row (i2; j3) as (i2; j2). We continue this process to get sequences
(ik), (jk), such that ik 6= ik+1, jk 6= jk+1, and such that the (ik; jk)th and
(ik; jk+1)th entries of A are all non-zero.
Our goal is to �nd a loop of an even number of distinct non-zero entries
in the matrix, connected by alternating horizontal and vertical moves. Once
we have such a loop, �nd the point (i; j) in the loop whose entry is minimal.
Suppose the (i; j)th entry is q. Decrease the (i; j)th entry by q to 0, increase
494
the next entry in the loop by q, decrease the next by q, and continue travel-
ling once around the entire loop, alternately adding and subtracting q in this
fashion. This will yield a magic matrix B with fewer non-zero entries thanA,
so by the induction hypothesis B contains a magic submatrix B0 of order 1.However, by the construction of B, the non-zero entries of B all correspond
to non-zero entries of A, so B0 is also a magic submatrix of A. We will thus
be done by induction.
Let us proceed with �nding this loop. As there are only a �nite num-
ber of entries of A, at some point in the sequence (i1; j1), (i1; j2), (i2; j2),(i2; j3), : : : there will be a repeated term. If the �rst repeated term is of the
form (il; jl), and the �rst appearance of this term is (ik; jk) (with k < l),
then the loop (ik; jk), (ik; jk+1), : : : , (il�1; jl), (il; jl) = (ik; jk) is exactlythe kind of loop we are looking for, and we are done. We are similarly �n-
ished if the �rst repeated term is of the form (il; jl+1) and the �rst occurrenceof that term is of the form (ik; jk+1).
Suppose instead that the �rst repeated term is of the form (il; jl+1) andthat the �rst appearance of the term is (ik; jk). Then, replacing jk by jl, we
obtain the loop (il; jl) = (ik; jl), (ik; jk+1), : : : , (il�1; jl), (il; jl), and we
are done. In the case when the �rst repeated term is of the form (il; jl) andthe �rst appearance of the term is (ik; jk+1), a similar trick works. Thus, we
have exhausted all cases, and the proof is complete.
Shreds and Slices
Erratum
There is a mistake in the 1987 Swedish Mathematical Olympiad, as
printed in CRUX and MAYHEM [1998:298]. The expression \�a+2b� 3c"in problem 2 of the Qualifying Round should read \�a + 2b+ 3c". Thanksto Solomon Golomb for pointing this out.
Goodbye, Richard!
We regret to inform our readership that Richard Hoshino, long time
Mayhem sta� member and High School Editor, will be leaving us after this
issue. We all thank Richard for his strong dedication and numerous contri-
butions to Mayhem over the years, and we hope that he has gained as much
in the experience as we have. We wish Richard the best of luck in his new
responsibilities, and we will name his successor in the next issue.
495
1998{1999 Olympiad Correspondence
Problems
Please mail your solutions to Professor E.J. Barbeau, Department of
Mathematics, University of Toronto, Toronto, ON M5S 3G3.
Set 1
1. ABC is an isosceles triangle with \A = 100� and AB = AC. The
bisector of angle B meets AC inD. Show that BD + AD = BC.
2. Let I be the incentre of triangle ABC. Let the lines AI, BI, and CI
produced intersect the circumcircle of triangle ABC at D, E, and F
respectively. Prove that EF is perpendicular to AD.
3. Let PQR be an arbitrary triangle. Points A, B, and C external to the
triangle are determined for which
\AQR = \ARQ = 15� ; \QPC = \RPB = 30� ;
\PQC = \PRB = 45� :
Prove that: (a) AC = AB ; (b) \BAC = 90� .
4. Let a and b be two positive real numbers. Suppose that ABC is a
triangle andD a point on side AC for which \BCA = 90�, jADj = a,
and jDCj = b. Let jBCj = x and \ABD = �. Determine the values
of x and � for the con�guration in which � assumes its maximum value.
5. Let C be a circle with centre O and radius k. For each point P 6= O, we
de�ne a mapping P ! P 0 where P 0 is that point on OP produced for
which jOP j � jOP 0j = k2:
In particular, each point on C remains �xed, and the mapping at other
points has period 2. This mapping is called inversion in the circle C with
centre O, and takes the union of the sets of circles and lines in the plane
to itself. (You might want to see why this is so. Analytic geometry is
one route.)
(a) Suppose that A and B are two points in the plane for which
jABj = d, jOAj = r, and jOBj = s, and let their respective im-
ages under the inversion beA0 andB0. Prove that jA0B0j =k2d
rs:
(b) Using (a), or otherwise, show that there exists a sequence fXngof distinct points in the plane with no three collinear for which all
distances between pairs of them are rational.
496
6. Solve each of the following two systems of equations:
(a) x+ xy + y = 2 + 3p2 ;
x2 + y2 = 6 :
(b) x2 + y2 +2xy
x+ y= 1 ;p
x+ y = x2 � y :
Set 2
7. For a positive integer n, let r(n) denote the sum of the remainders
when n is divided by 1, 2, : : : , n respectively.
(a) Prove that r(n) = r(n�1) for in�nitely many positive integers n.
(b) Prove thatn2
10< r(n) <
n2
4for each integer n � 7.
8. Counterfeit coins weigh a and genuine coins weigh b (a 6= b). You are
given two samples of three coins each and you know that each sam-
ple has exactly one counterfeit coin. What is the minimum number of
weighings required to be certain of isolating the two counterfeit coins
by means of an accurate scale (not a balance)?
(a) Solve the problem assuming a and b are known.
(b) Solve the problem assuming a and b are not known.
9. Similar isosceles triangles EBA, FCB, GDC, and HAD are erected
externally on the four sides of the planar quadrilateral ABCD with the
sides of the quadrilateral as their bases. Let M , N , P , and Q be the
respective midpoints of the segments EG,HF , AC, and BD. What is
the shape of PQMN?
10. Given two pointsA andB in the Euclidean plane, let C be free to move
on a circle with A as centre. Find the locus of P , the point of intersec-
tion of BC with the internal bisector of angle A of triangle ABC.
11. Let ABC be a triangle; let D be a point on AB and E a point on AC
such that DE and BC are parallel and DE is a tangent to the incircle
of the triangle ABC. Prove that
8DE � AB + BC +CA :
12. Suppose that n is a positive integer and that x+ y = 1. Prove that
xn+1
nXk=0
�n+ k
k
�yk + yn+1
nXk=0
�n+ k
k
�xk = 1 :
497
J.I.R. McKnight Problems Contest 1985
1. If Sn = 3+8x+15x2+24x3+ � � �+(n2+4n+3)xn�1, determine Snby �rst evaluating (1�x)Sn. Hence �nd the limit of Sn asn approaches
in�nity, given x = 13.
2. (a) P and Q are points (ap2; 2ap) and (aq2; 2aq) on the parabola
y2 = 4ax. Show that the equation of the chord PQ is
2x� (p+ q)y+ 2aq = 0.
(b) If O is the origin and the chords OP and OQ are perpendicular,
prove that the chord PQ cuts the x-axis in the same point for all
possible positions of P and Q.
3. In the �gure, angle A has a measure of 60�. At a distance of 10 cm
from the vertex, a perpendicular is erected and a square is constructed
on it with side s1. In toward the vertex of the angle a second square
of side s2 is formed. Then similarly a square of side s3, and so on
ad in�nitum. Find the sum of the areas of these squares in simplest
radical form and then give an approximation to the nearest hundredth
of a square centimetre.
A� -
10 cm
s3
s2
s1
4. A wire of length L is to be cut into two pieces, one of which is bent to
form a circle and the other to form a square. How should the wire be cut
if the sum of the areas enclosed by the two pieces is to be a maximum?
5. (a) Sketch the hyperbola represented by the equation
x2
a2�y2
b2= 1 ; a > b;
and draw its asymptotes.
(b) Draw a tangent to the hyperbola at any point on the hyperbola, and
prove that the portion of the tangent between the points where it
meets the asymptotes is bisected by the point of contact of the
tangent.
498
(c) Prove that the segment of the tangent in (b) forms with the asymp-
totes a triangle of constant area.
6. Prove that if cosx+ cos y = a and sinx� siny = b, then
cos(x� y) =a2 � b2
a2 + b2:
Swedish Mathematics Olympiad 1989
1989 Qualifying Round
1. Find the integer t and the hundreds digit a such that
(3(320+ t))2 = 492a04 :
2. Form all possible six digit numbers, each using the digits 1, 2, 3, 4, 5,6 exactly once. What is the sum of all these numbers?
3. Let ABC be an acute-angled triangle and let P be a point on the side
BC. Let P 0 be the re ection of P in the side AB, and let P 00 be the
re ection of P in the side CA. Show that the distance P 0P 00 is leastwhen P is the foot of the perpendicular from A to BC.
4. Show that if x, y, and z are positive real numbers and xy = yz = zx
then x = y = z.
5. The equations x2 + px+ q = 0 and qx2 +mqx+ 1 = 0, where m, p,
and q are real, and q > 0, have roots x1, x2, and x1, 1=x2 respectively.Show that mp � 4.
6. Assume that a1 < a2 < � � � < a995 are 995 real numbers. Form all
sums ai + aj , for 1 � i � j � 995. Show that at least 1989 di�erent
numbers are obtained. Show also that exactly 1989 di�erent numbers
are obtained if and only if a1, a2, : : : , a995 is an arithmetic progression.
The 1989 Final Round has already appeared in a previous issue of CRUX, in
Olympiad Corner #125, 1991.
499
Dividing Points Equally
Cyrus C. Hsiastudent, University of Toronto
We start o� with a simple problem and follow it through with some
related questions that can be reduced to this problem.
Problem 1. Given 2n points in the plane, is it possible to draw a straight
line so that there are an equal number of points on either side of the line?
Solution. The answer is yes. Since there are a �nite number of points,
namely 2n of them, there are a �nite number of lines that pass through a
pair of points, at most�2n
2
�. Each line passing through a pair of points has a
certain direction. So, the 2n points have at most�2n
2
�distinct directions. It
is not too hard then to pick a direction, call it �, di�erent from all of these.
Intuitively, it is possible to take a line with direction � and slide it
along (preserving its direction of course) until half the points are on either
side. This is possible since the line can pass through only one point at a time.
If it passed through more than one point, then the direction between the two
points must be the direction � of this line. This contradicts the choice of �
to be di�erent from any of those pairs from the 2n points. Hence, by sliding
the line with direction � along, we pass each point one at a time until we
have passed exactly half, and then we are done.
We follow this up with some generalizations and related problems for
the reader to solve.
Problem 1A. Prove that given mn points in the plane, we can �nd
m � 1 parallel lines that divide the plane into m regions with n points in
each region.
Problem 1B. Show that it is possible to divide 2n points in the plane by
two intersecting lines so that for each line, half the points lie on either side
of it. Show that it is possible to divide them by m concurrent lines so that
for each line, half the points lie on either side of it.
Problem 1C. Take 4n points in the plane and take any three distinct
lines, each pair of which divides the set of points into equal quarters. Show
that these three lines cannot be concurrent. (Hint: Show that the three lines
must form a triangle containing n of the points.)
Problem 1D. Is it possible to divide 2n points in the plane by a circle
so that half of the points are inside and half are outside? (Note: This does
not follow immediately from Problem 1 by inversion.)
Copyright c 1998 Canadian Mathematical Society
500
Problem 1E. Proposed by Loren C. Larson, St. Olaf College, North�eld,Minnesota.
Can every set of 4n points in the plane, no three of which are collinear,
be evenly quartered by two mutually perpendicular lines?
MathematicsMagazine, Problem 1513. Vol. 69, No. 5, December 1996,
p. 385.
In the next two problems, we consider three dimensions.
Problem2. There are 2n chocolate chips in a roll of frozen cookie dough.
Show that it is possible to divide them into two sets of n chips by a plane
cut.
Solution. The 2n chocolate chips can be thought of as points in space.
We want to roll the frozen cookie dough into a position where if the chocolate
chips fell straight down, they would not hit each other. Essentially, we want
to project the 2npoints onto a plane so that each point ismapped to a distinct
position. Then we are back to Problem 1 of dividing the points in the plane
by a line. (Why?) This problem then becomes one of showing that there is
such a projection.
In 3 dimensions, we need something that is similar to \slope". This
is where the concept of a direction vector comes into play. The direction
vector between two points from A = (a1; a2; a3) to B = (b1; b2; b3) is
given by (b1 � a1; b2 � a2; b3 � a3) (or any multiple of this). To reduce
two points having multiple direction vectors, we usually consider the unit
direction vector which is found simply by taking the given direction vec-
tor and dividing each coordinate by the vector's norm or length, namelyp(b1 � a1)2 + (b2 � a2)2 + (b3� a3)2. (The reader may wish to verify that
for any line, any two points on the line will give the same direction vector up
to a � sign.)
Since there are a �nite number of points, there are a �nite number of
unit direction vectors. We can then pick a unit direction vector di�erent from
all of them; call it � = (�1; �2; �3). By projecting each point in this direction
onto a plane perpendicular to this vector, each point projects to a di�erent
point in the plane.
Thus, take the dough so that the direction vector � points vertically
down on the kitchen table. Take a knife and cut along the line that would
divide the points in half in the plane.
Problem 3. Suppose the Earth has a population of 6 billion people in
the near future. Is it possible to draw an imaginary equator around the world
so that each hemisphere contains an equal number of humans?
Here is a lemma that we will need.
Lemma 3A. 2n points and a special point labelled S are given in the
plane. Any line passing through S passes through at most one of the 2npoints. It is then possible to divide the plane by a straight line that passes
501
through S so that n points are on either side of the line.
Solution. Consider the lines passing through S and one other point.
There are 2n of them. Pick any other line through S which is not one of
these. Say there are k points on one side and 2n � k points on the other.
If k = n then we are �nished. Otherwise, rotate this line counter-clockwise
about S. This line will go through all of the points one at a time as it rotates.
Let f(�) be the di�erence between the number of points on one side labelled
A and the number of points on the other side labelled B as shown when the
angle is � from the original line. Points on the line are to be ignored.
��������������1
�
qq A
Bqq
�
qS
A line with sides A and B.
Thus, at the beginning, f(0�) = (2n� k)� k = 2(n� k), and when it has
completed a rotation of 180�, we have f(180�) = k�(2n�k) = �2(n�k).Make sure that sides A and B stay in the same orientation.
��������������1
A
B
�
qS
A line with sides A and B reversed from the above �gure.
Thus these values have di�erent signs and if this were a problem of a contin-
uous function, then we could easily claim by the Intermediate Value Theorem
that there is a value 0� < � < 180� such that f(�) = 0. Nonetheless, we
can still conclude this, since we know that when the value of f changes, it
changes by a value of 1.
To see this, note that each time the line passes through a point, the
value on one side, say A decreases by 1 and the number of points on side
B stays the same. The value of f then decreases by one. Once the line
immediately passes by this point, the number of points on side B increases
by 1 and the number of points on side A remains the same. Again, the value
of f decreases by 1. This will happen for each point that the line crosses.
502
Note that when the line passes completely over a point the value of f changes
by a value of 2. Thus the parity of f changes when the line goes through a
point from the state when there are no points on the line and the parity stays
the same when the line passes by a point.
Since 2(n� k) and its negative are both integers, at some step in the
value of f , 0 is reached. Now we must check that this value is not obtained
when the line crosses through one of the 2n points. Since the values of 0and 2(n � k) have the same parity, when the value of 0 is achieved, it is
achieved in the state where the line does not pass through any of the 2npoints. Hence, for this line, there is an angle � counter-clockwise away from
the original line where the points are equally divided.
Solution to Problem 3.
We must assume, of course, that the people are points occupying a dis-
tinct and �xed location on a spherical Earth. Now for each pair of people
draw a great circle passing through them. That is, a circle given by the inter-
section of the sphere with a plane passing through the centre of the sphere;
this would be our de�nition of an equator. There are a �nite number of such
circles as there are a �nite number of pairs. Thus we may always choose an-
other point, call itN , not on any of the great circles. The point diametrically
opposite N , call it S, must not be on any of the great circles either. (Why?)
If a point was on a great circle then the point diametrically opposite it would
be also, by de�nition.
Now place the spherical world on a plane with the point S tangent to
it. From point N project each point onto the plane by drawing a line from
N through each point to intersect the plane { this projection is called the
stereographic map. Now consider lines through S in the plane. Any such line
can have at most one other point. If a line contains S and two other points
then that would mean that S was on the great circle through the preimage
of these points.
By Lemma 3A there is a line through S which cuts the plane into two
parts containing an equal number of points. This line projected back onto
the sphere is a great circle dividing the world into two equal populations.
Exercises
1. Give solutions to the problems listed above.
2. Is it possible to divide any 4n points in the plane by two intersecting
lines so that each of the four sectors contains n points?
3. Is it possible to divide the Earth into 4 quarters with an equal number
of people in each? (Assume a population of 6 billion.)
503
PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, De-partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.
In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.
To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81
2"�11" or A4 sheets of paper.
These may be typewritten or neatly hand-written, and should be mailedto the Editor-in-Chief, to arrive no later than 1 April 1999. They may alsobe sent by email to [email protected]. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication. Please note that we do not accept submissions sent by FAX.
2388. Proposed by Daniel Kupper, B �ullingen, Belgium.Suppose that for each k 2 N, the numbers ak; bk; zk 2 C. Suppose
that the polynomials
An(z) =
nXk=0
ak zk and Bn(z) =
nXk=0
bk zk
are related by An(zj) = Bn(z2j ) = 0 for j 2 f1; 2; : : : ; ng.
For each n 2 N, �nd an expression for bn in terms of a0; a1; : : : ; an.
2389. Proposed by Nikolaos Dergiades, Thessaloniki, Greece.Suppose that f is continuous on Rn and satis�es
f(a1; a2; a3; : : : ; an) � f
�a1 + a2
2;a2 + a3
2; a3; : : : ; an
�:
Let m =a1 + a2 + : : :+ an
2. Prove that
f(a1; a2; a3; : : : ; an) � f (m;m; : : : ;m) :
504
2390. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.
For � � 0 and p; q � 1, let Sn(�; p; q) :=
n��Xi=1
nXj=i+�
ip jq, where
n > �.
Given the statement: \Sn(�; p; q), understood as a polynomial in Q[n],is always divisible by (n� �)(n� �+ 1)(n� �+ 2)",
(a) give examples for � = 0; 1; 2; 3; 4;
(b)? prove the statement in general.
2391. Proposed by G. Tsintsifas, Thessaloniki, Greece.
Consider d + 1 points, B1, B2, : : : , Bd+1 in the unit sphere in Rd,
so that the simplex Sd(B) = B1B2 : : : Bd+1 includes the origin O.
Let P = fx j Bi � x � 1g for all i between 1 and d+ 1.
Prove that there is a point y 2 P such that jyj � d.
2392. Proposed by G. Tsintsifas, Thessaloniki, Greece.
Suppose that xi, yi, (1 � i � n) are positive real numbers. Let
An =
nXi=1
xiyi
xi + yi, Bn =
�Pn
i=1 xi� �Pn
i=1 yi�
Pni=1 (xi + yi)
,
Cn =
�Pn
i=1 xi�2
+�Pn
i=1 yi�2Pn
i=1 (xi + yi), Dn =
nXi=1
x2i + y2i
xi + yi.
Prove that
1. An � Cn,
2. Bn � Dn,
3. 2An � 2Bn � Cn � Dn.`
2393. Proposed by G. Tsintsifas, Thessaloniki, Greece.
Suppose that a, b, c and d are positive real numbers. Prove that
1.�(a+ b)(b+ c)(c+ d)(d+ a)
�3=2� 4abcd(a+ b+ c+ d)2;
2.�(a+ b)(b+ c)(c+ d)(d+ a)
�3� 16(abcd)2
Ya; b; c; dcyclic
(2a+ b+ c).
505
2394. Proposed by Vedula N. Murty, Visakhapatnam, India.
The inequality aabb ��a+ b
2
�a+b, where a; b > 0, is usually proved
using Calculus. Give a proof without the aid of Calculus.
2395. Proposed by Witold Janicki, Jagiellonian University, Krakow,Poland, Michael Sheard, St. Lawrence University, Canton, NY, USA, DanVelleman, Amherst College, Amherst, MA, USA, and StanWagon,MacalesterCollege, St. Paul, Minnesota, USA.
Let P be such that
(A) P (0) is true, and
(B) P (n) =) P (n+ 1).
Find an integer n > 106 such that P (n) can be proved without using
induction, but rather using
(L) the Law of Implication (that is, X and (X =) Y ) yield Y )
ten times only.
2396. Proposed by Jose Luis Diaz, Universitat Politecnica deCatalunya, Colum, Terrassa, Spain
Suppose that A(z) =
nXk=0
akzk is a complex polynomial with an = 1,
and let r = max0�k�n�1
njakj1=(n�k)
o. Prove that all the zeros of A lie in the
disk C =
�z 2 C : jzj �
r
21=n � 1
�.
2397. Proposed by Toshio Seimiya, Kawasaki, Japan.
Given a right-angled triangle ABC with \BAC = 90�. Let I be the
incentre, and let D and E be the intersections of BI and CI with AC and
AB respectively.
Prove thatBI2 + ID2
CI2 + IE2=
AB2
AC2.
2398. Proposed by Toshio Seimiya, Kawasaki, Japan.
Given a square ABCD with points E and F on sides BC and CD
respectively, let P and Q be the feet of the perpendiculars from C to AE
and AF respectively. Suppose thatCP
AE+CQ
AF= 1.
Prove that \EAF = 45�.
506
2399?. Proposed by David Singmaster, SouthBank University, Lon-don, England.
In James Dodson's The Mathematical Repository, 2nd ed., J. Nourse,
London, 1775, pp 19 and 31, are two variations on the classic \Ass andMule"
problem:
\What fraction is that, to the numerator of which 1 be added,
the value will be 1=3; but if 1 be added to the denominator, its
value is 1=4?"
This is easily done and it is easy to generalize to �nding x=y such that
(x + 1)=y = a=b and x=(y + 1) = c=d, giving x = c(a + b)=(ad � bc)and y = b(c+ d)=(ad � bc). We would normally take a=b > c=d, so that
ad � bc > 0, and we can also assume a=b and c=d are in lowest terms.
\A butcher being asked, what number of calves and sheep he
had bought, replied, `If I had bought four more of each, I should
have four sheep for every three calves; and if I had bought four
less of each, I should have had three sheep for every two calves'.
How many of each did he buy?"
That is, �nd x=y such that (x + 4)=(y + 4) = 4=3 and (x � 4)=(y � 4) =3=2. Again, this is easily done and it is easy to solve the generalisation,
(x + A)=(y + A) = a=b and (x � A)=(y � A) = c=d, getting
x = A(2ac � bc � ad)=(bc � ad) and y = A(ad + bc � 2bd)=(bc� ad).We would normally take a=b < c=d so that bc � ad > 0, and we can also
assume a=b and c=d are in lowest terms.
In either problem, given that a, b, c and d are integers, is there a con-
dition (simpler than computing x and y) to ensure that x and y are integers?
Alternatively, is there a way to generate all the integer quadruples a,
b, c, d, which produce integer x and y?
2400. Proposed by V�aclav Kone �cn �y, Ferris State University, BigRapids, Michigan, USA.
(a) Show that 1 + (� � 2)x <cos(�x)
1� 2x< 1 + 2x for 0 < x < 1=2.
[Proposed by Bruce Shawyer, Editor-in-Chief.]
(b)? Show thatcos(�x)
1� 2x<
�
2�2(�� 2)
�x�
1
2
�2
for 0 < x < 1=2.
507
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.
2090. [1995: 307, 1997: 433] Proposed by Peter Iv �ady, Budapest,Hungary.
For 0 < x < �=2 prove that
�sinx
x
�2
<�2 � x2
�2 + x2:
In the remarks following the solution, Walther Janous gave the exten-
sion to: what is the value of � that gives the best inequality of the type
�sinx
x
�2<
�2 � x2
�2 + x2
which is valid for all x 2 (0; �=2)?
Janous has now solved this question himself.
We prove that
�sinx
x
�2<
�2 � x2
�2 + x2(1)
is valid for all x 2 (0; �=2) if � �p6, and the optimal value is �opt =
p6.
(A) From the Taylor series expansion, we have
x2�2 � x2
�2 + x2� sin2 x = x2
�2 � x2
�2 + x2+
cos(2x)� 1
2
=
x2 +
1Xk=1
(�1)k2
�2kx2k+2
!
+
�x2 +
1Xk=1
(�1)k+122k+1
(2k+ 2)!x2k+2
!
= 2
1Xk=1
(�1)k
1
�2k�
22k
(2k+ 2)!
!x2k+2
= 2x4��2 � 6
4�2+ x2(: : : )
�> 0 :
Thus we have (as taking the limit as x ! 0 shows) that �2 � 6 � 0; that is� �p6.
508
(B) Suppose that x > 0. Since
�2 � x2
�2 + x2<
�2 � x2
�2 + x2
() �2�2 + �2x2 � �2x2 � x4 < �2�2 � �2x2 + �2x2 � x4
() �2x2 < �2x2 () �2 < �2 ;
it is su�cient to prove (1) for � = �opt =p6.
(C) Next, we prove the auxiliary inequality
sinx � x�x3
6+
x5
120(2)
is valid for all x > 0.
Indeed, let g(x) = x� x3
6+ x5
120� sinx. Then g(0) = 0. Furthermore,
we have
g0(x) = 1� x2
2+ x4
24� cosx ; g0(0) = 0 ;
g00(x) = �x+ x3
6+ sinx ; g00(0) = 0 ;
g(3)(x) = �1 + x2
2+ cosx ; g(3)(0) = 0 ;
g(4)(x) = x� sinx ; g(4)(0) = 0 ; and
g(5)(x) = 1� cosx � 0 :
Thus, g(4)(x) increases, so that g(4)(x) � g(4)(0) = 0.
Further, g(3)(x) increases, so that g(3)(x) � g(3)(0) = 0. And so on.
Thus, g(x) increases, so that g(x)� g(0) = 0, as claimed.
(D) Because of (C), and since sinx > 0 (since x 2 (0; �=2)), inequality (1)
with � = �opt =p6 [that is, x2(6�x2) > (6+x2) sin2 x] will follow from
x2(6� x2)� (6 + x2)
�x�
x3
6+
x5
120
�2
> 0 :
This inequality is equivalent to
x6
14400
�x6 � 34x4 + 400x2 � 960
�:=
x6
14400h(x) < 0 :
But, h0(x) = 6x5 � 136x3 + 800x = 2x�x4 � 68x2 + 400
�> 2x
�3x4 � 69x2 + 399
�= 6x
�x4 � 23x2 + 133
�= 6x
h�x2 � 23
2
�2+ 3
4
i> 0 ;
that is, h(x) increases as x > 0 increases. Since �
2< 1:6 andh(1:6) � �142,
the proof is complete.
509
2259. [1997: 301] Proposed by Paul Yiu, Florida Atlantic University,Boca Raton, Florida, USA.
Let X, Y , Z, be the projections of the incentre of 4ABC onto the
sidesBC,CA, AB respectively. LetX0, Y 0, Z0, be the points on the incirclediametrically opposite to X, Y , Z, respectively. Show that the lines AX0,BY 0, CZ0, are concurrent.
I. Solution by Christopher J. Bradley, Clifton College, Bristol, UK.The areal coordinates of the incentre I are 1
a+b+c(a; b; c), and those of
X are 12a(0; a+ b� c; a� b+ c).
Since�!XI =
��!IX0, the areal coordinates ofX0 are twice those of I minus
those of X, and hence are�2a
a+ b+ c;(c� a+ b)(c+ a� b)
2a(a+ b+ c);(b� a+ c)(b+ a� c)
2a(a+ b+ c)
�:
The equation of AX0 is thus
y(b+ a� c) = z(c+ a� b) :
The equations of BY 0 and CZ0 are obtained by cyclic changes, and all are
concurrent at the point P with coordinates
1
a+ b+ c(b+ c� a; c+ a� b; a+ b� c) :
II. Solution by Florian Herzig, student, Cambridge, UK.This problem is very similar to problem 2250 [1997:245, 1998:372].
I will use the following known theorem (a proof can be found in [1]):
In an inscriptable quadrilateral the incentre and
the midpoints of the diagonals are collinear.(1)
First let A0, B0, C0 be the midpoints of AX, BY and CZ. Because
of (1), in the limiting case of a triangle, the lines A0K, B0L, C0M pass
through I. Since XA0 : XA = XI : XX0 = 1 : 2, the lines AX0 and KI
are parallel. Let G be the centre of gravity of4ABC (and also of4KLM ).
A dilatation with centre G and factor �2 maps 4KLM onto 4ABC. If P
is the image of I under that mapping, then AX0, BY 0, CZ0 pass through Psince the lineKI is mapped onto the line AP but also onto the line AX0 asAX0 k KI. Moreover, because of the dilatation, P lies on IG and
IG : GP = 1 : 2 :
Reference
[1] H. D�orrie, Triumph derMathematik, Physica, W�urzburg 1958. (100 GreatProblems in Elementary Mathematics , Dover, N.Y. 1965.)
510
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;
FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CON AMORE
PROBLEM GROUP, Royal Danish School of Educational Studies, Copenhagen,Denmark; MICHAEL LAMBROU, University of Crete, Crete, Greece; GERRY
LEVERSHA, St. Paul's School, London, England; ISTV �AN REIMAN, Budapest, Hun-
gary; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Nether-lands; JOHN VLACHAKIS, Athens, Greece; and the proposer.
Bradley commented that if J is the centre of mass of the triangle thought
of as a uniform wire framework, then P lies on the extension of IJ and is such thatIJ = JP . The point P is the Nagel point. Also, G lies on this line, and
IG : GJ = 2 : 1. Further, G and P are the internal and external centres of
similitude of the incircles of triangles ABC and the median triangle (with verticesat the mid-points of the sides of triangle ABC).
Lambrou, who gave an indirect proof, gave a generalization and noted that it is
the converse of a result due to Rabinowotz (problem 1353, Mathematics Magazine,65 1992, p. 59).
Let X, Y , Z be the points of contact of the incircle of triangle ABC with
the sides BC, CA, AB respectively. Let I0 be any point within the incircle (not
necessarily the incentre). Let XI0, Y I0, ZI0 cut the incircle again at X0, Y 0, Z0
respectively. Then AX0, BY 0, CZ0 are concurrent.
Seimiya also noted that the point of concurrency is the Nagel point of4ABC.
2264. [1997: 364] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a right angled triangle with the right angle at A. PointsD and
E are on sides AB and AC respectively, such that DEkBC. Points F and
G are the feet of the perpendiculars from D and E to BC respectively.
Let I, I1, I2, I3 be the incentres of4ABC,4ADE,4BDF ,4CEGrespectively. Let P be the point such that I2PkI1I3, and I3PkI1I2.
Prove that the segment IP is bisected by the line BC.
Solution by Florian Herzig, student, Cambridge, UK.Let � = AD
AB= AE
AC: Also de�ne x; x1; x2; x3; y to be the lengths of
perpendiculars of I; I1; I2; I3; P to BC respectively. It is su�cient to prove
that x = y. First, (see, for example, Roger A. Johnson, Modern Geometry(1929), p. 189),
x =bc=2
(a+ b+ c)=2=
bc
a+ b+ c:
Then, by similarity,
DF = b �BD
a= (1� �)
bc
a;
and so,
x1 = DF + �x = (1� �)bc
a+ �
bc
a+ b+ c:
511
Moreover,
x2 =BD
BCx = (1� �)
bc2
a(a+ b+ c); x3 = (1� �)
b2c
a(a+ b+ c):
By de�nition of P ,
y = x1 � x2 � x3
= (1� �)bc
a+ �
bc
a+ b+ c� (1� �)
bc(b+ c)
a(a+ b+ c)= x
after an easy calculation, and that is what we wanted to prove.
[Editor's note: To see that y = x1 � x2 � x3; let Q = I1P ^ I2I3,R = I1P ^ BC, and z be the length of the perpendicular from Q to BC.
Then x2 + x3 = 2z and
I1R = I1Q+ QR = QP + QR = RP + 2QR = I1R �y
x1+ 2I1R �
z
x1;
so x1 = y+ 2z = y + x2 + x3.]
Solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid,
Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WALTHERJANOUS,
Ursulinengymnasium, Innsbruck, Austria; D.J. SMEENK, Zaltbommel, the Nether-
lands; and the proposer.
2267. [1997: 364] Proposed by Clark Kimberling, University ofEvansville, Evansville, IN, USA and Peter Y�, Ball State University, Muncie,IN, USA.
In the plane of 4ABC, let F be the Fermat point and F 0 its isogonalconjugate.
Prove that the circles through F 0 centred at A, B and C meet pairwise
in the vertices of an equilateral triangle having centre F .
Solution by Toshio Seimiya, Kawasaki, Japan.
Equilateral triangles BCD, CAE, ABF are erected externally on the
sides of4ABC. Then AD, BE, CF are concurrent at the Fermat point F ,
and \AFB = \BFC = \CFA = 120� (see, for example, H.S.M. Coxeter,
Introduction to Geometry (1961), x1.8).
Let the circles through F 0 centred at B and C meet at P . Since
BF 0 = BP and CF 0 = CP , P is the re ection of F 0 across BC. Simi-
larly Q is the re ection of F 0 across AC, and R0 is the re ection of F 0 acrossAB, where the circles through F 0 centred at A and C meet at Q and those
centred at A and B meet at R.
512
Since \CAQ = \F 0AC = \BAF , we get
\FAR = \FAC + \CAQ = \FAC + \BAF = \BAC :
Similarly we have \RAF = \BAC, so that \FAQ = \RAF .
SinceAQ = AF 0 = AR, AF is the perpendicular bisector ofQR. Sim-
ilarly CF is the perpendicular bisector of PQ. Thus we have
FR = FQ = FP , so that F is the circumcentre of4PQR.
Since AF ? QR and CF ? PQ, and \AFC = 120�, we have
\PQR = 60�. Similarly, we have \PRQ = 60�, so that 4PQR is equilat-
eral.
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and
MAR �IA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain
(2 solutions); FLORIAN HERZIG, student, Cambridge, UK; MICHAEL LAMBROU,
University of Crete, Crete, Greece; and the proposer.
2269. [1997: 365] Proposed by Crist �obal S �anchez{Rubio, I.B.Penyagolosa, Castell �on, Spain.
Let OABC be a given parallelogram with \AOB = � 2 (0; �=2].
A. Prove that there is a square inscribable in OABC if and only if
sin�� cos� �OA
OB� sin�+ cos�
and
sin�� cos� �OB
OA� sin�+ cos� :
B. Let the area of the inscribed square be Ss and the area of the given
parallelogram be Sp. Prove that
2Ss = tan2 ��OA2 + OB2 � 2Sp
�:
Editor's comment.
All those who sent in submissions on this problem noted that the pro-
poser had mislabeled the parallelogram OABC instead of OACB.
513
Solution by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg,Germany.
O
B C
DAQEP
S R
M
�
Suppose that the diagonals of parallelogram OACB meet at M . The centre
of the square must be M . If OB and � are �xed, then CD = OB sin� is
�xed. If the square rotates aboutM with one vertex on the lineOA, then the
other vertices lie on the lines QR, RS and SP . Thus, there is an inscribed
square if and only if the segment OB meets the segment SP .
Therefore OA � OB cos�+ OB sin� (from B = S and A = Q), and
OB sin��OB cos� � OA (from O = P andC = R). The same must hold
if we interchange the roles of A and B. So part A follows immediately.
O
S B R C
DQAEGP
HM
�
Let OB = b, OP = x, HP = y, PG = z. Then y = x tan� and
z = b sin� � y. Therefore OF = 12(a + b cos�) and x = OF � PF =
12(a+ b cos�� b sin�), giving 4Ss = 4(y2 + z2) and Sp = ab sin�.
514
It follows that
4Ss = 4y2 + 4z2
= (a+ b cos�� b sin�)2 tan2 �+ 4(b sin�� x tan�)2
= (a+ b cos�� b sin�)2 tan2 �+(2b sin�� (a+ b cos�� b sin�) tan�)
2
= tan2 ��2(a+ b cos�� b sin�)2 + 4b2 cos2�
�4b(a+ b cos�� b sin�) cos�) :
Therefore
2Ss = tan2��(a+ b cos�� b sin�)(a� b cos�� b sin�)
+2b2 cos2��
= tan2��(a� b sin�)2 � (b cos�)2 + 2b2 cos2 �
�= tan2�
�a2 + b2 � 2ab sin�
�= tan2�
�OA2 +OB2 � 2Sp
�;
as required.
Also solved by FLORIAN HERZIG, student, Cambridge, UK; and the proposer.
Walther Janous, Ursulinengymnasium, Innsbruck, Austria and Michael
Lambrou, University of Crete, Crete, Greece commented on the error in notation.
Janous gave a condition for the problem to be true.
2276. [1997: 430] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.
Quadrilateral ABCD is cyclic with circumcircle �(0; R).
Show that the nine-point (Feuerbach) circles of 4BCD, 4CDA,
4DAB and4ABC have a point in common, and characterize that point.
Solution by Francisco Bellot Rosado, I.B. Emilio Ferrari, and Mar��aAscensi �on L �opez Chamorro, I.B. Leopoldo Cano, Valladolid, Spain and asummary of their comments.
This problem is a classical result about cyclic quadrangles. We will
present a solution, then discuss some references. Let HA; BB ;HC ;HD be
the respective orthocentres of the triangles BCD; CDA;DAB, and ABC,
and let NA; NB ; NC ; ND be their nine-point centres.
Theorem 1. For any cyclic quadrangle, the centres of the nine-pointcircles of the four triangles formed, taking the vertices of the quadranglethree at a time, form a homothetic quadrangle.
Proof. The nine-point centres NA; NB ; NC ; ND are the midpoints of
the segments OHA; OHB ; OHC ; OHD; therefore the quadrangles
NANBNCND and HAHBHCHD are homothetic (with ratio 12). But
515
HAHBHCHD is also homothetic (and oppositely congruent) to ABCD.
Hence NANBNCND and ABCD are homothetic (with ratio �1
2).
As a consequence, the circumcentre M of NANBNCND is the mid-
point of the segment joining O to the circumcentre of HAHBHCHD, while
the circumradius is R2. In other words, M is the centre of symmetry of
ABCD and HAHBHCHD (the common midpoint of AHA, etc). Since our
four nine-point circles all have radius R2, they must all pass throughM , which
is the desired result.
Comments. It seems that this result is from Lemoine, 1869, NouvellsAnnals de Math �ematiques (pp. 174 and 317). M is sometimes called the
Mathot point, after Jules Mathot, Mathesis, 1901, p. 25. In [1]M is called
the anticentre of the cyclic quadrangle ABCD. Here is a summary of some
of the properties of this point:
Theorem 2. If ABCD is cyclic, the nine-point circles of the trianglesBCD,CDA,DAB,ABC,HBHCHD ;HCHDHA,HDHAHB;HAHBHC,and the Simson lines of these triangles (with respect to the fourth vertex ofeach quadrangle) all contain the Mathot point of ABCD, as do the linesperpendicular to a side of ABCD that pass through the midpoint of theopposite side.
Convenient references that present a solution to our problem include
1 Nathan Altshiller Court, College Geometry. Barnes and Noble, 1965
(theorem 263, p. 132.
2 Roger A. Johnson, Advanced Euclidean Geometry. Dover, 1960 (pp. 209
and 243).
Part of Theorem 1 was used in the 1984 Balkan Math. Olympiad. See [1985:
243] for related references.
Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece;
FLORIAN HERZIG, student, Cambridge, UK; WALTHER JANOUS, Ursulinen-
gymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete,
Greece; GERRY LEVERSHA, St. Paul's School, London, England; ISTV �AN REIMAN,
Budapest, Hungary (2 solutions); TOSHIO SEIMIYA, Kawasaki, Japan; and the
proposer.
2279. [1997: 431] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.
With the usual notation for a triangle, prove that
Xcyclic
sin3A cosB cosC =sr
4R4
�2R2 � s2 + (2R+ r)2
�:
516
Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.On [1997: 447{448] it was shown by Herzig that
Xsin3A cosB cosC =
�YsinA
��Xcos2A
�:
Using (see, for example, [1996: 130])
YsinA =
sr
2R2
and
Xcos2A = 3�
Xsin2A = 3�
s2 � 4Rr � r2
2R2
=6R2 � s2 + 4Rr + r2
2R2;
the desired identity easily follows.
Also solved by HAYO AHLBURG, Benidorm, Spain; FRANCISCO BELLOT
ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; THEODORE CHRONIS, student,
Aristotle University of Thessaloniki, Greece; GORAN CONAR, student, Gymna-sium Vara�zdin, Vara�zdin, Croatia; NIKOLAOS DERGIADES, Thessaloniki, Greece;
FLORIAN HERZIG, student, Cambridge, UK; RICHARD I. HESS, Rancho Palos
Verdes, California, USA; MURRAY S. KLAMKIN, University of Alberta, Edmonton,Alberta; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; SAI
C. KWOK, San Diego, California, USA; MICHAEL LAMBROU, University of Crete,
Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; VEDULAN.MURTY, Maharanipeta, India; ISAO NAOI, Seki-Shi, Gifu, Japan; BOB PRIELIPP,
University of Wisconsin{Oshkosh, Wisconsin, USA; D.J. SMEENK, Zaltbommel, the
Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; PAUL YIU, Florida At-lantic University, Boca Raton, Florida, USA; and the proposer.
Ahlburg and Prielipp gave the same solution as Sei�ert. Several other solvers
also used the identity from [1997: 447{448].
2280. [1997: 431] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle with incentre I. Let D be the second intersection of
AI with the circumcircle of4ABC. Let X, Y be the feet of the perpendic-
ulars from I to BD, CD respectively.
Suppose that IX + IY = 12AD. Find \BAC.
Combination of solutions by Michael Lambrou, University of Crete,Crete, Greece and by Florian Herzig, student, Perchtoldsdorf, Austria.
We show that
IX + IY = (sinA)AD :
The given condition then implies that sinA =1
2, so that A =
�
6or A =
5�
6.
517
Note �rst that 4IBD is isosceles with ID = BD:
\IBD = \IBC + \CBD =B
2+ \CAD
=B
2+A
2= \ABI + \BAI = \BID :
Thus IY = ID sin\CDI = BD sin\CDA. Let the diameter of the cir-
cumcircle of 4ABC be 1 (so that a chord equals the sine of either angle it
subtends on the circumference), in which case,
IY = BD �AC :
Similarly,
IX = CD� AB :
Thus (by Ptolemy's Theorem)
IX + IY = BD� AC +CD � AB = BC � AD = (sinA)AD :
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain;
FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; SABIN
CAUTIS, student, Earl Haig Secondary School, North York, Ontario; NIKOLAOS
DERGIADES, Thessaloniki, Greece; GERRY LEVERSHA, St. Paul's School, London,
England;VEDULA N. MURTY, Visakhapatnam, India; VICTOR OXMAN, University
of Haifa, Haifa, Israel; ISTV �AN REIMAN, Budapest, Hungary; D.J. SMEENK, Zalt-
bommel, the Netherlands; and the proposer.
2281. [1997: 431] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABC is a triangle, and D is a point on the side BC produced beyond
C, such that AC = CD. Let P be the second intersection of the circumcircle
of4ACD with the circle on diameter BC. Let E be the intersection of BP
with AC, and let F be the intersection of CP with AB.
Prove that D, E, F , are collinear.
Solution by Nikoloas Dergiades, Thessaloniki, Greece (with his nota-tion modi�ed to make use of directed line segments).
LetG be the point whereAP intersectsBD. The (convex) quadrilateral
APCD is cyclic and4ACD is isosceles so
\GPC = \CDA (cyclic)
= \CAD (isosceles)
= \CPD (cyclic):
Hence PC is the bisector of \GPD; since PB ? PC, PB is the exterior
bisector of the same angle. It follows that
GC
CD=
BG
BD; or
BG
GC= �
BD
DC:
518
Ceva's Theorem applied to 4ABC gives
BG
GC�CE
EA�AF
FB= 1 :
Thus
BD
DC�CE
EA�AF
FB= �1 ;
which, by Menelaus's Theorem, means that D;E; F are collinear.
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid,
Spain; FLORIAN HERZIG, student, Cambridge, UK; WALTHER JANOUS, Ursulinen-
gymnasium, Innsbruck, Austria; MICHAEL LAMBROU, University of Crete, Crete,
Greece; D.J. SMEENK, Zaltbommel, the Netherlands (2 solutions); and the proposer.
2282. [1997: 431] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.
A line, `, intersects the sides BC, CA, AB, of 4ABC at D, E, F
respectively such that D is the mid-point of EF .
Determine the minimum value of jEF j and express its length as ele-
ments of 4ABC.
Solution by Florian Herzig, student, Perchtoldsdorf, Austria.
Let M be the midpoint of BC. Drop perpendiculars from E, F onto
BC having lengths h1, h2 respectively. Observe that h1 = h2 by similar-
ity, as ED = DF . Hence the triangles BMF , CME are equal in area.
This implies that [ABC] = [AFME], using the common notation for areas.
Therefore
[ABC] = [AFME] = 1
2AM � EF � sin\(AM;EF ) � 1
2AM � EF
and so
EF �2[ABC]
AM
with equality if and only if EF ? AM . Equality is always possible by a
continuity argument: Consider all lines ` perpendicular to AM ; the lines
passing through B and C yield the extreme ratios FD : DE = 0 or in�nity,
whence there is a desired line in between.
[Ed: Dou gave three constructions for the pointsE0, F 0 that provide theminimum value of jEF j. Here is one of them: De�ne B0 as the point where
the perpendicular from B to AM meets the side AC; and D0 to be where
the line joiningA to the midpoint ofBB0 meets BC; the perpendicular from
D0 to AM meets AB and AC in the desired points F 0 and E0.]
519
Remark: The linesEF envelope a parabola (see [1991: 97]) that touches
BC in M;AC in Y and AB in Z where AC = CY , AB = BZ. Indeed
this problem shows that in this case the shortest segment of a tangent inter-
cepted by AB and AC comes from the tangent perpendicular to the axis of
the parabola. I will sketch a proof: Let the directrix of the parabola be d and
the focus be P . Drop perpendiculars from Y; Z on d which have feet K;L.
Then, as AY;AZ are tangents to the parabola, AY;AZ are perpendicular
bisectors of PK;PL using the tangent property (for parabolae). Hence A is
the circumcentre of4PLK and so the perpendicular from A to d is actually
the bisector of the linesKY;LZ; that is, it passes through M . Thus AM is
perpendicular to the directrix; that is, it is parallel to the axis of the parabola.
Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece; JORDI DOU,
Barcelona, Spain; RUSSELL EULER and JAWAD SADEK, NW Missouri State Uni-
versity, Maryville, Missouri, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,
USA; GERRY LEVERSHA, St. Paul's School, London, England: TOSHIO SEIMIYA,
Kawasaki, Japan; and the proposer.
Dou also proved the parabola property of the minimum EF . Several readers
interpreted the question as calling for the minimum value of EF in terms of the side
lengths of4ABC:
min jEF j =pa+ b+ c)(�a+ b+ c)(a� b+ c)(a+ b� c)
p2b2 + 2c2 � a2
:
2283. [1997: 432] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.
You are given triangle ABC with \C = 60�. Suppose that E is an
interior point of line segment AC such that CE < BC. Suppose that D is
an interior point of line segment BC such that
AE
BD=
BC
CE� 1 :
Suppose that AD and BE intersect in P , and the circumcircles of AEP and
BDP intersect in P and Q. Prove that QE k BC.
Solution by Michael Lambrou, University of Crete, Crete, Greece.
De�ne F to be the point on AC such that \CBF = 60�, and R to be
the point on BF such that ER is parallel to BC.
We shall show that the circumcircles of AEP , BDP intersect at R.
In other words R and Q coincide and the required conclusion QEjjBC will
follow.
520
If the projections of R and E on BC are R1 and E1 respectively,
then BR1 = CE1 = 1
2CE (as \C = 60�), so that RE = R1E1 =
BC � BR1 � CE1 = BC � CE. Hence from the stated condition we
have
AE
BD=
BC
CE� 1 ;
=RE
CE
so that
AE � CE = BD � RE : (1)
Suppose that the circumcircle of RAC cuts the extension of RE at G.
As RG, AC are intersecting chords we have AE � CE = EG � RE. Com-
paring with (1) we see EG = BD. Using this we see that triangles BRD,
ECG are congruent: they have BD = EG;BR = EC (clear) and
\B = 60� = \ECB = \CEG (alternate on parallels). Thus \EGC =\RDB and thus quadrilateral RGCD is cyclic. In other words C, G, A, R
and D are all on one circle. Hence
\RAD = \RCD (same arc)
= \CBE (symmetry)
= \BER (parallel lines);
that is, \RAP = \REP , showing that quadrilateral RAEP is cyclic, as
required.
It remains to verify that quadrilateral BDPR is cyclic:
\BPD = \APE
= \ARE (as triangle RAE is cyclic)
= \ACG (as quadrilateral RACG is cyclic)
= \BRD (as4BDR �= ECG):
This completes the proof.
Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece;
FLORIAN HERZIG, student, Cambridge, UK; TOSHIO SEIMIYA, Kawasaki, Japan;
D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.
Pompe shows more generally that QEjjBC still holds if the given condition
wereAE
BD=
BC
CE� 2 cosC :
Note that our featured solution accommodates this generalization: de�ne F to be
the point on AC for which \CBF = \C.
521
2284. [1997: 432] Proposed by Toshio Seimiya, Kawasaki, Japan.
ABCD is a rhombus with \A = 60�. Suppose that E, F , are points
on the sides AB, AD, respectively, and that CE, CF , meet BD [Ed: not
BC as was originally printed in error] at P , Q respectively. Suppose that
BE2 +DF 2 = EF 2.
Prove that BP 2 +DQ2 = PQ2.
Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.Apparently, there is a misprint in the published condition: \... meet BC
at ..." should read \... meet BD at ...". We solve the corrected version.
Without loss of generality, let AB = 1, and put EB = x, and FD = y.
Then DB = 1, AE = 1� x, and AF = 1� y.
By applying the Cosine Law to 4AEF (\EAF = 60�) we transform
the condition x2 + y2 = EF 2 to
x2 + y2 = (1� x)2 + (1� y)2� 2 �1
2� (1� x)(1� y);
which simpli�es to
1� x� y � xy = 0;
and so,
y =1� x1 + x
: (1)
Furthermore, BP is the angle-bisector of \EBC in 4EBC, so, by a
well-known formula,
BP =2EB � BC � cos 60�
EB + BC=
x
x+ 1:
Similarly, DQ = y
y+1. Using (1), DQ = 1�x
2. Now,
PQ = DB � BP �DQ = 1�x
x+ 1�
1� x
2=
1 + x2
2(1 + x):
Finally,
BP 2 +DQ2 =
�x
x+ 1
�2+
�1� x
2
�2
=1 + 2x2 + x4
4(1 + x)2=
(1 + x2)2
4(1 + x)2= PQ2 ;
which completes the proof.
522
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid,
Spain; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece;
GORAN CONAR, student, Gymnasium Vara�zdin, Vara�zdin, Croatia; NICOLAOSDERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, student, Cambridge, UK;
V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; MICHAEL
LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School,London, England; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; ANGEL
JOVAL ROQUET, I.E.S. Joan Brudieu, La Seu d'Urgell, Spain; D.J. SMEENK, Zalt-
bommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the pro-poser.
Most of the submitted solutions are similar to the above. All solvers have
noticed the misprint and solved the corrected problem.
2285. [1997: 432] Proposed by Richard I. Hess, Rancho PalosVerdes,California, USA.
An isosceles right triangle can be 100% covered by two congruent tiles.
Design a connected tile so that two of them maximally cover a non-
isosceles right triangle. (The two tiles must be identical in size and shape
and may be turned over so that one is the mirror image of the other. They
must not overlap each other or the border of the triangle.)
What coverage is achieved for a 30{60{90 right triangle?
Editorial comments by Bill Sands.
Two readers sent in solutions to this problem, but both seem to have
assumed that the tile must be a triangle, so the amount of the right triangle
their solutions cover is quite a bit less than optimal. The proposer's solution
is the best of the three received, although it has not been proved optimal
either, and this in fact may be quite di�cult to do! The proposer's solution
also had an error at one point, which slightly a�ected the end results. So
below is a corrected version of his solution. (The proposer has independently
veri�ed the numerical results given below.)
Let the given right triangle have angle � � 45�, with the side adja-
cent to this angle having length 1, so the other side has length tan � and the
hypotenuse has length sec �. The area of the triangle is (tan �)=2. The pro-
poser gives two methods for partially tiling this triangle with two congruent
tiles, each method better than the other for certain values of �.
523
Method A.
�� �
x tan �-� 1
tan �
x
x cot �
x csc �
ppppp
pppp
ppppp
pppp
pppppppppppp pp p pp p
From the �gure, x+ tan � + x cot � = 1, giving
x =1� tan �
1 + cot �:
The uncovered area is
x2 cot �
2=
(1� tan �)2 cot �
2(1 + cot �)2;
so the fraction of uncovered area is
Fa =
�1� tan �
1 + tan �
�2:
Note that this method is only valid for � � 45�, and for � = 45� gives
Fa = 0, the perfect tiling of the isosceles right triangle.
Method B.
-� 1
w
?
6
ppppppppppp
ppppppppp
pppppppp
ppppppp
pppppp
pppp
ppppp p
pppppppp
ppppppp
ppppp
pppp pp p
p p p pp pp p p p .�.A1
.�
. � = 2�
.�
v
.A2
v
�!A3!
524
Let � = 2�. From the �gure one can derive, for arbitrary v between 0 and
tan �=(1 + cos�),
w =v sin2 �+ sin�
cos�+ cos �
[this formula was given incorrectly by the proposer],
A1 =sin�(w cot�� 1)2
2; A2 =
v2 sin �
4;
and
A3 =sin�(w� v cos�)(tan� � v �w)
2;
where A1; A2; A3 are the uncovered areas of the triangle as shown. For a
given �, and hence �, vary v so that A = A1 +A2 +A3 is minimized. After
much calculation, one obtains: the minimum value of A occurs at
v =sin� cos�
cos3 �+ 2 cos2 �� 1;
whence
w =sin� cos�
cos2 �+ cos�� 1;
and the minimum value of A is
A =sin3 � cos�(3 cos2 �� cos4 �� 1)
(2 cos2�� 1)(cos3 �+ 2 cos2 �� 1)2:
Thus the uncovered fraction is
Fb =sin2 �(3 cos2 �� cos4 �� 1)
(cos3 �+ 2 cos2 �� 1)2:
When � = 30� (� = 15�), we get
Fb =15p3� 22
20p6 + 15
p3 + 36
p2 + 74
� :019915536;
that is, just under 2% of the 30{60{90 right triangle remains uncovered. Can
anyone do better?
One can also calculate that Method A and Method B give the same
uncovered area when � satis�es the remarkably simple equation
4 sin� cos4 � = 1;
namely for � � 17:648�, or � � 35:3�. Method B gives the better result
when � < 35:3�, and Method A when 35:3� < � � 45�. In particular,
Method B is the better way for the 30-60-90 triangle.
525
If anyone can �nd a reasonably short derivation of these relations, let
us know! And of course the entire problem is still open, in the sense that
no proof of minimality has been given, including for the 30{60{90 triangle.
Maybe some reader can �nd a third method that beats both of the proposer's
methods for some values of �.
2287. [1997: 501] Proposed by Victor Oxman, University of Haifa,Haifa, Israel.
Let G denote the point of intersection of the medians, and I denote
the point of intersection of the internal angle bisectors of a triangle. Using
only an unmarked straightedge, construct H, the point of intersection of the
altitudes.
Solution - see problem 2234 [1997:168, 1998: 247]
New solutions were sent in by FLORIAN HERZIG, student, Cambridge, UK;
and MICHAEL LAMBROU, University of Crete, Crete, Greece.
2288. [1997: 501] Proposed by Victor Oxman, University of Haifa,Haifa, Israel.
In the plane are a circle (without centre) and �ve points A, B, C, D,
E, on it such that arc AB = arc BC and arc CD = arc DE. Using only
an unmarked straightedge, construct the mid-point of arc AE.
Solution - see problem 2251 [1997: 300, 1998: 373]
New solutions were sent in by NIKOLAOS DERGIADES, Thessaloniki, Greece;
CYRUS HSIA, student, University of Toronto, Toronto, Ontario; and GOTTFRIED
PERZ, Pestalozzigymnasium, Graz, Austria.
2289?. [1997: 501] Proposed by Clark Kimberling, Evansville, IN,USA.
Use any sequence, fckg, of 0's and 1's to de�ne a repetition-resistantsequence s = fskg inductively as follows:
1. s1 = c1, s2 = 1� s1;
2. for n � 2, let
L = maxfi � 1 : (sm�i+2; : : : ; sm; sm+1)
= (sn�i+2; : : : ; sn; 0) for somem < ng;L0 = maxfi � 1 : (sm�i+2; : : : ; sm; sm+1)
= (sn�i+2; : : : ; sn; 1) for somem < ng:
(so thatL is themaximal length of the tail-sequence of (s1; s2; : : : ; sn; 0)that already occurs in (s1; s2; : : : ; sn), and similarly for L0), and
526
sn+1 =
8<:
0 if L < L0;1 if L > L0;cn if L = L0:�
For example, if ci = 0 for all i, then
s = (0; 1; 0; 0; 0; 1; 1; 0; 1; 0; 1; 1; 1; 0; 0; 1; 0; 0;
1; 1; 1; 1; 0; 1; 1; 0; 0; 0; 0; 0; 1; 0; 1; 0; : : : )�
Prove or disprove that s contains every binary word.
No solutions have been received | the problem remains open.
2290. [1997: 501] Proposed by Panos E. Tsaoussoglou, Athens,Greece.
For x; y; z � 0, prove that
�(x+ y)(y+ z)(z+ x)
�2 � xyz(2x+ y + z)(2y+ z + x)(2z + x+ y) :
Solution by Florian Herzig, student, Cambridge, UK.First note that we can assume that x, y, z are greater than zero since
otherwise the inequality becomes trivial. Now let a = 1x; b = 1
yand c = 1
z.
The inequality is therefore equivalent to
(a+ b)2(b+ c)2(c+ a)2 � (bc+m)(ca+m)(ab+m) ;
where m = bc + ca + ab. Since (a + b)(a + c) = a2 + m, etc., this is
equivalent to
(a2 +m)(b2 +m)(c2 +m) � (bc+m)(ca+m)(ab+m) ;
which is a consequence of Cauchy's Inequality:
(a2 +m)(b2 +m) � (ab+m)2
(the three such expressions are then multiplied together). Equality therefore
holds if and only if a = b = c; that is, x = y = z or two of the x, y, z are
zero.Also solved by MICHEL BATAILLE, Rouen, France; CHRISTOPHER
J. BRADLEY, Clifton College, Bristol, UK; THEODORE CHRONIS, Athens, Greece;
NIKOLAOS DERGIADES, Thessaloniki, Greece; RICHARD I. HESS, Rancho Palos
Verdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-tria; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; MICHAEL
LAMBROU, University of Crete, Crete, Greece (3 solutions); KEE-WAI LAU, Hong
527
Kong; VEDULA N. MURTY, Dover, PA, USA; GOTTFRIED PERZ, Pestalozzigymna-
sium, Graz, Austria; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DIGBY SMITH,
Mount Royal College, Calgary, Alberta; GEORGE TSAPAKIDIS, Agrinio, Greece;G. TSINTSIFAS, Thessaloniki, Greece; and the proposer.
2291. [1997: 501] Proposed by K.R.S. Sastry, Dodballapur, India.Let a, b, c denote the side lengths of a Pythagorean triangle. Suppose
that each side length is the sum of two positive integer squares. Prove that
360jabc.Solution by Florian Herzig, student, Cambridge, UK.More generally we show that this is even true for all right triangles
with each side being the sum of any two integer squares. We use the fol-
lowing well-known criterion (?) throughout the solution: a positive integern is the sum of two integer squares if and only if all prime factors p withp � 3 (mod 4) are contained in n to an even power. We may assume that
a, b, c have no common factor since if d > 1 is the highest common factor of
them and a = a1d, etc., then by (?) a1, etc. are again sums of two squares
(d has to contain each prime of the form 4k + 3 to an even power). Hence
we can write
a = m2 � n2
b = 2mn
c = m2 + n2
where m > n are relatively prime and m 6� n (mod 2).
For the power of 3 in abc we know that a2 + b2 = c2 and since
x2 � 1 (mod 3) for x not divisible by 3, it follows that at least one of a,
b, c is divisible by 3. By (?) this implies that this one is even divisible by 9,whence 9jabc.
For the power of 2 in abc notice thatm even, n odd is impossible since
it would then follow that a � 3 (mod 4). Thus m is odd and n is even.
As b=2 is the sum of two integer squares andm, n have no common factor, it
follows thatm, n are each the sum of two integer squares. The same follows
for m � n and m + n as a = (m � n)(m + n) is the sum of two integer
squares. Hence m and m � n are of the form 4k + 1 as both are odd. But
this implies that n is divisible by 4 and so 8jabc.Finally as x2 � 1 or 4 (mod 5) for x not divisible by 5, we conclude
(since a2 + b2 = c2) that at least one of a, b, c is divisible by 5 and hence
also abc. This shows that 360jabc.Moreover this is the best possible, even for side lengths being the sum
of two positive integer squares: just consider (a; b; c) = (225;272; 353),(153; 104; 185), and (65;72; 97).
Also solved by RICHARD I. HESS, Rancho Palos Verdes, California,
USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL
528
LAMBROU, University of Crete, Crete, Greece; JOEL SCHLOSBERG, student, Robert
Louis Stevenson School, New York, NY, USA; HEINZ-J�URGENSEIFFERT, Berlin, Ger-
many; KENNETHM. WILKE, Topeka, Kansas, USA; and the proposer.
There were three incorrect and two incomplete solutions submitted.
2292. [1997: 502] Proposed by K.R.S. Sastry, Dodballapur, India.
A convex quadrilateral Q has integer values for its angles, measured in
degrees, and the size of one angle is equal to the product of the sizes of the
other three.
Show that Q is either a parallelogram or an isosceles trapezium.
Solution by Florian Herzig, student, Cambridge, UK (slightly edited).
Let the angles be x, y, z, and xyz in degrees, which are positive inte-
gers. Then x + y + z + xyz = 360, and since the quadrilateral is convex,
xyz < 180. Without loss of generality, x � y � z. Hence
3z � x+ y+ z = 360� xyz > 180 ;
and so z > 60. But then 180 > xyz > 60xy, which gives xy � 2, andleaves us two cases to check.
If x = 1 and y = 2, then z = 119, which contradicts xyz < 180.
If x = y = 1 we obtain z = 179 and the last angle is also 179 degrees.
Depending on whether the angles of one degree are opposite or adjacent,
Q is a parallelogram or isosceles trapezium.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinen-
gymnasium, Innsbruck, Austria; RICHARD I. HESS, Rancho Palos Verdes, Cal-
ifornia, USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario;V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; MICHAEL
LAMBROU, University of Crete, Crete, Greece; GERRY LEVERSHA, St. Paul's School,
London, England; VICTOR OXMAN, University of Haifa, Haifa, Israel; D.J. SMEENK,Zaltbommel, the Netherlands; KENNETHM. WILKE, Topeka, Kansas, USA; and the
proposer.
529
2293. [1997: 502] Proposed by Claus Mazanti Sorensen, student,Aarhus University, Aarhus, Denmark.
A sequence, fxng, of positive integers has the properties:
1. for all n > 1, we have xn�1 < nxn;
2. for arbitrarily large n, we have x1x2 : : : xn�1 < nxn;
3. there are only �nitely many n dividing x1x2 : : : xn�1.
Prove that
1Xk=1
(�1)k
xkk!is irrational.
Solution by Michael Lambrou, University of Crete, Crete, Greece.We shall show the required irrationality without using conditions 1 and 3.
We argue by contradiction.
Suppose condition 2 is valid for all n � n0 for some �xed n0 2 N, andsuppose on the contrary that
1Xk=1
(�1)k
xkk!=
p
q
with p; q 2 N. By multiplying both terms of the fraction p=q by a positive
integer, if necessary, we may assume that q � n0. In particular we have by
condition 2 that
x1x2 : : : xq < (q+ 1)xq+1 (1)
and, for all k � q + 1, (since xi � 1 and 0 < xk�1 � x1x2 : : : xk�1 < kxk)
that 0 < xk�1 < kxk. So for k � q + 1 we have
1
xk�1(k� 1)!>
1
xkk!:
Thus the terms of1Xk=1
(�1)k
xkk!
are decreasing in absolute value. So by a well-known estimate of alternat-
ing series (see, for example, G.H. Hardy, A Course of Pure Mathematics,Cambridge University Press, Tenth Edition, 1952, p. 377) we have
0 <1
xq+1(q + 1)!�
1
xq+2(q+ 2)!
< (�1)q+1
1Xk=q+1
(�1)k
xkk!<
1
xq+1(q+ 1)!;
530
and so
0 < (�1)q+1
p
q�
qXk=1
(�1)k
xkk!
!<
1
xq+1(q + 1)!:
Multiplying across by q!(x1x2 : : : xq) we obtain
0 < (�1)q+1
"q!(x1x2 : : : xq)
p
q�
qXk=1
(�1)kq!(x1x2 : : : xq)xkk!
#
<x1x2 : : : xq
xq+1(q+ 1):
The middle term is an integer (as a sum of integers) but by inequality (1)
it is strictly between 0 and 1. This is absurd, and the required irrationality
follows.
Also solved by FLORIAN HERZIG, student, Cambridge, UK; and theproposer. Herzig also avoided use of condition (3).
2294. [1997: 502] Proposed by Zun Shan and Edward T.H. Wang,Wilfrid Laurier University, Waterloo, Ontario.
For the annual Sino-Japanese \Go" tournament, each country sends a
team of seven players, Ci's and Ji's, respectively. All players of each country
are of di�erent ranks (strengths), so that
C1 < C2 < : : : < C7 and J1 < J2 < : : : < J7:
Each match is determined by one game only, with no tie. The winner then
takes on the next higher ranked player of the opponent country. The tourna-
ment continues until all the seven players of one country are eliminated, and
the other country is then declared the winner. (For those who are not fa-
miliar with the ancient Chinese \Chess" game of \Go", a better and perhaps
more descriptive translation would be \the surrounding chess".)
(a) What is the total number of possible sequences of outcomes if each
country sends in n players?
(b)? What is the answer to the question in part (a) if there are three countries
participating with n players each, and the rule of the tournament is
modi�ed as follows:
The �rst match is between the weakest players of two countries
(determined by lot), and the winner of each match then plays the
weakest player of the third country who has not been eliminated (if
there are any left). The tournament continues until all the players
of two countries are eliminated.
531
Solution to (a) by Walther Janous, Ursulinengymnasium, Innsbruck,Austria and the proposers.
Let Tn denote the number sought. Suppose, for some k = 1, 2, : : : ,n, the winning country has lost k � 1 of its players before the tournament
is over; that is, the kth player of the winning country defeats the strongest
player of the losing country. Since each match eliminates exactly one player,
the total number of matches before the last game is n�1+k�1, or n+k�2,and among these n+k�2matches, k�1 of them resulted in the elimination
of a player from the winning country. Hence, for each k, the total number of
such sequences of outcomes isWk =�n+k�2k�1
�, and so
Tn = 2
nXk=1
Wk = 2
nXk=1
�n+ k� 2
k� 1
�
= 2
��n� 1
0
�+
�n
1
�+
�n+ 1
2
�+ : : :+
�2n� 2
n� 1
��
= 2
�2n� 1
n� 1
�=
2n
n
�2n� 1
n� 1
�=
�2n
n
�:
[Ed: The summation formula used above is a special case of the well-known
identityrX
k=0
�n+ k
k
�=
�n+ r + 1
r
�:
This identity can readily be proved by induction. See, for example, p. 207
of Applied Combinatorics (2nd Edition), John Wiley and Sons, Inc., 1984, by
Alan Tucker.]
Alternate solution to (a) by the proposers.
We represent the 2n players by a row of 2n white marbles. Choose
any n of the marbles and colour them black. Let the white (black) marbles
represent the players from China (Japan), and so they are listed (from left
to right) in order of their elimination, so that the colour of the last marble
indicates the winning country. Clearly, there is a one-to-one correspondence
between all such colourings and all the possible sequences of outcomes; for
example, the colouring depicted below corresponds to the following sequence
of outcomes:C1 defeats J1 and J2, but loses to J3;
C2 defeats J3 and J4, but loses to J5;
C3 defeats J5, but loses to J6;
C4 loses to J6;
C5 defeats J6, but loses to J7;
�nally, C6 defeats J7.
w w g w w g w g g w g w g g
532
Therefore Tn =
�2n
n
�as in solution I.
No other solutions were received to either (a) or (b)?. Hence, (b)? remains
open.
If one lets f(n) denote the number sought, then using brute-force computa-
tions on a tree diagram, the proposers found that f(1) = 12 and f(2) = 84. They
feel that, for general n, this is a di�cult problem.
2295. [1997: 503] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.
Find three positive integers a, b, c, in arithmetic progression (with pos-
itive common di�erence), such that a+b, b+c, c+a, are all perfect squares.
Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta.
We take the arithmetic progression as x � y, x, and x + y. Then we
must have
2x� y = u2; 2x = v2; 2x+ y = w2
so that x = (w2 + u2)=4, y = (w2� u2)=2, and u2 + w2 = 2v2.
The problem of �nding three squares in arithmetic progression is well-
known and is obtained easily from the identity
(1 + i)(p+ iq)2 = (p2 � q2� 2pq) + i(p2 � q2 + 2pq) :
On taking the square of the absolute values of each side, we have
2(p2 + q2)2 = (2pq+ p2 � q2)2 + (2pq+ p2 � q2)2:
Hence we can take
u = 2pq+q2�p2; w = 2pq+p2�q2; and v = p2+q2 (with p > q).
In order to havex > y, we must have 3u2 > w2 or (p2+q2)2 > 8pq(p2�q2).This can be satis�ed for an in�nite number of pairs (p; q) in which p is close
to q and not too small. For example, if we choose p = 10 and q = 9, weget x = (1992 + 1612)=4 and y = (1992 � 1612)=2. Since the Diophantine
equation is homogeneous, we can multiply x and y by 4, to give x = 65;522and y = 27;360, and then 2x� y = 3222, 2x = 3622, and 2x+ y = 3982.
Also solved by CHARLES ASHBACHER, Cedar Rapids, Iowa, USA; SAM
BAETHGE, Nordheim, Texas, USA; NIELS BEJLEGAARD, Stavanger, Norway;
CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORE CHRONIS,student, Aristotle University of Thessaloniki, Greece; CHARLES R. DIMINNIE, An-
gelo State University, San Angelo, TX, USA; FLORIAN HERZIG, student, Cam-
bridge, UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State
University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of Crete,
533
Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; KATHLEEN
E. LEWIS, SUNY Oswego, Oswego, NY, USA; DAVID E. MANES, SUNY at Oneonta,
Oneonta, NY, USA; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary,
Alberta; KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There was
one incorrect solution submitted.
Hess and Wilke both determine constraints on the ratio p=q which will guar-
antee that x > y above:
p
q< 2�
p3 + 2
q2�
p3 � 1:303225373
orp
q> 2 +
p3 + 2
q2 +
p3 � 7:595754113 :
Many solvers also pointed out that to get integers for x and y we need to have p and
q having the same parity, which explains why Klamkin needed to multiply his values
by 4 to get integers.
2296. [1997: 503] Proposed by Vedula N. Murty, Visakhapatnam,India.
Show that sin2�x
2>
2x2
1 + x2for 0 < x < 1.
Hence or otherwise, deduce that � <sin�x
x(1� x)< 4 for 0 < x < 1.
Solution by Florian Herzig, student, Cambridge, UK.
From the in�nite product formula for cosx [Ed.: See, for example,
1.431, #3 on p. 45 of Tables of Integrals, Series and Products (5th edition),
by I.S. Gradshteyn and I.M. Ryzhik, Academic Press, 1994], we get
cos
��x
2
�=
1Yk=0
�1�
x2
(2k+ 1)2
�� 1� x2 for 0 � x � 1 : (1)
Since 0 < (1� x2)(1 + x2) < 1 for 0 < x < 1, we have
cos2��x
2
�� (1� x2)2 <
1� x2
1 + x2;
or, equivalently,
sin2��x
2
�>
2x2
1 + x2:
From (1), we deduce that cos(�y) � 1 � 4y2 for 0 � y � 12. Hence, for
0 < x � 12and y = 1
2� x, we get
sin�x � 1� (1� 2x)2 = 4x(1� x) :
534
Dividing by x(1 � x) and noting that f(x) =sin(�x)
x(1�x) = f(1� x), we have
proved the right side of the double inequalities. Note that equality holds
when x = 12. [Ed.: Hence, the right inequality, as given, was incorrect.]
To prove the left inequality, note �rst that y = cos(�t) is concave on
[0; 12]. Since the chord joining (0;1) to (1
2; 0) has equation y = 1 � 2t, we
have, by integrating from 0 to x (0 < x � 12) that 1
�sin(�x) > x(1 � x),
orsin(�x)
x(1�x) > �. Again, since f(x) = f(�x), the inequality holds for all
x 2 (0; 1).
Also solved by PAUL BRACKEN, CRM, Universit �e de Montr �eal, Montr �eal,
Qu �ebec; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; NIKOLAOS
DERGIADES, Thessaloniki, Greece; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,
USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; HEINZ-J�URGEN
SEIFFERT, Berlin, Germany; and the proposer. There were also one incomplete andone incorrect solution.
The in�nite product formula was also used by Kone �cn �y, Lambrou and Sei�ert.
Besides Herzig, Bradley, Lambrou and Sei�ert also pointed out the error mentionedin the solution above.
Herzig remarked that the same problem appeared on p. 277 of G.H. Hardy'sA Course of Pure Mathematics, Cambridge University Press, and that it was also used
in the Mathematical Tripos 1993 for part IA (Cambridge exams after the �rst year),
paper 3, problem 7D. Janous asked the question:
Determine the set of all real numbers � such that sin�
��x2
�� 2x�
1 + x�
holds for all x 2 [0; 1].
Lambrou strengthened the �rst inequality to
sin2��x
2
�>
20x2
9(1 + x2)if 0 < x � 1
2;
sin2��x
2
�� x >
2x2
1 + x2if 1
2� x < 1 :
Both Kone �cn �y and Sei�ert pointed out that the left hand side of the double inequal-itiers is weaker than the following known inequality:
1� x2
1 + x2� sin(�x)
�xfor all x 6= 0 :
See CRUX [1993: 433]. According to Klamkin and Manes, this inequality is \not as
easy to prove".
535
2297. [1997: 503] Proposed by Bill Sands, University of Calgary,Calgary, Alberta.
Given is a circle of radius 1, centred at the origin. Starting from the
point P0 = (�1;�1), draw an in�nite polygonal path P0P1P2P3 : : : going
counterclockwise around the circle, where each PiPi+1 is a line segment tan-
gent to the circle at a point Qi, such that jPiQij = 2jQiPi+1j. Does this
path intersect the line y = x other than at the point (�1; 1)?Solution by Richard I. Hess, Rancho Palos Verdes, California, USA.
B
A
P0 Q0 P1
Q1
P2
Q2
P1
�1
1 12
12
14
14
q
q
q
q
q
q q
q
q
q
The total length of the in�nite
polygon P0P1P2P3 : : : is
1+1
2+1
2+1
4+1
4+1
8+1
8+� � � = 3 :
The total length of the arc from
A to B is � > 3. Therefore
P0P1P2 : : : will never reach B.
By calculation the limit point P1has angle �1 � 154:76 degrees.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; NIKOLAOS DERGIADES, Thessaloniki, Greece; FLORIAN HERZIG, student,
Cambridge, UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;
MICHAEL LAMBROU, University of Crete, Crete, Greece; VICTOR OXMAN, Uni-versity of Haifa, Haifa, Israel; and the proposer.
Janous and Oxman also gave the simple solution above.
2298. [1997: 503] Proposed by Bill Sands, University of Calgary,Calgary, Alberta.
The \Tickle Me" Feather Company ships its feathers in boxes which
cannot contain more than 1 kg of feathers each. The company has on hand
a number of assorted feathers, each of which weighs at most one gram, and
whose total weight is 1000001=1001 kg.
Show that the company can ship all the feathers using only 1000 boxes.
Solution by Kathleen E. Lewis, SUNY Oswego, Oswego, New York,USA.
Before packing the feathers in the boxes, �rst sort them by weight, from
heaviest to lightest. Now put 1000 feathers in each box, beginning with the
heaviest. Thus, the 1000000 heaviest feathers are now in the boxes with
no box containing more than 1 kg of feathers. For each remaining feather
(beginning still with the heaviest), put it in the �rst box that it will \�t" in;
that is, the �rst which it will not put over weight capacity. Continue this
process until all feathers are packed or until you reach a feather that will not
�t anywhere.
536
Suppose you �nd a feather that does not �t. Then in each box, adding
that feather would push the weight over 1 kg. Since there are at least 1000feathers already in each box, which are each the same weight as, or heavier
than, the new feather that will not �t, adding this feather would increase
the weight by a factor of at most 1001=1000. Thus the weight of each box
without this feather must be strictly greater than 1000=1001 kg. If we add
the weight of this feather to that of the feathers in any one box, it exceeds
1 kg. The total weight of the feathers in the other 999 boxes is more than
999 � 1000=1001 kg. Putting these weights together, we get a value strictly
greater than the 1000001=1001 kg which is supposed to be the total weight
of all the feathers. Therefore this situation cannot happen, so it must be
possible to pack all the feathers.
Also solved by NIKOLAOS DERGIADES, Thessaloniki, Greece; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; MICHAEL LAMBROU, Univer-
sity of Crete, Crete, Greece; DIGBY SMITH, Mount Royal College, Calgary, Alberta;
and the proposer.
Janous and the proposer note the more general result:
let a � 1 and let n be a positive integer, and let fw1; w2; : : : g bepositive real numbers satisfying
maxwi � 1
aand
Xwi � na+ 1
a + 1:
Then the wi's can be partitioned into at most n sets so that the sum
of each set is at most 1.
Readers may like to prove this. The given problem is the case a = n = 1000.
The proposer also notes that the result is best possible in two ways:
� if there are 1000001 feathers each of weight 1
1001+ � < 1
1000kg, for � su�-
ciently small, then the total weight of the feathers is just slightly greater than
1000001=1001 kg, but if only 1000 boxes are used, some box must contain at
least 1001 feathers which would weigh altogether more than 1 kg;
� if there are 999001 feathers each of weight 1
1000+ � kg, for � su�ciently small,
then each feather weighs just slightly more than 1 gm, and the total weight of
the feathers is still at most 1000001=1001 kg, but if only 1000 boxes are used,
some box must contain at least 1000 feathers which would weigh altogethermore than 1 kg.
Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell
Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer
Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil
Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,
J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia
537
YEAR END FINALE
Again, a year has own by! This year was owing smoothly until I had a
three month sick-leave break. Our Associate Editor, CLAYTON HALFYARD,
completed issue 5, and did issues 6 and 7. He did a splendid job. In par-
ticular, Memorial University student PAUL MARSHALL did almost all the
diagrams for issues 6 and 7. He too did a splendid job. I have resumed re-
sponsibilities with this issue. Thanks to all for their patience and assistance.
The online version of CRUX with MAYHEM continues to attract atten-
tion. There is a \new look" now. We recommend it highly to you. Thanks
are due to LOKI JORGENSON, NATHALIE SINCLAIR, FREDERIC TESSIER,
and the rest of the team at SFU who are responsible for this.
There are many people that I wish to thank most sincerely for par-
ticular contributions. Again, �rst and foremost is BILL SANDS. Bill is of
such value to me and to the continuance of CRUX with MAYHEM. As
well, I thank most sincerely, CATHY BAKER, ILIYA BLUSKOV, ROLAND
EDDY, CHRIS FISHER, BILL SANDS, JIM TOTTEN, and EDWARD WANG,
for their regular yeoman service in assessing the solutions; DENIS HANSON,
DIETER RUOFF, CHRIS FISHER, RICHARD MCINTOSH, DAIHACHIRO
SATO, DOUG FARENICK, HARLEY WESTON, for ensuring that we have qual-
ity articles; TED LEWIS, ANDY LIU, MAR�IA FALK de LOSADA, BILL SANDS,
JIM TIMOURIAN, for ensuring that we have quality book reviews, ROBERT
WOODROW, who carries the heavy load of two corners, and RICHARD GUY
for sage advice whenever necessary.
The editors of the MAYHEM section, NAOKI SATO, CYRUS HSIA,
ADRIAN CHAN, RICHARD HOSHINO, DAVID SAVITT and WAI LING YEE,
all do a sterling job.
I also thank two of our regulars who assist the editorial board with
proof reading; THEODORE CHRONIS and WALDEMAR POMPE. The quality
of these people is a vital part of what makes CRUX with MAYHEM what it
is. Thank you one and all.
As well, I would like to give special thanks to our Associate Editor,
CLAYTON HALFYARD, for guiding issues 5, 6 and 7 to fruition, and for
keeping me from printing too many typographical and mathematical er-
rors; and to my colleagues, PETER BOOTH, RICHARD CHARRON, ROLAND
EDDY, EDGAR GOODAIRE, ERIC JESPERS, MIKE PARMENTER, DONALD
RIDEOUT, NABIL SHALABY, in the Department of Mathematics and Statis-
tics at Memorial University, and to JOHN GRANT MCLOUGHLIN, Faculty
of Education, Memorial University, for their occasional sage advice. I have
also been helped by some Memorial University students, MIKE GILLARD,
DON HENDER, PAUL MARSHALL, JON MAUGER, SHANNON SULLIVAN,
TREVOR RODGERS, as well as a WISE Summer student, SHAWNA LEE.
538
The staff of the Department of Mathematics and Statistics at Memor-
ial University deserve special mention for their excellent work and support:
ROS ENGLISH, MENIE KAVANAGH, WANDA HEATH, and LEONCE
MORRISSEY; as well as the computer and networking expertise of RANDY
BOUZANE, DAVID VINCENT and PAUL DROVER.
Also the LATEX expertise of JOANNE LONGWORTH at the University of
Calgary, ELLEN WILSON at Mount Allison University, and the MAYHEM
sta�, is much appreciated.
GRAHAM WRIGHT, the Managing Editor, continues to be a tower of
strength and support. Graham has kept so much on the right track. He is
a pleasure to work with. The CMS's TEX Editor, MICHAEL DOOB has been
very helpful in ensuring that the printed master copies are up to the standard
required for the U of T Press who continue to print a �ne product.
Finally, I would like to express real and heartfelt thanks to the Head
of my Department, HERBERT GASKILL, and to the Acting Dean of Science
of Memorial University, WILLIE DAVIDSON. Without their support and un-
derstanding, I would not be able to do the job of Editor-in-Chief.
Last but not least, I send my thanks to you, the readers of CRUX with
MAYHEM. Without you, CRUX with MAYHEM would not be what it is.
Keep those contributions and letters coming in. We need your ARTICLES,
PROPOSALS and SOLUTIONS to keep CRUX with MAYHEM alive and well.
I do enjoy knowing you all.
In a letter from Ron Weedon, Reading, UK, concerning the reference
to F. G.-M., the author of the valuable 1912 book entitled Exercises de
G�eom�etrie, we learned that the author's full name is F. Gabriel-Marie (which
members of the Editorial Board had not known).
The letter made us wonder if there is a copy of the F. G.-M. book any-
where in Canada. Let us know if you have a copy. Apparently, the University
of Toronto does not have a copy, and the copy in the British Library is dam-
aged.
CRUX MAYHEM
u u
u
u u
?
539
INDEX TO VOLUME 24, 1998
Crux Articles
Cyclic ratio sums and products
Branko Gr �unbaum . ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. . 20
One Problem { Six Solutions
Georg Gunther ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... . 81
Sum of powers of a �nite sequence: a geometric approach
William O.J. Moser . .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 145
A Note on Special Numerals in Arbitrary Bases
Glenn Appleby, Peter Hilton and Jean Pedersen ... .. .. ... .. .. .. ... .. .. .. 210
Pythagoras Strikes Again!
K.R.S. Sastry ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... . 276
The Academy Corner Bruce Shawyer
February No. 16 ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 1
March No. 17 ..... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. 65
April No. 18 .... ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 129
May No. 19 ..... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 193
September No. 20 The Bernoulli Trials 1998 Questions
Ian Goulden and Christopher Small .. .. ... .. .. .. .. 257
December No. 21 The Bernoulli Trials 1998 Answers
Ian Goulden and Christopher Small .. .. ... .. .. .. .. 449
The Olympiad Corner R.E. Woodrow
February No. 187 ...... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 4
March No. 188 ...... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. .. 68
April No. 189 ...... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. . 131
May No. 190 .... ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. 196
September No. 191 ..... .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 261
October No. 192 ..... .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... . 322
November No. 193 .... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 385
December No. 194 ..... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 452
The Skoliad Corner R.E. Woodrow
February No. 27 ..... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. . 26
March No. 28 ..... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 88
April No. 29 .... ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 148
May No. 30 ..... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 215
September No. 31 .... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 281
October No. 32 .... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. . 334
November No. 33 ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. 398
December No. 34 .... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 477
Miscellaneous
Canadian Mathematical Society Award for
Contributions to Mathematical Education
Eric Muller, Chair, CMS Education Committee ..... .. .. ... .. .. .. ... .. .. .. 18
Letter From the Editors ... ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. . 321
Year End Finale .... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... 537
540
Book Reviews Andy Liu
Mini-Reviews Update
Reviewed by Andy Liu .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. . 12
From Erd }os to Kiev: Problems of Olympiad Caliber
by Ross Honsberger
Reviewed by Bill Sands . .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. . 78
Vita Mathematica
Edited by Ronald Calinger
Reviewed by Maria de Losada . .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... . 143
Mathematical Challenge & More Mathematical Challenges
by Tony Gardiner
Reviewed by Ted Lewis . .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 208
Models that Work
by Alan Tucker
Reviewed by Jim Timourian . .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. .. 275
Juegos y acertijos para la ense~nanza de las Matem�aticas
by Bernardo Recam�an Santos
Reviewed by Francisco Bellot Rosado .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 333
Dissections: Plane and Fancy
by Greg N. Frederickson
Reviewed by Andy Liu .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. 396
The Last Recreations
by Martin Gardiner
Reviewed by Andy Liu .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. 475
Problems
February 2301{2313 ..... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 45
March 2314{2325 ..... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 107
April 2306, 2326{2337 .... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... 175
May 2338{2350 ..... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. . 234
September 2326, 2329, 2351{2363 ..... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. 301
October 2364{2375 .... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 363
November 2374, 2376{2387 .... ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. 424
December 2388{2400..... .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... . 503
Solutions
February 2145, 2153, 2167, 2181, 2198{2207 ..... .. ... .. .. ... .. .. .. ... . 48
March 2208{2220 ..... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 110
April 2219{2222, 2224{2226, 2229{2230 .... .. ... .. .. .. ... .. .. ... . 178
May 2223, 2227{2228, 2231{2239 ..... .. ... .. .. ... .. .. .. ... .. .. .. . 237
September 2015, 220A, 2206, 2240{2246, 2248 ..... ... .. .. .. ... .. .. ... .. 305
October 2247, 2249{2256, 2258 ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. 366
November 1637, 2257, 2260{2263, 2265{2266, 2268,
2270{2275, 2277{2278 .... .. ... .. .. ... .. .. .. ... .. .. .. .. 427
December 2090, 2259, 2264, 2267, 2269, 2276,
2279{2285, 2287-2298 .... .. ... .. .. .. ... .. .. .. ... .. .. .. 507
541
Mathematical Mayhem
February .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. .. 30
March .... .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 92
April . .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... 158
May .... ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 220
September ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 286
October ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 342
November .... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 406
December ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 485
Mayhem Articles
Powers of Two
Wai Ling Yee ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 33
The Cantor Set and Cantor Function
Naoki Sato . .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 37
The Order of a Zero
Naoki Sato . .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 101
Tips on Inequalities
Naoki Sato . .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 161
Riveting Properties of Pascal's Triangle
Richard Hoshino . ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. 168
The Pentagon Theorem
Hiroshi Kotera .. .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... . 291
A Combinatorial Triad
Cyrus Hsia .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 296
Pushing the Envelope
Naoki Sato . .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 357
Bogus Arguments and Arcane Identities
Ravi Vakil . ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. 413
The Fibonacci Sequence
Wai Ling Yee . .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 415
Dividing Points Equally
Cyrus Hsia .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 499
Shreds and Slices
February .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. .. 30
March .... .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 92
May .... ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 220
September ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 286
November .... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 406
December ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 494
The K Method .... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 30
The Fibonacci Triangle ..... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... 92
From the Archives ..... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 220
Primitive Roots and Quadratic Residues, part 2 .... .. .. ... .. .. .. ... .. .. ... 286
An Algebraic Relation with a Geometric Twist .... .. .. ... .. .. ... .. .. .. ... .. 406
Miscellaneous
IMO Report Adrian Chan . ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. 412
1998{1999 Olympiad Correspondence Problems E.J. Barbeau .. .. .. .. ... .. 495
542
Mayhem Problems
February .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. .. 41
March .... .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 94
April . .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... 158
May .... ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. .. 222
September ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 288
October ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 342
November .... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 410
December ..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. . 485
High School Problems
February H223{H226 ..... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 42
April H237{H240 ..... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. .. 159
September H241{H244 ..... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. 289
November H245{H248 .... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. . 410
High School Solutions
March H217{H218 ..... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 95
May H221{H224 .... .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 222
October H224{H228 .... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. 342
December H229{H232.... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... . 485
Advanced Problems
February A209{A212 .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. 43
April A213{216 .... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... 159
September A217{A220 ...... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. 289
November A221{A224 .... ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. . 411
Advanced Solutions
March A193{A196 .... .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 98
May A197{A200 ..... .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... 226
October A201{A204 ..... ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 349
December A205{A208..... .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... . 489
Challenge Board Problems
February C75{C76 ...... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. .. .. 43
April C77{C78 ..... ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. . 160
September C79{C80 ..... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... . 290
November C81{C82 .... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. 411
Challenge Board Solutions
March C73 .... .. .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. . 100
May C74 .... .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. 228
October C75{C76 ..... .. .. .. ... .. .. ... .. .. .. ... .. .. .. ... .. .. ... .. .. .. .. 353
December C77{C78 ..... .. .. .. ... .. .. ... .. .. .. ... .. .. ... .. .. .. ... .. .. ... . 492
Contests
Swedish Mathematics Olympiad 1986 Qualifying Round ...... .. .. ... .. .. . 231
J.I.R. McKnight Problems Contest 1982 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 232
Swedish Mathematics Olympiad 1987 Qualifying Round ...... .. .. ... .. .. . 298
J.I.R. McKnight Problems Contest 1983 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 300
J.I.R. McKnight Problems Contest 1984 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 421
Swedish Mathematics Olympiad 1988 Qualifying Round ...... .. .. ... .. .. . 422
J.I.R. McKnight Problems Contest 1985 .... ... .. .. .. ... .. .. ... .. .. .. ... .. . 497
Swedish Mathematics Olympiad 1989 Qualifying Round ...... .. .. ... .. .. . 498
543
Proposers, solvers and commentators in the PROBLEMS and SOLUTIONS
sections for 1998 are:
Gerald Allen: 2207, 2208, 2228, 2249, 2256, 2260, 2273, 2275
Hayo Ahlburg: 2249, 2268, 2279
Miguel Amengual Covas: 2209, 2235, 2237, 2238, 2244, 2249,
2253, 2254, 2258, 2259, 2263, 2280
Claudio Arconcher: 2209, 2262
�Sefket Arslanagi �c: 2181, 2204, 2205, 2207, 2218, 2221, 2232,
2237, 2238, 2249
Charles Ashbacher: 2219, 2232, 2238, 2245, 2295
Sam Baethge: 2221, 2225, 2226, 2237, 2238, 2245, 2249, 2258,
2295
Edward J. Barbeau: 2219, 2219, 2278
Michel Bataille: 2207, 2232, 2249, 2290
Frank P. Battles: 2238, 2275
F.R. Baudert: 2271
Niels Bejlegaard: 2236, 2239, 2248, 2251, 2258, 2295
Francisco Bellot Rosado: 2181, 2203, 2204, 2209, 2221, 2224,
2226, 2235, 2237, 2239, 2240, 2241, 2242, 2246, 2249, 2250, 2252, 2253,
2254, 2258, 2259, 2263, 2264, 2266, 2267, 2276, 2279, 2280, 2281, 2284
Adrian Birka: 2229, 2235, 2239, 2249
Mansur Boase: 2199, 2209, 2211, 2218, 2219, 2220, 2221, 2222,
2224, 2225, 2226, 2227, 2229, 2235, 2236, 2237, 2238, 2239, 2248, 2254
Paul Bracken: 2206, 2245, 2256, 2258, 2260, 2268, 2296
Christopher J. Bradley: 2181, 2199, 2200, 2201, 2203, 2204,
2205, 2206, 2207, 2208, 2209, 2211, 2212, 2214, 2215, 2217, 2218, 2219,
2221, 2222, 2224, 2225, 2226, 2227, 2229, 2231, 2235, 2236, 2237, 2238,
2239, 2240, 2241, 2242, 2243, 2244, 2245, 2246, 2247, 2249, 2250, 2251,
2252, 2253, 2254, 2255, 2256, 2257, 2258, 2259, 2260, 2261, 2262, 2263,
2264, 2265, 2266, 2268, 2270, 2271, 2273, 2275, 2290, 2292, 2295, 2296,
2297
Jeremy T. Bradley: 2200
Adam Brown: 2181
James T. Bruening: 2245
Miguel Angel Cabez �on Ochoa: 2204, 2205, 2207, 2212, 2221,
2238, 2245, 2258, 2273
Sabin Cautis: 2280
Adrian Chan: 2204, 2209, 2218, 2221, 2224, 2226, 2229, 2232,
2235, 2239, 2249
Ji Chen: 220A, 2015
Denise Cheung: 2237
Jimmy Chui: 2229, 2235, 2237, 2239, 2240, 2249
Theodore Chronis: 2181, 2215, 2218, 2245, 2248, 2249, 2253,
2254, 2255, 2256, 2258, 2260, 2271, 2274, 2275, 2279, 2284, 2290, 2295,
, , 2443
Goran Conar: 2181, 2204, 2212, 2218, 2221, 2245, 2249, 2253,
2256, 2279, 2284
Curtis Cooper: 2225
Filip Crnogorac: 2235
Tim Cross: 2273
Luz M. DeAlba: 2256, 2275
Paul-Olivier Dehaye: 2199
Nikolaos Dergiades: 2268, 2276, 2278, 2279, 2280, 2281, 2282,
2283, 2284, 2288, 2290, 2292, 2296, 2297, 2298
Charles R. Diminnie: 2207, 2208, 2218, 2219, 2225, 2226, 2228,
2244, 2245, 2249, 2256, 2260, 2268, 2273, 2275, 2295
Angel Dorito: 2261
David Doster: 2199, 2207, 2217, 2221, 2222, 2227, 2232, 2237,
2245, 2248, 2249, 2256, 2268, 2275, 2203, 2209, 2220, 2234, 2235, 2237,
2244, 2255, 2282
Mayumi Dubree: 2238
Keith Ekblaw: 2207, 2211, 2212, 2238, 2260, 2261, 2275
Hans Engelhaupt: 2251, 2256, 2258, 2269, 2271, 2273, 2275
Russell Euler: 2254, 2256, 2260, 2268, 2275, 2282
Noel Evans: 2256, 2273
C. Festraets-Hamoir: 2201, 2203, 2204, 2208, 2209
F.J. Flanigan: 2181, 2199, 2275, 2443
Je�rey K. Floyd: 2238
Ian June L. Garces: 2212, 2224, 2226
Robert Geretschl�ager: 2145, 2199, 2211, 2212
Shawn Godin: 2222, 2248, 2251
Joaqu��n G �omez Rey: 2210, 2220, 2223, 2227, 2245, 2277, 2278
Douglass Grant: 2015
Herbert G�ulicher: 2231
David Hankin: 2226
Stergiou Haralampos: 2244
Yeo Keng Hee: 2203, 2214, 2217, 2221, 2225
G.P. Henderson: 2233, 2274
Florian Herzig: 2181, 2199, 2201, 2202, 2203, 2204, 2205, 2206,
2207, 2208, 2209, 2211, 2212, 2213, 2214, 2215, 2217, 2218, 2219, 2219,
2220, 2221, 2224, 2225, 2226, 2229, 2230, 2231, 2232, 2235, 2237, 2238,
2239, 2245, 2247, 2248, 2251, 2253, 2254, 2256, 2258, 2259, 2260, 2261,
2262, 2263, 2264, 2265, 2266, 2267, 2269, 2271, 2272, 2273, 2274, 2275,
2276, 2277, 2279, 2280, 2281, 2282, 2283, 2284, 2287, 2290, 2291, 2292,
2293, 2295, 2296, 2297
Richard I. Hess: 2181, 2198, 2199, 2200, 2207, 2208, 2211, 2212,
2217, 2218, 2219, 2220, 2221, 2222, 2225, 2227, 2229, 2231, 2233, 2237,
2238, 2244, 2245, 2247, 2248, 2256, 2258, 2260, 2261, 2262, 2273, 2275,
2279, 2290, 2291, 2292, 2295, 2297
John G. Heuver: 2226, 2231, 2235, 2238
Joe Howard: 2206, 2214, 2218, 2232, 2256, 2260, 2275
Cyrus Hsia: 2199, 2209, 2212, 2218, 2219, 2221, 2225, 2288, 2292
Robert B. Israel: 2208, 2211, 2213, 2217, 2218, 2221, 2222, 2225,
2226, 2227
Walther Janous: 2015, 2090, 2181, 2198, 2199, 2201, 2202, 2203,
2204, 2205, 2205, 2206, 2206, 2207, 2208, 2209, 2212, 2214, 2215, 2217,
2218, 2219, 2220, 2221, 2222, 2224, 2225, 2227, 2229, 2231, 2232, 2233,
2235, 2236, 2237, 2238, 2239, 2240, 2243. 2244, 2245, 2247, 2248, 2249,
2250, 2252, 2253, 2254, 2255, 2256, 2258, 2260, 2261, 2262, 2263, 2264,
2266, 2268, 2269, 2271, 2273, 2274, 2275, 2276, 2278, 2278, 2279, 2281,
2282, 2284, 2290, 2291, 2292, 2294, 2295, 2296, 2297, 2298
D. Kipp Johnson: 2181, 2226, 2229, 2232, 2237, 2238, 2254, 2255,
2258
Dag Jonsson: 2254
�Angel Joval Roquet: 2258
Geo�rey A. Kandall: 2254
Murray S. Klamkin: 2145, 2181, 2199, 2279, 2290, 2295
Clark Kimberling: 2267, 2289
Kenneth Kam Chiu Ko: 2225, 2229, 2239
V�aclav Kone�cn �y: 2181, 2199, 2200, 2203, 2204, 2205, 2206, 2209,
2212, 2214, 2215, 2217, 2218, 2219, 2222, 2224, 2225, 2236, 2237, 2238,
2240, 2242, 2245, 2249, 2251, 2253, 2254, 2256, 2258, 2261, 2266, 2271,
2274, 2275, 2278, 2279, 2282, 2284, 2292, 2295, 2296
Edward J. Koslowska: 2211
Sai C. Kwok: 2279
Michael Lambrou: 2181, 2198, 2199, 2201, 2202, 2203, 2204,
2205, 2205, 2206, 2207, 2208, 2209, 2211, 2212, 2213, 2214, 2215, 2215,
2217, 2218, 2219, 2220, 2221, 2222, 2224, 2225, 2226, 2227, 2228, 2229,
2231, 2232, 2233, 2235, 2236, 2237, 2238, 2239, 2240, 2241, 2242, 2243,
2244, 2245, 2246, 2247, 2248, 2250, 2251, 2253, 2254, 2255, 2256, 2257,
2258, 2259, 2260, 2261, 2262, 2263, 2265, 2266, 2267, 2268, 2269, 2271,
2273, 2274, 2275, 2276, 2278, 2278, 2279, 2280, 2281, 2283, 2284, 2287,
2290, 2291, 2292, 2293, 2295, 2296, 2297, 2298
Kee-Wai Lau: 220A, 2204, 2204, 2205, 2206, 2214, 2218, 2221,
2222, 2227, 2229, 2237, 2239, 2244, 2245, 2248, 2249, 2254, 2255, 2256,
2258, 2262, 2263, 2268, 2271, 2273, 2275, 2290
James Lee: 2235
Gerry Leversha: 2201, 2202, 2203, 2204, 2205, 2207, 2208, 2209,
2211, 2212, 2215, 2217, 2218, 2219, 2221, 2222, 2223, 2224, 2225, 2226,
2231, 2232, 2235, 2237, 2238, 2239, 2240, 2243, 2245, 2248, 2251, 2252,
2253, 2254, 2258, 2259, 2260, 2261, 2262, 2263, 2266, 2273, 2275, 2276,
2278, 2279, 2280, 2282, 2284, 2292, 2295
Kathleen E. Lewis: 2211, 2225, 2238, 2295, 2298
544
Alan Ling: 2239, 2245, 2248, 2249
Mar��a Ascensi �on L �opez Chamorro: 2204, 2221, 2224,
2226, 2253, 2265, 2267, 2276
Nick Lord: 2256, 2260
David E. Manes: 2295
Pavlos Maragoudakis: 2253, 2254, 2260
Anne Martin: 2232
Giovanni Mazzarello: 2212, 2224, 2226
J.A. McCallum: 2222, 2238, 2273
Phil McCartney: 2256, 2260
Sean McIlroy: 2211, 2229, 2232
Can Anh Minh: 2218, 2232
William Moser: 2212
Vedula N. Murty: 2198, 2206, 2245, 2249, 2253, 2260, 2275,
2279, 2280, 2290, 2296
Grady Mydlak: 2238
Isao Naoi: 2279
David Nicholson: 2235
Victor Oxman: 2213, 2218, 2234, 2236, 2240, 2244, 2245, 2249,
2251, 2254, 2255, 2256, 2258, 2260, 2262, 2263, 2268, 2280, 2287, 2288,
2292, 2297
Michael Parmenter: 2181, 2260, 2275
M. Perisastry: 2256, 2275
Gottfried Perz: 2199, 2209, 2211, 2212, 2221, 2225, 2226, 2232,
2235, 2237, 2240, 2244, 2284, 2288, 2290, 2295
Waldemar Pompe: 1637, 2015, 2224, 2228, 2230, 2238, 2257,
2258, 2265, 2266, 2283
Luis A. Ponce: 2254, 2258
Bob Prielipp: 2218, 2221, 2245, 2248, 2249, 2279
Efstratios Rappos: 2222
Istv �an Reiman: 2201, 2203, 2204, 2209, 2218, 2221, 2224, 2230,
2231, 2236, 2237, 2252, 2253, 2258, 2259, 2276, 2280
Renato Alberto Rodrigues: 2256
Juan-Bosco RomeroM�arquez: 2218, 2262, 2268
Angel Joval Roquet: 2284
Jawad Sadek: 2254, 2256, 2260, 2268, 2275, 2282
Rose Marie Saenz: 2211
Bill Sands: 2015, 2207, 2211, 2217, 2285, 2297, 2298
Jeevan Sandhya: 2244
Crist �obal S�anchez{Rubio: 2269
Christos Saragiotis: 2238, 2245, 2232
K.R.S. Sastry: 2226, 2237, 2242, 2244, 2249, 2252, 2291, 2292
Joel Schlosberg: 2211, 2217, 2225, 2229, 2238, 2245, 2248, 2273,
2278, 2291, 2443
Robert P. Sealy: 2232, 2245
Harry Sedinger: 2181
Heinz-J�urgen Sei�ert: 2181, 2198, 2199, 2202, 2204, 2205,
2206, 2206, 2207, 2211, 2212, 2214, 2221, 2222, 2227, 2229, 2232, 2233,
2236, 2237, 2238, 2240, 2245, 2248, 2256, 2258, 2260, 2262, 2268, 2273,
2275, 2278, 2279, 2290, 2291, 2295, 2296
Toshio Seimiya: 2201, 2203, 2209, 2224, 2230, 2231, 2234, 2235,
2236, 2237, 2240, 2241, 2244, 2246, 2250, 2251, 2252, 2253, 2254, 2255,
2257, 2258, 2259, 2262, 2263, 2264, 2265, 2266, 2267, 2276, 2280, 2281,
2282, 2282, 2283, 2284
Reza Shahidi: 2226, 2232, 2238, 2245, 2275
Zun Shan: 2202, 2204, 2206, 2207, 2208, 2211, 2212, 2215, 2217,
2218, 2219, 2221, 2225, 2245, 2294
Shi-Chang Shi: 2015
Trey Smith: 2207, 2228, 2244, 2249, 2256, 2260, 2273, 2275
D.J. Smeenk: 2181, 2201, 2203, 2209, 2221, 2224, 2226, 2231,
2232, 2235, 2236, 2237, 2238, 2240, 2241, 2242, 2244, 2245, 2246, 2250,
2251, 2253, 2254, 2257, 2258, 2259, 2263, 2264, 2265, 2266, 2268, 2270,
2275, 2276, 2279, 2280, 2281, 2283, 2284, 2292, 2295
Digby Smith: 2181, 2208, 2212, 2218, 2219, 2232, 2238, 2245,
2247, 2249, 2256, 2275, 2290, 2295, 2298
Christopher So: 2235, 2239, 2249
Claus Mazanti Sorensen: 2293
David R. Stone: 2181, 2199, 2207, 2211, 2221, 2225, 2226, 2228,
2232, 2238, 2243, 2245, 2247, 2248, 2249, 2256, 2260, 2268, 2273, 2275,
2278
Peter Tingley: 2272
Panos E. Tsaoussoglou: 2198, 2199, 2204, 2205, 2208, 2219,
2221, 2226, 2231, 2236, 2238, 2245, 2249, 2251, 2253, 2254, 2255, 2256,
2258, 2260, 2279, 2284, 2290
George Tsapakidis: 2262, 2275, 2290, 1637
G. Tsintsifas: 2290
Enrique Valeriano: 2254, 2255, 2258
Meletis D. Vasiliou: 2237
David Vella: 2249
John Vlachakis: 2254, 2258, 2259, 2260, 2262, 2266, 2275
Edward T.H. Wang: 2181, 2202, 2204, 2206, 2207, 2208, 2211,
2212, 2215, 2217, 2218, 2219, 2221, 2225, 2232, 2239, 2245, 2249, 2260,
2294
Kenneth M. Wilke: 2181, 2225, 2291, 2292, 2295
Yeo Keng Hee: 2209, 2215, 2218
Karen Yeats: 2235, 2244
Paul Yiu: 2181, 2226, 2244, 2254, 2258, 2258, 2259, 2273, 2279
Peter Y�: 2267
Vrej Zarikian: 2226, 2228, 2247, 2256, 2443
Roger Zarnowski: 2207, 2228, 2249, 2256, 2260, 2268, 2273,
2275
Con Amore Problem Group: 2227, 2238, 2239, 2242, 2244,
2252, 2254, 2255, 2256, 2257, 2258, 2259, 2261, 2268, 2275, 2443
Skidmore College Problem Group: 2238
University of Arizona Problem Solving Lab: 2256
Proposers, solvers and commentators in the MAYHEM PROBLEMS and
SOLUTIONS sections for 1997 are:
Mohammed Aassila: A218, A219, A223, C79
Miguel Carri �on �Alvarez: A202, A204, H225, H226
Evan Borenstein: H222
Adrian Chan: H224
Keon Choi: H229, H230, H231
Lino Demasi: H229, H230, H231
Deepee Khosla: A194, A195, A196
Christopher Long: C78
Vedula N. Murty: H228
Katya Permiakova: H229
Waldemar Pompe: A220, A224
Joel Schlosberg: H217, H222, H224
D.J. Smeenk: A195, A197, A207
Matt Szczesny: C73
Alexandre Trichtchenko: A202, A204, A207, A208, A217, H240
Hoe Teck Wee: H224
Wendy Yu: H229, H230, H231