1

LETTER FROM THE EDITORS

This issue sees a change in the Editorship of CRUXMATHEMATICORUM. Af-

ter many successful years under the accomplished leadership of Bill Sands and

Robert Woodrow, the editorial o�ce has moved from the University of Cal-

gary to Memorial University of Newfoundland. It is our intention to main-

tain the high standard that has led to CRUXMATHEMATICORUM having an

excellent international reputation as one of the world's top problem solving

journals. This high standard is a result of the submissions from the reader-

ship, and we encourage you all to continue to provide this. Please write to

the new editor if you have suggestions for improvements.

Readers will be delighted to know that Robert Woodrow has agreed to

continue as editor of the Olympiad Corner. However, he will soon be relin-

quishing the Skoliad Corner. Contributions intended for that section should

be sent directly to the editor. Bill Sands will remain \in the background",

and the new editor is indeed delighted to be able to call on Bill's sage advice

whenever necessary.

There are a few changes in the works. We shall continue with the

present format, which the readership appears to enjoy, but we shall change

from 10 issues of 36 pages (360 pages per volume) to 8 issues of 48 pages

(384 pages per volume). Although this has two fewer issues, the bonus is

that there are 24 extra pages per year. This allows increased e�ciency of

printing and helps to control mailing costs. The new schedule will have is-

sues for February, March, April, May, September, October, November and

December. We feel that this �ts in with the teaching year of the majority of

the readership.

Since postal rates for mailing outside of Canada are signi�cantly higher

than within Canada, and as is customary for many publications, the Society

has reluctantly adopted the policy that all subscription and other rates for

subscribers with non-Canadian addressesmust be paid in US funds. Although

this policy was instituted in 1995, the Society has, until the present time,

accepted payment in Canadian funds from any subscriber. However, e�ective

January 1996, the Society will require payment in US fundswhere applicable.

We trust that our non-Canadian subscribers will understand the necessity of

this change in payment policy.

We are also considering how to make CRUX available electronically to

its subscribers. One possibility would be to send postscript �les to sub-

scribers. Please communicate your thoughts to the Editor-in-Chief or to the

Managing Editor (email: gpwright@acadvm1.uottawa.ca).

We are now encouraging electronic submission ofmaterial. Please email

submissions to cruxeditor@cms.math.ca. We use LATEX2e. Those of you who

would like some of the technical details of the style use are asked to request

this by email.

Bruce Shawyer, Editor-in-Chief Graham Wright, Managing Editor

2

LETTRE DES R �EDACTEURS

Le pr �esent num�ero marque l'entr �ee en fonction d'une nouvelle �equipe �a la di-

rection de CRUXMATHEMATICORUM. Apr �es de nombreuses ann �ees sous la

direction experte de Bill Sands et de Robert Woodrow, le conseil de r �edaction

quitte l'Universit �e de Calgary pour s'installer �a l'Universit �e Memorial de

Terre-Neuve. Nous avons l'intention de maintenir le niveau de qualit �e qui

a fait de CRUX MATHEMATICORUM une publication reconnue mondiale-

ment comme l'une des meilleures en r �esolution de probl �emes. L'excellence

de notre p �eriodique tient �a votre apport, chers lecteurs, et nous vous encour-

ageons tous �a poursuivre dans la meme voie. Si vous avez des am�eliorations

�a proposer, veuillez en faire part au nouveau r �edacteur en chef.

Vous serez enchant �e d'apprendre que Robert Woodrow a accept �e de

conserver la direction de la chronique \Olympiade"; il d �elaissera toutefois

la chronique \Skoliad". Si vous d �esirez contribuer �a cette derni �ere, veuillez

vous adresser au r �edacteur en chef. Bill Sands, pour sa part, demeurera

\en veilleuse"; fort heureusement, nous pourrons compter sur ses judicieux

conseils au besoin.

Quelques changements sont �a venir. Nous conserverons la pr �esentation

actuelle de la revue puisqu'elle semble plaire �a nos lecteurs, mais nous pro-

duirons d �esormais huit num�eros de 48 pages (env. 384 pages par volume) au

lieu de dix num�eros de 36 pages (env. 360 pages par volume). Vous recevrez

donc deux num�eros de moins par ann �ee, mais vous aurez droit en prime �a 24

pages de plus. Cette formule s'av �ere plus avantageuse du point de vue de

l'impression et des frais d'exp �edition. Le nouveau calendrier de publication

pr �evoit la parution d'un num�ero en f �evrier, mars, avril, mai, septembre, oc-

tobre, novembre et d �ecembre. Nous pensons qu'il correspond au calendrier

scolaire de la majorit �e de nos lecteurs.

Comme il lui en coute beaucoup plus cher d'exp �edier ses publications

�a l' �etranger qu'au Canada, la Soci �et �e, �a l'instar de nombreux autres �editeurs

de revues, a �et �e contrainte d'adopter une nouvelle politique; dor �enavant,

tous les abonn �es dont l'adresse postale n'est pas au Canada devront r �egler

leurs frais d'abonnement et autres en devises am �ericaines. Meme si cette

politique avait �et �e introduite en 1995, la Soci �et �e a jusqu' �a pr �esent accept �e

les paiements en dollars canadiens qu'elle avait re�cus de ses abonn �es. Mais

�a partir de janvier 1996, elle exigera des paiements en dollars am �ericains, le

cas �ech �eant.

Nos abonn �es de l'ext �erieur du Canada comprendront qu'il nous �etait

devenu n �ecessaire de modi�er ainsi notre grille tarifaire.

Par ailleurs, nous songeons �a distribuer CRUX par voie �electronique.

L'une des options envisag �ees serait l'envoi de �chiers postscript aux abonn �es.

Veuillez transmettre vos commentaires au r �edacteur en chef ou au r �edaction-

g �erant (courrier �electronique : gpwright@acadvm1.uottawa.ca).

3

Nous encourageons d �esormais nos lecteurs �a soumettre leurs contri-

butions par courrier �electronique �a cruxeditor@cms.math.ca. Nous utilisonsLATEX2e. Si vous d �esirez obtenir certains d �etails techniques quant au style,

faites-en la demande par courrier �electronique.

Bruce Shawyer, R �edacteur en chef Graham P. Wright, R �edacteur-g �erant

CONGRATULATIONS

CRUX would like to extend its collective congratulations to Professor

Ron Dunkley, founder of the Canadian Mathematics Competitions, on his

appointment as a Member of the Order of Canada.

GUIDELINES FOR ARTICLES

Articles for this section of CRUX should satisfy the following:

� have length of two to four pages, ideally (we have allowed up to six

pages in exceptional circumstances);

� be of interest to advanced high school and �rst or second year university

students;

� contain some new material that leads to further interesting questions

(for this level);

� be well referenced as to origin of the problem and related material;

� not contain long involved formulas or expressions, that is, we like more

elegant mathematics as opposed to that which involves tedious calcu-

lations and attention to detail. We really want to emphasize ideas and

avoid many, many formulas. CRUX does not want to be too technical

in its appeal, rather we want to have wide general interest.

| Denis Hanson

4

UNITARY DIVISOR PROBLEMS

K.R.S. Sastry

Suppose d is a (positive integral) divisor of a (natural) number n. Then

d is a unitary divisor of n if and only if d and n=d are relatively prime, that

is (d; n=d) = 1.

For example, 4 is a unitary divisor of 28 because (4; 28=4) = (4; 7) = 1.

However, 2 is not a unitary divisor of 28 because (2; 28=2) = (2; 14) = 2

6= 1. Our aim is to consider

(i) the unitary analogue of a number theory result of Gauss, and

(ii) the unitary extension of super abundant numbers.

THE NUMBER d�(n) AND THE SUM ��(n) OF UNITARY DIVI-SORS of n.

To familiarize ourselves with the novel concept of unitary divisibility,

the following table, adapted from the one in [2], is presented. It consists of

n; 1 � n � 12; n's unitary divisors; d�(n), the number of unitary divisors of

n; and ��(n), the sum of unitary divisors of n.

TABLE

n unitary divisors of n d�(n) ��(n)

1 1 1 1

2 1, 2 2 3

3 1, 3 2 4

4 1, 4 2 5

5 1, 5 2 6

6 1, 2, 3, 6 4 12

7 1, 7 2 8

8 1, 8 2 9

9 1, 9 2 10

10 1, 2, 5, 10 4 18

11 1, 11 2 12

12 1, 3, 4, 12 4 20

From the above table it is clear that if n = p�11 is the prime decomposition

of n, then its unitary divisors are 1 and p�11 . So, d�(n) = 2 and ��(n) = 1+

p�11 . Ifn = p�11 p�22 is the prime decomposition of n, then its unitary divisors

are 1; p�11 ; p�22 ; p�11 p�22 . So d�(n) = 4 = 22 and��(n) = (1+p�11 )(1+p�22 ).

5

Now it is a simple matter to establish the following results:

Let n = p�11 p�22 � � � p�k

k denote the prime decomposition of n.

Then

d�(n) = 2k; (1)

��(n) = (1 + p�11 )(1 + p�22 ) � � � (1 + p�k

k ); (2)

If (m;n) = 1; then ��(mn) = ��(m)��(n): (3)

THE EULER � FUNCTION AND A RESULT OF GAUSS.

The Euler � function counts the number �(n)of positive integers that are less

than and relatively prime to n. Also, �(1) = 1 by de�nition. For example,

�(6) = 2 because 1 and 5 are the only positive integers that are less than

and relatively prime to 6. The following results are known [1]:

If n = p�11 ; then �(n) = (p1 � 1)p�1�11 : (4)

If n =

kYi=1

p�i

i ; then �(n) =

kYi=1

� (p�i

i ) : (5)

If (m;n) = 1; then �(mn) = �(m)�(n): (6)

LetD = fd : d is a divisor of ng. Then Gauss showed that

P�(d) = n.

For example, if n = 12, then D = f1; 2; 3; 4; 6; 12g and

P�(d) = �(1) +

�(2)+�(3)+�(4)+�(6)+�(12) = 1+1+2+2+2+4 = 12. Analogously,

ifD� = fd� : d� is a unitary divisor of ng, then what is

P�(d�)? The answer

is given by Theorem 1.

Theorem 1 Let n =

kYi=1

p�i

i denote the prime decomposition of n and D� =

fd� : d� is a unitary divisor of ng. Then

X�(d�) =

kYi=1

[1 + �(p�i

i )] :

Proof: The unitary divisors of n are the 2k elements, see (1), in the set

D� = f1; p�11 ; p�22 ; � � � ; p�k

k ; p�11 p�22 ; � � � ; p�k�1

k�1 p�k

k ; � � � ;kYi=1

p�i

i g:

6

Hence

X�(d�) = �(1) +

kXi=1

� (p�i

i ) +Xi<j

��p�i

i p�j

j

�+ � � �+ �

kYi=1

p�i

i

!:

On repeated applications of (5) and (6) we �nd that

X�(d�) = 1 +

kXi=1

� (p�i

i ) +Xi<j

� (p�i

i )��p�j

j

�+ � � �+

kYi=1

� (p�i

i ) :

But the right-hand side expression in the above equation is precisely the

expansion of

kYi=1

[1 + �(p�i

i )].

For a numerical illustration, let n = 108. Then D� = f1; 4; 27; 108g,P�(d�) = �(1)+�(4)+�(27)+�(108) = 1+2+18+36 = 57 by actual

count. Also n = 108 = 22 � 33.Therefore

P�(d�) = [1 + �(22)][1 + �

�33�] = [1 + 2][1 + 18] = 57, from

Theorem 1.

It is the converse problem that is more challenging to solve: Given a

positive integer m, �nd the set N such that N = fn : P�(d�) = mg.

There is a solution n if and only if m has the form

kYi=1

[1 + �(p�i

i )]. To see

this, let m = 4. Then there is no solution n such that

P�(d�) = m. This

follows because if n = p�, then 1 + �(p�) = 4 yields (p� 1)p��1 = 3. If

p = 2, then 2��1 = 3 has no solution for a positive integer �. If p is odd,

then (p� 1)p��1 is even and hence there is no solution of (p� 1)p��1 = 3

for a positive integer �. We leave it as an exercise to show that N = � if

m = 2� for � > 1. It is easy to verify that if m = p is an odd prime, then

N = fp; 2pg. As another exercise, the reader may �nd N whenm = 1995.

It is an open problem to determine the integers m for which N = �.

In the next section, we consider an extension of the concept of super-

abundant numbers in the context of unitary divisors.

UNITARY SUPER ABUNDANT NUMBERS.

We call a natural number n unitary abundant if ��(n) is greater than 2n. For

example, 150 is unitary abundant because ��(150) = 312 > 2(150).

To extend the work of Erd �os and Alaoglu [3], we call a natural number

n unitary super abundant if��(n)

n� ��(m)

mfor all natural numbers m � n.

Theorem 2 shows that the product of the �rst k primes, k = 1; 2; � � � is a

unitary super abundant number.

7

Theorem 2Let pk denote the kth

prime, k = 1; 2; � � � : Then n = p1; p2 � � � pkis a unitary super abundant number.

Proof: Consider the natural numbers of m � n. Then m belongs to one of

the three groups described below.

I. m is composed of powers of primes pj � pk.

II. m is composed of powers of some primes pj � pk and powers of some

primes p` > pk.

III. m is composed of powers of primes p` � pk.

First of all we note that the total number of primes composing m in any of

the three groups does not exceed k.

We now show that��(m)

m� ��(n)

nin all the above three cases.

Case I. Let m = p�11 p�22 � � � p�j

j where some �'s, except �j , may be

zero. From (2) we see that

��(m)

m=

jYi=1�i 6=0

�1 + p�i

i

p�i

i

�

=

jYi=1�i 6=0

�1 +

1

p�i

i

�

�jY

i=1�i 6=0

�1 +

1

pi

�

�kYi=1

�1 +

1

pi

�

=��(n)

n: (7)

Case II. In this case m = p�11 p�22 � � � p�j

j p�k+1

k+1 � � � p�`

` : Here too (7)

holds. Some �'s, except �j and �`, may be zero. Furthermore,

` > k) 1 +1

p`< 1 +

1

pk< 1 +

1

pk�1< � � � < 3

2: (8)

8

Again, from (2) and (3)

��(m)

m=

jYi=1�i 6=0

�1 + p�i

i

p�i

i

� `�kYi=1�i 6=0

1 + p

�k+i

k+i

p�k+i

k+i

!

=

jYi=1�i 6=0

�1 +

1

p�i

i

� `�kYi=1�i 6=0

1 +

1

p�k+i

k+i

!

�kYi=1

�1 +

1

p�i

i

�

�kYi=1

�1 +

1

pi

�

=��(n)

n;

on using (7) and (8).

Case III. In this case m = p�k

k � � � p�`

` . Here too (7) holds. Some �'s,

except �`, may be zero. As in earlier cases

��(m)

m=

`�kYi=0�i 6=0

1 +

1

p�k+i

k+i

!

�`�kYi=0

�1 +

1

pk+i

�

�kYi=1

�1 +

1

pi

�

=��(n)

non using (8).

We now give an example of a number n to show that there are values

of m less than n that can belong to any of the three groups described in

Theorem 2.

Let n = 210 = 2:3:5:7. Then��(n)

n= 96

35.

I. m = 120 = 23:3:5 is less than n and belongs to Group I.

Here

��(m)

m=

9

5� ��(n)

nholds.

II. m = 165 = 3:5:11 is less than n and belongs to Group II.

Here

��(m)

m=

96

55� ��(n)

nholds.

9

III. m = 143 = 11:13 is less than n and belongs to Group III.

Here too,

��(m)

m=

168

143� ��(n)

nholds.

If we write nk = p1p2 � � � pk; k = 1; 2; � � � , then we observe that

��(nk+1)

nk+1=

��(nk)

nk

�1 +

1

pk+1

�

>��(nk)

nk; k = 1; 2; � � � :

This above observation, coupled with the argument used in the proof of The-

orem 2, implies Theorem 3.

Theorem 3 The only unitary super abundant numbers are

nk; k = 1; 2; 3; � � � :

That is

2; 6; 30; 210; 2310; � � � :

Acknowledgement.

The author thanks the referees for their suggestions.

References

[1] L.E. Dickson, History of the Theory of Numbers, Vol. I, Chelsea Publi-

cations (1971), 113{114.

[2] R.T. Hanson and L.G. Swanson, Unitary Divisors, Mathematics Maga-zine, 52 (September 1979), 217{222.

[3] R. Honsberg, Mathematical Gems I, MAA, 112{113.

2943 Yelepet

Dodballapur 561203

Bangalore District

Karnataka, India

10

THE SKOLIAD CORNER

No. 11

R.E. Woodrow

This number marks the �rst anniversary of the Skoliad Corner. While

I have not received a lot of correspondence about the column per se, what Ihave received has been positive. A good source of more entry level problems

seems needed, and it would be nice to �nd a way of recognizing e�orts by

students at this level to solve the problems.

As a problem set this issue, we give the Sharp U.K. Intermediate Math-

ematical Challenge which was written by about 115,000 students February 2,

1995. Entrants had to be in the equivalent of School Year 11 or below for

England and Wales. Questions 1{15 were worth 5 marks each and Questions

16{25 were worth 6 marks each. The allotted time was one hour. The use

of calculators, rulers, and measuring instruments was forbidden. The con-

test was organized by the U.K.Mathematics Foundation with additional help

from The University of Birmingham. Many thanks go to Tony Gardiner, Uni-

versity of Birmingham, for sending me this contest and others sponsored by

the Foundation.

SHARP U.K. INTERMEDIATEMATHEMATICAL CHALLENGE

February 2, 1995

Time: 1 hour

1. Which of these divisions has a whole number answer?

A. 1234� 5 B. 12345� 6 C. 123456� 7

D. 1234567� 8 E. 12345678� 9

2. The word \thirty" has six letters, and 6 is a factor of 30. How many

of the numbers from one up to twenty have this curious property?

A. 14 B. 2 C. 3 D. 4 E. 5

3. Ren �e Descartes (1596{1650) published his most famous book Dis-course on Method in 1637. To illustrate his \method" he included an ap-

pendix on Geometry, in which he introduced the idea of cartesian coordinatesand showed how their use allowed one to solve many problems which had

previously been unsolved. Descartes is best known for a sentence on which

11

he based part of his philosophy. The famous sentence (in Latin) is \Cogito,

ergo sum". This is usually translated into the �ve English words given below

(in dictionary order). When they are written in the correct order to make

Descartes' saying, which word is in the middle?

A. am B. I C. I D. therefore E. think

4. What is the sum of the �rst nine prime numbers?

A. 45 B. 78 C. 81 D. 91 E. 100

5. Three hedgehogsRoland, Spike and Percival are having a race againsttwo tortoises Esiotrot and Orinoco. Spike (S) is 10 m behindOrinoco (O), who

is 25 m ahead of Roland (R). Roland is 5 m behind Esiotrol (E), who is 25 m

behind Percival (P). Which of the following words gives the order (�rst to

�fth) at this point in the race?

A. POSER B. SPORE C. ROPES D. PORES E. PROSE

6. Nabil was told to add 4 to a certain number and then divide the

answer by 5. Instead he �rst added 5 and then divided by 4. He came up

with the \answer" 54. What should his answer have been?

A. 34 B. 43 C. 45 D. 54 E. 211

7. Find the value of the angle BAD, where AB = AC, AD = BD,

and angleDAC = 39�.

�����@

@@@@

DDDDDD

A

B D C

A. 39� B. 45� C. 47� D. 51� E. 60�

8. A maths teacher who lives \opposite" (or is it \adjacent to"?) our

local hospital claims they recently hung up a banner saying: \Selly Oak Hos-

pital Can Always Handle the Odd Accident". In fact, between March 1993

and March 1994 the hospital handled 53; 744 accident and emergency cases.

Approximately how many cases was this per day?

A. 90 B. 150 C. 250 D. 500 E. 1000

9. The diagonal of a square has length 4 cm. What is its area (in cm2)?

A. 2 B. 4 C. 4p2 D. 8 E. 16

10. If x = 3, which expression has a di�erent value from the other

four?

A. 2x2 B. x2 + 9x C. 12x D. x2(x� 1)2 E. 2x2(x� 1)

12

11. The triangle formed by joining the points with coordinates (�2; 1),(2;�1) and (1; 2) is

A. scalene B. right-angled but not isosceles C. equilateral

D. isosceles but not right-angled E. right-angled and isosceles

12. The number of students who sat the 1992U.K. S.M.C.was 80; 000.

The number who sat the 1993 U.K. S.M.C. was 105; 000. Which calculation

gives the right percentage increase?

A.80;000

104;000� 100 B.

80;000

100� 105; 000 C.

105;000

80;000� 100

D.(105;000�80;000)

80;000� 100 E.

(105;000�80;000)

105;000� 100

13. Professor Hardsum was very absent-minded. He kept forgetting

his four-digit \PIN number"; without it he could not use his bank card to get

cash out of the bank cash machines. Then one day he noticed that none of

the digits is zero, the �rst two digits form a power of �ve, the last two digits

form a power of two, and the sum of all the digits is odd. Now he no longer

needs to remember the number because he can always work it out each time

he needs it! What is the product of the digits of his PIN number?

A. 60 B. 120 C. 240 D. 480 E. 960

14. A stopped clock may be useless, but it does at least show the

correct time twice a day. A \good" clock, which gains just one second each

day, shows the correct time far less often. Roughly how often?'

&

$

%t-6

A. once every 60 days B. once every 72 days C. once every 360 days

D. once every 12 years E. once every 120 years

15. What is the value of the fraction

1 +2

1 + 31+4

when written as a decimal?

A. 1:5 B. 2:25 C. 2:5 D. 2:6 E. 3:5

13

16. Each year around 750; 000 bottles of water are prepared for the

runners in the London Marathon. Each bottle holds 200 ml. George (who

�nished last year in 3hr 39min) reckoned that of each bottle used \one quar-

ter was more or less drunk, one half was sloshed all over the runner, and one

quarter went straight into the gutter". If four �fths of the bottles prepared

were actually used by the runners, roughly how many litres were \more or

less drunk"?

A. 30; 000 B. 36; 000 C. 120; 000 D. 150;000 E. 190; 000

17. Augustus Gloop eats x bars of chocolate every y days. How many

bars does he get through each week?

A. 7xy

B.7y

xC. 7xy D. 1

7xyE. x

7y

18. How many squares can be formed by

joining four dots in the diagram?

A. 4 B. 5 C. 9 D. 11 E. 13 q qq qq qq q qq q q

19. Timmy Riddle (no relation) never gives a straightforward answer.

When I asked him how old he was, he replied: \If I was twice as old as I was

eight years ago, I would be the same age as I will be in four years time".

How old is Timmy?

A. 12 B. 16 C. 20 D. 24 E. 28

20. Ivor Grasscutter's lawn, which is circular with radius 20 m, needs

retur�ng. Ivor buys the turf in 40 cm wide strips. What is the approximate

total length he needs?

A. 300 m B. 600 m C. 1500 m D. 3000 m E. 6000 m

21. Last year's carnival procession was 112km long. The last oat set

o�, and �nished, three quarters of an hour after the �rst oat. Just as the

�rst oat reached us, young Gill escaped. She trotted o� to the other end of

the procession and back in the time it took for half the procession to pass us.

Assuming Gill trotted at a constant speed, how fast did she go?

A. 3 km/h B. 4 km/h C. 5 km/h D. 6 km/h E. 7 km/h

22. In this 4 by 4 square you have to get from X to Y moving only

along black lines. How many di�erent shortest routes are there from X

to Y ?

r

r

X

Y

A. 18 B. 26 C. 28 D. 32 E. 34

14

23. When we throw two ordinary dice, the possible totals include all

the numbers from 2 to 12. Suppose we have two dice with blank faces. If we

mark the six faces of one dice 1; 2; 2; 3; 3; 4 how should we mark the faces of

the second dice so that all totals from 2 to 12 are possible, and each total has

exactly the same probability of occurring as with two ordinary dice?

A.

1; 2; 2; 7; 7; 8

B.

1; 3; 4; 5; 6; 8

C.

1; 4; 4; 5; 5; 8

D.

1; 3; 4; 5; 7; 8

E.

1; 2; 3; 6; 7; 8

24. A regular pentagon is inscribed in a circle of radius p cm, and the

vertices are joined up to form a pentagram. If AB has length 1 cm, �nd the

area inside the pentagram (in cm2).

BBBBBBZ

ZZ

ZZ �

��

��������BA

A. p2 B. 5p=2 C. 5p2=2 D. 5p E. 5p2

25. The six faces of a made-up cube are labelled F;H; I; N; X;Z.

Three views of the labelled cube are shown. The cube is then opened up to

form the net shown here (the net has been turned so that the F is upright).

What should be drawn (upright) in the shaded square?

XXX

XXX

��

��XXX��

XXX

XX

XX

XX

XXX

XXX

��

��XXX��

XXX

XX

XX

XX(((XX

XXX

XXX

��

��XXX��

XXX

XX��XX

(((

A. H B. I C. N D. X E. Z

Last month we gave the problems of the Saskatchewan Senior Mathe-

matics Contest 1994. This month we give the \o�cial" solutions.

My thanks go to Gareth Gri�th, University of Saskatchewan, longtime

organizer of the contest, for supplying the problems and answers.

15

Saskatchewan Senior Mathematics Contest 1994

1. Solve the equation

1 + 68x�4 = 21x�2:

Solution. The equation 1+68x�4 = 21x�2 is equivalent to u2�21u+

68 = 0 where u = x2. Since (u� 4)(u� 17) = 0, x2 = 4 or 17. Therefore

x = �2 or �p17.

2. Find the number of divisors of 16 128

(including 1 and 16 128).

Solution. To begin with, consider a simpler example: Find the number

of divisors of 12. The divisors are:

20 � 30; 21 � 30; 22 � 30;

20 � 31; 21 � 31; 22 � 31:

There are 6 divisors. We notice that we arrive at the number 6 as (a+1)(b+1)

where 12 = 2a � 3b. This results is true in general

16128 = 28 � 32 � 7:

The number of divisors is (8 + 1)(2 + 1)(1 + 1) = 54.

3. In the �gure, lines ABC and ADE intersect at A. The points

BCDE are chosen such that angles CBE and CDE are equal. Prove that

the rectangle whose sides have length AB and AC and the rectangle whose

sides have length AD and AE are equal in area.

�����������

����������������

QQQQQQQQQQQQQQ

@@@@@@@@@@@@

C

B

A

D

E

q q

qq q

Solution. Since angle CBE = angle CDE, the points BCDE are

concyclic. (Converse of the theorem \angles in the same segment are equal".)

(This is known as the theorem of Thales.) ABC andADE are secants of this

circle. Therefore AB � AC = AD � AE.

16

4. (a) State the domain and range of the functions

f(x) = tanx and g(x) = loga x; where a =5�

4:

(b) Determine the smallest value of x for which tanx = log5�=4 x.

Solution. (a) The domain of tanx is all the real numbers except for odd

multiplies of �=2. The range of tanx is all real numbers. The domain of

loga x is all positive real numbers. The range is the real numbers. This is

true for all a > 0, a 6= 1, and thus is true for a = 5�4.

(b) Consider the graphs of y = tanx, y = log(5�)=4 x for 0 < x < 3�2.

��=20��=2 3�=21

y = log5�=4(x)

log(5�)=2 x is not de�ned for x � 0.

The graphs indicate that the smallest value for x at which the graphs

intersect lies between � and3�2. But tan 5�

4= 1 and log(5�)=4

5�4

= 1.

Therefore this point is x = 5�4.

Note: for students who have studied calculus:

tanx 6= log(5�)=4 x for 0 < x <�

2;

since (i) tan 1 > tan�

4= 1, log(5�)=4 1 = 0 and

5�

4> 3 > e;

and (ii) sec2 x > 1 >1

x ln 5�

4

for 1 < x <�

2.

17

5. (a) Prove that the system of equations

x + y = 1

x2 + y2 = 2

x3 + y3 = 3

has no solution.

(b) Determine all values of k such that the system

x + y = 1

x2 + y2 = 2

x3 + y3 = k

has at least one solution.

Solution. (a)x + y = 1

x2 + y2 = 2

�.

Therefore 1 = (x+ y)2 = x2 + 2xy + y2 = 2 + 2xy. Therefore xy = �12.

If x3 + y3 = 3 then

(x+ y)(x2� xy + y2) = (1)(2� xy) = 3:

Therefore xy = �1. This is a contradiction and so the system has no solu-

tion.

(b) In order that the system admits at least one solution,

(x+ y)(x2� xy + y2) = (1)(2� xy) = k:

Therefore xy = 2� k and xy = �12so that 2� k = �1

2; k = 5

2.

6. (Contributed by Murray Bremner, University of Saskatchewan.)

This problem shows how we may �nd all solutions to the equation X2 +

Y 2 = Z2where X, Y and Z are positive integers. Such a solution (X;Y; Z)

is called a Pythagorean triple. If (X;Y; Z) have no common factor (other

than 1), we call (X;Y; Z) a primitive Pythagorean triple.

Part I

Remarks. Part I shows that if a, b are positive integers with no common

factor and a > b then X = a2 � b2, Y = 2ab, Z = a2 + b2 is a primitive

Pythagorean triple.

(a) Let a, b be two positive integers with a > b. Show thatX = a2�b2,Y = 2ab, Z = a2 + b2 is a Pythagorean triple.

Solution. Part I (a).

X2 + Y 2 = (a2 � b2)2 + 4a2b2

= a4 � 2a2b2 + b4 + 4a2b2

= a4 + 2a2b2 + b2 = Z2;

18

(b) Now assume that a, b have no common factor and not both are odd.

Show that (X;Y; Z) in (a) is a primitive Pythagorean triple. (Hint: Suppose

thatX, Y , Z have a common factor, p = some prime number. Then p divides

Z +X and Z �X. Note that Z + X = 2a2 and Z �X = 2b2. So what is

p? But Z must be odd (why?) so p can't be 2).

Solution. Part I (b). Since a, b have no common factor, they cannot

both be even. Since a, b are not both odd (given), one is odd and the other

even. Therefore Z = a2 + b2 is odd. (The square of an odd number is

odd; the square of an even number is even.) By the hint, if we suppose that

there exists a prime p which divides X, Y and Z, p divides Z + X = 2a2

and Z � X = 2b2. But p cannot divide a and b (given) and so p divides 2.

Therefore p = 2. But Z is odd. This contradiction implies that there does

not exist such a prime p. Therefore, (X;Y; Z) is a primitive Pythagorean

triple.

Part II

Remarks. Part II shows that every primitive Pythagorean triple arises

this way (for suitable choice of a, b).

(a) Let (X;Y; Z) be any Pythagorean triple. Show that the point (XZ; YZ)

lies on the unit circle x2 + y2 = 1.

Solution. Part II (a). The point (XZ; YZ) lies on the circle x2 + y2 = 1 if

and only if (XZ)2 + (Y

Z)2 = 1. This is true since (X;Y; Z) is a Pythagorean

triple.

(b) Let the slope of the line l which joins (�1; 0) to (XZ; YZ) be b=a

where a, b are positive integers with no common factor and a > b. Find the

points of intersection of the line l and the unit circle in terms of a, b to show

that

X

Z=a2 � b2

a2 + b2;

Y

Z=

2ab

a2 + b2:

Solution. Part II (b). The line l, through (�1;0) and with slopebahas

equation y = ba(x+ 1). l intersects the circle at those points for which

x2 +

�b

a(x+ 1)

�2= 1 or (a2 + b2)x2 + 2b2x+ b2 � a2 = 0

or x2 +2b2

a2 + b2x+

b2 � a2

a2 + b2= 0 since a2 + b2 6= 0:

(x+ 1)

�x+

b2 � a2

a2 + b2

�= 0:

Therefore x = �1 or x = a2�b2

a2+b2. So

XZ

= a2�b2

a2+b2.

Further y = ba

ha2�b2+a2+b2

a2+b2

i= 2ab

a2+b2and so

YZ= 2ab

a2+b2as required.

19

(c) If (X;Y; Z) is a primitive Pythagorean triple and if a, b are not both

odd, show that X = a2 � b2, Y = 2ab, Z = a2 + b2.

Solution. Part II (c). Since (X;Y; Z) is a Pythagorean triple, it follows

from parts (a) and (b) there exist a, b such that

X

Z=a2 � b2

a2 + b2;

Y

Z=

2ab

a2 + b2:

Further, since a, b are not both odd, a2 + b2 and a2 � b2 are both odd.

If a2� b2 and a2 + b2 have a common factor then that factor would be

a factor of their sum and di�erence, namely 2a2, 2b2. Note that 2 is not a

common factor and therefore since (X;Y; Z) is a primitive triple a, b have

no common factor. Thus

X = a2 � b2; Y = 2ab; Z = a2 + b2:

(d) If (X;Y; Z) is a primitive Pythagorean triple and if a, b are bothodd, a > b, we let r = 1

2(a+ b), s = 1

2(a� b).

(i) Prove that r, s are positive integers, that r > s, that r, s have no

common factor (other than 1) and that r, s are not both odd.

(ii) Let X0 = 2(2rs), Y 0 = 2(r2 � s2), Z0 = 2(r2 + s2). Show that

X = X0=2, Y = Y 0=2, Z = Z0=2. (Thus X = 2rs, Y = r2 � s2, Z =

r2 + s2).

Solution. Part II (d). (i) r = 12(a + b), s = 1

2(a � b) and a, b are

both odd. The sum and di�erence of two odd numbers is even and one half

of an even number is an integer. Since a > b > 0 (given), r, s are positive

integers.

r � s = b > 0 and therefore r > s:

Suppose that r, s have a common factor, p > 1. Then p divides r + s = a

and r � s = b. But a, b have no common factor. Therefore neither do r, s.

Suppose that r, s are both odd. Then r + s is even. But r + s = a which is

odd. Therefore, r, s are not both odd.

(ii) Note that

X0 = 2(2rs) = (a+ b)(a� b) = a2 � b2;

Y 0 = 2(r2 � s2) = 2�14(a2 + 2ab+ b2 � a2 + 2ab� b2)

�= 2ab;

Z0 = 2(r2 + s2) = 2�1

4(a2 + 2ab+ b2 + a2 � 2ab+ b2)

�= a2 + b2:

Therefore (X0; Y 0; Z0) is a Pythagorean triple, but it is not primitive since

2 divides X0, Y 0 and Z0. However X = X0

2, Y = Y 0

2, Z = Z0

2is primitive

(given) and thus X = 2rs, Y = r2 � s2, Z = r2 + s2.

For the sake of interest, we list the Pythagorean triples that result from

small values of a, b (a > b):

20

a b a2 � b2 2ab a2 + b2 Remarks

2 1 3 4 5 Surely, the most famous Pythagorean

triple

3 2 5 12 13

3 1 8 6 10 Both a; b are odd; see case d(ii)

4 3 7 24 25

4 2 12 16 20 Since a; b are both even, the triple

cannot be primitive. The correspond-

ing primitive triple is the case a = 2,

b = 1.

4 1 15 8 17

5 4 9 40 41

5 3 16 30 34 Both a; b are odd; see case d(ii)

5 2 21 20 29

5 1 24 10 26 Case d(ii) again

That completes the Skoliad Corner for this month. Please sendme your

pre-Olympiad contests and suggestions for future directions for this feature

of Crux.

Historical Titbit

Taken from a 1894 Ontario Public School textbook.

Show how to dissect a rhombus, and re-assemble the pieces to make a rect-angle.

Now this is quite easy, so we shall restate the problem:

Given a rhombus with diagonals of length a and b, show how to dissect the

rhombus, and re-assemble the pieces to make a rectangle of maximum pos-

sible area.

Call this maximum area �. Let 0 < � < �. Show how to dissect the

rhombus, and re-assemble the pieces to make a rectangle of area �.

In each case, the dissection should be with the minimum possible number of

pieces.

21

THE OLYMPIAD CORNER

No. 171

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, and we begin the 1996 volume of CruxMathe-maticorum with a change in the number of issues as well as in the Editorship.

This is the �rst number of the Corner which I will submit that does not fall

under the critical gaze of Bill Sands who stepped down as Editor-in-Chief at

the end of December. I want to express my particular gratitude to Bill for his

many hours of hard work and devotion to the publication over ten years. It

is also time to welcome the new Editor-in-Chief, with whom I look forward

to working over the next years.

Before launching into new material, let us pause to thank those who

contributed to the Corner in 1995. My particular thanks go to Joanne Long-

worth whose layout skills with LATEX enhance the readability of the copy. It

is also time to thank those who have contributed problem sets, solutions,

comments, corrections, and criticisms over the past year. Among our con-

tributors we have:

Miquel Amenguel Covas Federico Ardila M. �Sefket Arslanagi�c

Seung-Jin Bang Gerd Baron Bruce Bauslaugh

Francisco Bellot Rosado Christopher Bradley Himadri Choudhury

Tim Cross George Evagelopoulos Tony Gardiner

Murray Grant Gareth Gri�th Georg Gunther

Walther Janous Geo�rey Kandall Derek Kisman

Murray Klamkin Joseph Ling Andy Liu

Sam Maltby Beatriz Margolis Stewart Metchette

John Morvay Waldemar Pompe Bob Prielipp

Toshio Seimiya Michael Selby Bruce Shawyer

D.J. Smeenk Daryl Tingley Jim Totten

Panos E. Tsaoussoglou Stan Wagon Edward T.H. Wang

Martin White Chris Wildhagen C.S. Yogananda

Thank you all (and anyone I've left out by accident).

As an Olympiad set this number, we give the problems of the 29th

Spanish Mathematical Olympiad (National Round). My thanks go to Georg

22

Gunther, Sir Wilfred Grenfell College and Canadian Team leader at the 34th

I.M.O. at Istanbul, Turkey; and to Francisco Bellot Rosada, Valladolid, Spain,

for collecting and sending me a copy of the contest.

29th SPANISH MATHEMATICAL OLYMPIAD

(National Round)

Madrid, February 26{27, 1992

FIRST DAY (Time: 4.5 hours)

1. At a party there are 201 people of 5 di�erent nationalities. In each

group of six, at least two people have the same age. Show that there are at

least 5 people of the same country, of the same age and of the same sex.

2. Given the number triangle

0 1 2 3 4 : : : : : : 1991 1992 1993

1 3 5 7 : : : : : : 3983 3985

4 8 12 : : : : : : 7968

in which each number equals the sum of the two above it, show that the last

number is a multiple of 1993.

3. Show that in any triangle, the diameter of the incircle is not bigger

than the circumradius.

SECOND DAY (Time: 4.5 hours)

4. Show that each prime number p (di�erent from 2 and from 5) has

an in�nity of multiples which can be written as 1 1 1 1 : : : 1.

5. Given 16 points as in the �gure

r b b bb b b bb b b bb b b r

A

D

(lattice points)

in which the points A and D are marked, determine, for all the possible

manners, two other points B, C given that the six distances between these

four points should be di�erent. In this set of 4 points,

(a) Howmany �gures of four points are there (with the condition above)?

(b) How many of them are non-congruent?

23

(c) If each point is represented by a pair of integers (Xi; Yi), show that

the sum jXi�Xj j+ jYi�Yj j, extended to the six pairs AB, AC, AD, BC,

BD, CD, is constant.

6. A game-machine has a screen in which the �gure below is showed.

At the beginning of the game, the ball is in the point S.

S

A

C

B

G

D����

����

��������

��������� bbbbb

����

����

�HHHHH

With each impulse from the player, the ball moves up to one of the neigh-

bouring circles, with the same probability for each. The game is over when

one of the following events occurs:

(1) The ball goes back to S, and the player loses.

(2) The ball reaches G, and the player wins.

Determine:

(a) The probability for the player to win the game.

(b) The mean time for each game.

Next I'd like to give six Klamkin Quickies. Many thanks to Murray Klamkin,

University of Alberta, for creating them for us. The answers will appear in

the next number.

SIX KLAMKIN QUICKIES

1. Which is larger

(3p2� 1)1=3 or

3p1=9� 3

p2=9 + 3

p4=9?

2. Prove that

3�min

�a

b+b

c+c

a;b

a+c

b+a

c

�� (a+ b+ c)

�1

a+

1

b+

1

c

�

where a, b, c are sides of a triangle.

24

3. Let ! = ei�=13. Express1

1�!as a polynomial in ! with integral

coe�cients.

4. Determine all integral solutions of the simultaneous Diophantine

equations x2 + y2 + z2 = 2w2and x4 + y4 + z4 = 2w4

.

5. Prove that if the line joining the incentre to the centroid of a triangle

is parallel to one of the sides of the triangle, then the sides are in arithmetic

progression and, conversely, if the sides of a triangle are in arithmetic pro-

gression then the line joining the incentre to the centroid is parallel to one

of the sides of the triangle.

6. Determine integral solutions of the Diophantine equation

x� y

x+ y+y� z

y+ z+z �w

z +w+w � x

w + x= 0

(joint problem with Emeric Deutsch, Polytechnic University of Brooklyn).

To �nish this number of the Corner, we turn to readers' solutions to

problems from the May 1994 number of the Corner and the Final Round of

the 43rd Mathematical Olympiad (1991{92) in Poland [1994: 129{130].

1. Segments AC and BD intersect in point P so that PA = PD,

PB = PC. Let O be the circumcentre of triangle PAB. Prove that lines

OP and CD are perpendicular.

Solutions by Joseph Ling, University of Calgary; by Toshio Seimiya,Kawasaki, Japan; and by Chris Wildhagen, Rotterdam, The Netherlands. Wegive Seimiya's solution and comment.

O

B C

E

DA

P

Because PA = PD, PB = PC,

and \APB = \DPC,

we get �PAB � �PDC, so that

\BAP = \CDP: (1)

At least one of \PAB and \PBA is acute, so we may assume without

loss of generality that \PAB is acute. SinceO is the circumcentre of�PAB

we get OB = OP and \BOP = 2\BAP , so that

\OPB = 90� � 1

2\BOP = 90� � \BAP: (2)

Let E be the intersection of OP with CD. Then

\EPD = \OPB: (3)

25

From (1), (2) and (3) we have

\EPD = 90� � \CDP:

Thus \EPD+ \EDP = \EPD+ \CDP = 90�. Therefore OP ? CD.

Comment: Generally if A, B, C, D are concyclic, we have OP ? CD and

this theorem is an extension of Brahmagupta's theorem.

2. Determine all functions f de�ned on the set of positive rational

numbers, taking values in the same set, which satisfy for every positive ra-

tional number x the conditions

f(x+ 1) = f(x) + 1 and f(x3) = (f(x))3:

Solution by Edward T.H. Wang and by Siming Zhan, Wilfrid LaurierUniversity, Waterloo, Ontario.

Let N andQ+denote the set of positive integers. and the set of positive

rational numbers, respectively. We show that f(x) = x, for all x 2 Q+, is

the only function satisfying the given conditions. First of all, by the �rst

condition and an easy induction we see that f(x + n) = f(x) + n, for all

x 2 Q+, and for all n 2 N. Now for arbitrary

p

q2 Q+

, where p; q 2 N, wehave

f

�p

q= q2

�3!= f

�p3

q3+ 3p2 + 3pq3 + q6

�

= f

�p

q

�3!+ 3p2 + 3pq3 + q6: (1)

On the other hand

f

�p

q+ q2

�3!= f

��p

q+ q2

��3=

�f

�p

q

�+ q2

�3

=

�f

�p

q

��3+3

�f

�p

q

��2q2 +3

�f

�p

q

��2� q4+q6: (2)

Letting t = f(pq) and comparing (1) and (2), we get, since f(p

3

q3) = (f(p

q))3,

p(p+q3) = q2t2+q4t or q2t2+q4t�p(p+q3) = 0 or (qt�p)(qt+p+q3) = 0.

Since qt+p+ q3 > 0, we must have t = p

q, i.e. f(p

q) = p

q, and we are done.

3. Prove that the inequality

rXn=1

rX

m=1

aman

m+ n

!� 0

holds for any real numbers a1; a2; : : : ; ar. Find conditions for equality.

26

Solutions by Seung-Jin Bang, Seoul, Korea; by Joseph Ling, Universityof Calgary; and by Chris Wildhagen, Rotterdam, The Netherlands. We giveLing's solution.

Consider the polynomial

p(x) =

rXn=1

rX

m=1

amanxm+n�1

!:

Then

xp(x) =

rXn=1

rXm=1

amanxm+n =

rX

m=1

amxm

! rX

n=1

anxn

!

=

rXi=1

aixi

!2� 0;

for all x 2 R.In particular p(x) � 0 for all x � 0. Hence

0 �Z 1

0

p(x)dx =

rXn=1

rX

m=1

aman

m+ nxm+n

�10

!

=

rXn=1

rXm=1

aman

m+ n

The inequality is strict unless xp(x) � 0, that is a1 = a2 = � � � = ar = 0.

[Editor's Remark. All three solutions involved the integral. Can you furnish

a nice solution avoiding the calculus?]

4. De�ne the sequence of functions f0; f1; f2; : : : by

f0(x) = 8 for all x 2 R;fn+1(x) =

px2 + 6fn(x) for n = 0; 1; 2; : : : and for all x 2 R:

For every positive integer n, solve the equation fn(x) = 2x.

Solutions by Seung-Jin Bang, Seoul, Korea; and by Chris Wildhagen,Rotterdam, The Netherlands. We use Bang's comment and solution.

This problem is the same as problem 2 of the 19th Annual U.S.A.Math-

ematical Olympiad (which appeared inMath. Magazine 64, 3 (1991), pp. 211{213.)

Since fn(x) is positive, fn(x) = 2x has only positive solutions. We

show that, for each n, fn(x) = 2x has a solution x = 4. Since f1(x) =px2 + 48, x = 4 is a solution of f2(x) = 2x. Now fn+1(4) =

p42 + 6fn(4)

=p42 + 6 � 8 = 8 = 2 �4, which completes the inductive step. Next, induc-

tion on n gives us that for each n,fn(x)

xdecreases as x increases in (0;1).

It follows that fn(x) = 2x has the unique solution x = 4.

27

6. Prove that, for every natural k, the number (k3)! is divisible by

(k!)k2+k+1

.

Solution by Chris Wildhagen, Rotterdam, The Netherlands.Applying the well known fact that (ab)! is divisible by (a!)b � b! yields

(k3)! = (k � k2)! is divisible by (k!)k2 � (k2)! and (k2)! = (k � k)! is divisibleby (k!)k+1 from which the required result follows immediately.

The astute reader will notice we did not have a solution on �le from the

readership to problem 5. There's your challenge!

That completes the Olympiad Corner for this issue. The Olympiad season is

fast approaching. Please send me your national and regional contests.

Introducing the new Editor-in-Chief

This issue of CRUX marks the change of the Editor-in- Chief from Bill

Sands and Robert Woodrow to Bruce Shawyer. For those of you who do not

know Bruce, here is a short pro�le:

Born: Kirkcaldy, Scotland1

Educated Kirkcaldy High School

University of St. Andrews, Scotland

Employment University of Nottingham, England

University of Western Ontario, Canada

Memorial University of Newfoundland, Canada

(Head of Department, 1985{91)

Visiting Positions �Universit�at Ulm, Germany

University of St. Andrews

Service Canadian Mathematical Society

Team Leader, Team Canada at IMO 1987 and 1988

IMO95, Principal Organiser

Research Summability of Series and Integrals

Approximation Theory

LATEX macros for the picture environment

Home Page http://www.math.mun.ca/~bshawyer

1 Kirkcaldy, an industrial town north of Edinburgh, is the birthplace of Sir Sandford

Fleming, who invented time zones, and of Adam Smith, the Father of economics.

28

THE ACADEMY CORNER

No. 1

Bruce Shawyer

All communications about this column should be sent to ProfessorBruce Shawyer, Department of Mathematics and Statistics, Memorial Uni-versity of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

With this issue, we start a new corner, which, for want of a better name,

is being called \the Academy Corner". It will be concerned, in particular,

with problem solving at the undergraduate level. This can take the form of

a competition, a \problem of the week", a course in problem solving, etc.

Your submissions and comments are welcome!

Many Universities hold their own undergraduate Mathematics Compe-

titions. We invite subscribers to send us information about them for publi-

cation here. We also solicit nice solutions to these problems. Please send

them to the Editor-in-Chief.

We shall begin with the competition with which the editor is most fa-

miliar: the Undergraduate Mathematics Competition at Memorial Univer-

sity of Newfoundland. This competition is used as primary information for

choosing Memorial's team for the Atlantic Provinces Council on the Sciences

Mathematics Competition (more on that in a later issue).

Memorial University Undergraduate

Mathematics Competition 1995

Time allowed | 3 hours

1. Find all integer solutions of the equation x4 = y2 + 71.

2. (a) Show that x2 + y2 � 2xy for all real numbers x, y.

(b) Show that a2 + b2 + c2 � ab+ bc+ ca for all real numbers a, b, c.

3. Find the sum of the series

1

2!+

2

3!+

3

4!+

4

5!+ : : :+

99

100!:

29

4. If a, b, c, d are positive integers such that ad = bc, prove that a2 +

b2 + c2 + d2 is never a prime number.

5. Determine all functions f : R! Rwhich satisfy

(x� y)f(x+ y)� (x+ y)f(x� y) = 4xy(x2 � y2)

for all real numbers x, y.

6. Assume that when a snooker ball strikes a cushion, the angle of inci-

dence equals the angle of re ection.

For any position of a ball A, a point P on the cushion is determined as

shown.

Prove that if the ball A is shot at point P , it will go into the pocket B.

i i i

i i i

wr

r

rrr

P

A

B

This concludes the �rst Academy Corner.

Mathematical Literacy

1. Who thought that the binary system would convince the Emperor of

China to abandon Buddhism in favour of Christianity?

2. Who asked which king for one grain of wheat for the �rst square of a

chess board, two grains for the second square, four grains for the third

square, and so on?

3. In which well known painting, by whom, does a Magic Square appear?

4. Where was bread cut into \Cones, Cylinders, Parallelograms, and sev-

eral other Mathematical Figures"?

5. Which mathematician said: \Philosophers count about two-hundred

and eighty eight views of the sovereign good"?

Answers will be given in a subsequent issue.

30

BOOK REVIEWS

Edited by ANDY LIU

The Monkey and the Calculator (Le singe et la calculatrice), softcover,ISBN 2-909737-07-1, 128 pages, 48 French francs (plus mailing).

Aladdin's Sword (Le sabre d'Aladin), softcover, ISBN 2-909737-08-X,

128 pages, 48 French francs (plus mailing).

Both published in 1995 by Production et Organisation du Loisir Educatif

(POLE), 31 avenue des Gobelins, 75013 Paris, France.

Reviewed by Claude La amme, University of Calgary.

These two delightful little pocket books each contain a selection of

almost 100 puzzles from the eighth International Championship of Mathe-

matical Puzzles. This is an annual competition in four stages and in seven

categories from elementary to advanced, open to anyone and administered

by FFJM, the French Federation of Mathematical Games. These books are

numbers 14 and 15 respectively in a series taken from these competitions

and published by POLE, and most of these books are still in print.

The �rst book is from the Junior High level and the second from the

Grand Public competition, slightlymore sophisticated. Nevertheless, no spe-

cialized knowledge is required for the competitions and appropriately the

statements of these puzzles are all very elementary; but they puzzle your

puzzler to the point that once you have read one of the problems, you cannot

leave the book until you �gure out a solution. Now this is a serious com-

petition and a mere solution is usually not enough, the most elegant and

complete one is the favourite!

These puzzles should be of interest to anyone looking for some \fun

brain gymnastics" (gymnastique intellectuelle), or \neurobics" as I have seensomewhere in the books. I can see here a great resource for teachers trying

to interest students with some challenging but elementary and accessible

problems. Here is a sample from the second book, called \A Numismatic

Coincidence":

By multiplying the number of pieces in her coin collection by 1994,

the clever Miss Maths arrives at a product the sum of the digits

of which is exactly equal to the number of pieces in her coin col-

lection.

How many pieces does Miss Maths have in her coin collection?

All the problems are translated into English, a job very well done even

if this means that some problems require a new answer, such as the problem

asking us to �ll in the blank with a number (written out in words) making the

following sentence true (a hyphen doesn't count as a letter):

31

\This sentence has letters."

There are two answers in English, and only one (and it is di�erent) for the

French version.

The problem solutions are in French only but you will not be puzzled

here, even a small diagram or a few numbers and the solution jumps at you.

Apart from the puzzles themselves, you will �nd a few advertisements for

other mathematics magazines, even a puzzling one on music and mathemat-

ics, as well as for camps and Summer Schools. You will also �nd details

regarding the competition itself.

By the way, the titles refer to some of the more puzzling puzzles in the

books!

Tournament of the Towns, 1980{1984, Questions and Solutions; pub-lished in 1993, softcover, 117+ pages, Australian $18.

Tournament of the Towns, 1989{1993, Questions and Solutions; pub-lished in 1994, softcover, 209+ pages, Australian $22.

Both edited by Peter J. Taylor, Australian Mathematics Trust, Univer-

sity of Canberra, P. O. Box 1, Belconnen, A.C.T. 2616, Australia.

Reviewed byMurray S. Klamkin, University of Alberta.

In my previous review [1992: 172] of the �rst book of this series, which

are the problems and solutions for 1984{1989 (andwhichwas published�rst),

I had given a short description of the competition and references for further

information on the competition. More complete information on this is given

in the prefaces of the books.

This competition, as I said before, is one of the premier mathematics

competitions in the world for secondary school students and contains many

wonderful challenging problems (even to professional mathematicians). In

view of my previous review, I just give another sampling of these problems.

Junior Questions

1. Construct a quadrilateral given its side lengths and the length joining

the midpoints of its diagonal. [1983]

2. (a) A regular 4k-gon is cut into parallelograms. Prove that among these

there are at least k rectangles.

(b) Find the total area of the rectangles in (a) if the lengths of the sides

of the 4k-gon equal a. [1983]

3. Prove that in any set of 17 distinct natural numbers one can either �nd

�ve numbers so that four of them are divisible into the other or �ve

numbers none of which is divisible into any other. [1983]

32

4. Given the continued fractions

a =1

2 + 1

3+ 1

���+ 199

and b =1

2 + 1

3+ 1

���+ 199+ 1

100

;

prove that ja� bj < 1

99! � 100! : [1990]

5. The numerical sequence fxng satis�es the conditionxn+1 = jxnj � xn�1

for all n > 1. Prove that the sequence is periodic with period 9, i.e.,

for any n � 1, we have xn = xn+9. [1990]

6. There are 16 boxers in a tournament. Each boxer can �ght nomore often

than once a day. It is known that the boxers are of di�erent strengths,

and the stronger man always wins. Prove that a 10 day tournament can

be organized so as to determine their classi�cation (put them in order

of strength). The schedule of �ghts for each day is �xed on the evening

before and cannot be changed during the day. [1991]

Senior Questions

1. We are given 30 non-zero vectors in 3 dimensional space. Prove that

among these there are two such that the angle between them is less

than 45�. [1980]

2. Prove that every real positive number may be represented as a sum

of nine numbers whose decimal representation consists of the digits 0

and 7. [1981]

3. A polynomial P (x) has unity as coe�cient of its highest power and has

the property that with natural number arguments, it can take all values

of form 2m, where m is a natural number. Prove that the polynomial

is of degree 1. [1982]

4. A square is subdivided into K2equal small squares. We are given a

broken line which passes through the centres of all the smaller squares

(such a broken line may intersect itself). Find the minimum number of

links in this broken line. [1982]

5. Do there exist 1; 000; 000 distinct positive numbers such that the sum

of any collection of these numbers is never an exact square? [1989]

6. There are 20 points in the plane and no three of them are collinear. Of

these points 10 are red while the other 10 are blue. Prove that there

exists a straight line such that there are 5 red points and 5 blue points

on either side of this line. [1990]

33

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 that 1 September 1996. They may alsobe sent by email to cruxeditor@cms.math.ca. (It would be appreciated ifemail proposals and solutions were written in LATEX, preferably in LATEX2e).Solutions received after the above date will also be considered if there issu�cient time before the date of publication.

2101. Proposed by Ji Chen, Ningbo University, China.Let a; b; c be the sides and A;B;C the angles of a triangle. Prove that

for any k � 1, X ak

A� 3

�

Xak;

where the sums are cyclic. [The case k = 1 is known; see item 4.11, page 170

of Mitrinovi�c, Pe�cari�c, Volenec, Recent Advances in Geometric Inequalities.]

2102. Proposed by Toshio Seimiya, Kawasaki, Japan.ABC is a triangle with incentre I. Let P and Q be the feet of the

perpendiculars from A to BI and CI respectively. Prove that

AP

BI+AQ

CI= cot

A

2:

2103. Proposed by Toshio Seimiya, Kawasaki, Japan.ABC is a triangle. Let D be the point on side BC produced beyond

B such that BD = BA, and let M be the midpoint of AC. The bisector of

\ABC meets DM at P . Prove that \BAP = \ACB.

34

2104. Proposed by K. R. S. Sastry, Dodballapur, India.In how many ways can 111 be written as a sum of three integers in geometric

progression?

2105. Proposed by Christopher J. Bradley, Clifton College, Bristol,U. K.

Find all values of � for which the inequality

2(x3 + y3 + z3) + 3(1 + 3�)xyz � (1 + �)(x+ y + z)(yz+ zx+ xy)

holds for all positive real numbers x; y; z.

2106. Proposedby Yang Kechang, Yueyang University, Hunan, China.A quadrilateral has sides a; b; c; d (in that order) and area F . Prove that

2a2 + 5b2 + 8c2 � d2 � 4F:

When does equality hold?

2107. Proposed by D. J. Smeenk, Zaltbommel, The Netherlands.Triangle ABC is not isosceles nor equilateral, and has sides a; b; c. D1

and E1 are points of BA and CA or their productions so that

BD1 = CE1 = a. D2 andE2 are points ofCB andAB or their productions

so that CD2 = AE2 = b. Show that D1E1 k D2E2.

2108. Proposed by Vedula N. Murty, Dover, Pennsylvania.Prove that

a+ b+ c

3� 1

4

3

s(b+ c)2(c+ a)2(a+ b)2

abc;

where a; b; c > 0. Equality holds if a = b = c.

2109. Proposed by Victor Oxman, Haifa, Israel.In the plane are given a triangle and a circle passing through two of the

vertices of the triangle and also through the incentre of the triangle. (The

incentre and the centre of the circle are not given.) Construct, using only an

unmarked ruler, the incentre.

35

2110. Proposed by Jordi Dou, Barcelona, Spain.Let S be the curved Reuleaux triangle whose sides AB, BC and CA

are arcs of unit circles centred at C, A andB respectively. Choose at random

(and uniformly) a point M in the interior and let C(M) be a chord of S for

which M is the midpoint. Find the length l such that the probability that

C(M) > l is 1=2.

2111. Proposed by Hoe Teck Wee, student, Hwa Chong Junior Col-lege, Singapore.

Does there exist a function f : N �! N (where N is the set of positive

integers) satisfying the three conditions:

(i) f(1996) = 1;

(ii) for all primes p, every prime occurs in the sequence

f(p), f(2p), f(3p); : : : ; f(kp); : : : in�nitely often; and

(iii) f(f(n)) = 1 for all n 2 N ?

2112. Proposed by Shawn Godin, St. Joseph Scollard Hall, NorthBay, Ontario.

Find a four-digit base-ten number abcd (with a 6= 0) which is equal to

aa + bb + cc + dd.

2113. Proposed by Marcin E. Kuczma, Warszawa, Poland.Prove the inequality

nXi=1

ai

! nXi=1

bi

!�

nXi=1

(ai + bi)

! nXi=1

aibi

ai + bi

!

for any positive numbers a1; : : : ; an; b1; : : : ; bn.

36

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

1827. [1993: 78; 1994: 57; 1995: 54] Proposed by �Sefket Arslanagi�c,Trebinje, Yugoslavia, and D. M. Milo�sevi�c, Pranjani, Yugoslavia.

Let a; b; c be the sides, A;B; C the angles (measured in radians), and

s the semi-perimeter of a triangle.

(i) Prove that X bc

A(s� a)� 12s

�;

where the sums here and below are cyclic.

(ii)�It follows easily from the proof of Crux 1611 (see [1992: 62] and

the correction on [1993: 79]) that also

X b+ c

A� 12s

�:

Do the two summations above compare in general?

IV. Comment by Waldemar Pompe, student, University of Warsaw,Poland.

Here we give a short demonstration of the equality

X bc

s� a= s+

(4R+ r)2

s; (1)

which has appeared on [1995: 55]. SinceXa = 2s;

Xbc = s2 + r2 + 4Rr; abc = 4sRr ;

we get X(s� b)(s� c) = 3s2 � 4s2 +

Xbc = r2 + 4Rr :

Using the above equality we obtain

X 1

s� a=

r2 + 4Rr

(s� a)(s� b)(s� c)=r2 + 4Rr

sr2=r + 4R

sr:

ThereforeX bc

s� a= abc

X 1

a(s� a)= 4Rr

X�1

a+

1

s� a

�

= 4Rr

�s2 + r2 + 4Rr

4sRr+r + 4R

sr

�

= s+r2 + 4Rr

s+

4Rr + 16R2

s= s+

(4R+ r)2

s:

37

A (not quite as) short demonstration of (1) has also been sent in byToshio Seimiya, Kawasaki, Japan.

2006. [1995: 20] Proposed by John Duncan, University of Arkansas,Fayetteville; Dan Velleman, Amherst College, Amherst, Massachusetts; andStan Wagon, Macalester College, St. Paul, Minnesota.

Suppose we are given n � 3 disks, of radii a1 � a2 � � � � � an. We

wish to place them in some order around an interior disk so that each given

disk touches the interior disk and its two immediate neighbours. If the given

disks are of widely di�erent sizes (such as 100, 100, 100, 100, 1), we allow a

disk to overlap other given disks that are not immediate neighbours. In what

order should the given disks be arranged so as to maximize the radius of the

interior disk? [Editor's note. Readers may assume that for any ordering of

the given disks the con�guration of the problem exists and that the radius of

the interior disk is unique, though, as the proposers point out, this requires

a proof (which they supply).]

Solution by the proposers.Let r be the radius of the central disk. First look at a single tangent

con�guration made up of the central disk, and two disks of radii x and y.

The three centres form a triangle with sides r + x, r + y, and x + y; let

� = �r(x; y) be the angle at the centre of the disk with radius r. Applying

the law of cosines to this angle and simplifying gives

�r(x; y) = arccos

�1� 2xy

r2 + ry + rx+ xy

�:

A routine calculation shows that the mixed partial derivative �12 is given by

�12 =r2

2pxy(r2 + rx + ry)3=2

;

therefore �12 > 0 (for x; y > 0). Integrating from a to a + s and b to b+ t

(where s; t > 0) yields:

�(a+ s; b+ t) + �(a; b) > �(a+ s; b) + �(a; b+ t): (1)

Note that equality occurs if and only if s = 0 or t = 0.

We may assume n � 4 and a1 � a2 � � � � � an; letDi denote the disk

with radius ai. And let S(r) denotePn

i=1 �r(ai; ai+1), the total angle made

by the con�guration; when S < 2�, then the disk of radius r is too large and

the ring is not yet closed up.

THEOREM. The largest inner radius r occurs by placingD2 and D3 on

either side ofD1, thenD4 alongsideD2,D5 alongsideD3, and so on around

the ring.

38

Proof. By induction on n. We actually prove the stronger assertion that

for any radius r, S(r) for the con�guration of the statement of the theorem

is not less than S(r) for any other con�guration. This su�ces, for if r is such

that S(r) = 2�, then S(r) for any other con�guration is not greater than

2�.

For n = 4 there are only three arrangements:

&%'$D4

D3

D1

D2

Arrangement I

&%'$D3

D4

D1

D2

Arrangement II

&%'$D2

D4

D1

D3

Arrangement III

for which (suppressing the subscripts r in the �'s)

SI(r) = �(a1; a2) + �(a2; a4) + �(a3; a4) + �(a1; a3);

SII(r) = �(a1; a2) + �(a2; a3) + �(a3; a4) + �(a1; a4);

SIII(r) = �(a1; a3) + �(a2; a3) + �(a2; a4) + �(a1; a4):

By (1), SI(r) � SII(r) � SIII (r), which proves our assertion for this case.

For the general case, supposeE2; : : : ; En is an arrangement ofD2; : : : ;

Dn, with bi denoting the radius ofEi. Then a2 � b2 and wemay also assume

a3 � b3 (otherwise ip and relabel). Now it is su�cient, by the induction

hypothesis, to show that

�(a1; a2) + �(a1; a3)� �(a2; a3) � �(a1; b2) + �(a1; b3)� �(b2; b3):

But (1) implies that

�(a1; a2) + �(a1; a3) + �(b2; b3) � �(a1; a3) + �(a1; b2) + �(a2; b3);

and this means it is su�cient to prove:

�(a1; a3) + �(a2; b3) � �(a1; b3) + �(a2; a3):

But this too is a consequence of (1).

Note. A similar argument shows that the smallest inner radius occursfor the con�guration: : : : D5Dn�3D3Dn�1D1DnD2Dn�2D4 : : : .

There were no other solutions sent in.

The problem was motivated by the special case of three pennies andtwo nickels, which was in (the late) Joe Konhauser's collection of problems.

39

2007. [1995: 20] Proposed by Pieter Moree, Macquarie University,Sydney, Australia.

Find two primes p and q such that, for all su�ciently large positive real

numbers r, the interval [r; 16r=13] contains an integer of the form

2n; 2np; 2nq; or 2npq

for some nonnegative integer n.

Solution by Peter Dukes, student, University of Victoria, B.C.The prime numbers p = 3 and q = 13 solve the problem. To prove

this, it su�ces to �nd an increasing sequence of positive integers s1; s2; : : :

such that each si is of one of the forms stated in the problem, and si+1=si �16=13 for all i = 1; 2; : : : . Then, given any real number r � s1, the integer

sk (where k = minfi 2 Z+ : r � sig), is no more than 16r=13. For if

sk =2 [r; 16r=13], then sk=sk�1 � 16=13, contrary to construction. Of course

if k = 1 then r = s1; so s1 2 [r; 16r=13]. Now, consider the sequence

fsig : 24; 26; 32; 39; 48; 52; 64; 78; : : :

: : : ; 3 � 2m+3; 13 � 2m+1; 2m+5; 3 � 13 � 2m; : : : :It is a simple matter to verify that the ratio of consecutive terms si+1=sidoes not exceed 16=13 for any i 2 Z+. Thus, for all r � 24, the interval

[r; 16r=13] contains an integer of the form 2n, 2np, 2nq, or 2npq.

Also solved by FEDERICO ARDILA, student,Massachusetts Institute ofTechnology, Cambridge; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium,Bamberg, Germany; RICHARD I. HESS, Rancho Palos Verdes, California;PETER HURTHIG, Columbia College, Burnaby, British Columbia; ROBERTB. ISRAEL, University of British Columbia; DAVID E. MANES, State Univer-sity of New York, Oneonta; GOTTFRIED PERZ, Pestalozzigymnasium, Graz,Austria; CHRIS WILDHAGEN, Rotterdam, The Netherlands; and the pro-poser. One other solution contained a correct pair of primes, but the editorcould not decipher the proof that they worked.

Most solvers simply found a single pair of primes, some of which allowthe constant 16=13 to be replaced by a slightly smaller constant. As Engel-haupt, Hess and the proposer point out, the best you can do in this directionis that 16=13 can be replaced by any real number greater than 4

p2. Hess says

how: �nd primes \near" 2k1+1=4 and 2k2+1=2 for integers k1 and k2 (and hegives the example p = 4871, q = 181, which gets you within :00013 of 4

p2).

More precisely, if

p = 2k1+1=4 � t1 and q = 2k2+1=2 � t2 (1)

where k1; k2 are positive integers and t1; t2 are real numbers close to 1, then2k1+k2 < p � 2k2 < q � 2k1 < pq, and the ratios of consecutive terms are

4p2 � t1; 4

p2 � t2

t1;

4p2 � t1; 4

p2 � 1

t1t2;

40

all of which can be made arbitrarily close to 4p2. The reason (1) is possible

is because (by the Prime Number Theorem) there is always a prime betweenn and nt for any given t > 1, if n is big enough. By increasing the number ofprimes allowed, with an analogous change in the kind of integers you wantthe interval to contain, the proposer can get the interval down to [r; �r] forany given � > 1.

Subsequent to submitting the problem the proposer was able to gen-eralize it. He writes:

For given � > 1 we claim that there exists squarefree D such that

the consecutive integers n1; n2; n3; : : : of the form 2�d, with djD satisfy

ni+1=ni � � for all i su�ciently large. It su�ces to �nd an in�nite sub-

sequence fmig1i=1 such that mi+1=mi � � for all i su�ciently large. Let

3 = p1 < p2 < p3 < : : : be the sequence of odd consecutive primes. For

n � 1 let pin denote the smallest prime exceeding qn and let pjn denote

the greatest prime less than qn+1. Using the Prime Number Theorem it fol-

lows that limi!1pi+1pi

= 1, so there exists n such that pi=pi+1 � � for

i = in; : : : ; jn + 1. Put q0 = 2n, q1 = pin; q2 = p1+in; : : : ; qs = pjn and

D =Qsi=1 qi. Put ma+r(s+1) = 2rqa, for 0 � a � s, r � 0. Notice that

1 < mi+1=mi � � for every i � 1. Thus the claim holds withD =Qs

i=1 qi.

Remark. A similar result holds with 2 replaced by an arbitrary prime.

2008. [1995: 20] Proposed by Jun-hua Huang, The Middle SchoolAttached To Hunan Normal University, Changsha, China.

Let I be the incentre of triangle ABC, and suppose there is a circle

with centre I which is tangent to each of the excircles of�ABC. Prove that

ABC is equilateral.

Solution by Waldemar Pompe, student, University of Warsaw, Poland.We solve the problem assuming that the circle with centre I is tangent

either externally or internally to all the excircles. Without this assumption

the problem is not true; below we give a counterexample. (The triangleABC

in the �gure has sides 2; 9; 9, and of course is not equilateral.)

Assume �rst there is a circle C with centre I tangent externally to all

the excircles [i.e., C does not contain the excircles | Ed.]. Then according to

Feuerbach's theorem the circle C is the nine-point circle of ABC. Therefore

C is also tangent to the incircle of ABC, but since I is the centre of C, thecircle C and the incircle of ABC must coincide. It follows that the triangle

ABC is equilateral.

Now let C be the circle with centre I and tangent internally to all the

excircles. Let DE and FG be the chords of C containing the segments AB

and AC respectively. Since C and the incircle of ABC are concentric, the

lines DE and FG are symmetric to each other with respect to the line AI.

Therefore, since the excircles lying opposite to B and C are uniquely deter-

mined by the chordsDE, FG and the circle C, they also have to be symmetric

41

q

A

ICB

to each other with respect to the line AI. Thus rB = rC. Analogously we

show that rC = rA, which implies that ABC has to be equilateral, as we

wished to show.

Also solved by FEDERICO ARDILA, student, Massachusetts Instituteof Technology, Cambridge; FRANCISCO BELLOT ROSADO, I.B. Emilio Fer-rari, andMARIA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Val-ladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, U. K.;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; TOSHIOSEIMIYA, Kawasaki, Japan; ASHISH KR. SINGH, Kanpur, India; D. J.SMEENK, Zaltbommel, The Netherlands; HOE TECK WEE, student, HwaChong Junior College, Singapore; CHRIS WILDHAGEN, Rotterdam, TheNetherlands; and the proposer. There was one incorrect solution received.

Pompe was the only solver to �nd the counterexample. Everyone elseeither ignored the possibility, or (in a couple of cases at least) believed thatit could not occur! Half the solvers considered both cases of tangency, thatis, where the circle is tangent externally to all the excircles or internally toall of them, and the others considered only one.

2009. [1995: 20] Proposed by Bill Sands, University of Calgary.Sarah got a good grade at school, so I gave her N two{dollar bills.

Then, since Tim got a better grade, I gave him just enough �ve{dollar bills

so that he got more money than Sarah. Finally, since Ursula got the best

grade, I gave her just enough ten{dollar bills so that she got more money

than Tim. What is the maximum amount of money that Ursula could have

received? (This is a variation of problem 11 on the 1994 Alberta High School

Mathematics Contest, First Part; see the January 1995 Skoliad Corner [1995:

6].)

42

Solutionby Shawn Godin, St. Joseph ScollardHall, North Bay, Ontario.The most Ursula could receive is 2N + 14 dollars.

Sarah received 2N dollars and Tim received 2N + 1, 2N +2, 2N +3,

2N + 4 or 2N + 5 dollars depending on what residue class N belongs to

modulo 5. But since Tim gets $5 bills his amount is divisible by 5. Ursula

will then receive either $5more than Tim (if Tim's amount is not divisible by

10) or $10more than Tim (if Tim's amount is divisible by 10). Clearly 2N+5

is odd, thus not divisible by 10. So the maximum occurs when Tim receives

2N + 4 dollars, which means 2N � 1 mod 5, i.e. N � 3 mod 5, and that

maximum is 2N + 14.

Also solved byHAYO AHLBURG, Benidorm, Spain; FEDERICO ARDILA,student, Massachusetts Institute of Technology, Cambridge; CARLBOSLEY, student, Washburn Rural High School, Topeka, Kansas;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, U. K.; PAUL COLUCCI,student, University of Illinois; PETER DUKES, student, University of Victo-ria, B.C.; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Ger-many; J.K. FLOYD,Newnan, Georgia; TOBY GEE, student, The John of GauntSchool, Trowbridge, England; RICHARD K. GUY, University of Calgary;DAVID HANKIN, John Dewey High School, Brooklyn, New York; RICHARDI. HESS, Rancho Palos Verdes, California; CYRUS HSIA, student, Universityof Toronto, Toronto, Ontario; J. A. MCCALLUM, Medicine Hat, Alberta;ROBERT P. SEALY, Mount Allison University, Sackville, New Brunswick;HEINZ-J �URGEN SEIFFERT, Berlin, Germany; and the proposer. Three otherreaders sent in solutions which the editor judges are not precise enough.

Here Ursula gets at most $14 more than Sarah, which is 14=15 of theobvious maximum di�erence $5+$10 = $15. How small can this ratio be, ifwe replace the denominations $2, $5, $10 by three other positive integers?(ChoosingN; 2N � 1; 2N for large N gets it down arbitrarily close to 3=4.)

2010. [1995: 20] Proposed byMarcin E. Kuczma, Warszawa, Poland.In triangle ABC with \C = 2\A, line CD is the internal angle bisec-

tor (with D on AB). Let S be the centre of the circle tangent to line CA

(produced beyond A) and externally to the circumcircles of triangles ACD

and BCD. Prove that CS ? AB.

Composite of solutions by Hoe Teck Wee, student, Hwa Chong JuniorCollege, Singapore and Roland H. Eddy, Memorial University, St. John's,Newfoundland.

First, note that \BCD = \CAD, so that the line BC is tangent to

the circumcircle of �ACD. Next, perform an inversion with respect to C.

The line BC inverts into the line B0C, so that the circumcircle of

�ACD inverts into the line A0D0, which is parallel to B0C. The circum-

circle of �BCD inverts into the line B0D0. The line AD inverts into the

circumcircle of the quadrilateral CA0D0B0. Finally, the circle tangent to the

43

line CA, and externally tangent to the circumcircles of �ACD and �BCD

inverts into a circle �, which is tangent to the line segments A0C, A0D0and

B0D0as shown in the diagram.

A0 D0

C B0

Now, \BCD = \ACD, so that \B0CD0 = \A0CD0. Thus B0D0 =

A0D0. Since A0D0

and B0C are parallel, we have A0D0 = B0D0 = A0C.

Hence, the circumcentre of the quadrilateral A0D0B0C lies on the angle bi-

sectors of \CA0D0and \A0D0B0. These two angle bisectors intersect at the

centre of �. Thus, quadrilateral A0D0B0C and � are concentric. Let their

common centre be O.

Thus, C, O and S0 (the inverse of S) are collinear, and so CS is per-

pendicular to AB.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,U. K.; KEE-WAI LAU, Hong Kong; D. J. SMEENK, Zaltbommel, The Nether-lands; and the proposer.

2012. [1995: 52] Proposed by K. R. S. Sastry, Dodballapur, India.Prove that the number of primitive Pythagorean triangles (integer-sided

right triangles with relatively prime sides) with �xed inradius is always a

power of 2.

Solution by Carl Bosley, student, Topeka, KansasThe formula for the inradius of a right triangle, r = (a + b � c)=2,

where a, b are the legs and c is the hypotenuse, together with the formula

a = 2mn, b = m2 � n2, c = m2 + n2, for the sides of the triangle, where

m and n are relatively prime and not both odd, gives r = n(m� n).

Let p be a prime which divides r. If p = 2, p can divide n but not

m � n, since then m and n would either be both even or both odd. If p

is not 2, then p must divide n or m � n, but cannot divide both, since m

and n would have a common factor. Since each prime p other than 2 which

divides r divides n orm�n, but not both, and each combination of choices

produces a pair m, n which generates a right triangle, so if r has k distinct

prime factors greater than 2, there are 2k primitive Pythagorean triangles

with inradius r.

Also solved by CLAUDIO ARCONCHER, Jundia��, Brazil; CHRISTOPHERJ. BRADLEY, Clifton College, Bristol, U. K.; DAVID DOSTER, Choate Rose-

44

mary Hall, Wallingford, Connecticut;H. ENGELHAUPT, Franz{Ludwig{Gym-nasium, Bamberg, Germany; TOBY GEE, student, The John of Gaunt School,Trowbridge, England; RICHARD I. HESS, Rancho Palos Verdes, California;PETER HURTHIG, Columbia College, Burnaby, British Columbia; JAMSHIDKHOLDI, New York, N. Y.; V �ACLAV KONE �CN �Y, Ferris State University, BigRapids, Michigan; HEINZ{J �URGEN SEIFFERT, Berlin, Germany; LAWRENCESOMER, Catholic University of America, Washington, D. C.; CHRISWILDHAGEN, Rotterdam, The Netherlands: and the proposer. There wereseven incorrect solutions received, most of which did not handle the casep = 2 correctly.

Kholdi points out that the problem is solved on page 43 of Sierpi �nski'sPythagorean Triangles, published by the Graduate School of Science, YeshivaUniversity, 1962.

2013. [1995: 52] Proposed by Waldemar Pompe, student, Universityof Warsaw, Poland.

Given is a convexn-gonA1A2 : : : An (n � 3) and a pointP in its plane.

Assume that the feet of the perpendiculars from P to the linesA1A2; A2A3;

: : : ; AnA1 all lie on a circle with centre O.

(a) Prove that if P belongs to the interior of the n-gon, so does O.

(b) Is the converse to (a) true?

(c) Is (a) still valid for nonconvex n-gons?

Solution by Jerzy Bednarczuk, Warszawa, Poland.(a) We show that (a) is true for a polygon that is not necessarily convex

(so we will have treated (a) and (c) at the same time). Let us call the given

circle C. We �rst see that when P is inside the n-gon, then it must also be

insideC. The half-line onOP starting at P and going away fromO intersects

some side of the n-gon, say AiAi+1, at a point S. If P were exterior to C,

then the circle with diameter PS would lie in the exterior of C. Since the

latter circle is the locus of points B with \PBS = 90�, the foot of the

perpendicular from P to AiAi+1 would then lie outside of C, contrary to the

de�nition of C. Also by de�nition, P cannot lie on C so we conclude that P

must lie inside C as claimed. By way of contradiction we now assume that

O lies outside of the given n-gon. Since P is inside, the segment OP would

intersect some side of the n-gon, say AjAj+1, at some point T . Since P is

also inside C, the circle whose diameter is PT would be contained inside

C, which would force the foot of the perpendicular from P to AjAj+1 to lie

inside C contrary to its de�nition. We conclude that O lies inside the given

n-gon as desired.

(b) No. We can �nd a counterexample for any n > 2 ! First draw a

circle C and then choose any point P lying OUTSIDE of this circle. Next take

n points B1; B2; : : : ; Bn on the circle C and through each Bi draw the line

perpendicular to the line PBi. Those lines will form an n-gon which will

45

contain O if you choose each Bi so that the half-plane determined by the

line perpendicular to PBi will contain all other Bj together with O.

Also solved by the proposer.

2014. [1995: 52] Proposed by Murray S. Klamkin, University ofAlberta.

(a) Show that the polynomial

2�x7 + y7 + z7

�� 7xyz

�x4 + y4 + z4

�has x+ y + z as a factor.

(b)�Is the remaining factor irreducible (over the complex numbers)?

I. Solution to (a) by Jayabrata Das, Calcutta, India.Let f(x; y; z) = 2

�x7 + y7 + z7

�� 7xyz

�x4 + y4 + z4

�. If we can show

that f(x; y; z) = z when x+ y + z = 0, we are done.

We know, for x+ y + z = 0, that x3 + y3 + z3 = 3xyz. Thus

x7 + y7 + z7 + x3y4 + x3z4 + y3z4 + y3x4 + z3y4 + z3x4

=�x3 + y3 + z3

� �x4 + y4 + z4

�= 3xyz

�x4 + y4 + z4

�so that

x7 + y7 + z7 = 3xyz�x4 + y4 + z4

��x3y4 � x3z4 � y3z4 � y3x4 � z3y4 � z3x4

Therefore

f(x; y; z) = 2�x7 + y7 + z7

�� 7xyz�x4 + y4 + z4

�= 6xyz

�x4 + y4 + z4

��2 �x3y4 + x3z4 + y3z4 + y3x4 + z3y4 + z3x4

��7xyz �x4 + y4 + z4

�= �xyz �x4 + y4 + z4

��2x3y3(x+ y)� 2y3z3(y + z)� 2z3x3(z + x)

= �xyz�x4 + y4 + z4

�+ 2x3y3z + 2xy3z2 + 2x3yz3

= �xyz�x4 + y4 + z4 � 2x2y2 � 2y2z2 � 2z2x2

�= �xyz

��x2 + y2 + z2

�2 � 4�x2y2 + y2z2 + z2x2

��:

Since x2 + y2 + z2 = �2(xy+ yz + zx), we now have that

f(x; y; z) = �xyz�4 (xy + yz+ zx)

2 � 4�x2y2 + y2z2 + z2x2

��= �4xyz

�2x2yz+ 2xy2z + 2xyz2

�= �8xyz

�xyz(x+ y+ z)

�= 0 :

46

II. Solution to (a) by Cyrus Hsia, student, University of Toronto, Toron-to, Ontario.Consider the sequence an = xn+ yn+ zn. The characteristic equation with

roots x, y, z, is

a3 �Aa2 + Ba� C = 0;

where A = x+ y + z, B = xy + yz + zx and C = xyz.

The sequence fang follows the recurrence relation:

an+3 = Aan+2 �B an+1 + C an :

Now, we have

a0 = x0 + y0 + z0 = 3 ;

a1 = x1 + y1 + z1 = A ;

a2 = x2 + y2 + z2 = (x+ y + z)2 � 2(xy+ yz + zx) = A2 � 2B

From the recurrence relation, we see:

a3 = Aa2 �B a1 + C a0

= A3 � 2AB �AB + 3C

= Ak3 + 3C; where k3 is some term in x, y and z

Similarly

a4 = Ak4 + 2B2; where k4 is some term in x, y and z,

a5 = Ak5 � 5BC; where k5 is some term in x, y and z,

a6 = Ak6 � 2B3 + 3C2; where k6 is some term in x, y and z,

a7 = Ak7 + 7B2C; where k7 is some term in x, y and z.

Thus,

2�x7 + y7 + z7

�� 7xyz

�x4 + y4 + z4

�= 2a7 � 7C a4

= a�Ak7 + 7B2C

�� 7C

�Ak4 + 2B2

�= Ak

where k is some term in x, y and z; that is,

x+ y + z divides 2�x7 + y7 + z7

�� 7xyz

�x4 + y4 + z4

�.

Part (a) was also solved by �SEFKET ARSLANAGI �C, Berlin, Germany;CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol, U. K.; MIGUEL ANGELCABEZ �ON OCHOA, Logro ~no, Spain; ADRIAN CHAN, Grade 8 student, Up-per Canada College, Toronto, Ontario; TIM CROSS, Wolverley High School,Kidderminster, U. K.; RICHARD I. HESS, Rancho Palos Verdes, Califor-nia; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan; J. A. MCCALLUM,

47

Medicine Hat, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRISWILDHAGEN, Rotterdam, The Netherlands; and the proposer. One incor-rect solution to part (b) was received.

Other solvers of part (a) made use of computer algebra, properties ofroots, the substitutionx = �y� z, or direct division. Some solvers pointedout that computer algebra failed to factorize the remaining factor. The edi-tor (Shawyer) tried using DERIVE on a MSDOS 486 66MHZ computer. Thefactorisation stopped with no factors after 155 seconds. MAPLE also failedto �nd any factors. McCallum commented \An asterisk on a question ofKlamkin's is equivalent to a DO NOT ENTER sign!"

2015. [1995: 53 and 129 (Corrected)] Proposed by Shi-Chang Chiand Ji Chen, Ningbo University, China.

Prove that

�sin(A) + sin(B) + sin(C)

�� 1

A+

1

B+

1

C

�� 27

p3

2�;

where A, B, C are the angles (in radians) of a triangle.

I. Solution by Douglass L. Grant, University College of Cape Breton,Sydney, Nova Scotia, Canada.If A = B = C = �=3, equality obtains. It then su�ces to show that each

factor has an absolute minimum at the point. Note that C = � � (A+ B).

Let S = f(A;B) : A > 0; B > 0 A+ B < �g.Let f(A;B) =

1

A+

1

B+

1

� � (A+ B). Then f is unbounded on S. So,

if there is a unique critical point for f on S, it must be an absolute minimum.

Now, FA(A;B) = � 1

A2+

1�� � (A+ B)

�2 = 0 implies that ��(A+

B) = A. Similarly, FB(A;B) = 0 implies that � � (A + B) = B, and so

that A = B. Hence A = B = � � (A+ B) = �3.

Let g(A;B) = sin(A) + sin(B) + sin�� � (A+ B)

�= sin(A) +

sin(B)+sin(A+B). We now obtain that 0 = gA(A;B) = cos(B)+cos(A+

B), 0 = gB(A;B) = cos(A)+ cos(A+B), so that cos(A) = cos(B). Since

no two distinct angles in (0; �) have equal cosines, we have that A = B.

Then 0 = cos(A)+cos(2A) = 2 cos2(A)+cos(A)�1=�2 cos(A)� 1

��cos(A) + 1

�: Since cos(A) cannot have the value �1, it must then have

value12, and so we have A = B = C = �

3.

48

II. Solution by the proposers.Let y(x) = x�1=3 cos(x) for 0 < x � �

2. Di�erentiating twice yields

x7=3 sec(x):y00(x) =2x tan(x)

3+

4

9� x2

>2x

3

�x+

x3

3

�+

4

9� x2

=2

9

�x2 � 3

4

�2+

23

72> 0 :

By the AM{GM inequality and the Jensen inequality, we haveXsin(A)�

X 1

A= 4

Ycos(A=2)�

X 1

A

�6Y

cos(A=2)

Y�A

2

�1=3

� 6

0BBB@cos

��

6

���

6

�1=31CCCA3

=27p3

2�:

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; FRANCISCOBELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; and BOB PRIELIPP, University ofWisconsin{Oshkosh. Their solutions were based on known geometric in-equalities. Also some readers wrote in after the initial publication of theproblem, pointing out that the original result could not be true, and sug-gesting possible corrections. Thank you.

XXhh

EE

�� BB

��

((��

hh

DD ����

((

49

On The Generalized PtolemyTheorem

Shailesh ShiraliRishi Valley School, Rishi Valley 517 352,

Chittoor Dt., A.P, INDIA

Introduction. The following note describes a few uses of a relatively less

known result in plane geometry, the Generalized Ptolemy Theorem (GPT, for

short), also known as Casey's Theorem (see Johnson[1]). Featured will be

two proofs of the problem proposed by India for the 33rd IMO in Moscow

[1993: 255; 1995: 86].

Theorem 1. Circles 1 and 2 are externally tangent at a point I, and both

are enclosed by and tangent to a third circle . One common tangent to 1

and 2 meets in B and C, while the common tangent at I meets in A

on the same side of BC as I. Then I is the incentre of triangle ABC.

r

r

r

r

12

B C

A

I

Figure 1.

Proof 1.

Notation: Let tij refer to the length of the external common tangent to circles

i and j (thus the two circles lie on the same side of the tangent). We use the

GPT, which we state in the following manner.

(The GPT) Let circles�, �, , � all touch the circle �, the contacts

being all internal or all external and in the cyclical order �, �, ,

�. Then:

t�� � t � + t� � t�� = t� � t�� :

50

Moreover, a converse also holds: if circles �, �, , � are located

such that

� t�� � t � � t�� � t� � t� � t�� = 0

for some combination of +, � signs, then there exists a circle

that touches all four circles, the contacts being all internal or all

external.

For a proof of the GPT and its converse, please refer to [1].

r

r

r

r

12

B C

A

I

r

rr

X YD

c z bBC = a

BX = x

CY = y

DX =DI

=DY

= �

Figure 2.

Consider the con�guration shown in Figure 2, where x and y are, re-

spectively, the lengths of the tangents from B and C to 1 and 2; D is

AI \ BC; z = jAI j; u = jIDj; and a, b, c are the sides of4ABC.We apply the GPT to the two 4-tuples of circles (A; 1; B; C) and

(A; 2; C; B). We obtain:

az + bx = c(2u+ y) (1)

az + cy = b(2u+ x) : (2)

Subtracting (2) from (1) yields bx � cy = u(c � b), so (x + u)=(y + u) =

c=b, that is, BD=DC = AB=AC, which implies that AI bisects \BAC

and that BD = ac=(b + c). Adding (1) and (2) yields az = u(b + c), so

z=u = (b+c)=a, that is, AI=ID = AB=BD, which implies that BI bisects

\ABC. This proves the result.

Proof 2.

Lemma. Let BC be a chord of a circle �, and let S1, S2 be the two arcs of �

cut o� byBC. LetM be the midpoint of S2, and consider all possible circles

that touch S1 and BC. Then the length tM of the tangent from M to

is constant for all such . (See Figure 3.)

51

r

r

r rr

M

S2 �

B CS

R

S1

tM

Figure 3.

Proof of Lemma. Let \ � = R, \ BC = S. Applying the GPT to the

4-tuple (B;; C;M), we �nd: BS � CM + CS � BM = tM � BC. SinceBM = CM , we obtain: tM = BM , a constant.

Proof of Theorem 1. (See �gure 4.) Let S1, S2 be the two arcs of � cut o� by

chord BC, S1 being the one containing A, and let M denote the midpoint

of S2. Using the above lemma,

tM1=MB =MI =MC = tM2

:

Therefore M has equal powers with respect to 1 and 2 and lies on their

radical axis, namely AI. It follows that AI bisects \A of4ABC.

r

r

r

r

12

B C

A

I

Mr

Figure 4.

52

Next, 4IBM is isosceles, so \IBM = �=2� C=2. Also, \CBM = A=2,

so \IBC = �=2� C=2� A=2 = B=2, that is, IB bisects \B of 4ABC.It follows that I is the incentre of4ABC.

} } } } }For non-believers, here are two more illustrations of the power and

economy of the GPT.

Theorem 2. Let4ABC have circumcircle �, and let be a circle lying within

� and tangent to it and to the sides AB (at P ) and AC (at Q). Then the

midpoint of PQ is the incentre of 4ABC. (See Figure 5.)

r

r

r

r

rr

r

r

r

D

C

B

R

A

�

S

I

P Q

Figure 5.

Proof. Let the GPT be applied to the 4-tuple of circles (A;B;; C). Let

AP = x = AQ. Then:

tAB = c; tA = AP = x; tAC = b ;

tB = BP = c� x; tBC = a; tC = CQ = b� x :

The GPT now gives: c(b � x) + (c � x)b = ax, so x = bc=s, where s =

(a+b+c)=2 is the semi-perimeter of4ABC. Let I denote the midpoint of

PQ; then IP = x sinA=2 = (bc=s) sinA=2, and the perpendicular distance

from I to AB is IP cosA=2, which equals (bc=s)�sinA=2

��cosA=2

�=�

(1=2)bc sinA�=s. But this is just the radius of the incircle of4ABC. Since

I is equidistant from AB and AC, it follows that I is the incentre of the

triangle.

The next illustration concerns one of the most celebrated discoveries in

elementary geometry made during the last two centuries.

53

Theorem 3. (Feuerbach's Theorem) The incircle and nine-point circle of a

triangle are tangent to one another.

Proof. Let the sides BC, CA, AB of 4ABC have midpoints D, E, F

respectively, and let be the incircle of the triangle. Let a, b, c be the sides

of 4ABC, and let s be its semi-perimeter. We now consider the 4-tuple of

circles (D;E; F;). Here is what we �nd:

tDE =c

2; tDF =

b

2; tEF =

a

2;

tD =

����a2 � (s� b)

���� =����b� c

2

���� ;tE =

����b2 � (s� c)

���� =����a� c

2

���� ;tF =

���� c2 � (s� a)

���� =����b� a

2

���� :We need to check whether, for some combination of +, � signs, we have

� c(b� a) � a(b� c) � b(a� c) = 0 :

But this is immediate! It follows from the converse to the GPT that there

exists a circle that touches each of D, E, F and . Since the circle passing

throughD, E, F is the nine-point circle of the triangle, it follows that and

the nine-point circle are tangent to one another.

One would surmise that the GPT should provide a neat proof of the

following theorem due to Victor Thebault:

Let 4ABC have circumcircle �, let D be a point on BC, and let

1 and 2 be the two circles lying within � that are tangent to

� and also to AD and BC. Then the centres of 1 and 2 are

collinear with the incentre of4ABC.

I have, however, not been able to �nd such a proof, and I leave the problem to

the interested reader. We note in passing that Thebault's theorem provides

yet another proof of Theorem 1.

References:

[1] R.A. Johnson, Advanced Euclidean Geometry, Dover, 1960.

Acknowledgements:

I thank the referee for making several valuable comments that helped

tidy up the presentation of the paper.

54

THE SKOLIAD CORNERNo. 12

R.E. Woodrow

This issue we feature Part I of the 1995{96 Alberta High School Mathe-

matics Competition, written November 21, 1995. My thanks go to Professor

T. Lewis, the University of Alberta for forwarding me a copy. Students have

90 minutes to complete their contest. It is mostly written by students in

Grade XII, but has often been won by students in earlier grades.

ALBERTA HIGH SCHOOL MATHEMATICSCOMPETITION

Part I

November 21, 1995 (Time: 90 minutes)

1. A circle and a parabola are drawn on a piece of paper. The number

of regions they divide the paper into is at most

A. 3 B. 4 C. 5 D. 6 E. 7.

2. The number of di�erent primes p > 2 such that p divides 712 �372 � 51 is

A. 0 B. 1 C. 2 D. 3 E. 4.

3. Suppose that your height this year is 10% more than it was last year,

and last year your height was 20% more than it was the year before. By what

percentage has your height increased during the last two years?

A. 30 B. 31 C. 32 D. 33 E. none of these.

4. Multiply the consecutive even positive integers together until the

product 2 � 4 � 6 � 8 � � � becomes divisible by 1995. The largest even integer

you use is

A. between 1 and 21 B. between 21 and 31

C. between 31 and 41 D. bigger than 41

E. non-existent, since the product never becomes divisible by 1995.

5. A rectangle contains three circles as in

the diagram, all tangent to the rectangle and to

each other. If the height of the rectangle is 4,

then the width of the rectangle is

A. 3 + 2p2 B. 4 + 4

p2

3C. 5 + 2

p2

3

D. 6 E. 5 +p10.

4

55

6. Mary Lou works a full day and gets her usual pay. Then she works

some overtime hours, each at 150% of her usual hourly salary. Her total pay

that day is equivalent to 12 hours at her usual hourly salary. The number of

hours that she usually works each day is

A. 6 B. 7:5 C. 8

D. 9 E. not uniquely determined by the given information.

7. A fair coin is tossed 10; 000 times. The probability p of obtaining

at least three heads in a row satis�es

A. 0 � p < 14

B. 14� p < 1

2C. 1

2� p < 3

4D. 3

4� p < 1 E. p = 1:

8. In the plane, the angles of a regular polygon with n sides add up to

less than n2 degrees. The smallest possible value of n satis�es:

A. n < 40 B. 40 � n < 80 C. 80 � n < 120

D. 120 � n < 160 E. n � 160.

9. A cubic polynomial P is such that P (1) = 1, P (2) = 2, P (3) = 3

and P (4) = 5. The value of P (6) is

A. 7 B. 10 C. 13 D. 16 E. 19.

10. The positive numbers x and y satisfy xy = 1. The minimum value

of 1

x4+ 1

4y4is

A. 12

B. 58

C. 1 D. 54

E. no minimum.

11. Of the points (0; 0), (2;0), (3; 1), (1;2), (3; 3), (4; 3) and (2; 4),

at most how many can lie on a circle?

A. 3 B. 4 C. 5 D. 6 E. 7.

12. The number of di�erent positive integer triples (x; y; z) satisfying

the equations

x2 + y � z = 100 and x+ y2� z = 124 is:

A. 0 B. 1 C. 2 D. 3 E. none of these.

13. Which of the following conditions does not guarantee that the

convex quadrilateral ABCD is a parallelogram?

A. AB = CD and AD = BC B. \A = \C and \B = \D

C. AB = CD and \A = \C D. AB = CD and AB is parallel to CD

E. none of these.

56

14. How many of the expressions

x3 + y4; x4 + y3; x3 + y3; and x4 � y4;

are positive for all possible numbers x and y for which x > y?

A. 0 B. 1 C. 2 D. 3 E. 4.

15. In triangle ABC, the altitude from A to BC meets BC at D,

and the altitude from B to CA meets AD at H. If AD = 4, BD = 3 and

CD = 2, then the length ofHD is

A.p52

B. 32

C.p5 D. 5

2E. 3

p5

2.

16. Which of the folowing is the best approximation to

(23 � 1)(33� 1)(43� 1)

(23 + 1)(33+ 1)(43+ 1)� � � (100

3� 1)

(1003+ 1)?

A. 3

5B. 33

50C. 333

500D. 3;333

5;000E. 33;333

50;000.

Last issue we gave the SHARP U.K. Intermediate Mathematical Chal-

lenge, written February 2, 1995. Here are answers.

1. E 2. C 3. D 4. E 5. A

6. B 7. C 8. B 9. D 10. A

11. E 12. D 13. C 14. D 15. B

16. A 17. A 18. C 19. C 20. D

21. D 22. E 23. B 24. B 25. A

That completes this month's Skoliad Corner. I need materials of a suit-

able level to build up a bank of contests. Please send me suitable materials

as well as comments, criticisms, and suggestions. I would like to have some

feedback too about how your students do with these materials.

57

THE OLYMPIAD CORNERNo. 172

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.

Because we are now publishing eight numbers of the Corner rather than

ten, I am giving two Olympiad Sets for your pleasure. Besides, here in Canada

it is winter, and quite cold, so having a stock of problems to contemplate

in a warm spot is a good idea. Both of the sets we give this number were

collected by Georg Gunther, Sir Wilfred Grenfell College, Corner Brook, when

he was Canadian Team Leader to the IMO at Istanbul. Many thanks to him

for gathering a wide sample of contests. We begin with the Telecom 1993

Australian Mathematical Olympiad.

TELECOM 1993 AUSTRALIANMATHEMATICAL OLYMPIAD

Paper 1

Tuesday, 9th February, 1993

(Time: 4 hours)

1. In triangleABC, the angleACB is greater than 90�. PointD is the

foot of the perpendicular from C to AB;M is the midpoint of AB; E is the

point on AC extended such that EM = BM ; F is the point of intersection

of BC and DE; moreover BE = BF . Prove that \CBE = 2\ABC.

2. For each function f which is de�ned for all real numbers and satis-

�es

f(x; y) = x � f(y)+ f(x) � y (1)

and

f(x+ y) = f(x1993) + f(y1993) (2)

determine the value f(p5753).

3. Determine all triples (a1; a2; a3), a1 � a2 � a3, of positive inte-

gers in which each number divides the sum of the other two numbers.

4. For each positive integer n, let

f(n) = [2pn]� [

pn� 1 +

pn+ 1]:

Determine all values n for which f(n) = 1.

Note: If x is a real number, then [x] is the largest integer not exceeding x.

58

Paper 2

Wednesday, 10th February, 1993

(Time: 4 hours)

5. Determine all integers x and y that satisfy

(x+ 2)4 � x4 = y3:

6. In the acute-angled triangle ABC, let D, E, F be the feet of alti-

tudes through A, B, C, respectively, andH the orthocentre. Prove that

AH

AD+BH

BE+CH

CF= 2:

7. Let n be a positive integer, a1; a2; : : : ; an positive real numbers

and s = a1 + a2 + � � �+ an. Prove that

nXi=1

ai

s� ai� n

n� 1and

nXi=1

s� ai

ai� n(n� 1):

8. The vertices of triangle ABC in the x�y plane have integer coor-

dinates, and its sides do not contain any other points having integer coor-

dinates. The interior of ABC contains only one point, G, that has integer

coordinates. Prove that G is the centroid of ABC.

Next we give the Final Round of the Japan Mathematical Olympiad.

JAPANMATHEMATICAL OLYMPIAD

Final Round | 11 February, 1993

(Time: 4.5 hours)

1. Suppose that two words A and B have the same length n > 1

and that the �rst letters of them are di�erent while the others are the same.

Prove that A or B is not periodic.

2. Let d(n)be the largest odd number which divides a givennumbern.

Suppose that D(n) and T (n) are de�ned by

D(n) = d(1) + d(2) + � � �+ d(n);

T (n) = 1 + 2 + � � �+ n:

Prove that there exist in�nitely many positive numbers n such that

3D(n) = 2T (n).

59

3. In a contest, x students took part, and y problems were posed. Each

student solved y=2 problems. For every problem, the number of students

who solved it was the same. For each pair of students, just three problems

were solved by both of them. Determine all possible pairs (x; y). Moreover,

for each (x; y), give an example of the matrix (aij) de�ned by aij = 1 if the

ith student solved the jth problem and aij = 0 if not.

4. Five radii of a sphere are given so that no three of them are in a

common plane. Among the 32 possible choices of an end point from each

segment, �nd out the number of choices for which the 5 points are in a hemi-

sphere.

5. Prove that there exists a positive constant C (independent of n, aj)

which satis�es the inequality

max0�x�2

nYj=1

jx� aj j � Cn max0�x�1

nYj=1

jx � aj j

for any positive integer n and any real numbers a1; : : : ; an.

Last issue we gave a set of six Klamkin Quickies. Here are his \quick"

solutions. Many thanks go to Murray Klamkin, the University of Alberta, for

sending them to me.

SIX KLAMKIN QUICKIES

1. Which is larger

(3p2� 1)1=3 or 3

p1=9� 3

p2=9 + 3

p4=9?

Solution. That they are equal is an identity of Ramanujan.

Letting x = 3

p1=3 and y = 3

p2=3, it su�ces to show that

(x+ y)(3p2� 1)1=3 = x3 + y3 = 1;

or equivalently that

(3p2 + 1)3(

3p2� 1) = 3;

which follows by expanding out the left hand side.

For other related radical identities of Ramanujan, see Susan Landau,

How to tangle with a nested radical, Math. Intelligencer, 16 (1994),

pp. 49{54.

2. Prove that

3min

�a

b+b

c+c

a;b

a+c

b+a

c

�� (a+ b+ c)

�1

a+

1

b+

1

c

�;

60

where a, b, c are sides of a triangle.

Solution. Each of the inequalities

3

�a

b+b

c+c

a

�� (a+ b+ c)

�1

a+

1

b+

1

c

�;

3

�b

a+c

b+a

c

�� (a+ b+ c)

�1

a+

1

b+

1

c

�;

follow from their equivalent forms (which follow by expansion):

(b+ a� c)(c� a)2 + (c+ b� a)(a� b)2 + (a+ c� b)(b� c)2 � 0;

(b+ c� a)(c� a)2 + (c+ a� b)(a� b)2 + (a+ b� c)(b� c)2 � 0:

3. Let ! = ei�=13. Express 11�! as a polynomial in ! with integral

coe�cients.

Solution. We have

2

(1� !)=

(1� !13)

(1� !)= 1 + ! + !2+ � � �+ !12;

0 =(1 + !13)

(1 + !)= 1� ! + !2� � � �+ !12:

Adding or subtracting, we get

1

(1� !)= 1 + !2 + !4+ � � �+ !12

= ! + !3+ � � �+ !11:

More generally, if ! = ei�=(2n+1),

1

(1� !)= 1 + !2+ !4 + � � �+ !2n:

4. Determine all integral solutions of the simultaneous Diophantine

equations x2 + y2 + z2 = 2w2 and x4 + y4 + z4 = 2w4.

Solution. Eliminating w we get

2y2z2 + 2z2x2 + 2x2y2� x4 � y4� z4 = 0

or

(x+ y + z)(y+ z � x)(z + x� y)(x+ y � z) = 0;

so that in general we can take z = x+y. Note that if (x; y; z; w) is a solution,

so is (�x;�y;�z;�w) and permutations of the x, y, z. Substituting back,

we get

x2 + xy + y2 = w2:

61

Since (x; y;w) = (1;�1; 1) is one solution, the general solution is obtained

by the method of Desboves, that is, we set x = r + p, y = �r + q and

w = r. This gives r =(p2+pq+q2)

(q�p) . On rationalizing the solutions (since the

equation is homogeneous), we get

x = p2 + pq+ q2 + p(q � p) = q2+ 2pq;

�y = p2 + pq+ q2 � q(q� p) = p2 + 2pq;

w = p2 + pq+ q2;

z = q2� p2:

5. Prove that if the line joining the incentre to the centroid of a triangle

is parallel to one of the sides of the triangle, then the sides are in arithmetic

progression and, conversely, if the sides of a triangle are in arithmetic pro-

gression then the line joining the incentre to the centroid is parallel to one

of the sides of the triangle.

Solution. Let A, B, C denote vectors to the respective vertices A, B,

C of the triangle from a point outside the plane of the triangle. Then the

incentre I and the centroid G have the respective vector representations I

and G, where

I =(aA+ bB+ cC)

(a+ b+ c); G =

(A+ B+ C)

3;

(where a, b, c are sides of the triangle). IfG�I = k(A�B), then by expanding

out

(b+ c� 2a� k0)A+ (a+ c� 2b+ k0)B+ (a+ b� 2c)C = 0;

where k0 = 3k(a + b + c). Since A, B, C are linearly independent, the

coe�cient of C must vanish so that the sides are in arithmetic progression.

Also then k0 = b+ c� 2a = 2b� a� c.

Conversely, if 2c = a + b, then G� I =3(A�B)(b�a)6(a+b+c)

, so that GI is parallel

to the side AB.

6. Determine integral solutions of the Diophantine equation

x� y

x+ y+y� z

y+ z+z �w

z +w+w � x

w + x= 0

(joint problem with Emeric Deutsch, Polytechnic University of Brooklyn).

Solution. It follows by inspection that x = z and y = w are two

solutions. To �nd the remaining solution(s), we multiply the given equation

by the least common denominator to give

P (x; y; z; w) = 0;

62

where P is the 4th degree polynomial in x, y, z, w which is skew symmetric

in x and z and also in y and w. Hence,

P (x; y; z; w) = (x� z)(y� w)Q(x; y; z; w);

where Q is a quadratic polynomial. On calculating the coe�cient of x2 in P ,

we get 2z(y�w). Similarly the coe�cient of y2 is�2w(x� z), so that

P (x; y; z; w) = 2(x� z)(y� w)(xz� yw):

Hence, the third and remaining solution is given by xz = yw.

Next we turn to the readers' solutions of problems from earlier numbers

of the Corner. Let me begin by thanking Beatriz Margolis, Paris France; Bob

Prielipp, University of Wisconsin-Oshkosh, USA; Toshio Seimiya, Kawasaki,

Japan; D.J. Smeenk, Zaltbommel, the Netherlands and ChrisWildhagen, Rot-

terdam, the Netherlands, for sending in nice solutions to some of the prob-

lems of the 1994 Canadian and 1994 U.S.A. Mathematical Olympiads. As

we publish \o�cial" solutions to the former, and refer readers to an MAA

publication for the latter, I normally do not publish these solutions.

Next, correspondence from Murray Klamkin, the University of Alberta,

�led under the September number of the Corner, contains comments about

problems from several numbers.

3. [1993: 5, 1994: 69] 1991 British Mathematical Olympiad.

ABCD is a quadrilateral inscribed in a circle of radius r. The diagonals

AC, BD meet at E. Prove that if AC is perpendicular to BD then

EA2 + EB2 + EC2 + ED2 = 4r2: (�)Is it true that, if (�) holds, then AC is perpendicular to BD? Give a reason

for your answer.

Klamkin's Comment. A simpler solution than that published, plus a

generalization, is given in Crux, 1989, p. 243, #1.

3. [1994: 184] 1st Mathematical Olympiad of the Republic of China

(Taiwan).

If x1; x2; : : : ; xn are n non-negative numbers, n � 3 and x1 + x2 +

� � �+ xn = 1 prove that x21x2 + x22x3 + � � �+ x2nx1 � 4=27.

Klamkin's Comment. This problem appeared as problem 1292 in The

Math. Magazine, April 1988.

Now we turn to readers' solutions of problems proposed to the Jury,

but not used, at the 34th International Mathematical Olympiad at Istanbul.

A booklet of \o�cial solutions" was issued by the organizers. Because I do

not have formal permission to reproduce these solutions, I will only discuss

readers' solutions that are di�erent.

63

2. [1994: 216] Proposed by Canada.

Let triangle ABC be such that its circumradius R = 1. Let r be the

inradius of ABC and let p be the inradius of the orthic triangle A0B0C0 oftriangle ABC. Prove that p � 1� 1=3(1 + r)2.

Solution by D.J. Smeenk, Zaltbommel, The Netherlands.

In the �rst instance I misread the problem: I read1

3(1 + r)2instead of

13(1 + r)2!

B C

A

A0

C0B0

�

� ��

�

�

Now r�ABC = R(cos�+ cos�+ cos � 1). From R = 1, we obtain

r = cos�+ cos� + cos � 1. Similarly,

p =1

2[cos(� � 2�) + cos(� � 2�) + cos(� � 2 )� 1]

= 2 cos� cos� cos :

We are to show that p+ 13(1 + r)2 � 1. (1)

Equivalently,

2 cos� cos� cos +1

3(cos�+ cos� + cos )2 � 1: (2)

Now

�1 < cos� cos� cos � 18;

1 < cos�+ cos� + cos � 32:

)(3)

[Bothema, Geom. Inequalities, 2.24, resp. 2.16]

Now 2 � 18+ 1

3(32)2 = 1, so it is clear that (1) holds.

3. [194: 216] Proposed by Spain.

Consider the triangleABC, its circumcircle k of centre O and radiusR,

and its incircle of centre I and radius r. Another circle Kc is tangent to the

sides CA, CB at D, E, respectively, and it is internally tangent to k. Show

that I is the midpoint of DE.

64

Solution by D.J. Smeenk, Zaltbommel, The Netherlands.

Assume (see �gure) that � > . Let F be the centre of Kc and � its

radius. Then F lies on the production of AI. Now FD = �, so

AF =�

sin �2

:

Also AO = R, OF = R� �, and \FAO = ' = 12(� � ).

A

B C

E

D

'

G I0

F

O

R

R�p

�=2

�=2

�=2

Apply the Law of Cosines to �AFO:

OF 2 = AF 2+ AO2 � 2AF � AO cos';

or

(R� �)2 =�2

sin2 �2

+ R2 �2R� cos

��� 2

�2 sin �

2

; � 6= 0:

From this we obtain

�2R+ � =�

sin2 �2

�2R cos

��� 2

�cos

��+

2

� ;

or� cos2 �

2

sin2 �=2=

4R sin �

2sin

2

sin�=2:

Thus

� =4R sin �

2sin �

2sin

2

cos2 �

2

=r

cos2 �

2

= FD:

65

Now let DE intersect AF at I0. Then DI0 = DF cos �2= r

cos �2

.

Let G be the foot of the perpendicular from I0 to AB.

Then I0G = I0D cos �2= r

cos �2

cos �2= r. So I0 coincides with I.

7. [1994: 217] Proposed by Israel.

The vertices D, E, F of an equilateral triangle lie on the sides BC,

CA, AB respectively of a triangle ABC. If a, b, c are the respective lengths

of these sides and S is the area of ABC, prove that

DE � 2p2S � fa2 + b2 + c2 + 4

p3Sg�1=2:

Comment by Murray S. Klamkin, The University of Alberta.

This problem is equivalent to problem #624 of Crux [1982: 109{110].

11. [1994: 241] Proposed by Spain.

Given the triangle ABC, let D, E be points on the side BC such that

\BAD = \CAE. IfM andN are respectively, the points of tangency with

BC of the incircles of the triangles ABD and ACE, show that

1

BM+

1

MD=

1

NC+

1

NE:

Solution by D.J. Smeenk, Zaltbommel, The Netherlands.

We are to show 1MB

+ 1MD

= 1NC

+ 1NE

, or equivalently

BD �NC �NE = CE �MB �MD: (1)

We denote \BAD = \CAE = '.

q

q

qq

q

q q

q

q

B C

A

M D NE

I1 I2

' '

+'� �+'

Applying the law of sines to �ABD we obtain

BD =c sin'

sin(� + '); AD =

c sin�

sin(� + '): (2)

The sine law for �ACE gives

CE =b sin'

sin( + '); AE =

b sin

sin( + '): (20)

66

From �ABD:

2MB = AB + BD � AD

2MD = �AB + BD + AD;

and from �ACE:

2NC = AC + CE � AE

2NE = �AC +CE +AE:

From this we see that to show (1) we must show

BD(AC +CE � AE)(�AC + CE + AE)

= CE(AB +BD � AD)(�AB + BD +AD);

or equivalently

BD(CE2 � AC2 � AE2 + 2AC � AE)= CE(BD2 �AB2 � AD2 + 2AB � AD): (3)

Now use the law of cosines in �ACE, and in�ABD to obtain

CE2 � AC2 � AE2 = �2ACAE cos'

BD2 �AB2 � AD2 = �2ABAD cos':

�(4)

Combining (3) and (4) we see that we must show

BD �AC � AE(1� cos') = CE � AB �AD(1� cos'):

As 1� cos' 6= 0, we �nd with (2) and (2') that we must verify that

c sin'

sin(� + )� b � b sin

sin( + '=

b sin'

sin( + ')� c � c sin�

sin(� + ');

and as b sin = c sin�, this holds.

13. [1994: 241] Proposed by India.

A natural number n is said to have the property P if, whenver n divides

an � 1 for some integer a, n2 also necessarily divides an � 1.

(a) Show that every prime number has property P .

(b) Show there are in�nitely many composite numbers n that possess

property P .

Solution by E.T.H. Wang, Sir Wilfrid Laurier University, Waterloo, On-

tario.

(a) Suppose n = p is a prime such that ap � 1 mod p. Since ap �a mod p by Fermat's Little Theorem, we have a � 1 mod p. Thus a = kp+1

for some integer k. Hence, by the Binomial Theorem ap�1 = (kp+1)p�1 =

(kp)p +�p

1

�(kp)p�1 + � � �+

�p

p�1�kp. Since

�p

p�1�= p and p � 2, it follows

that p2 j ap � 1.

(b) We show that all composite numbers of the form n = 2p, where p

is an odd prime have property P . Suppose 2p j a2p � 1. Then p j (a2)p � 1

67

which, in view of (a), implies p2 j a2p � 1. On the other hand, 2 j a2p � 1

implies 2 j (ap � 1)(ap+ 1). Since ap � 1 and ap + 1 have the same parity,

they must both be even, and hence 4 j (ap�1)(ap+1). Since gcd(4; p2) = 1,

4p2 j a2p � 1 follows.

Remarks. (1) The fact that all prime numbers satisfy property P is well

known. In fact, using exactly the same argument as in the proof of (a) above,

one can show easily that if ap � bp mod p, then ap � bp mod p2. (See e.g.

Ex. 13 on page 190 of Elementary Number Theory and its Applications by

Kenneth H. Rosen, 3rd Edition).

(2) n = 4 also has property P . For suppose that 4 j a4 � 1. Then

4 j (a2 � 1)(a2 + 1), which implies that a is odd. Since any odd square is

congruent to 1modulo 8, we have 8 j a2�1, which, together with 2 j a2+1,

yields 16 j (a2 � 1)(a2 + 1).

(3) In view of (2) and our proof above, n = 8 is the �rst natural number

which need not possess property P . Indeed, it does not since 38� 1 = 6560

is divisible by 8 but not by 64.

(4) It might be of interest to characterize all natural numbers with prop-

erty P .

We �nish this number of the Corner with a solution to one of the IMO

problems from the 35th IMO in Hong Kong.

2. [1994: 244]

ABC is an isosceles triangle with AB = AC. Suppose that

(i) M is the midoint of BC and D is the point on the line AM such that

OB is perpendicular to AB;

(ii) Q is an arbitrary point on the segment BC di�erent from B and C;

(iii) E lies on the line AB and F on the line AC such that E, Q and F are

distinct and collinear.

Prove that OQ is perpendicular to EF if and only if QE = QF .

Solution by D. J. Smeenk, Zaltbommel, the Netherlands.

(=)). See Figure 1. Assume OQ ? EF . We want to showQE = QF .

Note that quadrilateral OBEQ is inscribed on the circle with diameter

OE. Thus \OEQ = \OBQ = \OQB = �2. Also OFCQ is cyclic and thus

\OFQ = \OCQ = \OAC = �

2. Together these give \OEQ = \OFQ

and OQ ? EF , so QE = QF .

68

�=2 �=2

�=2

�=2

�=2�=2

r

r

r rr r

r

r

A

B C

O

E

M

F

Q

Figure 1.

((=). See Figure 2. Assume that QE = QF . We want to show that

OQ ? EF .

Let D lie on AC with QD parallel to AB. As QE = QF we see that

D is the midpoint of AF . As AB = AC we have DQ = DC. (1)

Also,

QE = 2DQ = 2CD (2)

BE = AB � AE: (3)

From (2) and (3)

BE = AC � 2CD = (AC � CD)� CD

= AD � CD = DF � CD:

Thus BE = CF . (4)

Draw OB, OE, OC, and OF .

69

r

r

r rr r

r

r

A

B C

O

E

M

F

Q

Dr

Figure 2.

Then

�OBE �= �COF [BE = CF; OB = OC; \OBE = \OCF = n=2]:

(5)

Now this implies that OE = OF , QE = QF giving OQ ? EF .

Remark. This direction could be shortened by using the law of sines in tri-

angles BQF and CQF , but the given solution is, I think, more elementary,

and therefore more elegant.

That completes the material we have available for this number. The

Olympiad Season is fast approaching. Please collect your contests and send

them to me. Also send me your nice solutions to problems posed in the

Corner.

THE ACADEMY CORNERNo. 2

This will appear in a future issue.

70

IMO95 PUZZLES

In the September 1995 issue of CRUX, two puzzles were given about the

logo for the 1995 IMO. The second puzzle, attributed there to Mike Dawes

of the University of Western Ontario, asked:

Suppose that you take a physical model of this logo, and ma-

nipulate it to turn the in�nity sign into a circle. What does the

circle turn into?

It turns out that the same question occurred to several other people at the

same time, including some members of the Netherlands team at IMO 95.

Less than half an hour prior to the start of the competition, the

1995 IMO logo inspired one of the contestants (RvL) to pose the

following problem:

Consider the logo as a knot composed of a blue string

forming the in�nity-sign and a red string forming the

zero. Can the knot be rearranged so that the blue string

forms the zero and the red string forms the in�nity-sign

(without using scissors)?

The contestant solved the problem shortly after the competi-

tion and, upon returning home, wrote a program together with

one of the other members of his team (DG).

Ronald van Luijk (rmluijk@cs.ruu.nl) and Dion Gijswijt.

The program was sent by the Leader of the Netherlands team (Johannes

Notenboom) to the Editor-in-Chief (in his role as Chief Operating O�cer of

the 1995 IMO). We were so impressed by the program that we have had it

made available on the IMO95 WWW site. You may download this program

(zipped) by going to: http://camel.math.ca/IMO/IMO95/.

Ronald has pointed out the the IMO 95 logo is, in fact, the Whitehead

Link. See, for example, L.H. Kau�man's book: On Knots, Annals of Mathe-

matics Studies, 115, Princeton University Press, 1987, p. 14.

71

BOOK REVIEWS

Edited by ANDY LIU

Assessing Calculus Reform E�orts : A Report to the Community, edited

by James Leitzel and Alan C. Tucker, 1995. Paperback, 100+ pages, US$18.00,

ISBN 0-88385-093-1.

Preparing for a New Calculus, edited by Anita Solow, 1994. Paperback,

250+ pages, US$24.00, ISBN 0-88385-092-3.

Both published by The Mathematical Association of America, Washington,

DC 20036. Reviewed by Jack Macki, University of Alberta.

Anyone who attended the January 1995 joint meetings in San Francisco

could not help but be struck by the contrast between the evident malaise on

the research side of the meeting and the enthusiasm, energy and high quality

of the discourse on the educational side.

These two books, very di�erent in nature and content, together rep-

resent enormously persuasive documentation of the on-going revolution in

teaching and curriculumdesign at the University/College level. They describe

the past and present of reform in undergraduate calculus, and evaluate the

situation in a thorough and realistic manner.

Assessing Calculus Reform E�orts is one of the best presentations ever

of the history, present activities and future plans of the movement for reform

in the design and teaching of the undergraduate mathematics curriculum.

The history begins with the founding of the Committee on the Under-

graduate Program (CUP) in 1953. The authors describe the debate between

Anthony Ralston and Ron Douglas which led to the famous Tulane workshop

in 1986 (with 25 participants, four of whom were research mathematicians).

They follow the entire development of the reform movement, carefully de-

scribing the key players, and the roles of NRC and NSF. They present all

kinds of surprising (at least to this reviewer) information along the way |

did you know that in 1985 the IEEE, deeply dissatis�ed with existing texts,

published a calculus book? In the nine years since that seminal Tulane con-

ference, reform has clearly moved into a central position in the mathematical

life of North America.

The e�ort of key individuals to convert the research community to be-

lievers (and activists) makes fascinating reading, beginningwith a widely dis-

seminated article by Peter Lax (UME Trends, May, 1990) which was highly

critical of the attitudes of the majority of research mathematicians. Very

determined department chairs at Stony Brook and Michigan convinced their

departments to adopt reform texts. Said Don Lewis, chair at Michigan, \An

NSF grant providing two months of summer salary is icing on the cake. The

cake is calculus."

The editors are very tactful in their presentation, but the lack of interest

in education of the majority of research mathematicians in the United States

72

(and Canada?) comes through quite clearly. They present the results of four

surveys carried out to date, and discuss the issue of \cosmetic" versus \real"

change. The bottom line:

� By Spring 1994, three-quarters of responding departments (1048 re-

spondents) had some reform underway, one-quarter of the respondents

were conducting major reforms.

� The Graduate Record Exam has been changed to re ect reform.

� The NSF continues to fund reform initiatives at the level of about $2.7

million per year.

� Sales of reform texts have increased to 108,700 annually (552 institu-

tions) in Fall, 1994 (not counting the 375 high schools which use them).

The main part of this book has only 44 pages of text, and then con-

cludes with 50 pages of important and highly readable appendices: data on

enrolments, copies of surveys, text-by-text description of reform materials

presently available, and a very detailed list of NSF-funded projects.

If you have colleagues who pooh-pooh the reform movement (and who

doesn't?), get them started on the path to rightness with this book. If they

can read it and remain unwilling to look at change, order them co�ns, they

probably died a long time ago!

Preparing for a New Calculus is the proceedings of a conference held

at the University of Illinois in April, 1993. Eighty individuals attended, 40

of them associated with projects at colleges or universities, 25 of them asso-

ciated with projects at the high school level, seven from community college

projects, and eight from organizations like the MAA, NSF, NCTM, etc., and

publishers.

Part I contains seven background papers, which were provided to par-

ticipants before the conference. The �rst paper cites some impressive �gures:

well over 100 schools use the Harvard Consortium Project materials; 27% of

all college-level calculus students in the State of Washington are in a reform

course. All seven papers o�er several important observations based on

experience, for example:

� There is no correct way to do a reform course, there are many ways to

treat calculus e�ectively.

� Reform courses have fewer topics with much richer content.

� The \rule of three" (graphical, numerical, analytical) should be replaced

by the \rule of four" (writing) or \�ve" (oral).

� It is mindless to ask a student who has access to a calculator to approxi-

matep17 using a di�erential, but one can instead study Euler's method

for solving the logistic di�erential equation before the students know

any integration theory.

73

� Reform makes appropriate use of technology; technology does not equal

reform.

� While most mathematics professors believe that learning is transmitted

(I call this the \I am God" theory of learning), most evidence points to

the correctness of the constructive theory of learning, which emphasizes

that most learning is constructed by the learner in response to chal-

lenges to re�ne or revise what he/she already knows in order to cope

with new situations. Sally Berenson, North Carolina State University:

\Telling is not teaching, listening is not learning".

� Never underestimate the importance and di�culty of educating faculty

for real change. Teaching reform courses is very hard work.

� Creating a cooperative learning environment is not easy, but pays o�

for all students, especially minorities.

� It is very useful to carry open-ended long-term applied problems

through one or, even better, several courses. At the U.S. Military

Academy at West Point, they begin with a few core applied problems

and proceed to attack them with non-calculus methods, leading to a

semester's study of di�erence equations and the associated linear al-

gebra. Calculus starts in term two! K. Stroyan of the University of

Iowa starts his course by asking \Why is it we can eradicate polio and

smallpox, but not measles and rubella?"

You get the picture, there is just a tremendous amount of experience

and insight available in these seven articles | and wonderful quotes: The

chairman of physics at Duke, Larry Evans, when asked to comment on changes

in the teaching of calculus:

\There is nowhere to go but up."

Other important themes: downgrading of exams, strong emphasis on

(and evaluation of) writing, emphasis on cooperative learning. And don't ex-

pect all students to love the changes | students who are successful in stan-

dard courses often react negatively to being saddled with lab partners and

being evaluated in unfamiliar ways. In fact, several authors emphasize that

student questionnaires cannot be the sole means of teaching evaluation |

one needs to talk to students who have gone down the road a way in order

to get a fair picture.

The second part of Preparing for a New Calculus reports on the work-

shops, one each on the topics of content, teaching strategies, and institutional

context. Each report has a long and useful list of suggestions and observa-

tions. Among them:

� Characterize and reward that which constitutes e�ective teaching.

74

� Allocate time as a resource to support involved faculty.

� Every person involved with evaluating sta� should read Ernest Boyer's

Carnegie Foundation Report \Scholarship Reconsidered".

The third section consists of �ve contributed papers of very high quality

on topics ranging from calculator courses, to the gateway exam at Michigan.

The �nal piece is a thoughtful article by Peter Renz of Academic Press on

Publishers, Innovation and Technology, which should be required reading

for all university mathematics professors. Many of the articles in this book

have extensive lists of very timely and useful references.

All in all, Preparing for a New Calculus is an excellent detailed introduc-

tion to calculus reform, to be read after Assessing Calculus Reform E�orts.

Anyone who can read them both and not be interested in and excited about

reform doesn't need a co�n, they've been dead too long!

Introducing the new Associate Editor-in-Chief

For those of you who do not know Colin, here is a short pro�le:

Born: Bedford, England 1

Educated Bedford Modern School

University of Birmingham, England

University of London, England

Employment Queen's College, Nassau, Bahamas

Sir John Talbot's Grammar School,

Whitchurch, England

Memorial University of Newfoundland, Canada

Mathematical Mathematical Education

Interests History and Philosophy of Science

Nineteenth century Astrophysics

1 Bedford is halfway between Oxford and Cambridge (just an average place).

75

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 812"�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 that 1 October 1996. They may also be

sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email

proposals and solutionswere written in LATEX, preferably in LATEX2e). Graph-

ics �les should be in epic format, or plain postscript. Solutions received after

the above date will also be considered if there is su�cient time before the

date of publication.

2114. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABCD is a square with incircle �. A tangent ` to � meets the sides

AB and AD and the diagonal AC at P , Q and R respectively. Prove that

AP

PB+AR

RC+AQ

QD= 1:

2115. Proposed by Toby Gee, student, The John of Gaunt School,

Trowbridge, England.

Find all polynomials f such that f(p) is a prime for every prime p.

2116. Proposedby Yang Kechang, Yueyang University, Hunan, China.

A triangle has sides a; b; c and area F . Prove that

a3b4c5 � 25p5(2F )6

27:

When does equality hold?

76

2117. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle withAB > AC, and the bisector of\AmeetsBC at

D. Let P be an interior point of the sideAC. Prove that \BPD < \DPC.

2118. Proposed by Paul Yiu, Florida Atlantic University, Boca Raton,Florida, USA.

The primitive Pythagorean triangle with sides 2547 and 40004 and hy-

potenuse 40085 has area 50945094, which is an 8-digit number of the form

abcdabcd. Find another primitive Pythagorean triangle whose area is of this

form.

2119. Proposed by Hoe Teck Wee, student, Hwa Chong Junior Col-

lege, Singapore.

(a) Show that for any positive integer m � 3, there is a permutation of m

1's,m 2's andm 3's such that

(i) no block of consecutive terms of the permutation (other than the

entire permutation) contains equal numbers of 1's, 2's and 3's; and

(ii) there is no block ofm consecutive terms of the permutation which

are all equal.

(b) For m = 3, how many such permutations are there?

2120. Proposed by Marcin E. Kuczma, Warszawa, Poland.

LetA1A3A5 andA2A4A6 be nondegenerate triangles in the plane. For

i = 1; : : : ; 6 let `i be the perpendicular from Ai to line Ai�1Ai+1 (where of

course A0 = A6 and A7 = A1). If `1; `3; `5 concur, prove that `2; `4; `6 also

concur.

2121. Proposed by Krzysztof Chelmi �nski, Technische Hochschule

Darmstadt, Germany; and Waldemar Pompe, student, University of War-

saw, Poland.

Let k � 2 be an integer. The sequence (xn) is de�ned by x0 = x1 = 1

and

xn+1 =xkn + 1

xn�1for n � 1:

(a) Prove that for each positive integer k � 2 the sequence (xn) is a se-

quence of integers.

(b) If k = 2, show that xn+1 = 3xn � xn�1 for n � 1.

(c)� Note that for k = 2, part (a) follows immediately from (b). Is there an

analogous recurrence relation to the one in (b), not necessarily linear,

which would give an immediate proof of (a) for k � 3?

77

2122. Proposed by Shawn Godin, St. Joseph Scollard Hall, North

Bay, Ontario.

Little Sam is a unique child and his math marks show it. On four tests

this year his scores out of 100 were all two-digit numbers made up of eight

di�erent non-zero digits. What's more, the average of these scores is the

same as the average if each score is reversed (so 94 becomes 49, for example),

and this average is an integer none of whose digits is equal to any of the digits

in the scores. What is Sam's average?

2123. Proposed by Sydney Bulman{Fleming and Edward T. H. Wang,

Wilfrid Laurier University, Waterloo, Ontario.

It is known (e.g., exercise 23, page 78 of Kenneth H. Rosen's Elemen-

tary Number Theory and its Applications, Third Edition) that every natu-

ral number greater than 6 is the sum of two relatively prime integers, each

greater than 1. Find all natural numbers which can be expressed as the sum

of three pairwise relatively prime integers, each greater than 1.

2124. Proposed by Catherine Shevlin, Wallsend, England.

Suppose that ABCD is a quadrilateral with \CDB = \CBD = 50�

and \CAB = \ABD = \BCD. Prove that AD ? BC.

A B

C

D

Mathematical Literacy

1. Who said: \To be able to read the great book of the universe, one must

�rst understand its language, which is that of mathematics".

2. In referring to \the unreasonable e�ectiveness of mathematics in the

natural sciences", who wrote: \The miracle of the appropriateness of

the language of mathematics for the formulation of the laws of physics

is a wonderful gift which we neither understand nor deserve".

78

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

1827. [1993: 78; 1994:57] Proposed by �Sefket Arslanagi�c, Berlin,

Germany, and D.M. Milo�sevi�c, Pranjani, Yugoslavia.

In commenting on the solutions submitted, the editor asked for a short

proof of X bc

s� a= s+

(4R+ r)2

s:

Two very nice and very di�erent solutions have been received.

I. Solution by TOSHIO SEIMIYA, Kawasaki, Japan.

Since tan(A=2) = r1=s, and using a result in R.A. Johnson, Advanced

Euclidean Geometry, p. 189, we obtain

Xtan(A=2) =

Xr1=s = (4R+ r)=s:

Since A + B + C = �, we then haveX

tan(B=2)tan(C=2) = 1, which

leads to �Xtan(A=2)

�2= �1 +

Xsec2(A=2):

Together, we now haveX

sec2(A=2) = 1 +

�4R+ r

s

�2.

Sincebc

s� a= s

�bc

s(s� a)

�= s sec2(C=2), we have

X bc

s� a= s

�Xsec2(A=2)

�= s+

(4R+ r)2

s:

II. Solution by Waldemar Pompe, student, University of Warsaw, Pol-

and.

SinceX

a = 2s,Pbc = s2 + r2 + 4Rr, and abc = 4sRr, we get:

X(s� b)(s� c) = 3s2 � 4s2 +

Xbc = r2 + 4Rr:

Using this, we obtain

X 1

s� a=

r2 + 4Rr

(s� a)(s� b)(s� c)=r2 + 4Rr

sr2=r+ 4R

sr:

79

Therefore

X bc

s� a= abc

X 1

a(s� a)= 4Rr

X�1

a+

1

s� a

�

= 4Rr

�s2 + r2 + 4Rr

4sRr+r + 4R

sr

�

= s+r2 + 4Rr

s+

4Rr + 16R2

s= s+

(4R+ r)2

s:

2000. [1994: 286]Proposed byMarcin E. Kuczma, Warszawa, Poland.

A 1000{element set is randomly chosen from f1; 2; : : : ; 2000g. Let pbe the probability that the sum of the chosen numbers is divisible by 5. Is p

greater than, smaller than, or equal to 1=5?

Comment by Stan Wagon, Macalester College, St. Paul, Minnesota,

USA.

The answer is that

p =1

5+

4

5

�400

200

��2000

1000

� =1

5+ 410�482:

See Stan Wagon and Herbert S. Wilf, When are subset sums equidistributed

modulom?, Electronic Journal of Combinatorics 1 (1994).

In particular, we obtained a necessary and su�cient condition that the t{

subsets of [n] be equidistributed mod m. That condition is:

t � > n � mod d for all d dividing m (except d = 1), where �refers to the least non-negative residue.

In the case at hand, m = 5, so only d = 5 need be considered, and then

t� = n� = 0, so equidistribution fails. While this does not directly answer

problem 2000 (since it gives no information about the speci�c remainder 0

mod 5), the paper does discussmany intriguingopen questions related to the

mod m distribution of subset sums. Thus it strikes me that, because some

of your readers were successful at generalizing the problem as stated, they

would be interested in this reference. Indeed, perhaps some characterization

of the quadruples (i; t; n;m), such that the set of t{subsets of [n] whose

mod m sum is i has less than average size, is possible.

The problem arose from lottery considerations. When are the tickets in a

lottery equidistributed with respect to the mod m value of their sums?

80

2011. [1995: 52] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle with incentre I. BI and CI meet AC and AB at

D and E respectively. P is the foot of the perpendicular from I to DE, and

IP meets BC at Q. Suppose that IQ = 2IP . Find angle A.

Solution by the proposer.

Let X and Y be the feet of perpendiculars from I to BC and AB, then

IX = IY = r; where r is the inradius of4ABC.We put \ABI = \IBC = �, \ACI = \ICB = , \EIB = �,

\IDE = x, and \IED = y. Then we have � + = � and x + y = �.

As \IQC = \IBQ + \BIQ = \IBQ + \DIP = � + �2� x, we get

r = IX = IQ sin\IQC = IQ sin(� + �2� x), so that

r = IQ cos(� � x) (1)

As \AEI = \EBI + \EIB = �+ �, we get

r = IY = IE sin\AEI = IE sin(�+ �): (2)

Because PI = IE siny; and IQ = 2PI, we have from (1)

r = 2PI cos(� � x) = 2IE siny cos(� � x): (3)

From (2) and (3) we have (cancelling IE),

sin(�+ �) = 2 siny cos(� � x) = sin(� + y � x) + sin(x+ y� �)

= sin(� + y � x) + sin(�� �):

Therefore sin(� + y � x) = sin(�+ �) � sin(�� �) = 2 cos� sin�; thus

we obtain

sin� cos(y� x) + cos� sin(y� x) = 2 cos� sin�: (4)

Similarly we have (exchanging � for , and x for y, y for x, simultaneously)

sin cos(x� y) + cos sin(x� y) = 2 cos� sin , or

sin cos(y� x)� cos sin(y � x) = 2 cos� sin : (5)

Multiplying (4) by sin and (5) by sin�, we get

(sin cos� + cos sin�) sin(y� x) = 0

or sin(� + ) sin(y� x) = 0, that is sin� sin(y� x) = 0:

Since sin� > 0, we have sin(y � x) = 0, therefore y = x, and

cos(y � x) = 1: Hence we get from (4) sin� = 2 cos� sin�: Because

sin� > 0 we get cos� = 12: Thus we have � = 60�. As � = 90�� 1

2\A; we

obtain \A = 60�:

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UKand KEE-WAI LAU, Hong Kong. There was one incorrect solution.

81

2016. [1995: 53] Proposed by N. Kildonan, Winnipeg, Manitoba.

Recall that 0:19 stands for the repeating decimal 0:19191919 : : : , for

example, and that the period of a repeating decimal is the number of digits

in the repeating part. What is the period of

(a) 0:19 + 0:199 ; (b) 0:19� 0:199 ?

Solution by Christopher J. Bradley, Clifton College, Bristol, UK

(a) We have

0:19 + 0:199 =19

99+

199

999=

19� 10101 + 199� 1001

999999=

391118

999999

= 0:391118;

so the period is 6.

(b) Since

0:19� 0:199 =19� 199

99� 999;

we look for a number with all nines which is a multiple of 999�99. In order

to be a multiple of 999 it must have 3k nines for some integer k. When this

number is divided by 999, the quotient has k ones interspersed with pairs

of zeros: 1001001 : : : 1001. In order that this quotient be divisible by 99

as well, k must be divisible by 9 and must be even (using the well known

rules for divisibility by 9 and 11). Hence the required number will have

3 � 9 � 2 = 54 nines, so the period is 54. That the number has the full

period of 54 results from the fact that 19 and 199 are both coprime to 99 and

999 ensuring no fortuitous cancellations.

Also solved by HAYO AHLBURG, Benidorm, Spain; �SEFKET

ARSLANAGI �C, Berlin, Germany; NIELS BEJLEGAARD, Stavanger, Norway;

FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, andMARIA ASCENSI �ON

L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CARL BOSLEY,

student, Washburn Rural High School, Topeka, Kansas, USA; MIGUEL

ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; ADRIAN CHAN, student, Upper

Canada College, Toronto, Ontario; TIM CROSS, Wolverley High School, Kid-

derminster, UK; KEITH EKBLAW, Walla Walla, Washington, USA; JEFFREY

K. FLOYD, Newnan, Georgia, USA; RICHARD I. HESS, Rancho Palos Verdes,

California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-

tria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,USA;

DAVID E. MANES, State University of New York, Oneonta, NY, USA; J. A.

MCCALLUM, Medicine Hat, Alberta; P. PENNING, Delft, the Netherlands;

GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; HEINZ{J �URGEN

SEIFFERT, Berlin, Germany; DAVID R. STONE, Georgia SouthernUniversity,

Statesboro; PANOS E. TSAOUSSOGLOU, Athens, Greece; EDWARD T. H.

WANG, Wilfrid Laurier University, Waterloo, Ontario; CHRIS WILDHAGEN,

Rotterdam, the Netherlands; and the proposer.

82

Part (a) only was solved by CHARLES ASHBACHER, Cedar Rapids,

Iowa, USA; TOBY GEE, student, The John of Gaunt School, Trowbridge, Eng-

land; and JOHN S. VLACHAKIS, Athens, Greece. Part (b) only was solved by

KEE-WAI LAU, Hong Kong.

Kone�cn �y and Perz had solutionswhich were comparable to Bradley's in

their simplicity and minimal dependence on calculation.

Cross notes that the product in (b) equals

3781

98901=

3780

98901+

1

98901;

where the �rst fraction is 140=3663 = 0:038220, and the second fraction is

the \rather interesting" recurring decimal

0:000010 111121 222232 333343 444454 555565 666676 777787 888899 !

Bellot and L �opez mention the following general result of J. E. Oliver

which is quoted (for an arbitrary number of fractions) on page 166 of Dick-

son's History of the Theory of Numbers, Volume 1: if x0=x and z0=z are

periodic fractions, with periods a and b respectively, then (x0=x)(z0=z) hasa period of

xz

[x; z]� [a; b];

where [a; b] is the least commonmultiple of a and b. This formula, which can

be more simply written as (x; z)[a; b] where (x; z) is the greatest common

divisor of x and z, gives precisely 54 in our case. However, perhaps \pe-

riod" here was intended to mean any period rather than just the smallest pe-

riod, because the formula fails in many cases. For example, for the fractions

11=3 and 1=11, which have periods 1 and 2 respectively, the formula gives

(3; 11)[1; 2] = 1 � 2 = 2 as the period for the product (11=3)(1=11) = 1=3,

which has period only 1 of course. Unfortunately, the reference given for

Oliver's result is page 295 of Math. Monthly, Vol. 1, 1859; this cannot be

the familiar American Math.Monthly, whichonly started publishing in 1894.

Can any reader supply us with more information?

2017. [1995: 53] Proposed by D. J. Smeenk, Zaltbommel, the Neth-

erlands.

We are given a �xed circle � and two �xed points A and B not lying

on �. A variable circle through A and B intersects � in C and D. Show that

the ratioAC � ADBC � BD

is constant. [This is not a new problem. A reference will be given when the

solution is published.]

83

Solution by Toshio Seimiya, Kawasaki, Japan.

Let be a �xed circle passing through A;B and intersecting � at X

and Y . Then XY , CD, and AB are all parallel or concurrent at the radical

centre of the three circles.

Case 1. XY kAB:If XY kAB then CDkAB, so AC = BD and AD = BC: Hence

AC � ADBC � BD = 1 (which is a constant).

Case 2. XY 6 k AB.

Let P be the intersection of XY with AB, then P is a �xed point (the

radical centre) and CD passes through P .

Since either \CAD = \CBD, or \CAD+ \CBD = 180�, we get

[ACD]

[BCD]=

AC � ADBC � BD

; (1)

where [UVW ] denotes the area of triangle UVW .

Let A0, B0 be the feet of perpendiculars from A, B to CD, then

AA0kBB0; and

AA0

BB0=AP

BP: (2)

Because[ACD]

[BCD]=

AA0

BB0, we have from (1) and (2)

AC � ADBC � BD

=AP

BP= constant.

The proposer tells us that the problem comes from a 1939 book by

Dr. P. Molenbroek.

(Note that \CAD and \DBC might be supplementary rather than

equal, so sin\CAD = sin\DBC still holds, but the triangles CAD and

DBC are not necessarily similar.)

Also solved by: CLAUDIO ARCONCHER, Jundia��, Brazil;

CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; DAN PEDOE,Minneapo-

lis,Minnesota, USA; P. PENNING, Delft, theNetherlands; GOTTFRIED PERZ,

Pestalozzigymnasium, Graz, Austria; there were four incorrect solutions.

84

2018. [1995: 53] Proposed byMarcin E. Kuczma, Warszawa, Poland.

Howmany permutations (x1; : : : ; xn) of f1; : : : ; 2g are there such thatthe cyclic sum

Pn

i=1 jx1 � xi+1j (with xn+1 = x1) is (a) a minimum, (b)

a maximum?

Solutionby Carl Bosley, student,Washburn Rural High School, Topeka,

Kansas, USA.

(a) Let j and k be such that xj = 1 and kk = n. For all m, de�ne xn+m =

xm. Then we have

nXi=1

jxi � xi+1j =

k�1Xi=j

jxi � xi+1j +j�1Xi=k

jxi � xi+1j

� jxk � xj j+ jxj � xkj by the triangle inequality

= 2n� 2:

Equality holds if and only if both sequences

fxk = n; xk+1; : : : ; xj�1; xj = 1gfxk; xk�1; : : : ; xj+1; xj = 1g

are monotonic decreasing. Each element of fn � 1; n � 2; : : : ; 3; 2gmay be in either the �rst or second sequence, but not both. Choosing which

elements are in these sequences determines the positions of all of fn � 1;

n�2; : : : ; 2; 1g uniquely. Thus there are n2n�2 permutations with minimal

sum 2n� 2.

(b) Consider the e�ect of a single element xi on the sum. If xi > xi+1and xi > xi�1, then the two terms in the sum involving xi are (xi � xi+1)

and (xi � xi�1). Thus the term xi contributes 2xi to the sum. Similarly, if

xi < xi+1 and xi < xi�1, then the term xi contributes �2xi to the sum.

Further, if xi is less than one of xi�1, xi+1, and greater than the other, then

xi has no contribution to the sum.

Suppose that j elements have no contribution to the sum. There are n

pairs of elements xi, xi+1. In each pair, one number is greater and one is

less than the other. Thus there are n�j2

numbers xi which contribute 2xi to

the sum, and n�j2

numbers xi which contribute �2xi to the sum. If we can

arrange the numbers xi which contribute 2xi to the sum to be the largestn�j2

numbers, and the numbers xi which contribute �2xi to the sum to be

the smallest n�j2

numbers, then a maximum value is clearly attained with j

as small as possible.

This is indeed possible. If n is even (n = 2k) and j = 0, we must have

that fxi; xi+2; : : : ; xi�4; xi�2g is a permutation of fk+1; k+2; : : : ; 2kgfor i = 0 or i = 1, and the other k numbers must be a permutation of

f1; 2; : : : ; kg. Otherwise, for some xi, we would have xi < xi+1 < xi+2which is a contradiction. This gives a same maximal sum of 2(k+ 1 + k +

2 + : : :+ 2k)� 2(1 + 2 + : : :+ k) = 2(k2) = n2=2, for any permutation

85

of fk + 1; k + 2 : : : ; 2kg and f1; 2; : : : ; kg. So there are 2(k!)2 possible

permutations with maximal sum if n = 2k.

If n is odd (n = 2k + 1), then j must be odd, so j is at least 1.

Placing the middle element k+1 in one of the 2k+1 possible positions gives

permutations of f1; 2; : : : ; kg and fk+2; k+3; : : : ; 2k+1g in alternatingpositions. This gives the same maximal sum, (n2 � 1)=2, for every such

permutation. Hence there are 2(2k + 1)(k!)2 possible permutations with

maximal sum of n = 2k + 1.

Also solved by NEILS BEJLEGAARD, Stavanger, Norway; TOBY GEE,

student, The John of Gaunt School, Trowbridge, England, part (a) only;

P. PENNING, Delft, the Netherlands; and the proposer.

2019. [1995: 53] Proposed by P. Penning, Delft, the Netherlands.

In a plane are given a circle C with diameter `, and a point P within C

but not on `. Construct the equilateral triangles that have one vertex at P ,

one on C, and one on `.

Solutionby Carl Bosley, student,Washburn Rural High School, Topeka,

Kansas, USA.

Assume that we have a solution triangle, and consider the e�ect of ro-

tating the circleC by 60� about P . We can do this in two directions, clockwise

or counterclockwise. The vertex of the equilateral triangle on C is rotated

to the vertex on the original diameter. Since each rotated circle intersects

the diameter only once (or twice, if ` is to be interpreted as the line con-

taining the given diameter), this vertex is now known and we can easily �nd

the vertex on C of the equilateral triangle. We see that there are two such

equilateral triangles (four if the extension of the diameter is allowed). To

construct the pair of rotated circles, �nd their centres as the third vertices

of the equilateral triangles that have PO as base, where O is the centre C;

their radius is half the length of the given diameter.

Also solved by CLAUDIO ARCONCHER, Jundia��, Brazil; NIELS

BEJLEGAARD, Stavanger, Norway; CHRISTOPHER J. BRADLEY, CliftonCol-

lege, Bristol, UK; PETER HURTHIG, Columbia College, Burnaby, BC; D. J.

SMEENK, Zaltbommel, the Netherlands; and the proposer. One incorrect

solution was received.

Most of the solvers used an argument similar to the featured solution.

Bradley remarked that he has seen the rotation technique used to solve the

analogous problem in which a line parallel to ` is given instead of the cir-

cle C. Bejlegaard went even further, pointing out that the given triangle

could have any prescribed shape (instead of equilateral) and that C and `

could be any two curves for which the points of intersection of ` with the

rotated image of C could be constructed.

86

2020. [1995: 53] Proposed by Christopher J. Bradley, Clifton Col-

lege, Bristol, UK.

Let a, b, c, d be distinct real numbers such that

a

b+b

c+c

d+d

a= 4 and ac = bd:

Find the maximum value of

a

c+b

d+c

a+d

b:

Solutionby Carl Bosley, student,Washburn Rural High School, Topeka,

Kansas, USA.

Let x =a

band y =

b

c. Then by ac = bd we have

c

d=

b

a=

1

x,

d

a=c

b=

1

y,a

c= xy and

b

d=b

c� cd=y

x. Thus we are to �nd the maximum

value of xy+y

x+

1

xy+x

ysubject to the conditions that a, b, c, d are distinct

and

(�) x+ y +1

x+

1

y= 4:

Let e = x+1

xand f = y+

1

y. Then xy +

z

x+

1

xy+x

y= ef .

By the Arithmetic Mean-Geometric Mean Inequality, we have t + 1

t� 2 if

t > 0 and t+1

t� �2 if t < 0.

By (�), we have that x and y cannot both be negative. If both are positive,

then (�) implies x = y = 1 or a = b = c, a contradiction. Hence exactly

one of x and y is negative.

Assume, without loss of generality, that x > 0, y < 0. Then we get f ��2; e = 4 � f � 6 and so ef � �12. Equality holds, for example, when

a = 3 + 2p2; b = 1; c = �1 and d = �(3 + 2

p2).

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; NIELS

BEJLEGAARD, Stavanger, Norway; ADRIAN CHAN, student, Upper Canada

College, Toronto, Ontario; F.J. FLANIGAN, San Jose State University, San

Jose, California, USA; SHAWN GODIN, St. Joseph Scollard Hall, North Bay,

Ontario; PETER HURTHIG, Columbia College, Burnaby, BC; V �ACLAV

KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; KEE-WAI

LAU, Hong Kong; DAVID E. MANES, SUNY at Oneonta, Oneonta, New York,

USA; BEATRIZ MARGOLIS, Paris, France; P. PENNING, Delft, the Nether-

lands; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; JOHN VLACHAKIS,

Athens, Greece; and the proposer. Ten incorrect or incomplete solutions

were also received. (Is this a record?)

87

Many of them showed thata

c+b

d+c

a+d

b� 4 and then claimed erroneously

that 4 is actually the maximum value.

From Bosley's solutions above, it is not di�cult to show that the upper

bound�12 is attained if and only if (a; b; c; d) equals (k; (3�2p2)k;�(3�

2p2)k;�k) or (k;�k;�(3�2

p2)k; (3�2

p2)k) for some k 6= 0. This was

shown by Flanigan and Mane.

Arslanagi�c conjectured that the minimum value of the given sum is�27661900

!

2021. [1995: 89] Proposed by Toshio Seimiya, Kawasaki, Japan.

P is a variable interior point of a triangle ABC, and AP , BP ,

CP meet BC, CA, AB at D, E, F respectively. Find the locus of P so

that

[PAF ] + [PBD] + [PCE] =1

2[ABC];

where [XY Z] denotes the area of triangle XY Z.

I. Solution by P. Penning, Delft, the Netherlands.

First, note that [PAE] + [PFB] + [PDC] = 12[ABC].

A B

C

DE

F

P

Let AF=AB = 12� x, BD=BC = 1

2� y, CE=CA = 1

2� z.

From Ceva's Theorem, we get:�12� x

� �12� y

� �12� z

�=�12+ x

� �12+ y

� �12+ z

�or

x+ y + z = 4xyz: (1)

From the �gure, we can see that�12� x

�[ABC] = [PAF ]+[PAE]+[PEC] = [PAE]+1

2[ABC]�[PBD] :

Thus

x[ABC] = [PBD]� [PAE]: (2)

88

Similarly, we have

y[ABC] = [PCE]� [PFB]; (3)

z[ABC] = [PAF ]� [PDC]: (4)

Adding (2), (3), and (4) gives

(x+ y + z)[ABC] = 12[ABC]� 1

2[ABC];

so that x+ y+ z = 0. This, together with (1), gives xyz = 0, so that x = 0

or y = 0 or z = 0.

Thus the locus consists of the three medians of the triangle (excluding the

end points).

II. Solution by the Austin Academy Problem Solvers, Austin, Texas,

USA.

Assign masses a, b, c to the vertices A, B, C, respectively, so that P

is the centre of mass and a+ b+ c = 2.

Then[PAF ]

[ABC]=FP

CF� AFAB

=c

(a+ b+ c)� b

(a+ b).

There are similar expression for PBD and PCE.

So we need to �nd the locus of P such that

bc

(a+ b)+

ac

(b+ c)+

ab

(a+ c)=

(a+ b+ c)

2( = 1):

Cross multiplying and simplifying leads to

(a� b)(b� c)(c� a)(a+ b+ c) = 0:

But a+ b+ c = 2, so we must have at least one of a = b, b = c and c = a.

Now, this is exactly when P lies on a median of triangle ABC (excluding the

end points).

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; CARL BOSLEY,

student,Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER

J. BRADLEY, CliftonCollege, Bristol, UK;MIGUEL AMENGUAL COVAS, Cala

Figuera, Mallora, Spain; PETER HURTHIG, Columbia College, Burnaby, BC;

KEE-WAI LAU, Hong Kong; ASHISH KR. SINGH, Student, Kanpur, India;

D. J. SMEENK, Zaltbommel, the Netherlands; HOE TECK WEE, student,

Hwa Chong Junior College, Singapore; and the proposer.

89

2022. [1995: 89] Proposed by K. R. S. Sastry, Dodballapur, India.

Find the smallest integer of the form

A ? B

B;

whereA andB are three-digit integers andA?B denotes the six-digit integer

formed by placing A and B side by side.

Solution by M. Parmenter, Memorial University of Newfoundland,

St. John's, Newfoundland.

I claim that the answer to the problem is 121.

Note that when A = 114 and B = 950, thenA ? B

B= 121, so this number

is actually obtained. We must therefore show that in any other case when

D =A ? B

Bis an integer, then D � 121.

Observe thatA ? B

B=

(1000A+ B)

B=

1000A

B+ 1, so we must now show

that whenever E =1000A

Bis an integer, then E � 120.

Let B = GX where G = gcd(B; 1000). We will show that wheneverA

Xis

an integer, then F =

�1000

G

��A

X

�� 120. Note that since B < 1000, we

have X <1000

G. Also, recall that we are only interested in cases when

A

Xis

an integer.

If G = 1, 2, 4, 5 or 8, then1000

G> 120, and we are done.

If G = 10, then X < 100, soA

X� 2 (since A � 100). In this case,

F � 200, and we are done.

IfG = 20, then X < 50 andA

X� 3. Thus F � 150, and we are done.

IfG = 25, then X < 40 andA

X� 3. Thus F � 120, and we are done.

IfG = 40, then X < 25 andA

X� 5. Thus F � 125, and we are done.

IfG = 50, then X < 20 andA

X� 6. Thus F � 120, and we are done.

If G = 100, then X < 10 andA

X� 12. (Note that x � 9). Thus

F � 120, and we are done.

90

If G = 125, then X < 8 andA

X� 15. Thus F � 120, and we are

done.

If G = 200, then X < 5 andA

X� 25. (Note that x � 4). Thus

F � 125, and we are done.

If G = 250, then X < 4 andA

X� 34. Thus F � 136, and we are

done.

If G = 500, then B = 500 and1000A

B= 2A � 200, and we are now

all done.

Note that I am assuming that A is an \honest" three-digit integer, that is

100 � A. Otherwise, there is a trivial solution: A = 001, B = 500, and

A ? B

B= 3.

Also solved by CARL BOSLEY, student, Washburn Rural High School,

Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK;MIGUELANGEL CABEZ �ONOCHOA, Logro ~no, Spain; JEFFREY R. FLOYD,

Newnan, Georgia, USA; TOBY GEE, student, the John of Gaunt School, Trow-

bridge, England; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, On-

tario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, Hong

Kong; J. A. MCCALLUM, Medicine Hat, Alberta; STEWART METCHETTE,

Culver City, California, USA; P. PENNING, Delft, the Netherlands; the SCI-

ENCE ACADEMY PROBLEM SOLVERS; Austin, Texas, USA; DAVID STONE

and BILL MEISEL, Georgia Southern University, Statesboro, Georgia, USA;

PANOS E. TSAOUSSOGLOU, Athens, Greece; HOE TECK WEE, student, Hwa

Chong Junior College, Singapore; CHRISWILDHAGEN, Rotterdam, theNeth-

erlands; and the proposer. Some solvers made use of computers to search

for all possible solutions. One incorrect solution was submitted. Janous

suggested an extension to:

Let � 2 N. Determine the integer-minimum, �n of�A

B, where A

and B are n-digit numbers.

91

2023. [1995: 89]Proposed byWaldemar Pompe, student, University

of Warsaw, Poland.

Let a, b, c, d, e be positive numbers with abcde = 1.

(a) Prove that

a+ abc

1 + ab+ abcd+

b+ bdc

1 + bc+ bcde+

c+ cde

1 + cd+ cdea

+d+ dea

1 + de+ deab+

e+ eab

1 + ea+ eabc� 10

3:

(b) Find a generalization!

Solution to (a) by Carl Bosley, student, Washburn Rural High School,

Topeka, Kansas, USA.

Let x1 = a, x2 = ab, x3 = abc, x4 = abcd and x5 = abcde = 1.

Multiply the second, the third, the fourth and the �fth fractions on the left bya

a,ab

ab,abc

abcand

abcd

abcdrespectively. Then the expression on the left becomes:

x1 + x3

x5 + x2 + x4+

x2 + x4

x1 + x3 + x5+

x3 + x5

x2 + x4 + x1

+x4 + x1

x3 + x5 + x2+

x5 + x2

x4 + x1 + x3:

Add 1 to each of the �ve fractions. Then the desired inequality is equivalent

to: 5Xi=1

xi

!�1

x5 + x2 + x4+

1

x1 + x3 + x5+

1

x2 + x4 + x1

+1

x3 + x5 + x2+

1

x4 + x1 + x3

�� 25

3;

which follows by applying the Arithmetic Mean { Harmonic Mean inequality

to the second factor on the left. Equality holds if and only if the �ve denom-

inators are all equal, which happens if and only if x1 = x2 = x3 = x4 = x5,

which is true if and only if a = b = c = d = e = 1.

Solution to (b) by Hoe Teck Wee, student, Hwa Chong Junior College,

Singapore.

Suppose that a1; a2; : : : ; an are positive real numbers such that

a1a2 : : : an = 1, where n > 1. For each i = 1; 2; : : : ; n, de�ne ai+n = ai,

and for i; j = 1; 2; : : : ; n, let Ai;j =

i+j�1Yr=i

ar. Further, for each i =

1; 2; : : : ; n, de�ne Ai;j+n = Ai;j.

Let I be any non-empty subset of S = f0; 1; 2; : : : ; n � 1g. Then the

92

generalized inequality is:

nXi=1

0BB@

Xj2S�I

Ai;j

Xk2I

Ai;k

1CCA � n (n� jIj)

jIj :

The proof is as follows: let u =

n�1Xr=0

A1;r. Then

Xj2I

Ai;j

n�1Xk=0

Ai;k

=

Xj2I

A1;i�1Ai;j

n�1Xk=0

A1;i�1Ai;k

=

Xj2I

A1;j+i�1

n�1Xk=0

A1;k+i�1

=1

u

Xj2I

A1;j+i�1;

so that

nXi=1

0BBBBB@

Xj2I

Ai;j

n�1Xk=0

Ai;k

1CCCCCA =

nXi=1

1

u

Xj2I

A1;j+i�1 =1

u

Xj2I

nXi=1

A1;j+i�1

=1

u

Xj2I

u = jIj:

By the Arithmetic Mean { Harmonic Mean inequality, we have:

nXi=1

0BB@

Xj2S

Ai;j

Xk2I

Ai;k

1CCA � n2

nXi=1

Xj2I

Ai;j

Xk2S

Ai;k

=n2

jIj ;

and thus

nXi=1

0BB@

Xj2S�I

Ai;j

Xk2I

Ai;k

1CCA � n2

jIj� n =

n (n� jIj)jIj

:

Note that part (a) is the special case when n = 5 and I = f0; 2; 4g.

93

Besides Bosley and Wee, both parts were also solved by VEDULA N.

MURTY, Andhra University, Visakhapatnam, India; and the proposer.

Part (a) only was solved by SABIN CAUTIS, Earl Haig Secondary School,

North York, Ontario.

2024. [1995: 90] Proposed by Murray S. Klamkin, University of

Alberta, Edmonton, Alberta.

It is a known result that if P is any point on the circumcircle of a given

triangle ABC with orthocentre H, then (PA)2+(PB)2+(PC)2� (PH)2

is a constant. Generalize this result to an n-dimensional simplex.

Solution by Christopher J. Bradley, Clifton College, Bristol, UK.

Let O be the centre of a circumhypersphere, let R be the radius, and

let A1, A2, : : : ; An be the vertices. De�neH by the vector expression:

OH =

nXi=1

OAi :

Also, let P be any point on the surface of the circumhypersphere such that

jOP j2 = R2.

Then we claim that

nXi=1

(PAi)2 � (PH)2 = constant.

Let OP = ~x. Then

nXi=1

(PAi)2 � (PH)2 =

nXi=1

(~x� OAi) : (~x� OAi)

� ~x�

nXi=1

OAi

!:

~x�

nXi=1

OAi

!

= n j~xj2 � 2~x:

nXi=i

OAi +

nXi=1

OA2i

� j~xj2 + 2~x:

nXi=1

OAi �nXi=1

nXj=1

OAi:OAj

= (n� 1)R2 �nXi=1

nXj=1

OAi:OAj

= constant.

CARL BOSLEY, student, Washburn Rural High School, Topeka, Kansas,

USA, WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria, and the

proposer had essentially the same solution.

94

2027. [1995: 90] Proposed by D. J. Smeenk, Zaltbommel, the Neth-

erlands.

Quadrilateral ABCD is inscribed in a circle �, and has an incircle as

well. EF is a diameter of � with EF ? BD. BD intersects EF inM and

AC in S. Show that AS : SC = EM :MF .

Solution by Hoe Teck Wee, student, Hwa Chong Junior College, Singa-

pore.

Note: It is necessary that A and E lie on the same side of the lineBD,

otherwise the result is false.

Since ABCD has an incircle, we have AB + CD = BC + AD, or,

equivalently, BC � CD = AB � AD. Thus

EF

B

D

A

C

M

S

AS

AC=

Area�ABD

Area�BCD=

ABAD sin(\BAD)

BC CD sin(\BCD)

=ABAD

BC CD(since \BAD + \BCD = 180�)

=2ABAD

�(BC �CD)2 � BD2

�2BC CD ((AB � AD)2 � BD2)

=ABAD

�BC2 +CD2 � 2BCCD � BD2

�2BC CD (AB2 + AD2 � 2ABAD � BD2)

=

BC2 + CD2 � BD2 � 2BCCD

2BCCDAB2 + AD2 � BD2 � 2ABAD

2ABAD

=cos(\BCD)� 1

cos(\BAD)� 1;

using the cosine rule, applied to triangles ABD and BCD.

95

Since EF is a diameter, we have BE + DF = DE + BF , and so, by

an argument similar to the above, we get

EM

MF=

cos(\BFD)� 1

cos(\BED)� 1:

Since ABCD is a cyclic quadrilateral, we have \BED = \BAD and

\BFD = \BCD, and hence that AS : SC = EM : MF .

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI

LAU, Hong Kong; P. PENNING, Delft, the Netherlands; SCIENCE

ACADEMY PROBLEM SOLVERS, Austin, Texas, USA; TOSHIO SEIMIYA,

Kawasaki, Japan; ASHISH KR. SINGH, student, Kanpur, India; and the pro-

poser. These solvers assumed implicitly that A and E lay on the same side

of the line BD.

2028. [1995: 90] Proposed byMarcin E. Kuczma, Warszawa, Poland.

If n � m � k � 0 are integers such that n + m � k + 1 is a power

of 2, prove that the sum�n

k

�+�m

k

�is even.

Solution by Hoe Teck Wee, student, Hwa Chong Junior College, Singa-

pore.

For any non-negative integer t, let the binary representation of t be

(arar�1 � � � a0)2, where ai 2 f0; 1g, i = 0; 1; : : : ; r. Let f(t) be the largest

integer � such that 2� divides t. Let g(t) be the sum of the digits in the

binary representation of t. Then

f(t!) =

1Xj=1

�t

2j

�=

rXj=1

(arar�1 � � � aj)2

=

rXj=1

j�1Xs=0

2s =

rXj=1

aj(2j � 1) = t� g(t):

From this we get

f

��n

k

��= g(n� k) + g(k)� g(n):

Similarly, f��m

k

��= g(m�k)+ g(k)� g(m). Let n+m�k+1 = 2�, that

is, n+m� k = 2� � 1. Then, by considering the binary representations of

n, m� k, 2� � 1,m, and n� k, it is clear that

g(n) + g(m� k) = � = g(n� k) + g(m):

Thus g(n � k) + g(k) � g(n) = g(m � k) + g(k) � g(m), that is,

f��n

k

��= f

��m

k

��. Hence 2j

�n

k

�, 2j

�m

k

�, so

�n

k

�+�m

k

�is even.

96

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck,

Austria; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; SOFIYA VASINA, stu-

dent, University of Arizona, Tucson, USA; and the proposer. Kuczma notes

that the problem can be solved without any computation as follows:

Look at Pascal's triangle, rows 0 through 2�, modulo 2, and vi-

sualize it geometrically as an equilateral triangle with black and

white spots (zeros and ones). It has three (geometric) symmetry

axes, and each one of these symmetries preserves the spot de-

sign. This fact can be considered as known. And, if not, it su�ces

to draw the \row 0 through 7 triangle" and notice that it consists

of three copies of the (twice) smaller triangle formed by rows 0

through 3, the remaining quarter in the centre of the �gure being

�lled with zeros. The base row is all ones. These observations

provide a scheme for an induction proof of the claim for rows 0

through 2� � 1. The symmetry with respect to one of the oblique

axes is precisely the contents of the problem, expressed in alge-

braic terms.

Comment by the Editor-in-Chief

Choosing a solution to print is not an easy task. Each collection of

submitted solutions is assigned to a member of the Editorial Board, and these

editors work independently of one another. The choice of which solution to

highlight is left entirely to that editor. Sometimes it happens, as in this

issue, that several solutions are chosen from the same solver. I would like

to assure all subscribers that every submission is very important to CRUX,

and that we encourage everyone to submit solutions, as well as proposals

for problems, articles for publication, and contributions to the other corners.

Every subscriber is very important to us and we really value all contributions.

And while I am on the subject of contributions, please continue to send

in proposals for problems. We publish 100 per year, and we do not have too

many in reserve at this time. Without your contributions, there would be no

CRUX.

97

Dissecting Triangles

into Isosceles Triangles

Daniel Robbinsstudent, �Ecole Secondaire Beaumont, Beaumont

Sudhakar Sivapalanstudent, Harry Ainlay Composite High School, Edmonton

Matthew Wongstudent, University of Alberta, Edmonton

Problem 2 of Part II of the 1993-1994 Alberta High SchoolMathematicsCompetition (see CRUX [20:65]) goes as follows:

An isosceles triangle is called an amoeba if it can be divided into

two isosceles triangles by a straight cut. Howmany di�erent (that

is, not similar) amoebas are there?

All three authors wrote that contest. Afterwards, they felt that the problemwould have been more meaningful had they been asked to cut non-isoscelestriangles into isosceles ones.

We say that a triangle is n-dissectible if it can be dissected inton isosce-les triangles where n is a positive integer. Since we are primarily interestedin the minimum value of n, we also say that a triangle is n-critical if it isn-dissectible but not m-dissectible for any m < n. The isosceles trianglesthemselves are the only ones that are 1-dissectible, and of course 1-critical.

Note that, in the second de�nition, we should not replace \not m-dissectible for anym < n" by \not (n� 1)-dissectible". It may appear thatif a triangle is n-dissectible, then it must also bem-dissectible for allm > n.However, there are two exceptions. The solution to the contest problem,which motivated this study, shows that almost all 1-dissectible triangles arenot 2-dissectible. We will point out later that some 2-dissectible ones arenot 3-dissectible.

On the other hand, it is easy to see that all 1-dissectible triangles are3-dissectible. Figure 1 illustrates the three cases where the vertical angle isacute, right, and obtuse respectively.

������AAAAAA�

��ZZZ

��

�� @

@@@

��

������ HHHH

HH���

SSS

(a) (b) (c)

Figure 1.

98

We now �nd all 2-dissectible triangles. Clearly, such a triangle can onlybe cut into two triangles by drawing a line from a vertex to the opposite side,as illustrated in Figure 2.

SSSSSSSS��

����

����

����

�����������

�

2� 2�

A

BD

C

Figure 2.

Note that at least one of \ADB and \ADC is non-acute. We mayassume that \ADB � 90�. In order for BAD to be an isosceles triangle,we must have \BAD = \ABD. Denote their common value by �. By theExterior Angle Theorem, \ADC = 2�. There are three ways in which CADmay become an isosceles triangle.

Case 1.

Here \ACD = \ADC = 2�, as illustrated in Figure 2. Then \CAD =

180��4� > 0�. This class consists of all triangles in which two of the anglesare in the ratio 1:2, where the smaller angle � satis�es 0� < � < 45�. Ofthese, only the (36�; 72�; 72�) triangle is 1-dissectible, but it turns out thatevery triangle here is 3-dissectible.

Case 2.

Here, \CAD = \ADC = 2�. Then \CAB = 3� and \ACD = 180� �4� > 0�. This class consists of all triangles in which two of the angles arein the ratio 1:3, where the smaller angle � satis�es 0� < � < 45�. Of

these, only the (36�; 36�; 108�) and the (1807

�; 540

7

�; 540

7

�) triangles are 1-

dissectible. It turns out that those triangles for which 30� < � < 45�, witha few exceptions, are not 3-dissectible.

Case 3.

Here, \ACD = \CAD. Then their common value is 90� � � so that\CAB = 90�. This class consists of all right triangles. Of these, only the(45�; 45�; 90�) triangle is 1-dissectible, and it turns out that every trianglehere is 3-dissectible.

We should point out that while our three classes of 2-dissectible tri-angles are exhaustive, they are not mutually exclusive. For instance, the(30�; 60�; 90�) triangle appears in all three classes, with the same dissection.The (20�; 40�; 120�) triangle is the only one with two di�erent dissections.

We now consider 3-dissectible triangles. Suppose one of the cuts doesnot pass through any vertices. Then it divides the triangle into a triangle anda quadrilateral. The latter must then be cut into two triangles, and this cutmust pass through a vertex. Hence at least one cut passes through a vertex.

99

Suppose no cut goes from a vertex to the opposite side. The only pos-sible con�guration is the one illustrated in Figure 1(a). Since the three anglesat this point sum to 360�, at least two of them must be obtuse. It followsthat the three arms have equal length and this point is the circumcentre ofthe original triangle. Since it is an interior point, the triangle is acute. Thusall acute triangles are 3-dissectible.

In all other cases, one of the cuts go from a vertex to the opposite side,dividing the triangle into an isosceles one and a 2-dissectible one. There arequite a number of cases, but the argument is essentially an elaboration ofthat used to determine all 2-dissectible triangles. We leave the details to thereader, and will just summarize our �ndings in the following statement.

Theorem.

A triangle is 3-dissectible if and only if it satis�es at least one of the followingconditions:

1. It is an isosceles triangle.

2. It is an acute triangle.

3. It is a right triangle.

4. It has a 45� angle.

5. It has one of the following forms:

(a) (�, 90� � 2�, 90� + �), 0� < � < 45�;

(b) (�, 90� � 3�2, 90� + �

2), 0� < � < 60�;

(c) (�, 360� � 7�, 6� � 180�), 30� < � < 45�;

(d) (2�, 90� � 3�2, 90� � �

2), 0� < � < 60�;

(e) (3�, 90� � 2�, 90� � �), 0� < � < 45�;

(f) (180� � 2�, 180� � �, 3� � 180�), 60� < � < 90�;

(g) (180� � 4�, 180� � 3�, 7� � 180�), 30� < � < 45�.

6. Two of its angles are in the ratio p:q, with the smaller angle strictlybetween 0� and r�, for the following values of p, q and r:

p 1 1 1 1 1 1 2 3 3q 2 3 4 5 6 7 3 4 5r 60 30 22.5 30 22.5 22.5 60 67.5 67.5

100

The fact that all right triangles are 2-dissectible is important becauseevery triangle can be divided into two right triangles by cutting along thealtitude to its longest side. Each can then be cut into two isosceles trianglesby cutting along the median from the right angle to the hypotenuse, as il-lustrated in Figure 3. It follows that all triangles are 4-dissectible, and thatthere are no n-critical triangles for n � 5.

SSSSSSSS��

����

����

����

��

HHHH

HHHH

����

Figure 3.

We can prove by mathematical induction on n that all triangles are n-dissectible for all n � 4. We have established this for n = 4. Assume that alltriangles are n-dissectible for some n � 4. Consider any triangle. Divide itby a line through a vertex into an isosceles triangle and another one. By theinduction hypothesis, the second can be dissected into n triangles. Hencethe original triangle is (n+ 1)-dissectible.

Announcement

For the information of readers, we are saddened to note the deaths oftwo mathematicians who are well-known to readers of CRUX.

Leroy F. Meyers died last November. He was a long time contribu-tor to CRUX. See the February 1996 issue of the Notices of the American

Mathematical Society.

D.S. Mitrinovi�c died last March. He is well-known for his work on Ge-ometric Inequalities, in particular, being co-author of Geometric Inequalities

and Recent Advances in Geometric Inequalities. See the July 1995 issue ofthe Notices of the American Mathematical Society.

101

THE SKOLIAD CORNER

No. 13

R.E. Woodrow

This issue we feature the Eleventh W.J. Blundon Contest, writtenFebruary 23, 1994. This contest is sponsored by the Department of Math-ematics and Statistics of Memorial University of Newfoundland and is oneof the \Provincial" contests that receives support from the Canadian Math-ematical Society

THE ELEVENTH W.J. BLUNDON CONTESTFebruary 23, 1994

1. (a) The lesser of two consecutive integers equals 5 more than threetimes the larger integer. Find the two integers.

(b) If 4 � x � 6 and 2 � y � 3, �nd the minimum values of(x� y)(x+ y).

2. A geometric sequence is a sequence of numbers in which each term,after the �rst, can be obtained from the previous term, by multiplying bythe same �xed constant, called the common ratio. If the second term of ageometric sequence is 12 and the �fth term is 81=2, �nd the �rst term andthe common ratio.

3. A square is inscribed in an equilateral triangle. Find the ratio of thearea of the square to the area of the triangle.

4. ABCD is a square. Three parallel lines l1, l2 and l3 pass throughA, B andC respectively. The distance between l1 and l2 is 5 and the distancebetween l2 and l3 is 7. Find the area of ABCD.

5. The sum of the lengths of the three sides of a right triangle is 18.The sum of the squares of the lengths of the three sides is 128. Find the areaof the triangle.

6. A palindrome is a word or number that reads the same backwardsand forwards. For example, 1991 is a palindromic number. How many palin-dromic numbers are there between 1 and 99; 999 inclusive?

7. A graph of x2�2xy+y2�x+y = 12 and y2�y�6 = 0will producefour lines whose points of intersection are the vertices of a parallelogram.Find the area of the parallelogram.

8. Determine the possible values of c so that the two lines x� y = 2

and cx+ y = 3 intersect in the �rst quadrant.

9. Consider the function f(x) = cx2x+3

, x 6= �3=2. Find all values

of c, if any, for which f(f(x)) = x.

102

10. Two numbers are such that the sum of their cubes is 5 and thesum of their squares is 3. Find the sum of the two numbers.

Last issue we gave the problems of Part I of the Alberta High SchoolMathematics Competition, which was written Tuesday November 21, 1995.This month we give the solutions. How well did you do?

ALBERTA HIGH SCHOOL MATHEMATICSCOMPETITIONPart I: Solutions

November 21, 1995 (Time: 90 minutes)

1. A circle and a parabola are drawn on a piece of paper. The numberof regions they divide the paper into is at most

A. 3 B. 4 C. 5 D. 6 E. 7.

Solution. (D) The parabola divides the plane into two regions. Thecircle intersects the parabola in at most four points, so that it is divided bythe parabola into at most four arcs. Each arc carves an existing region intotwo.

2. The number of di�erent primes p > 2 such that p divides 712 �372 � 51 is

A. 0 B. 1 C. 2 D. 3 E. 4.

Solution. (D) We have 712 � 362 � 51 = (71 + 37)(71� 37)� 51 =

3 � 17(36 � 2� 1).

3. Suppose that your height this year is 10% more than it was last year,and last year your height was 20% more than it was the year before. By whatpercentage has your height increased during the last two years?

A. 30 B. 31 C. 32 D. 33 E. none of these.

Solution. (C) Suppose the height was 100 two years ago. Then it was120 a year ago and 132 now.

4. Multiply the consecutive even positive integers together until theproduct 2 � 4 � 6 � 8 � � � becomes divisible by 1995. The largest even integeryou use is

A. between 1 and 21 B. between 21 and 31

C. between 31 and 41 D. bigger than 41

E. non-existent, since the product never becomes divisible by 1995.

103

Solution. (C) All factors of 1995 are distinct and odd, with the largestone being 19. Hence the last even number used is 38.

5. A rectangle contains three circles as inthe diagram, all tangent to the rectangle and toeach other. If the height of the rectangle is 4,then the width of the rectangle is

A. 3 + 2p2 B. 4 + 4

p2

3C. 5 + 2

p2

3

D. 6 E. 5 +p10.

4

Solution. (A) Let O be the centre of the large circle, P that of oneof the small circles, and Q the point of tangency of the small circles. Then\PQO = 180�, PQ = 1 and OP = 1 + 2. By Pythagoras' Theorem,OQ = 2

p2.

6. Mary Lou works a full day and gets her usual pay. Then she workssome overtime hours, each at 150% of her usual hourly salary. Her total paythat day is equivalent to 12 hours at her usual hourly salary. The number ofhours that she usually works each day is

A. 6 B. 7:5 C. 8

D. 9 E. not uniquely determined by the given information.

Solution. (E) Suppose Mary Lou usually works x hours per day but yon that day. All we know is x+ 3

2(y� x) = 12 or x+ 3y = 24.

7. A fair coin is tossed 10; 000 times. The probability p of obtainingat least three heads in a row satis�es

A. 0 � p < 1

4B. 1

4� p < 1

2C. 1

2� p < 3

4D. 3

4� p < 1 E. p = 1:

Solution. (D) Partition the tosses into consecutive groups of three, dis-carding the last one. If we never get 3 heads in a row, none of the 3333

groups can consist of 3 heads. The probability of this is (78)3333, which is

clearly less than 1

4. In fact,

�7

8

�12=

1176492

812<

1310722

236=

1

8:

8. In the plane, the angles of a regular polygon with n sides add up toless than n2 degrees. The smallest possible value of n satis�es:

A. n < 40 B. 40 � n < 80 C. 80 � n < 120

D. 120 � n < 160 E. n � 160.

104

Solution. (E) The sum of the angles is exactly (n� 2)180�. From

180 <n2

n� 2= n+ 2 +

4

n� 2< n+ 3;

we have n > 177.

9. A cubic polynomial P is such that P (1) = 1, P (2) = 2, P (3) = 3

and P (4) = 5. The value of P (6) is

A. 7 B. 10 C. 13 D. 16 E. 19.

Solution. (C) By the Binomial Theorem, P (5) = 4P (4) � 6P (3) +

4P (2) � P (1) = 9. It follows that P (6) = 5P (5)� 10P (4) + 10P (3) �5P (2) + P (1) = 16.

10. The positive numbers x and y satisfy xy = 1. The minimum valueof 1

x4+ 1

4y4is

A. 1

2B. 5

8C. 1 D. 5

4E. no minimum.

Solution. (C) We have

1

x4+

1

4y4=

�1

x2�

1

2y2

�2+

1

x2y2� 1;

with equality if and only if x2 = 2y2.

11. Of the points (0; 0), (2;0), (3; 1), (1;2), (3; 3), (4; 3) and (2; 4),at most how many can lie on a circle?

A. 3 B. 4 C. 5 D. 6 E. 7.

Solution. (C) Since (0; 0), (1; 2) and (2;4) are collinear, a circle passesthrough at most two of them. Since (2;0), (3; 1), (3; 3) and (4; 3) are notconcyclic, a circle passes through at most three of them. The circle with centre(1; 2) and passing through (0; 0) also passes through (2;0), (3;1), (3;3) and(2; 4).

12. The number of di�erent positive integer triples (x; y; z) satisfyingthe equations

x2 + y � z = 100 and x+ y2� z = 124 is:

A. 0 B. 1 C. 2 D. 3 E. none of these.

Solution. (B) Subtraction yields 24 = x+ y2 � x2 � y = (y� x)(y +

x � 1). Note that one factor is odd and the other even, and that the �rst issmaller than the second. Hence either y � x = 1 and y + x � 1 = 24, ory � x = 3 and y + x � 1 = 8. They lead to (x; y; z) = (12;13; 57) and(3; 6;�85) and (3; 6;�85) respectively. However, we must have z > 0.

105

13. Which of the following conditions does not guarantee that theconvex quadrilateral ABCD is a parallelogram?

A. AB = CD and AD = BC B. \A = \C and \B = \D

C. AB = CD and \A = \C D. AB = CD and AB is parallel to CDE. none of these.

Solution. (C) Let \ABD = \BAD = \DCB = 40� and \CBD =

80�. Then ABCD is not a parallelogram. Let E on BC be such that\CDE = 40�. Then triangles BAD and CDE are congruent, so thatAB = CD. It is easy to see that the other three conditions do guaranteeparallelograms.

14. How many of the expressions

x3 + y4; x4 + y3; x3 + y3; and x4 � y4;

are positive for all possible numbers x and y for which x > y?

A. 0 B. 1 C. 2 D. 3 E. 4.

Solution. (A) We have x3 + y4 < 0 for x = �1

4and y = �1

3. Each of

the other three expressions is negative if x = 0 and y < 0.

15. In triangle ABC, the altitude from A to BC meets BC at D,and the altitude from B to CA meets AD at H. If AD = 4, BD = 3 andCD = 2, then the length ofHD is

A.p5

2B. 3

2C.p5 D. 5

2E. 3

p5

2.

Solution. (B) Note that \CAD = 180� � \BCA = \CBE. Hencetriangles CAD andHBD are similar, so that

HD

BD=CD

AD:

16. Which of the following is the best approximation to

(23 � 1)(33� 1)(43� 1)

(23 + 1)(33 + 1)(43 + 1)� � �

(1003� 1)

(1003 + 1)?

A. 3

5B. 33

50C. 333

500D. 3;333

5;000E. 33;333

50;000.

106

Solution. (C) The given expression factors into

(2� 1)(22 + 2 + 1)(3� 1)(32 + 3+ 1) � � � (100� 1)(1002 + 100 + 1)

(2 + 1)(22� 2 + 1)(3 + 1)(32� 3 + 1) � � � (100 + 1)(1002� 100 + 1):

Since ((n + 2) � 1) = n + 1 and (n + 1)2 � (n + 1) + 1 = n2 + n + 1,cancellations yield

(2� 1)(3� 1)(1002 + 100 + 1)

(22 � 2 + 1)(99+ 1)(100+ 1)=

10101

15150:

That completes the Skoliad Corner for this issue. Send me contest ma-terials, as well as your comments, suggestions, and desires for future direc-tions for the Skoliad Corner.

Citation

As was announced in the February 1996 issue of CRUX, Professor Ron Dunk-ley was appointed to the Order of Canada. This honour was bestowed onRon by the Governor-General of Canada, the His Excellency The Right Hon-ourable Rom�eo LeBlanc, in mid-February, and we are pleased to publish acopy of the o�cial citation:

Professor Ronald Dunkley, OC

A professor at the University of Waterloo and founding member ofthe Canadian Mathematics Competition, he has dedicated hiscareer to encouraging excellence in students. He has trainedCanadian teams for the International Mathematics Olympiad,

authored six secondary school texts and chaired two foundationsthat administer signi�cant scholarship programs. An inspiringteacher, he has stimulated interest and achievement among

students at all levels, and provided leadership and developmentprograms for teachers across the country.

107

THE OLYMPIAD CORNER

No. 173

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.

The �rst Olympiad problems that we give in this issue are the prob-lems of the Selection Tests for the Romanian Team to the 34th InternationalMathematical Olympiad. My thanks go to Georg Gunther, Sir Wilfred Gren-fell College for collecting them and sending them to us while he was Canadianteam leader at the IMO at Istanbul, Turkey.

SELECTION TESTS FOR THE ROMANIAN TEAM,34th IMO.

Part I | Selection Test for Balkan Olympic Team

1. Prove that the sequence Im (zn), n � 1, of the imaginary parts ofthe complex numbers zn = (1+ i)(2+ i) � � � (n+ i) contains in�nitely manypositive and in�nitely many negative numbers.

2. Let ABC be a triangle inscribed in the circle O(O;R) and cir-cumscribed to the circle J (I; r). Denote d = Rr

R+r. Show that there exists

a triangle DEF such that for any interior point M in ABC there exists apoint X on the sides of DEF such that MX � d.

3. Show that the set f1; 2; : : : ; 2ng can be partitioned in two classessuch that none of them contains an arithmetic progression with 2n terms.

4. Prove that the equation xn + yn = (x+ y)m has a unique integersolution withm > 1, n > 1, x > y > 0.

Part II | First Contest for IMO Team1st June, 1993

1. Find the greatest real number a such that

xpy2 + z2

+y

pz2 + x2

+zp

x2 + y2> a

is true for all positive real numbers x, y, z.

2. Show that if x, y, z are positive integers such that x2 + y2 + z2 =

1993, then x+ y + z is not a perfect square.

108

3. Each of the diagonals AD, BE and CF of a convex hexagonABCDEF determine a partition of the hexagon into quadrilaterals havingthe same area and the same perimeter. Does the hexagon necessarily have acentre of symmetry?

4. Show that for any function f : P(f1;2; : : : ; ng) ! f1; 2; : : : ; ngthere exist two subsets, A and B, of the set f1; 2; : : : ; ng, such that A 6= B

and f(A) = f(B) = maxfi j i 2 A \Bg.

Part III | Second Contest for IMO Team2nd June, 1993

1. Let f : (0;1)! R be a strict convex and strictly increasing func-tion. Show that the sequence ff(n)gn�1, does not contain an in�nite arith-metic progression.

2. Given integer numbers m and n, with m > n > 1 and (m;n) =

1, �nd the gcd of the polynomials f(X) = Xm+n � Xm+1 � X + 1 andg(X) = Xm+n +Xn+1 �X + 1.

3. Prove that for all integer numbers n, with n � 6, there exists ann-point set M in the plane such that every point P in M has at least threeother points inM at unit distance to P .

4. For all ordered 4-tuples (n1; n2; n3; n4) of positive integer numberswith ni � 1 and n1 + n2 + n3 + n4 = n, �nd the 4-tuples for which the

numbern!

n1!n2!n3!n4!2l, where

l =

�n1

2

�+

�n2

2

�+

�n3

2

�+

�n4

2

�+ n1n2 + n2n3 + n3n4;

has a maximum value.

Part IV | Third Contest for IMO Team3rd June, 1993

1. The sequence of positive integers fxngn�1 is de�ned as follows:x1 = 1, the next two terms are the even numbers 2 and 4, the next threeterms are the three odd numbers 5, 7, 9, the next four terms are the evennumbers 10, 12, 14, 16 and so on. Find a formula for xn.

2. The triangle ABC is given and let D, E, F be three points such

that D 2 (BC), E 2 (CA), F 2 (AB), BD = CE = AF and \BAD =

\CBE = \ACF . Show that ABC is equilateral.

109

3. Let p be a prime number, p � 5, andZ�p = f1; 2; : : : ; p�1g. Provethat for any partition with three subsets of Z�p there exists a solution of theequation

x+ y � z mod p;

each term belonging to a distinct member of the partition.

As a national Olympiad, we have the Final Round of the CzechoslovakMathematical Olympiad, 1993.

CZECHOSLOVAK MATHEMATICAL OLYMPIAD 1993Final Round

1. Find all natural numbers n for which 7n� 1 is a multiple of 6n� 1.

2. A 19� 19 table contains integers so that any two of them lying onneighbouring �elds di�er at most by 2. Find the greatest possible numberof mutually di�erent integers in such a table. (Two �elds of the table areconsidered neighbouring if they have a common side.)

3. A triangle AKL is given in a plane such that j\ALKj > 90� +

j\LAKj. Construct an equilateral trapezoid ABCD, AB ? CD, such thatK lies on the side BC, L on the diagonal AC and the outer section S ofAK and BL coincides with the centre of the circle circumscribed around thetrapezoid ABCD.

4. A sequence fang1n=1 of natural numbers is de�ned recursively bya1 = 2 and an+1 = the sum of 10th powers of the digits of an, for all n � 1.Decide whether some numbers can appear twice in the sequence fang1n=1.

5. Find all functions f :Z!Zsuch that f(�1) = f(1) and

f(x) + f(y) = f(x+ 2xy) + f(y� 2xy)

for all integers x, y.

6. Show that there exists a tetrahedron which can be partitioned intoeight congruent tetrahedra, each of which is similar to the original one.

Ah, the �ling demons are at it again. When I attacked a rather sus-picious looking pile of what I thought were as yet un�led solutions to 1995problems from the Corner, I found a small treasure-trove of solutions to var-ious problem sets from 1994, and some comments of Murray Klamkin aboutearlier material that he submitted at the same time. The remainder of thiscolumn will be devoted to catching up on this backlog in an attempt to bringthings up to the November 1994 issue. First two comments about solutionsfrom 1992 numbers of the corner.

110

2. [1990: 257; 1992: 40] 1990 Asian Paci�c Mathematical Olympiad.

Let a1; a2; : : : ; an be positive real numbers, and let Sk be the sum ofthe products of a1; a2; : : : ; an taken k at a time. Show that

SkSn�k ��n

k

�2a1a2 : : : an; for k = 1; 2; : : : ; n� 1:

Comment by Murray S. Klamkin, University of Alberta. It should benoted that the given inequality, as well as other stronger ones, follows fromthe known and useful Maclaurin inequalities, and that

"Sk�nk

�#1=k

is a non-increasing sequence in k, and with equality i� all the ai's are equal.

2. [1992: 197] 1992 Canadian Mathematical Olympiad.For x; y; z � 0, establish the inequality

x(x� z)2 + y(y� z)2 � (x� z)(y� z)(x+ y� z);

and determine when equality holds.

Comment by Murray S. Klamkin, University of Alberta. It should benoted that the given inequality is the special case � = 1 of Schur's inequality

x�(x� y)(x� z)� y�(x� z)(x� y) + z�(z � x)(z � y) � 0:

For a proof, since the inequality is symmetric, we may assume that x � y �z, or that x � z � y. Assuming the former case, we have

x�(x� y)(x� z)� y�(y� z)(x� y) + z�(z � x)(z � y)

� x�(x� y)(y� z)� y�(x� z)(x� y) + z�(z � x)(z � y) � 0:

We have assumed � � 0. For � < 0, we have

(yz)��(x� y)(x� z)� (zx)��(y� z)(x� y) + (xy)��(x� z)(y� z) �

(yz)��(x�y)(x� z)� (zx)��(y� z)(x� z)+(xy)��(x� z)(y� z) � 0:

The case x � z � y goes through in a similar way.Note that if � is an even integer, x, y, z can be any real numbers.

111

Next a comment about a solution from the February 1994 number.

6. [1994: 43; 1992: 297] Vietnamese National Olympiad.

Let x, y, z be positive real numbers with x � y � z. Prove that

x2

y+y2z

x+z2x

y� x2 + y2 + z2:

Comment byMurray S. Klamkin, University of Alberta. We give a moredirect solution than the previous ones and which applies in many cases wherethe constraint conditions are x � y � z.

Let z = a, y = a+ b, x = a+ b+ cwhere a > 0, and b; c � 0. Substi-tuting back in the inequality, multiplying by the least common denominatorand combining the cubic terms, we get

(a+ b+ c)3(ab+ b2) + a3c(a+ b+ c) � (a+ b)3(ab+ ac):

On inspection, for every term in the expansion of the right hand side there isa corresponding term on the left hand side, which establishes the inequality.

Amongst the solutions sent in were three solutions by Klamkin to prob-lems 3, 4 and 7 of the 1992 Austrian-Polish Mathematics Competition. Wediscussed reader's solutions to these in the December number [1995: 336{340]. My apologies for not mentioning his solutions there. He also sent in acomplete set of solutions to the 43rd Mathematical Olympiad in Poland, forwhich we discussed solutions to most of the problems in the February 1996Corner [1996: 24{27]. One problem solution was not covered there and wenext give his solution to �ll that gap.

5. [1994: 130] 43rd Mathematical Olympiad in Poland.

The rectangular 2n-gon is the base of a regular pyramid with vertexS. A sphere passing through S cuts the lateral edges SAi in the respectivepoints Bi (i = 1; 2; : : : ; 2n). Show that

nXi=1

SB2i�1 =nXi=1

SB2i:

Solution by Murray S. Klamkin, University of Alberta. Let S be theorigin (0;0; 0) of a rectangular coordinate system and let the coordinatesof the vertices Ak of the regular 2n-gon be given by (r cos �k; r sin �k; a),k = 1; 2; : : : ; 2n where �k = �k=n. A general sphere through S is given by

(x� h)2 + (y� k)2 + (z � l)2 = h2 + k2 + l2:

Since the parametric equation of the line SAk is given by

112

x = tr cos �k; y = tr sin �k; z = ta;

its intersection with the sphere is given by

(tr cos �k � h)2 + (tr sin �k � k)2 + (ta� l)2 = h2 + k2 + l2:

Solving for t, t = 0 corresponding to point S and

t =(hr cos �k + kr sin �k + al)

(r2 + a2):

Since SBk = tpr2 + a2, the desired result will follow if

Xcos �2k�1 =

Xcos �2k and

Xsin�2k�1 =

Xsin �2k

(where the sums are over k = 1; 2; : : : ; n). Since in the plane

(cos �2k�1; sin�2k�1); k = 1; 2; : : : ; n;

are the vertices of a regular n-gon, bothP

cos �2k�1 andP

sin �2k�1 vanishand the same for

Pcos �2k and

Psin �2k.

Next we give a comment and alternative solution to a problem dis-cussed in the May 1994 Corner.

7. [1994: 133; 1993: 66-67] 14th Austrian-Polish Mathematical

Olympiad.

For a given integer n � 1 determine the maximum of the function

f(x) =x+ x2 + � � � x2n�1

(1 + xn)2

over x 2 (0;1) and �nd all x > 0 for which this maximum is attained.

Comment byMurray S. Klamkin, University of Alberta. Here is a morecompact solution than the previously published one. We show that the max-imum value of f(x) is attained for x = 1, by establishing the inequality

4(x+ x2 + � � �x2n�1) � (2n� 1)(1 + 2xn + x2n):

This is a consequence of the majorization inequality [1], i.e., ifF (x) is convexand the vector (x1; x2; : : : ; xn) majorizes the vector (y1; y2; : : : ; yn), then

F (y1) + F (y2) + � � �+ F (yn) � F (x1) + F (x2) + � � �+ F (xn):

Note that for a vector X of n components xi in non-increasing order to ma-jorize a vector Y of n components yi in non-increasing order, we must have

113

pXi=1

xi �pX

i=1

yi for p = 1; 2; : : : ; n� 1 andX

xi =X

yi

and we write X � Y . Since xt is convex in t, we need only show thatX = (2n;2n; : : : ; 2n; n; n; : : : ; n; 0; 0; : : : ; 0) � Y = (2n � 1; : : : ; 2n �1; 2n�2; : : : ; 2n�2; : : : ; 1; : : : ; 1)where forX there are 2n�1 componentsof 2n; 2(2n� 1) components of n; (2n� 1) components of 0, while for Ythere are 4 components of each of (2n� 1); (2n� 2); : : : ; 1.

It follows easily that X and Y have the same number of componentsand that the sums of their components are equal. As for the rest, actually allwe really need show is that

4[(2n� 1) + (2n� 2) + � � �+ (n+ 1)]

= 6n2 � 12n < (2n� 1)(2n)+ (2n� 3)(n)

= 6n2 � 5n:

Reference.

1. A.W. Marshall, I. Olkin, Inequalities: Theory of Majorization and Its

Applications, Academic Press, NY, 1979.

Now we turn to some more solutions to problems proposed to the jurybut not used at the IMO at Istanbul. Last number we gave solutions to someof these. My \found mail" includes another solution to 2 [1994: 216] byMurray Klamkin, University of Alberta, and solutions to 13 [1994: 241] byBob Prielipp, University of Wisconsin-Oshkosh and by Cyrus C. Hsia, stu-dent, Woburn Collegiate Institute, Scarborough, Ontario.

1. [1994: 216] Proposed by Brazil.

Show that there exists a �nite set A � R2 such that for every X 2 A

there are points Y1; Y2; : : : ; Y1993 in A such that the distance between X

and Yi is equal to 1, for every i.

Solution by Cyrus C. Hsia.

We will prove the following proposition Pn: There exists a �nite setA � R2 such that for every point X in A there are n points Y1; Y2; : : : ; Ynin A such that the distance between X and Yi is equal to 1, for every i, andn is an integer greater than 1.

(By mathematical induction on n). For the case n = 2 just take anytwo points a unit distance apart. For n = 3 just take any three points a unitdistance apart from each other (i.e. any equilateral triangle of side 1 hasvertices with this property). The proposition is satis�ed in these two cases.

114

Suppose that the proposition Pn is true for n = k points. Let there beL (�nite) points that satisfy the proposition. Since there are a �nite numberof points inAk there are a �nite number of unit vectors formed between anytwo points of distance 1 unit apart. Now choose any unit vector, ~v, di�erentfrom any of the previous ones. Consider the set of 2L points formed by theoriginal L points translated by ~v and the original L points. (No point canbe translated onto another point by our choice of ~v). Now by the inductionhypothesis, every point in the original L points was a unit distance fromk other points. But the translation produced two such sets with each pointfrom one set a unit distance from its translated point in the other. Thus everypoint of the 2L points are at least a unit distance from k + 1 other points,which means the proposition is true for n = k + 1. Induction complete.

6. [1994: 217] Proposed by Ireland.

Let n, k be positive integers with k � n and let S be a set containingdistinct real numbers. Let T be the set of all real numbers of the form x1 +

x2 + � � �xk where x1; x2; : : : ; xk are distinct elements of S. Prove that Tcontains at least k(n� k) + 1 distinct elements.

Solution by Cyrus C. Hsia.

The problem should say: Let n, k be positive integers with k � n andlet S be a set containing n distinct real numbers. Let T be the set of all realnumbers of the form x1 + x2 + � � � + xk where x1; x2; : : : ; xk are distinctelements of S. Prove that T contains at least k(n�k)+1 distinct elements.

WOLOG let x1 < x2; � � � < xn since all xi are distinct.

Then consider the k(n� k) + 1 increasing numbers

x1 + x2 + x3 + � � �+ xk�1 + xk < x1 + x2 + x3 + � � �

+xk�1 + xk+1 < � � � < x1 + x2 + x3 + � � �+ xk�1 + xn (n� k + 1)

< x1 + x2 + x3 + � � �+ xk�2 + xk + xn < x1 + x2 + x3 + � � �

+xk+1 + xn < � � � < x1 + x2 + x3 + � � �+ xn�1 + xn (n� k)

< x1 + x2 + � � �+ xk�1 + xn�1 + xn < x1 + x2 + � � �+ xk�1

+xn�1 + xn < � � � < x1 + x2 + � � �+ xn�2 + xn�1 + xn (n� k)

.

.

....

< x2 + xn�k+2 + � � �+ xn�1 + xn < x3 + xn�k+2 + � � �

+xn�1 + xn < � � � < xn�k+1 + xn�k+2 + � � �+ xn�1 + xn (n� k)

There are at least (n � k + 1) + (k � 1)(n� k) = k(n� k) + 1 distinctnumbers.

9. [1994: 217] Proposed by Poland.

Let Sn be the number of sequences (a1; a2; : : : an), where ai 2 f0; 1g,in which no six consecutive blocks are equal. Prove that Sn ! 1 whenn!1.

115

Solution by Cyrus C. Hsia.

\No six consecutive blocks are equal" interpretation: There is no se-quence with the consecutive numbers 0; 0; 0; 0; 0; 0 or 1; 1; 1; 1; 1; 1 anywhere.

Consider the blocks 0; 1 and 1; 0. If the sequences were made onlywith these then we cannot have six consecutive blocks equal. Let Tn be thenumber of such sequences for n even (and ending with 0 or 1 for n odd). Forexample

T2 = 2 f0; 1 or 1; 0gT3 = 4 f0; 1; 0 or 0; 1; 1 or 1; 0; 0 or 1; 0; 1g

Thus Tn = 2dn2e. Now Tn < Sn since any 1 in a Tn sequence can be

changed to a 0 to form a new sequence in Sn which was not counted in Tn.

E.g. 1; 0; 0; 1; 0; 1 in Tn changes to 1; 0; 0; 0; 0; 1 which is in Sn.

Note: this change cannot produce six consecutive blocks equal.

Thus as n!1, Tn = 2dn2e !1. And since Sn > Tn, Sn !1.

18. [1994: 242] Proposed by the U.S.A.

Prove that

a

b+ 2c+ 3d+

b

c+ 2d+ 3a+

c

d+ 2a+ 3b+

d

a+ 2b+ 3c�

2

3

for all positive real numbers a, b, c, d.

Solution by Cyrus C. Hsia.

Using the Cauchy-Schwartz-Buniakowski inequality, we have�a

b+ 2c+ 3d+

b

c+ 2d+ 3a+

c

d+ 2a+ 3b+

d

a+ 2b+ 3c

��(a(b+ 2c+ 3d) + b(c+ 2d+ 3a) + � � � ) � (a+ b+ c+ d)2

) S(4)(ab+ a+ ad + bc+ bd+ cd) � (a+ b+ c+ d)2;

where

S =

�a

b+ 2c+ 3d+

b

c+ 2d+ 3a+

c

d+ 2a+ 3b+

d

a+ 2b+ 3c

�:

Now a2+ b2 � 2ab from the AM-GM inequality. Likewise for the other �vepairs we have the same inequality. Adding all six pairs gives

3(a2 + b2 + c2 + d2) � 2(ab+ ac + � � � )) 3(a+ b+ c+ d)2 � 8(ab+ ac+ � � � ):

Therefore

S �(a+ b+ c+ d)2

4(ab+ ac+ � � � )�

(83(ab+ ac+ � � � )

4(ab+ ac+ � � � )�

2

3;

as required.

116

Comment by Murray Klamkin, University of Alberta.

A generalization of this problem appeared recently in School Science &

Mathematics as problem #4499. The generalization is the following:

Let n be a natural number greater than one. Show that, for all positivenumbers a1; a2; : : : ; an,

X ai

ai+1 + 2ai+2 + � � �+ (n� 1)ai+n�1�

2

n� 1;

with equality if and only if a1 = a2 = � � � an. Here all the subscripts greaterthan n are reduced modulo n. For n = 2 and n = 3, the stated inequalityreduces to

a1

a2+a2

a1� 2

and

a1

a2 + 2a3+

a2

a3 + 2a1+

a3

a1 + 2a2� 1;

respectively. The special case n = 4 was proposed to the jury for the 34thInternational Mathematical Olympiad.

That's all the space I have this issue. Send me your nice solutions, andyour regional and national Olympiads!

Mathematical Literacy

1. Which Victorian physicist characterised nonnumerical knowledge as\meagre and unsatisfactory".

2. Who said, in 1692, \There are very few things which we know; whichare not capable of being reduc'd to a Mathematical Reasoning".

3. Which mathematician is reputed to have examined (and passed)Napoleon at the Ecole Militaire in 1785.

4. Which mathematician was said to have: \frequented low company, withwhom he used to guzzle porter and gin".

117

THE ACADEMY CORNER

No. 2

Bruce Shawyer

All communications about this column should be sent to Professor

Bruce Shawyer, Department of Mathematics and Statistics, Memorial Uni-

versity of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

In the �rst corner, I mentioned the APICS (Atlantic Provinces Councilon the Sciences) annual undergraduate contest. This contest is held at the endof October each year during the Fall APICSMathematics Meeting. As well asthe competition, there is a regular mathematics meeting with presentations,mostly from mathematicians in the Atlantic Provinces. An e�ort is made toensure that most of the talks are accessible to undergraduate mathematicsstudents. In fact, a key talk is the annual W.J. Blundon Lecture. This isnamed in honour of Jack Blundon, who wasHead of Department atMemorialUniversity of Newfoundland for the 27 years leading up to 1976. Jack was agreat support of CRUX, and long time subscribers will remember his manycontributions to this journal.

APICS Mathematics Contest 1995

Time allowed: three hours

1. Given the functions g : R! R and h : R! R, with g(g(x)) = x forevery x 2 R, and � a real number such that j�j 6= 1, prove that thereexists exactly one function f : R! R such that

�f(x) + f(g(x)) = h(x) for every x 2 R.

2. A solid fence encloses a square �eld with sides of length L. A cow is ina meadow surrounding the �eld for a large distance on all sides, and istied to a rope of length R attached to a corner of the fence.

What area of the meadow is available for the cow to use?

3. Find all solutions to

(x2 + y)(x+ y2) = (x� y)3

where x and y are integers di�erent from zero.

4. For what positive integers n, is the nth Catalan number,

�2n

n

�

n+ 1, odd?

118

5. N pairs of diametrically opposite points are chosen on a circle of ra-dius 1. Every line segment joining two of the 2N points, whether inthe same pair or not, is called a diagonal.

Show that the sum of the squares of the lengths of the diagonals de-pends only on N , and �nd that value.

6. A �nite pattern of checkers is placed on an in�nite checkerboard, atmost one checker to a square; this is Generation 0.

Generation N is generated from Generation (N � 1), for N = 1, 2,3; : : : , by the following process:

if a cell has an odd number of immediate horizontal of vertical neigh-bours in Generation (N � 1), it contains a checker in GenerationN ;otherwise it is vacant.

Show that there exists an X such that GenerationX consists of at least1995 copies of the original pattern, each separated from the rest of thepattern by an empty region at least 1995 cells wide.

7. A and B play a game. First A chooses a sequence of three tosses of acoin, and tells it to B;

then B chooses a di�erence sequence of three tosses and tells it to B.

Then they throw a fair coin repeatedly until one sequence or the othershows up as three consecutive tosses.

For instance, A might choose (head, tail, head); then B might choose(tail, head, tail). If the sequence of tosses is (head, tail, tail, head, tail),then B would win.

If both players play rationally (make their best possible choice), whatis the probability that A wins?

119

BOOK REVIEWS

Edited by ANDY LIU

e The Story of a Number by Eli Maor.Published by Princeton University Press, Princeton NJ, 1994,ISBN 0-691-03390-0, xiv+223 pages, US$24.95.Reviewed by Richard Guy, University of Calgary.

As the author says in his Preface:

The story of � has been extensively told, no doubt because itshistory goes back to ancient times, but also because much of itcan be grasped without a knowledge of advanced mathematics.Perhaps no book did better than Petr Beckmann's A History of �,a model of popular yet clear and precise exposition. The numbere fared less well.

It continues to fare less well. Many of us are searching for ways to bringmathematics to a much broader audience. This book is a good opportunitymissed.

There are too many formulas, and they appear too early; even in thePreface there are many. Moreover, they are rarely displayed (only one outof more than a dozen in the Preface), which makes the text hard to read |particularly for the layman.

Several chapters are nomore than excerpts from standard calculus texts.Chap. 4, Limits. Chap. 10, dy=dx = y, Chap. 12, Catenary. Chaps. 13 & 14,Complex Numbers; even including a treatment of the Cauchy-Riemann andLaplace equations.

This is a history book, but the history is often invented, either con-sciously or unconsciously (Pythagoras experimenting with strings, bells andglasses of water; Archimedes using parabolic mirrors to set the Roman eetablaze; Descartes watching a y on the ceiling; `quotations' of Johann andDaniel Bernoulli which are �gments of E. T. Bell's fertile imagination), orjust plain wrong.

Pascal's dates (1623-1662) are given, and an illustration of \his" trian-gle from a work published nearly a century before Pascal was born, and an-other from a Japanese work more than a century after he died. The author,elsewhere, mentions that it appeared in 1544 in Michael Stifel's Arithmetica

integra; all of which leaves the reader bewildered. The triangle was knownto the Japanese some centuries before, to the Chinese some centuries beforethat, and to Omar Khayyam before that.

After saying that it is unlikely that a member of the Bach family metone of the Bernoullis, the author gives a conversation between J. S. Bachand Johann Bernoulli. This piece of �ction distracts the reader from the realtopic, which is the mathematics of musical scales. No connection with e ismentioned.

120

One of the various examples that is given of the occurrence of the ex-ponential function, or of its inverse, the natural logarithm, is the Weber-Fechner law, which purports to measure the human response to physicalstimuli. This is probably also regarded as a piece of �ction by most mod-ern scientists. It has far less plausibility than the gas laws, say.

There are many missed opportunities: more examples of damping, thedistribution of temperature and pressure in the earth's atmosphere, alter-nating currents, radioactive decay and carbon dating; and why not mentionthe distribution of the prime numbers | they are as relevant and probablyof more interest to the person-in-the-street than many of the topics covered.

Let us look at some of the examples that are given.

Newton's law of cooling says that the temperature (di�erence) is pro-portional to its derivative. There is no mention of its eclipse by the work ofDulong & Petit and the Stefan-Boltzmann law [1], presumably because theyhave little connexion with e.

The parachutist whose air resistance is proportional to his velocity; inpractice the resistance is at least quadratic and probably more complicated[2].

Yet another example is the occurrence of the logarithmic spiral in artand in nature, without any hint of the controversial aspects of the subject.There is the oft-quoted and illustrated capitulum of a sun ower; though thecurrent wisdom [3] is that Fermat's spiral, r = a

p�, which does not involve

e, provides a more realistic model. Incidentally, this spiral is essentiallyidentical to the orbit of a non-relativistic charged particle in a cyclotron; thereare area, (resp. energy), considerations which make this the `right' spiral forthese two applications.

Population growth gets only a one-sentence mention, whereas thereare important mathematical and social lessons to be learnt from this topic.The author is not afraid of quoting whole pages of calculus, so why not thefollowing? There are many good examples of exponential growth, or decay,in everyday life. But none of them is perfect. Look at a perturbation ofthe equation for the exponential function, a particular case of Bernoulli'sequation:

dp

dt= kp� �p2

which, after multiplying by ekt=p2, we may write as

d

dt

ekt

p

!= �ekt

(Yes! Integrating factors and even characteristic equations occur on pp. 104{105.) Integrate and divide by ekt:

121

1

p= Ce�kt +

�

k

As the time tends to in�nity, the population tends to k=�. If � is small, thisis large, but at least it is �nite, and there is some hope for our planet. Whatis the �p2 term? It is roughly proportional to the number of pairs of people,and the term represents the e�ect of competition. But what if � is negative?What if we substitute cooperation for competition? As we approach the �nitetime (1=k) ln(Ck=(��)) the population tends to in�nity! No wonder thatcapitalism is more successful than communism.

There is a good deal of irrelevant padding, presumably in an attemptto make the book `popular': the cycloid, the lemniscate, Euler's formula forpolyhedra, the Newton-Leibniz controversy, and even Fermat's Last Theo-rem, though Andrew Wiles has a wrong given name in the Index.

There are misprints: `Appolonius' (p. 66; he does not appear in theIndex), `audibile' (p. 112); | no mathematics book is free of misprints. Butthere are worse than misprints: | `each additional term brings us closerto the limit (this is not so with a series whose terms have alternating signs)'(p. 36); `can be thought of as the sum of in�nitelymany small triangles' (p. 54)and Note 1 on p. 93 might lead many readers to infer that a continuousfunction has a derivative.

Let us hope that the book will spark the interest of non-mathematicians,but the fear is that it will con�rm their suspicions that we mathematiciansjust juggle symbols which the rest of the world has little hope of compre-hending.

REFERENCES

1. CharlesH. Draper, Heat and the Principles of Thermodynamics, Blackie& Son, London and Glasgow, 1893, pp. 221{222.

2. Horace Lamb, Dynamics, Cambridge Univ. Press, 1914, p. 290.

3. Helmut Vogel, A better way to construct the sun ower head, Math.

Biosciences, 44 (1979), pp. 179{189.

122

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 812"�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 that 1 November 1996. They may also

be sent by email to cruxeditor@cms.math.ca. (It would be appreciated if

email proposals and solutions were written in LATEX, preferably in LATEX2e).Graphics �les should be in epic format, or plain postscript. Solutions re-

ceived after the above date will also be considered if there is su�cient time

before the date of publication.

2125 Proposed by Bill Sands, University of Calgary, Calgary, Alberta.

At Lake West Collegiate, the lockers are in a long rectangular array,with three rows of N lockers each. The lockers in the top row are numbered1 to N , the middle row N +1 to 2N , and the bottom row 2N +1 to 3N , allfrom left to right. Ann, Beth, and Carol are three friends whose lockers arelocated as follows:

: : : : : :

��@@

��@@

��@@

By the way, the three girls are not only friends, but also next-doorneighbours, with Ann's, Beth's, and Carol's houses next to each other (in thatorder) on the same street. So the girls are intrigued when they notice thatBeth's house number divides into all three of their locker numbers. What isBeth's house number?

123

2126 Proposed by Bill Sands, University of Calgary, Calgary, Alberta.

At Lake West Collegiate, the lockers are in a long rectangular array, withthree rows ofN lockers each, whereN is some positive integer between 400

and 450. The lockers in the top row were originally numbered 1 to N , themiddle row N + 1 to 2N , and the bottom row 2N + 1 to 3N , all from leftto right. However, one evening the school administration changed aroundthe locker numbers so that the �rst column on the left is now numbered 1

to 3, the next column 4 to 6, and so forth, all from top to bottom. Threefriends, whose lockers are located one in each row, come in the next morningto discover that each of them now has the locker number that used to belongto one of the others! What are (were) their locker numbers, assuming thatall are three-digit numbers?

2127 Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is an acute triangle with circumcentre O, andD is a point on theminor arc AC of the circumcircle (D 6= A;C). Let P be a point on the sideAB such that \ADP = \OBC, and let Q be a point on the side BC suchthat \CDQ = \OBA. Prove that\DPQ = \DOC and\DQP = \DOA.

2128 Proposed by Toshio Seimiya, Kawasaki, Japan.

ABCD is a square. Let P and Q be interior points on the sides BCand CD respectively, and let E and F be the intersections of PQ with ABand AD respectively. Prove that

� � \PAQ+ \ECF <5�

4:

2129? Proposed by Stanley Rabinowitz, Westford, Massachusetts,

USA.

For n > 1 and i =p�1, prove or disprove that

1

4i

4nXk=1

gcd (k;n)=1

ik tan

�k�

4n

�

is an integer.

2130 Proposed by D. J. Smeenk, Zaltbommel, the Netherlands.

A and B are �xed points, and ` is a �xed line passing through A. C is avariable point on `, staying on one side of A. The incircle of�ABC touchesBC at D and AC at E. Show that lineDE passes through a �xed point.

124

2131 Proposed by Hoe Teck Wee, student, Hwa Chong Junior Col-

lege, Singapore.

Find all positive integers n > 1 such that there exists a cyclic permuta-tion of (1; 1; 2; 2; : : : ; n; n) satisfying:

(i) no two adjacent terms of the permutation (including the last and �rstterm) are equal; and

(ii) no block of n consecutive terms consists of n distinct integers.

2132 Proposed by �Sefket Arslanagi�c, Berlin, Germany. Let n be aneven number and z a complex number.Prove that the polynomial P (z) = (z + 1)n � zn � n is not divisible byz2 + z + n.

2133 Proposed by K. R. S. Sastry, Dodballapur, India.

Similar non-square rectangles are placed outwardly on the sides of aparallelogram �. Prove that the centres of these rectangles also form a non-square rectangle if and only if � is a non-square rhombus.

2134? Proposed by Waldemar Pompe, student, University of War-

saw, Poland.

Let fxng be an increasing sequence of positive integers such that thesequence fxn+1 � xng is bounded. Prove or disprove that, for each inte-ger m � 3, there exist positive integers k1 < k2 < : : : < km, such thatxk1 ; xk2 ; : : : ; xkm are in arithmetic progression.

2135 Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,

Madrid, Spain.

Let n be a positive integer. Find the value of the sum

bn=2cXk=1

(�1)k(2n� 2k)!

(k+ 1)!(n� k)!(n� 2k)!:

2136 Proposed by G. P. Henderson, Campbellcroft, Ontario.

Let a; b; c be the lengths of the sides of a triangle. Given the values ofp =

Pa and q =

Pab, prove that r = abc can be estimated with an error

of at most r=26.

2137 Proposed by Aram A. Yagubyants, Rostov na Donu, Russia.

Three circles of (equal) radius t pass through a point T , and are eachinside triangle ABC and tangent to two of its sides. Prove that:

(i) t =2R

R+ 2, (ii) T lies on the line segment joining the centres

of the circumcircle and the incircle of �ABC.

125

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

2015. [1995: 53 and 129 (Corrected)] Proposed by Shi-Chang Chi

and Ji Chen, Ningbo University, China.

Prove that

�sin(A) + sin(B) + sin(C)

�� 1

A+

1

B+

1

C

��

27p3

2�;

where A, B, C are the angles (in radians) of a triangle.

Editor's comment on the featured solution by Douglass L. Grant, Uni-

versity College of Cape Breton, Sydney, Nova Scotia, Canada. [1996: 47]There is a slight and very subtle aw in the published solution. To

correct this, all that is required is to replace the open domain S by a closed

domain.

The error is a very natural one, and has been made in the past by manyothers. We refer readers to Mathematics Magazine, 58 (1985), pp. 146{150,for several examples illustrating this subtle point.

2025. [1995: 90] Proposed by Federico Ardila, student, Massachu-

setts Institute of Technology, Cambridge, Massachusetts, USA.

(a) An equilateral triangle ABC is drawn on a sheet of paper. Provethat you can repeatedly fold the paper along the lines containing the sides ofthe triangle, so that the entire sheet of paper has been folded into a wad withthe original triangle as its boundary. More precisely, let fa be the functionfrom the plane of the sheet of paper to itself de�ned by

fa(x) =

�x if, x is on the same side of BC as A is,the re ection of x about line BC, otherwise,

(fa describes the result of folding the paper along the line BC), and analo-gously de�ne fb and fc. Prove that there is a �nite sequence fi1; fi2; : : : ; fin,with each fij = fa, fb or fc, such that fin(: : : (fi2(fi1(x))) : : : ) lies in or onthe triangle for every point x on the paper.

(b)� Is the result true for arbitrary triangles ABC?

126

Solution to (a) by Catherine Shevlin, Wallsend, England.

We shall show that it is possible by reversing the problem: we startwith the triangleABC and unfold three copies of it along the lines containingthe edges of the original triangle, to create a larger shape. This process ofunfolding shapes is repeated as illustrated in the diagram. We see how theplane is covered by the sequence of shapes, fSkg. The areas of the shapes

increase: the area of Sk is 1+ 3+ 6+ : : :+3(k� 1) =3k2 � 3k+ 2

2!1

as k!1.

: : :

S1 S2 S3 S4 S5 : : :

We now show that the minimum distance from a point on the boundary ofthe shape to the centre increases without bound.

Let O be the centre of the original triangle. Then, proceeding away fromtriangleABC in each of the six principal directions is a sequence of triangles,as illustrated below. For convenience, we rename triangleABC asA1A2A3.

A2 A4 A6 : : :

q

: : :

A1 A3 A5 : : :

O

The distance OAk is the minimum distance of the kth repetition from O. Thisis easy to calculate in terms of the side of the original triangle. It is clear thatOAk !1 as k!1.

It is easy to see that the area covered includes a circle of radius OAk.Hence, the area covered by the repetitions tends to in�nity, and so it willcover any �nite area. Thus the sheet of paper will be covered by the foldings.

Part (a) was also solved by the proposer. One incorrect attempt was

received. No-one sent in anything on part (b), so this problem remains

open.

127

2026. [1995: 90] Proposed by Hiroshi Kotera, Nara City, Japan.

One white square is surrounded by four black squares:

Two white squares are surrounded by six black squares:

Three white squares are surrounded by seven or eight black squares:

What is the largest possible number of white squares surrounded by n blacksquares? [According to the proposer, this problem was on the entrance ex-amination of the junior high school where he teaches!]

Solution by Carl Bosley, student, Washburn Rural High School, Top-

eka, Kansas, USA.

The largest possible number of white squares is8>><>>:

2k2 � 2k+ 1 if n = 4k;

2k2 � k if n = 4k+ 1;

2k2 if n = 4k+ 2;

2k2 + k if n = 4k+ 3:

Consider an arrangement of black squares surrounding some white re-gion.

Three black squares that are horizontally or vertically adjacent can bechanged as follows to increase the number of white squares surrounded byone as shown below.

-

Suppose there are two pairs of black squares that are horizontally ad-jacent. Then we can shift all the black squares between these pairs downas shown and still keep [at least] the same number of white squares sur-rounded, as shown below.

-

128

[Similarly we can assume there are not two pairs of vertically adjacent blacksquares.] Thus the maximum number of white squares surrounded can beobtained when there are either one horizontal and one vertical pair of ad-jacent black squares or when there are no horizontally or vertically adjacentblack squares. [For example, it is impossible to have one horizontal pair andno vertical pair of adjacent black squares: just consider the usual chessboardcolouring of the squares. | Ed.]

Label a set of diagonally adjacent squares as shown below.

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

Then if all the black squares are diagonally adjacent, we see that the squaresalternate from type 1 to type 2 and that there must be an even number ofthem. Thus, if there is an odd number of black squares, we have two blacksquares that are vertically adjacent and two black squares that are horizon-tally adjacent.

Connect the centres of adjacent squares. Then for a maximal numberof white squares surrounded, this polygon must be convex, since if it is notconvex, we can increase the number of white squares surrounded, as shownbelow.

-

Thus the black squares must form a rectangular formation, as in the secondpart of the �gure above, if the number of black squares is even. If there are2a + 2b black squares, a + 1 along two opposite sides and b + 1 along theother two sides, the number of white squares enclosed can be found to beab+(a�1)(b�1) = 2ab�a�b+1, so the maximum value is 2k2�2k+1

if there are 4k black squares and 2k2 if there are 4k + 2 black squares, foreach k > 0.

129

[Editor's note. With 2a+2b = 4k the number of white squares is2ab�2k+1,which is maximized when a = b = k; when 2a+2b = 4k+2 the number ofwhite squares is 2ab� 2k, which is maximized when fa; bg = fk; k+ 1g.]

If the number of black squares is odd, there are two squares that arehorizontally adjacent and two squares that are vertically adjacent. We canremove one of these squares and shift the others to form a closed �gure asshown below.

-

This second �gure must be a rectangle, so we can add on at most a � 1

squares, where a+ 1 is the number of squares in one of the largest sides ofthe rectangle. Therefore if there are 4k + 1 black squares, we can surround2k2 � k white squares; with 4k+ 3 black squares, we can surround 2k2 + k

white squares. [Editor's note. Here are some details. Letting the resultingblack rectangle have a+1 squares on one side and b+1 on the other, wherea � b, there are 2a + 2b black squares surrounding 2ab� a � b+ 1 whitesquares as before. Thus in the original �gure there are 2a + 2b + 1 blacksquares surrounding 2ab � b white squares. Now if the number of blacksquares is 4k+1, then a+b = 2k, and 2ab�b is maximized when a = b = k;thus the number of white squares surrounded is 2k2 � k. Similarly, if thenumber of black squares is 4k + 3, then a + b = 2k + 1, and 2ab � b ismaximized when a = k + 1, b = k; and this time the number of whitesquares surrounded is 2(k+ 1)k� k = 2k2 + k.]

Also solved by HAYO AHLBURG, Benidorm, Spain; RICHARD I. HESS,

Rancho Palos Verdes, California, USA; R. DANIEL HURWITZ, Skidmore Col-

lege, Saratoga Springs, New York; WALTHER JANOUS, Ursulinengymnasium,

Innsbruck, Austria; KOJI SHODA, Nemuro City, Japan; UNIVERSITY OF

ARIZONA PROBLEM SOLVING LAB, Tucson; and the proposer. One incor-

rect and one incomplete solution were also received.

Bosley and Janous actually made similar minor errors at the end of

their solutions. Bosley's has been corrected in the above writeup.

2029?. [1995: 91] Proposed by Jun-hua Huang, the Middle School

Attached To Hunan Normal University, Changsha, China.

ABC is a triangle with area F and internal angle bisectors wa, wb, wc.Prove or disprove that

wbwc + wcwa +wawb � 3p3F:

130

Solution by Kee-Wai Lau, Hong Kong.

The inequality is true.

Denote, as usual, the semi-perimeter, inradius and circumradius by s, r andR respectively. In the following, sums and products are cyclic over A, B, C,and/or over a, b, c, as is appropriate.

We need the following identities:

wa =2ab

b+ ccos(A=2); etc., (1)

F = 1

2ab sin(C); etc., (2)X

sin2(A) =s2 � 4Rr � r2

2R2; (3)X

sin(A) =s

R; (4)

Xsin(A) sin(B) =

s2 + 4Rr + r2

4R2; (5)Y

sin(A) =sr

2R2; (6)

Xcos(A) =

R+ r

R; (7)

Xcos(A) cos(B) =

r2 + s2 � 4R2

4R2; (8)

Ycos(A) =

s2 � 4R2 � 4Rr � r2

4R2; (9)

Ycos

�A� B

2

�=

s2 + 2Rr + r2

8R2: (10)

We also need the inequality:

16Rr � 5r2 � s2 � 4R2 + 4Rr + 3r2: (11)

Identities (1) and (2) are well-known. Identities (3) through (9) can be foundon pages 55{56 of [1]. Identity (10) can be obtained from (5) and (8) by

rewriting cos�A�B2

�as 1

4(1 +

Psin(A) sin(B) +

Pcos(A) cos(B)).

Inequality (11) is due to J.C. Gerretsen, and can be found on page 45of [1].

131

From (1) and (2), we havePwawb

F= 8

X�c2 cos(A=2) cos(B=2)

(c+ a)(c+ b) sin(C)

�

=X�

2 sin(C)

cos((B � C)=2)cos((A� C)=2)

�:

Thus, it is su�cient to prove that

Xsin(A) cos((B� C)=2) �

3p3

2

Ycos((B� C)=2): (12)

By squaring both sides, we see that (12) is equivalent to

4X

sin2(A) cos2((B� C)=2)

+8X

sin(A) sin(B) cos((B � C)=2)cos((A� C)=2)

� 27Y

cos2((B � C)=2): (13)

To prove (13), it su�ces to show that

4X

sin2(A) cos2((B � C)=2)

+8Y

cos((B� C)=2)X

sin(A) sin(B)

�27Y

cos2((B �C)=2) � 0: (14)

Since

4X

sin2(A) cos2((B �C=2))

= 2�X

sin2(A) +X

cos(B) cos(C)

�Y

cos(A)X

cos(A) +Y

sin(A)X

sin(A)�;

we use (3) { (10), to obtain that (14) is equivalent to

11s4 + (22r2 � 20rR� 64R2)s2 � 148r2R2 � 20r3R+ 11r4 � 0:

(15)

Let x = s2 and denote the left side of (15) by f(x). Then f(x) is a convexfunction of x. In view of (11), in order to prove (15), it is su�cient to showboth

f(16Rr� 5r2) � 0; (16)

f(4R2 + 4Rr + 3r2) � 0: (17)

Now

f(16Rr� 5r2) = �4r(R� 2r)(256R2 � 155rR+ 22r2);

and

f(4R2 + 4Rr + 3r2) = �4(R� 2r)(20R3 + 36rR2 + 45r2R+ 22r3):

Since R � 2r, both (16) and (17) hold, and the desired inequality is proved.

132

Reference

[1.] D.S. Mitrinovi�c, J.E. Pe�cari�c and V. Volenec, Recent Advances in Geo-

metric Inequalities, Kluwer Academic Publishers, 1989.

No other solutions were received.

2030. Proposed by Jan Ciach, Ostrowiec �Swi�etokrzyski, Poland.

For which complex numbers s does the polynomial z3 � sz2 + sz � 1

possess exactly three distinct zeros having modulus 1?

I. Combination of solutions by F.J. Flanigan, San Jose State University,

and the late John B. Wilker, University of Toronto.

Denote by S the set of such complex numbers s. We o�er two para-metrizations of S (neither injective), the second of these yielding a descrip-tion of S as the interior of a certain curvilinear triangle (hypocycloid or del-toid) inscribed in the disc jzj � 3.

Since the three zeros have modulus 1 and product equal to unity, theymay be written as ei�; ei�; e�i(�+�) with 0 � �; � < 2�. It follows that

s = ei� + ei� + e�i(�+�): (1)

Moreover, the three zeros will be distinct if and only if � 6= � and neither2�+� nor 2�+� is an integer multiple of 2�. (We note in passing that thefact the coe�cient of z in the cubic is s does not impose a further restriction.)

From (1) we see that jsj � 3 with equality if and only if � = � =

�(�+ �) mod 2�.

We improve (1) by noting that ei� + ei� = ei(�+�)=2(ei(���)=2 +

e�i(���)=2) = 2 cos �ei� where � = (�+ �)=2 and � = (�� �)=2. Thuswe have

s = 2 cos �ei� + e�2i�: (2)

Equation (2) enables us to visualize the parameter set S as follows. Fix� and let cos � vary through its full range: �1 � cos � � 1. Then the complexnumbers 2 cos �ei� lie on a line segment of length 4 centred at the origin inthe direction of the vector ei�. Thus the points s for this �xed � lie on a linesegment of length 4 joining P = e�2i� + 2ei� to P 0 = e�2i� � 2ei�. Thissegment is centred at the point e�2i� and makes an angle � with the x-axis.The set S is the union of all these (overlapping) segments PP 0.

As � varies, the path of either endpoint of the segment is the curvilineartriangle with vertices at 3; 3e�i=3, and 3e��i=3, namely,

133

z(t) = 2eit + e�2it

(where P = z(�) and P 0 = z(� + �)). One veri�es easily that this curveis a hypocycloid with three cusps: it is the locus of a point P �xed to thecircumference of a circle (whose diameter is PP 0) that is rolling clockwisearound the inside of the circle jzj = 3 as � runs from 0 to 2�. This curve iscalled a deltoid because its shape resembles the Greek letter � ([2], pp 73{79).

De�ne P 00 = 2 cos(3�)ei� + e�2i�. Note that P 00 lies on PP 0, andsince

P 00 = (e3i� + e�3i�)ei� + e�2i� = e4i� + 2e�2i� = z(�2�);

it lies on the deltoid as well. Moreover, P 00 is the only point of the deltoidin the interior of the segment PP 0 (as is easily veri�ed, or see [2], p. 75),so that PP 0 is tangent to the deltoid at P 00. It follows that all points insidethe deltoid are in S (which is the union of these diameters). Note furtherthat P (the point where � = 0) corresponds to � = �, P 0 (where � = �) to� = � + 2�, and P 00 (where � = 3�) to � = �2�; consequently, the valuesof s on the deltoid correspond to multiple zeros of the given cubic (ei� = ei�

or ei� = e�i(�+�)). Since the proposal excludes multiple zeros we concludethat S is precisely the interior of the region bounded by the deltoid.

Editor's comment (by Chris Fisher). It is certainly clear (from (2)) thatall points between P and P 0 on the segment PP 0 lie in S. Many of thosewhose solutions described S geometrically concluded that the interior of thesegment joining the points P and P 0 automatically lies inside the deltoid,since P and P 0 lie on the boundary. However, further argument seems tobe required: since S is not a convex region we have no guarantee against aninterval of points on PP 0 that (like P 00) lie outside the interior of the deltoid.

II. Solution by Kurt Fink and Jawad Sadek, Northwest Missouri State

University.

Let P (z) = z3 � sz2 + sz � 1 and let P 0(z) = 3z2 � 2sz + s beits derivative. A result of A. Cohn, specialized to P (z) (see [3], p. 206,Exercise 3), states that the zeros of P (z) lie on the unit circle and are simpleif and only if the zeros of P 0(z) lie in jzj < 1 or, equivalently, the zeros of thepolynomial sz2 � 2sz + 3 lie in jzj > 1. An application of Theorem 6.8b onp. 493 of [1], shows that (we omit the elementary calculations) this occurs ifand only if jsj < 3 and jsj4 + 18jsj2 � 8<(s3)� 27 < 0. If s = x+ iy, thecondition becomes

134

(x2 + y2)2 + 18(x2 + y2)� 8(x3 � 3xy2)� 27 < 0:

This is the interior of a deltoid whose closure is contained in the disk ofradius 3 centred at the origin; it touches the boundary circle at the points(3; 0); (�3=2;3

p3=2), and (�3=2;�3

p3=2). The closed deltoid region con-

tains the unit circle and boundaries of these curves meet at (�1; 0),(1=2;

p3=2) and (1=2;�

p3=2).

References

1. Peter Henrici, Applied and Computational Complex Analysis, Vol. 1,Wiley, New York, 1974.

2. E.H. Lockwood, A Book of Curves, Cambridge Univ. Press, Cambridge,1963.

3. Morris Marden, Geometry of Polynomials, Amer. Math. Society,Princeton, 1985.

Also solved by ED BARBEAU, University of Toronto, Toronto, Ontario;

CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; KEITH EKBLAW,

Walla Walla, Washington, USA; JEFFREY K. FLOYD, Newnan, Georgia, USA;

RICHARD I. HESS, Rancho Palos Verdes, California, USA; MURRAY S.

KLAMKIN, University of Alberta, Edmonton, Alberta; WALTHER JANOUS,

Ursulinengymnasium, Innsbruck, Austria; P. PENNING, Delft, the Nether-

lands; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the proposer.

Janous showed that if the polynomial

p(z) = zn � szn�1 + � � �+ (�1)n

has n zeros of modulus 1, then s =

nXk=1

ei�k , where �1; : : : ; �n 2 [0; 2�) and

�1 + � � � + �n = 0 (mod 2�). Furthermore, if ak is the coe�cient of zk

then, necessarily, (�1)n ak is the coe�cient of zn�k. The proposer outlinedan argument that if the polynomial

p(z) = zn � szn�1 + sz � 1

hasn zeros of modulus 1, then s belongs to the interior of the region boundedby the hypocycloid

z(t) =n� 1

n� 2eit +

1

n� 2e�i(n�1)t:

135

2031. [1995: 129] Proposed by Toshio Seimiya, Kawasaki, Japan.

Suppose that �, �, are acute angles such that

sin(�� �)

sin(�+ �)+

sin(� � )

sin(� + )+

sin( � �)

sin( + �)= 0:

Prove that at least two of �, �, are equal.

All solvers had the same idea, so we present a composite solution.By dividing the �rst term, top and bottom, by cos� cos�, and the

other two terms similarly, the given condition is equivalent to

tan�� tan�

tan�+ tan�+

tan� � tan

tan� + tan +

tan � tan�

tan + tan�= 0:

By multiplying by the common denominator, this reduces to

(tan�� tan�)(tan� � tan )(tan � tan�) = 0:

Hence, at least two of tan�, tan�, tan are equal, and since �, �, areacute angles, at least two of them must be equal.

Solved by �SEFKET ARSLANAGI �C, Berlin, Germany; CARL BOSLEY, stu-

dent, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J.

BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGEL CABEZ �ON OCHOA,

Logro ~no, Spain; SABIN CAUTIS, student, Earl Haig Secondary School, North

York, Ontario; ADRIAN CHAN, student, Upper Canada College, Toronto,

Ontario; THEODORE CHRONIS, student, Aristotle University of Thessa-

lonika, Thessalonika, Greece; DAVID DOSTER, Choate Rosemary Hall, Wall-

ingford, Connecticut, USA; CYRUS HSIA, student, University of Toronto,

Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck,

Austria; KEE-WAI LAU, Hong Kong; VICTOR OXMAN, Haifa University,

Haifa, Israel; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; D. J. SMEENK,

Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece;

HOE TECK WEE, student, Hwa Chong Junior College, Singapore; and the

proposer.

2032. [1995: 129] Proposed by Tim Cross, Wolverley High School,

Kidderminster, UK.

Prove that, for nonnegative real numbers x, y and z,px2 + 1+

py2 + 1+

pz2 + 1 �

p6(x+ y+ z) :

When does equality hold?

This problem attracted 25 correct solutions. Six solvers (marked y inthe list of solvers) gave the generalization highlighted below. So we present

a composite solution based on the submissions of several solvers.

136

We prove the generalization: for nonnegative real numbers fxkg,

nXk=1

qx2k + 1 �

vuut2n

nXk=1

xk:

From (xkxj � 1)2 � 0, we get

x2kx2j + x2k + x2j + 1 � x2k + 2xkxj + x2j :

Hence

(x2k + 1)(x2j + 1) � (xk + xj)2:

Taking the square root, and since the xk are nonnegative, we have

2

q(x2k + 1)

q(x2j + 1) � 2(xk + xj):

Hence

nXk=1

nXj=1

q(x2k + 1)

q(x2j + 1) � 2(n� 1)

nXk=1

xk:

We now add the nonnegative quantities (xk � 1)2, to get

nXk=1

(xk � 1)2 +

nXk=1

nXj=1

q(x2k + 1)

q(x2j + 1) � 2(n� 1)

nXk=1

xk:

Thus

nXk=1

x2k + n

nXk=1

nXj=1

q(x2k + 1)

q(x2j + 1) � 2n

nXk=1

xk:

But the left side is the square of

nXk=1

q(x2k + 1);

and so the result follows.

Equality holds when xk = 1 for all k such that 1 � k � n.

137

Solved by y �SEFKET ARSLANAGI �C, Berlin, Germany; yNIELSBEJLEGAARD, Stavanger, Norway; CARL BOSLEY, student, Washburn Ru-

ral High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton

College, Bristol, UK; yMIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain;

SABIN CAUTIS, student, Earl Haig Secondary School, North York Ontario;

ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;

yTHEODORE CHRONIS, student, Aristotle University of Thessalonika,

Greece; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut,

USA; TOBY GEE, student, the John of Gaunt School, Trowbridge, England;

CYRUS HSIA, student, University of Toronto, Toronto, Ontario; PETER

HURTHIG, Columbia College, Burnaby, BC; yWALTHER JANOUS, Ursul-

inengymnasium, Innsbruck, Austria; yV �ACLAV KONE �CN �Y, Ferris State Uni-

versity, Big Rapids, Michigan, USA; SAI C. KWOK, Boulder, Colorado, USA;

KEE-WAI LAU, Hong Kong; VICTOR OXMAN, Haifa, Israel; GOTTFRIED

PERZ, Pestalozzigymnasium, Graz, Austria; SCIENCE ACADEMY PROBLEM

SOLVERS, Austin, Texas, USA; yHEINZ-J�URGEN SEIFFERT, Berlin, Germany;

DIGBY SMITH, Mount Royal College, Calgary, Alberta; PANOS E.

TSAOUSSOGLOU, Athens, Greece; EDWARD T. H. WANG, Wilfrid Laurier

University, Waterloo, Ontario; HOE TECK WEE, student, Hwa Chong Junior

College, Singapore; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and

the proposer.

Four incorrect solutions were received. They all used calculus. Those

solvers are referred to the comment on the solution to problem 2015, earlier

in this issue.

2033. [1995: 129] Proposed by K. R. S. Sastry, Dodballapur, India.

The sides AB, BC, CD, DA of a convex quadrilateral ABCD areextended in that order to the points P , Q, R, S such that BP = CQ =

DR = AS. If PQRS is a square, prove that ABCD is also a square.

Solution by Cyrus Hsia, student, University of Toronto, Toronto, On-

tario.

Suppose that ABCD is not a rectangle. Then it must contain an inte-rior angle greater than �

2. Without loss of generality, let \A > �

2.

Since PQRS is a square, PQ = QR = RS = SP . Rotate QP aboutP clockwise untilQ coincides with S. Let B0 be the new position of B. Now

\B0PA = \B0PS + \SPA

= \BPQ+ \SPA (from rotation)

= \SPQ =�

2:

(Editor's note: Since PQRS is a square, this is just a rotation through3�2

radians and the fact that \B0PA = �2follows directly.)

138

Since \A > �2;\SAP < �

2which implies that PB0 = AS > h, where

h is an altitude from S to AP . Then�

2> \SB0P = \QBP;

so \ABC > �2.

Similar arguments show that \BCD > �2and \CDA > �

2. Therefore,

the sum of the interior angles of ABCD > 2�. Impossible!

Therefore, ABCD is a rectangle.

Since \SAP = \PBQ = \QCR = \RDS = �2, the triangles

ASP; BPQ; CQR;DRS are congruent and AB = BC = CD = DA.Therefore, ABCD is a square.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK; TOSHIO SEIMIYA, Kawasaki, Japan; D. J. SMEENK, Zaltbommel, the

Netherlands; and the proposer. There was one incorrect solution. The pro-

poser notes that his starting point was the analogous theorem for triangles.

Bradley notes that the problem was set in the Second Selection examina-

tion for the 36th IMO in Bucharest in April 1995 and is attributed to L.

PANAITOPOL. He also notes that two solutions are given, one trigonomet-

rical, which is correct, the other, a pure solution, which is in error.

Finally, both Bradley and Smeenk note that a simple modi�cation of the

proof works in the case of convex polygons of any number of sides.

2034. [1995: 130, 157] Proposed by Murray S. Klamkin and M. V.

Subbarao, University of Alberta.

(a) Find all sequences p1 < p2 < � � � < pn of distinct prime numberssuch that �

1 +1

p1

��1 +

1

p2

�� � ��1 +

1

pn

�is an integer.

(b) Can �1 +

1

a21

��1 +

1

a22

�� � ��1 +

1

a2n

�be an integer, where a1; a2; : : : are distinct integers greater than 1?

Solution to (a), by Heinz-J �urgen Sei�ert, Berlin, Germany.If the considered product is an integer, then pnj(pi + 1) for some

i 2 f1; 2; : : : ; n� 1g. Since pi + 1 � pn, it then follows that pn = pi + 1,which implies pi = 2 and pn = 3. Thus, n = 2, p1 = 2, p2 = 3. This isindeed a solution since

�1 + 1

2

� �1 + 1

3

�= 2.

Solution to (b). In the following, we present four di�erent solutionssubmitted by eight solvers and the proposers. In all of them, p denotes thegiven product, and it is shown that 1 < p < 2 and thus, p cannot be aninteger. Clearly one may assume, without loss of generality, that 1 < a1 <

a2 < : : : < an.

139

Solution I, by Carl Bosley, student, Washburn Rural High School, Top-eka, Kansas, USA; Kee-Wai Lau, Hong Kong; and Kathleen E. Lewis, SUNY,Oswego, New York.

Since 1 + x < ex for x > 0, we have

1 < p < exp

�1

a21+

1

a22+ : : :+

1

a2n

�

< exp

�1

22+

1

32+ : : :

�

= exp

��2

6� 1

�� 1:90586 < 2:

Solution II, by Toby Gee, student, the John of Gaunt School, Trow-bridge, England.

Since a1 � 2, we have

1 < p �n+1Yk=2

�1 +

1

k2

�=

n+1Yk=2

k2 + 1

k2<

n+1Yk=2

k2

k2 � 1

=

n+1Yk=2

�kk + 1

! n+1Yk=2

k

k� 1

!=

2(n+ 1)

n+ 2< 2:

Solution III, by Walther Janous, Ursulinengymnasium, Innsbruck, Aus-tria; V�aclav Kone �cn �y, Ferris State University, Big Rapids, Michigan, USA;Heinz-J �urgen Sei�ert, Berlin, Germany; and the proposers.

1 < p <

1Yt=2

�1 +

1

t2

�=

1

2

1Yt=1

�1 +

1

t2

�=

sinh�

2�� 1:838 < 2.

Solution IV, Richard I. Hess, Rancho Palos Verdes, California, USA.

Let p1 =

1Yk=2

�1 +

1

h2

�. Then clearly 1 < p < p1 = Q � R where

Q =

10Yk=2

�1 +

1

h2

�� 1:6714 and R =

1Yk=11

�1 +

1

k2

�.

Now, lnR =

1Xk=11

ln

�1 +

1

k2

�<

Z 1

10

ln

�1 +

1

x2

�dx <

Z 1

10

dx

x2=

1

10.

(Ed: This is because f(x) = ln

�1 +

1

x2

�is strictly decreasing on (0;1)

and ln(1 + t) < t for t > 0).Therefore, R < e0:1 � 1:1052, and so p1 < (1:68)(1:11) = 1:8648 < 2.

Part (a) was also solved by SABIN CAUTIS, student, Earl Haig Sec-

ondary School, North York, Ontario; SHAWN GODIN, St. Joseph Scollard

Hall,

North Bay, Ontario; CYRUS HSIA, student, University of Toronto, Toron-

to, Ontario; and ASHISH KR. SINGH, student, Kanpur, India.

140

2036. [1995: 130] Proposed by Victor Oxman, Haifa University, Is-

rael.

You are given a circle cut out of paper, and a pair of scissors. Showhow, by cutting only along folds, to cut from the circle a �gure which hasarea between 27% and 28% of the area of the circle.

Solution. The response to this problem was 10 di�erent solutions in therange [27:2535 : : : ; 27:8834 : : : ]. Except for the proposer's solution, all in-volved straight line cuts. The proposer's solution involves cutting along fourcurved lines, outlined by parts of the circle that have been folded inwards.

The minimum number of (single) folds was four { Hurthig wondered if a prizewas available for the least!

Rather than give the details of any one, we shall present ten diagrams thatillustrate the ingenuity of our readership, and leave you, the reader, to dothe exact calculation for each.

@@@@@@@@

Gee, Godwin, Hsai

In the above diagram, each line bisects an angle formed previously, until anangle of 35�=64 is generated.

Bosley

������@@@@@@�

�����@@@@@@

������

������

������

Godwin Grant, Hurthig

Hess Hurthig

��

��

��

�� @

@@

@@

@@

@

Jackson

141

��

��

��

�� @

@@

@@

@@

@

Perz

������@@@@@@�

�����

Tsaoussoglou

������@@@@@@�

�����@@@@@@

The proposer

Solved by CARL BOSLEY, student, Washburn Rural High School, Top-

eka, Kansas, USA; TOBY GEE; student, the John of Gaunt School, Trow-

bridge, England; DOUGLASS GRANT, University College of Cape Breton;

Sydney, Nova Scotia; RICHARD I. HESS, Rancho Palos Verdes, California,

USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario; PETER

HURTHIG, Columbia College, Burnaby, BC; DOUGLAS E. JACKSON, East-

ern New Mexico University, Portales, New Mexico, USA; GOTTFRIED PERZ,

Pestalozzigymnasium, Graz, Austria; PANOS E. TSAOUSSOGLOU, Athens,

Greece; and the proposer.

2037. [1995: 130] Proposed by Paul Yiu, Florida Atlantic University,

Boca Raton, Florida, USA.

The lengths of the base and a slant side of an isosceles triangle areintegers without common divisors. If the lengths of the angle bisectors ofthe triangle are all rational numbers, show that the length of the slant sideis an odd perfect square.

Solution by David Doster, Choate Rosemary Hall, Wallingford, Con-

necticut, USA.

Let �ABC be isosceles, with b = c, and let the angle bisectors beAD, BE and CF . It is given that a and c are relatively prime integers.Using Stewart's Theorem, for example, one can show that the length of theangle bisector to AB (and hence to AC) is

CF =

pac(a+ b+ c)(a+ b� c)

a+ b=apc(2c+ a)

a+ c:

[This formula can also be derived from the formula for the angle bisectorgiven in CRUX on [1995: 321]. | Ed.] Also [by the Pythagorean Theorem]

AD =

p4c2 � a2

2:

SinceCF andAD are both rational, wemust have, for some positive integersu and v,

142

c(2c+ a) = u2 and 4c2 � a2 = v2:

Since gcd(a; c) = 1, it follows that gcd(a; 2c + a) = 1. Hence both c and2c+ a are perfect squares. Now, if c is even, then a, being relatively primeto c, is odd. From the equation 4c2�a2 = v2 we see that v too is odd. Thusa2 + v2 � 2 mod 4. But 4c2 � 0 mod 4, a contradiction. Thus c is an oddperfect square.

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,

Spain; CARL BOSLEY, student,WashburnRural High School, Topeka, Kansas,

USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL

ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; ADRIAN CHAN, student, Upper

Canada College, Toronto, Ontario; RICHARD K. GUY, University of Calgary,

Calgary, Alberta; RICHARD I. HESS, Rancho Palos Verdes, California, USA;

CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, Hong

Kong; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the proposer. Four

slightly incorrect solutionswere also received, most of these assuming, with-

out proof, that the altitude from the apex of the triangle is an integer.

Amengual notes that the perimeter of the triangle is also a perfect

square. This is contained in the above proof, since 2c + a is the perime-

ter.

The proposer gave several examples of triangles satisfying the given

property. The smallest has base 14 and slant sides 25.

2038. [1995: 130] Proposed by Neven Juri�c, Zagreb, Croatia.

Show that, for any positive integersm and n, there is a positive integerk so that �p

m+pm� 1

�n=pk +

pk � 1:

Solution by Francisco Bellot Rosado, I.B. Emilio Ferrari, Valladolid,

Spain andMaria Ascensi �on L �opez Chamorro, I.B. LeopoldoCano, Valladolid,

Spain. We prove a more general result:

Let a � b be non-negative real numbers, and let n 2 N. Let

p =(a+ b)2 � (a� b)n

2:

Then

p2 +�a2 � b2

�n=

1

4

�(a+ b)n + (a� b)n

�2;

giving

143

qp2 + (a2 � b2)

n=

1

2

�(a+ b)n + (a� b)n

�= (a+ b)n � p:

This given

qp2 + (a2 � b2)

n+pp2 = (a+ b)n: (1)

In (1), set a =ps, b =

ps� k, where s; k 2 N; s � k, so that p2 is a

positive integer, and we have

�ps+

ps� k

�n=pp2 +

pp2 � k2: (2)

Letting k = 1 in (2), gives the result of the original problem here:

�ps+

ps� 1

�n=pp2 +

pp2 � 1: (3)

If we let s = 2 in (3), we get�p2 + 1

�n=pp2 +

pp2 � 1;

which was a problem proposed by Hong Kong (but not used) at the 1989IMO at Braunschweig, Germany. This version also appears as ElementaryProblem E 950 (proposed by W.R. Ransom, Tufts College) in the AmericanMathematical Monthly 58 (1951) p. 566, on which we have based our solu-tion.

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; CARL BOSLEY,

student,Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER

J. BRADLEY, CliftonCollege, Bristol,UK;MIGUELANGEL CABEZ �ONOCHOA,

Logro ~no, Spain; ADRIAN CHAN, student, Upper Canada College, Toronto,

Ontario; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut,

USA; TOBY GEE, the John of Gaunt School, Trowbridge, England; RICHARD

I. HESS, Rancho Palos Verdes, California, USA; JOHN G. HEUVER, Grande

Prairie Composite High School, Grande Prairie, Alberta; CYRUS HSIA, stu-

dent, University of Toronto, Toronto, Ontario; PETER HURTHIG, Columbia

College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck,

Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,

USA; VICTOR OXMAN, Haifa, Israel; P. PENNING, Delft, the Netherlands;

GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; CRIST �OBAL

S �ANCHEZ{RUBIO, I.B. Penyagolosa, Castell �on, Spain; SCIENCE ACADEMY

PROBLEM SOLVERS; Austin, Texas, USA; HEINZ-J�URGEN SEIFFERT, Berlin,

Germany; TOSHIO SEIMIYA, Kawasaki, Japan; PANOS E. TSAOUSSOGLOU,

Athens, Greece; HOE TECK WEE, student, Hwa Chong Junior College, Singa-

pore; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer.

One incorrect submission was received.

144

Chan states that the problem is in the 1980 Romanian Mathematical

Olympiad. Perz observes that the special casem = 10 occurs in CRUX [1977:190], and, along with Hsia, notes that the case m = 2 was in the 1994

Canadian Mathematical Olympiad (see CRUX [1994: 151], [1994: 186])

2039?. [1995: 130] Proposed by Dong Zhou, FudanUniversity, Shang-hai, China, and Ji Chen, Ningbo University, China.

Prove or disprove that

sinA

B+

sinB

C+

sinC

A�

9p3

2�;

where A,B, C are the angles (in radians) of a triangle. [Compare with CRUX1216 [1988: 120] and this issue! ]

Solution by Douglass L. Grant, University College of Cape Breton, Syd-

ney, Nova Scotia.

In problem 2015 [1995: 53 and 129 (Corrected), 1996: 47], it was shownthat

K = (sinA+ sinB + sinC)

�1

A+

1

B+

1

C

��

27p3

2�;

where A, B, C are the angle of a triangle measured in radians, and that thelower bound is attained in the equilateral triangle.

Let

F =sinA

A+

sinB

B+

sinC

C;

G =sinA

B+

sinB

C+

sinC

A;

H =sinA

C+

sinB

A+

sinC

B:

Then K = F + G +H. By problem 1216 [1987: 53, 1988: 120], we have

2 < F � 9p3

2�, so that we have G+H � 18

p3

2�.

But G can be obtained fromH by a permutation of symbols. Thus each ofG

and H has the same minimum, which will be at least 9p3

2�.

But, in the equilateral triangle,G (andH) achieves this value, so the requiredinequality is true and sharp.

No other submission was received.

145

On the Sum of n Dice

J.B. KlerleinDepartment of Mathematics, Western Carolina University,

Cullowhee, North Carolina, USA.

Introduction

An interesting problem can be found in Tucker [2, p. 421] which we rephrase

as follows.

Suppose n distinct fair dice are rolled and S is the sum of their

faces. Show that the probability that 2 divides S is 12.

There are several ways to solve this problem. One, which we will use, lends

itself to generalization.

Die Two

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

Die 3 4 5 6 7 8 9

One 4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

Table 1. Possible Sums of Two Dice

The result is clear for one die, and easily veri�ed for two dice. See Table 1.

The thirty-six entries in Table 1 each have the same probability. Since there

are eighteen even sums among the thirty-six entries, the result follows for

n = 2. A sub-problem is, how to represent the sum of three, or more gen-

erally n, dice while maintaining the di�erent frequencies for di�erent sums.

That is, the sum 3 is only obtained in one way using three dice, while the

sum 15 can be obtained in ten ways with three dice. Fortunately, these fre-

quencies are unnecessary for this solution.

In Table 2, we label the rows by the possible sums of (n� 1) dice and the

six columns by the possible values of the nth die. We note that, unlike the

entries in Table 1, the entries in Table 2 are not equi-probable. However,

the table does allow us to count the number of even sums among the 6n

possible outcomes from rolling n dice. For suppose there are fn+2 ways to

obtain the sum (n + 2) when (n � 1) dice are rolled. Then by looking at

the row labelled by (n + 2), we can account for 3fn+2 even sums among

the 6n possible outcomes for n dice, since there are three even entries in this

row. Since each row of the table consists of six consecutive integers, there

146

Die n

1 2 3 4 5 6

n� 1 n n+ 1 n+ 2 n+ 3 n+ 4 n+ 5

n n+ 1 n+ 2 n+ 3 n+ 4 n+ 5 n+ 6

Sum of n+ 1 n+ 2 n+ 3 n+ 4 n+ 5 n+ 6 n+ 7

(n� 1) n+ 2 n+ 3 n+ 4 n+ 5 n+ 6 n+ 7 n+ 8

dice

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

6n� 7 6n� 6 6n� 5 6n� 4 6n� 3 6n� 2 6n� 1

6n� 6 6n� 5 6n� 4 6n� 3 6n� 2 6n� 1 6n

Table 2. Possible Sums of n Dice

are exactly three even entries in each row. Thus, we can sum the frequencies

(fn�1+ fn+ fn+2+ : : :+ f6n�6) and multiply this sum by 3 to obtain the

number of even sums among the 6n possible outcomes. As the sum of the

frequencies is 6n�1, we have the probability that 2 divides S is

3 � 6n�16n

=1

2:

As mentioned above, this method generalizes easily. Let us �rst state two

problems. Suppose n dice are rolled, and let Sn be the sum of their faces.

Let P (kjSn) be the probability that k divides Sn.

1. Find values of k such that P (kjSn) = 1

kfor all n.

2. Find values of k and n for which P (kjSn) = 1

k.

Problem 1. Find values of k such that P (kjSn) = 1kfor all n.

As mentioned, each row in Table 2 consists of six consecutive integers. Thus,

in each row, two entries are divisible by 3, and one entry is divisible by 6.

It follows, in exactly the same manner as above, that P (3jSn) = 13and

P (6jSn) = 16. Clearly, P (1jSn) = 1. Combining these observations with

our �rst result, we have that, for all n; P (kjSn) = 1kwhen k = 1; 2; 3; and 6.

By considering n = 1 (or n = 3, if we wish to avoid the trivial case), we see

that the only values of k that hold for all n are precisely k = 1; 2; 3; and 6.

Problem 2. Find values of k and n for which P (kjSn) = 1k.

Let n and k satisfy P (kjSn) = 1k. Since 6nP (kjSn) is an integer, it fol-

lows that k divides 6n, that is, k is of the form 2s3t for some non-negative

integers s and t.

For 0 � s; t � 1 we have k = 1; 2; 3; or 6. These values of k, as we have

seen, are precisely the solutions to Problem 1. Let us consider k = 4.

147

We shall show that P (4jSn) = 14if and only if n � 2 (mod 4). We note

that in Table 2 each row does not contain the same number of multiples of 4.

For this reason, we are led to consider another approach, in particular, the

congruence classes depicted in Table 3.

1 2 3 4 5 6

0 1 2 3 0 1 2

1 2 3 0 1 2 3

2 3 0 1 2 3 0

3 0 1 2 3 0 1

Table 3. Congruence Classes Modulo 4 for n Dice

We construct Table 3 in a similar fashion to Table 2, except that the congru-

ence classes 0; 1; 2; and 3 (modulo 4) label the rows. The columns are still

labelled by the face values of thenth die. The entry in the ith row and the jth

column is the congruence class of the sum of an element from the congruence

class of the ith row and the face value of the jth column.

Let P (n;� i) be the probability that the sum of n dice is congruent

to i (mod 4) for i = 0; 1; 2; 3: Note that P (4jSn) is precisely P (n;� 0).

The following recurrence relations are a consequence of Table 3.

6P (n;� 0) = 2P (n� 1;� 3) + 2P (n� 1;� 2)

+P (n� 1;� 1) + P (n� 1;� 0)

= P (n� 1;� 3) + P (n� 1;� 2) + 1

6P (n;� 1) = 2P (n� 1;� 3) + 2P (n� 1;� 0)

+P (n� 1;� 1) + P (n� 1;� 2)

= P (n� 1;� 3) + P (n� 1;� 0) + 1

6P (n;� 2) = 2P (n� 1;� 0) + 2P (n� 1;� 1)

+P (n� 1;� 2) + P (n� 1;� 3)

= P (n� 1;� 0) + P (n� 1;� 1) + 1

6P (n;� 3) = 2P (n� 1;� 1) + 2P (n� 1;� 2)

+P (n� 1;� 3) + P (n� 1;� 0)

= P (n� 1;� 1) + P (n� 1;� 2) + 1

From these recurrence relations, we observe that

P (n;� 0) + P (n;� 2) = 12

and P (n;� 1) + P (n;� 3) = 12:

With a little iteration, we obtain P (n;� 0) = 3231296

+ 41296

P (n� 4;� 2).

148

Replacing the last term by�12� P (n� 4;� 0)

�we have,

P (n;� 0) = 14+�

11296

� 41296

P (n� 4;� 0)�:

It follows that P (4jSn) = 14if and only ifP (4jSn�4) = 1

4. NowP (4jS2) = 1

4,

but P (4jSm) 6= 14ifm = 1; 3; or 4.

Thus, P (4jSn) = 14if and only if n � 2 (mod 4).

Conclusion

What about other possible k's (k = 2s � 3t) for which it may be that

P (kjSn) = 1kfor some n? To be honest, we don't know. We have checked

those k's which are less than 54 for n � 55. For k = 8 and 9 respectively,

the only values of n � 55 for which P (kjSn) = 1kare n = 3 and 2 respec-

tively. For n � 55; P (kjSn) = 1knever holds for 12 < k < 54. The last

value for which we could report is k = 12. But rather than spoil the fun of

the interested reader, we say no more.

References

1. Ne�, J.D. Dice tossing and Pascal's triangle. Two-Year College Math-ematics Journal, 13 (1982), pp. 311-314.

2. Tucker, A. Applied Combinatorics, 2nd Ed., John Wiley and Sons, 1984.

Here are three curves that divide a circle into four equal areas.

Any other nice examples?

149

THE SKOLIAD CORNER

No. 14

R.E. Woodrow

First this month, an observation by an astute reader, Derek Kisman,

student, Queen Elizabeth High School, Calgary, who noticed that the �gure

we gave for question 25 of the Sharp U.K. Intermediate Mathematical Chal-

lenge, 1995, [1996: 14] is missing the crucial upright F in the net. It should

be in the top right corner!

As a contest this month we give the European \Kangaroo" Mathemat-

ical Challenge, written Thursday 23 March, 1995. It is organized by the

U.K. Mathematics Foundation and L'Association Europ �eennes �Kangarou

des Math �ematiques�. The contest is for students at about school year nine

or below. This was written by about 5000 students in the U.K. My thanks

go to Tony Gardiner of the U.K. Mathematics Foundation for sending the

materials to Crux.

EUROPEAN \KANGAROO" MATHEMATICALCHALLENGE23 March 1995

1. 1� 9� 9� 5� (1 + 9 + 9 + 5) makes:

A. 0 B. 381 C. 481 D. 429 E. 995

2. Which shape does not appear in this �gure?

A. circle

B. square

C. right-angled triangle

D. isosceles triangle

E. equilateral triangle&%'$����@

@@@

3. The whole numbers from 1 to 1995 are alternately added and sub-

tracted thus: 1� 2+3� 4+5� 6+ � � �+1993� 1994+1995. What is the

result?

A. 997 B. 1995 C. 998 D. 0 E. �9974. What is the angle between the hour hand and the minute hand of a

clock at 1.30?

A. 180� B. 120� C. 130� D. 150� E. 135�

5. C1 is a circle of radius 6cm, C2 is a circle of radius 8cm. Jos �e wants

the two circles to touch tangentially. He knows that there are two possibili-

ties for the distance between their centres. What are these two distances?

A. 3 and 4cm B. 2 and 8cm C. 2 and 14cm D. 6 and 8cm E. 6 and 14cm

150

6. A train 1km long is restricted to travel through a tunnel of length

1km at 1km/h. How long does it take the train to pass through the tunnel?

A. 1 hour B. 1h 30m C. 2 hours D. 3 hours E. 1=2 hour

7. The angle x in the �gure is equal to:

A. 20�

B. 22�

C. 24�

D. 26�

E. 28�

64�22�

2x

x

22�

8. Which number is smallest?

A. 1995 B. 19� 95 C. 1995 D. 1995 E. 1995

9. What is the ratio of the perimeter of the

shaded region to the circumference of the circle?

A. 3

4B. 4+�

4�C. 2�

4+�D. 4+�

2�E. 1

4

10. After two successive 20% reductions, a coat costs $320. What was

its original price?

A. $204 B. $400 C. $448 D. $500 E. $533

11. Nine people are sitting in a room; their average age is 25. In

another room eleven people are gathered their average age is 45. If the two

groups were to combine, what would their average age be?

A. 70 B. 36 C. 35 D. 32 E. 20

12. How many squares are there in this �gure?

A. 13

B. 14

C. 19

D. 21

E. 23

151

13. The quadrilateral PQRS is a rectangle; M is any point on the

diagonal SQ. What can one say for sure about the two shaded regions?

A. the upper area is larger

B. the lower area is larger

C. they always have equal areas

D. the two areas are equal only ifM is the

midpoint of SQ

E. insu�cient information ��������

P

M

Q

RS

14. A metallic disc of diameter 20cm weighs 2:4kg. From it one cuts

out a disc of diameter 10cm. What does it weigh?

A. 1:2kg B. 0:8kg C. 0:6kg D. 0:5kg E. 0:4kg

15. What is the area (in unit squares) of the quadrilateral MNPQ?

A. 9

B. 10

C. 11

D. 12

E. 13

���������HH

HH

Q

MN

P

16. Six hundred and twenty-�ve students enter a 100m competition.

The track has �ve lanes. At the end of each race the winner survives, while

the other four are eliminated. How many races are needed to determine the

champion sprinter?

A. 98 B. 106 C. 125 D. 126 E. 156

17. In the �gure O is the centre of

the circle and OA = BC. Which of the

following relations holds?

A. 2x = 3y B. x = 2y C. x = y

D. x+ y = 90� E. x+ 2y = 180�

O

A

B

Cx

y

18. What is the sum of all the digits in the number 1095 � 95?

A. 6 B. 29 C. 108 D. 663 E. 842

19. In a group of pupils, 40% of the pupils have poor eyesight; 70% of

these wear glasses, while the other 30% wear contact lenses. If there are 21

pairs of glasses, which of these assertions is true?

A. 45 pupils have poor eyesight

B. 30 pupils have good eyesight

C. there are 100 pupils in the group

D. 10 pupils wear contact lenses

E. none of the other assertions is true

152

20. One has to arrange four pawns in this �g-

ure so that each column contains one pawn, and each

row contains at most one pawn. How many di�erent

arrangements are possible?

A. 64 B. 28 C. 16 D. 8 E. 4

21. Let X = X4 and X j Y = X + Y . Then 2 j 2 equals:

A. 3� 24 B. 29 C. 212 D. 220 E. 23

22. Bruce buys three ostriches, seven koalas and one kangaroo. Dar-

ren buys four ostriches, ten koalas and one kangaroo. Sheila buys one ostrich,

one koala and one kangaroo. Bruce pays 3150 Australian dollars and Darren

pays 4200 Australian dollars. How many Australian dollars must Sheila pay?

A. 1700 B. 1650 C. 1200 D. 1050 E. 950

23. The diagonal of an isosceles

trapezium is 16cm long, and makes an an-

gle of 45� with the base. What is the area

of the trapezium?

A. 64cm2 B. 96cm2 C. 128cm2

D. more information required E. 256cm2

45�

16cm

24. Each positive whole number which (in base 10) can be written as

a string of 1's and 2's only is called simple. For example 22121 and 2222 are

simple; 1021 is not simple. How many simple numbers are there less than

one million?

A. 63 B. 62 C. 127 D. 128 E. 126

25. What is the maximum possible number of points of intersection

one can get with eight circles?

A. 16 B. 32 C. 38 D. 44 E. 56

Next we give solutions to the contest given last issue, The EleventhW.J.

Blundon Contest.

THE ELEVENTH W.J. BLUNDON CONTESTFebruary 23, 1994

1. (a) The lesser of two consecutive integers equals �ve more than

three times the larger integer. Find the two integers.

Solution. Let the integers be n, n + 1. Then n = 3(n + 1) + 5 so

n = �4.

153

(b) If 4 � x � 6 and 2 � y � 3, �nd the minimum values of

(x� y)(x+ y).

Solution. If 4 � x � 6 and 2 � y � 3, then (x�y)(x+y) = x2�y2 �42 � 32 = 16� 9 = 5.

2. A geometric sequence is a sequence of numbers in which each term

after the �rst can be obtained from the previous term by multiplying by the

same �xed constant, called the common ratio. If the second term of a geo-

metric sequence is 12 and the �fth term is 81=2, �nd the �rst term and the

common ratio.

Solution. Let the �rst term be a, and the ratio r. Then the sequence

is a; ar; arr = ar2; ar3; ar4; ar5; : : : , with the nth term being arn�1. So

ar = 12 and ar4 = 81=2. Dividing gives

r3 =ar4

ar=

81=2

12=

81

24=

27

8;

so r = 32. Since ar = a � 3

2= 12, a = 8.

3. A square is inscribed in an equilateral triangle. Find the ratio of the

area of the square to the area of the triangle.

Solution. Let the equilateral triangleABC and the squareDEFG be as shown.

Now \ADG = \ABC = \AGD =

\ACB = 60� so ADG is an equilateral

triangle with side length s, the length of

the side of the square. If DE is superim-

posed over GF a third equilateral triangle

is formed with height s.

s

s=2

A

E CF

GD

B

Now the area of an equilateral triangle with side length s isp34s2, while

the area of an equilateral triangle with height s is 1p3s2. So the area of the

square is s2 and the area of ABC is

p3

4s2 + s2 +

s2p3=

1 +

p3

4+

1p3

!s2:

The ratio of the area of the square to the area of the triangle is then

1

1 + 7

4p3

=4p3

4p3 + 7

=4p3

�48 + 49(�4

p3 + 7) = 28

p3� 48:

154

4. ABCD is a square. Three parallel lines l1, l2 and l3 pass through

A, B andC respectively. The distance between l1 and l2 is 5 and the distance

between l2 and l3 is 7. Find the area of ABCD.

Solution. Let the line perpendicular

to l2 through B meet l1 at E and

l2 at F respectively. Let l2 meet

AD at G. Then \BAE = \CBF

and \AEB = 90� = \BFC so

�BAE is similar to �BCF . If the

side length is s this gives

sps2 � 52

=s

7

so s2 = 52 + 72 = 25 + 49 = 74.

A B

C

E

F

s

s

5

7

l1

l3

l2

D

G

5. The sum of the lengths of the three sides of a right triangle is 18.

The sum of the squares of the lengths of the three sides is 128. Find the area

of the triangle.Solution.

a2 + b2 = c2

a+ b+ c = 18

and a2 + b2 + c2 = 128: a

b c

So 2c2 = 128, c2 = 64 and c = 8. Thus a+ b = 10, a2 + b2 = 64, so

2ab = (a+ b)2 � (a2 + b2)

= 100� 64 = 36

The area 12ab = 2ab

4= 36

4= 9.

6. A palindrome is a word or number that reads the same backwards

and forwards. For example, 1991 is a palindromic number. How many palin-

dromic numbers are there between 1 and 99; 999 inclusive?

Solution. Let us �nd the number nk of the properly k digit palindromic

numbers. To be a k digit number the �rst digit must be one of 1; 2; 3; : : : ; 9.

The answer now depends whether k is even or odd.

Now n1 = 9. If k > 1 is odd, k = 2l + 1, l � 1 then nk = 9� 10l. If

k is even, k = 2l, l � 1 then nk = 9� 10l�1. For the answer we want

n1 + n2 + n3 + n4 + n5 = 9+ 9 + 90 + 90 + 900 = 1098:

7. A graph of x2�2xy+y2�x+y = 12 and y2�y�6 = 0will produce

four lines whose points of intersection are the vertices of a parallelogram.

Find the area of the parallelogram.

Solution. Note that x2 � 2xy + y2� x+ y = 12 is equivalent to

(x� y)2 � (x� y)� 12 = 0 or

155

((x� y)� 4)(x� y+ 3) = 0

Thus its graph is the two parallel lines x� y � 4 = 0 and x� y + 3 = 0.

Also y2 � y � 6 = 0 is equivalent to (y � 3)(y + 2) = 0, giving two

horizontal lines y � 3 = 0 and y + 2 = 0. The sides of the parallelogram

lie on these lines. It has base 7 and height 5. (The distance between the

horizontal lines.)

8. Determine the possible values of c so that the two lines x� y = 2

and cx+ y = 3 intersect in the �rst quadrant.

Solution. For the curves to intersect in the �rst quadrant we need x �0, y � 0. Adding the equations gives (c+1)x = 5, so we require c+1 > 0,

or c > �1. Substitution of x = 5c+1

into x� y = 2 gives

y =5

c+ 1� 2 =

3� 2c

c+ 1

and we need 3� 2c � 0, so c � 32. The intersection is in the �rst quadrant

for �1 < c � 32.

9. Consider the function f(x) = cx2x+3

, x 6= �3=2. Find all values of

c, if any, for which f(f(x)) = x.

Solution.

f(f(x)) =c�

cx2x+3

�2�

cx2x+3

�+ 3

for x 6= �32

and f(x) 6= �3=2, i.e. cx=(2x + 3) 6= �3=2, x 6= �9=(6 + 2c). For these

excluded values f(f(x)) = x gives

c�

cx2x+3

�2�

cx2x+3

�+ 3

= x

so

c2x = x(2cx+ 3(2x+ 3))

c2x = 2cx2 + 6x2 + 9x

so equating coe�cients of x, x2, etc. (since the polynomials are equal for

in�nitely many values of x) gives c2 = 9 and 2c = 6, so c = 3.

10. Two numbers are such that the sum of their cubes is 5 and the

sum of their squares is 3. Find the sum of the two numbers.

Solution. Let the numbers be x and y. We are given x3 + y3 = 5 and

x2 + y2 = 3.

Let x+ y = A.

From (x+ y)3 = x3+ 3x2y+3xy2+ y3 = x3 + y3+ 3xy(x+ y) we

have A3 = 6+ 3xyA.

156

From (x + y)2 = x2 + y2 + 2xy we have A2 = 3 + 2xy or xy =1

2(A2 � 3).

So we get A3 = 5+ 3

2(A2 � 3)A or

1

2A3 +

9

2A+ 5 = 0 and A3 + 9A+ 10 = 0

this gives A = �1.

That completes the Skoliad Corner for this issue. Send me your contest

materials, comments, suggestions, and solutions.

Historical Titbit

Taken from a 1950's University Scholarship Paper.

A box contains forty-eight chocolates all exactly similar in appearance.

There are four of each of twelve di�erent sorts.

How many must you take out to be certain of having at least one of each of

�ve di�erent sorts?

Suppose that, having taken this number out, you eat four of them, and �nd

that they are all of one sort.

How many must you now put back to be certain that the box contains at least

two of each of eight di�erent sorts?

157

THE OLYMPIAD CORNER

No. 174

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 set this number we give the third Team Compe-

tition, the Baltic Way 1994. The contest was written in Vilnius, Lithuania.

Teams from Denmark, St. Petersburg, Poland, Latvia, Iceland, Lithuania, Es-

tonia and Sweden participated. My thanks go to Georg Gunther for collecting

this problem set when he was Canadian Team Leader to the IMO at Istanbul.

MATHEMATICAL TEAM CONTEST\BALTIC WAY | 92"

Vilnius, 1992 | November 5{8

1. Let p, q be two consecutive odd prime numbers. Prove that p+ q

is a product of at least 3 positive integers > 1 (not necessarily di�erent).

2. Denote by d(n) the number of all positive divisors of a positive

integer n (including 1 and n). Prove that there are in�nitely many n such

that nd(n)

is an integer.

3. Find an in�nite non-constant arithmetic progression of positive in-

tegers such that each term is neither a sum of two squares, nor a sum of two

cubes (of positive integers).

4. Is it possible to draw a hexagon with vertices in the knots of an in-

teger lattice so that the squares of the lengths of the sides are six consecutive

positive integers?

5. Given that a2 + b2 + (a + b)2 = c2 + d2 + (c + d)2, prove that

a4 + b4 + (a+ b)4 = c4 + d4 + (c+ d)4.

6. Prove that the product of the 99 numbers k3�1k3+1

, k = 2; 3; : : : ; 100,

is greater than 23.

7. Let a =1992p1992. Which number is greater:

ap p pa

a

a

9>>>>=>>>>;1992 or 1992?

158

8. Find all integers satisfying the equation

2x � (4� x) = 2x+ 4:

9. A polynomial f(x) = x3 + ax2 + bx + c is such that b < 0 and

ab = 9c. Prove that the polynomial has three di�erent real roots.

10. Find all fourth degree polynomials p(x) such that the following

four conditions are satis�ed:

(i) p(x) = p(�x), for all x,(ii) p(x) � 0, for all x,

(iii) p(0) = 1,

(iv) p(x) has exactly two local minimum points x1 and x2 such that

jx1 � x2j = 2.

11. Let Q+ denote the set of positive rational numbers. Show that

there exists one and only one function f : Q+! Q+ satisfying the following

conditions:

(i) If 0 < q < 12then f(q) = 1 + f

�q

1�2q

�.

(ii) If 1 < q � 2 then f(q) = 1 + f(q� 1).

(iii) f(q) � f(1q) = 1 for all q 2 Q+.

12. Let N denote the set of positive integers. Let ' : N ! N be a

bijective function and assume that there exists a �nite limit

limn!1

'(n)

n= L:

What are the possible values of L?

13. Prove that for any positive x1; x2; : : : ; xn, y1; y2; : : : ; yn the in-

equalitynXi=1

1

xiyi� 4n2Pn

i=1(xi + yi)2

holds.

14. There is a �nite number of towns in a country. They are connected

by one direction roads. It is known that, for any two towns, one of them can

be reached from the other one. Prove that there is a town such that all the

remaining towns can be reached from it.

15. Noah has 8 species of animals to �t into 4 cages of the ark. He

plans to put species in each cage. It turns out that, for each species, there

are at most 3 other species with which it cannot share the accommodation.

Prove that there is a way to assign the animals to their cages so that each

species shares with compatible species.

16. All faces of a convex polyhedron are parallelograms. Can the poly-

hedron have exactly 1992 faces?

159

17. Quadrangle ABCD is inscribed in a circle with radius 1 in such

a way that one diagonal, AC, is a diameter of the circle, while the other

diagonal, BD, is as long as AB. The diagonals intersect in P . It is known

that the length of PC is 2

5. How long is the side CD?

18. Show that in a non-obtuse triangle the perimeter of the triangle

is always greater than two times the diameter of the circumcircle.

19. Let C be a circle in the plane. Let C1 and C2 be nonintersecting

circles touching C internally at points A and B respectively. Let t be a com-

mon tangent of C1 and C2, touching them at points D and E respectively,

such that both C1 and C2 are on the same side of t. Let F be the point of

intersection of AD and BE. Show that F lies on C.

20. Let a � b � c be the sides of a right triangle, and let 2p be

its perimeter. Show that p(p � c) = (p � a)(p � b) = S (the area of the

triangle).

As a second Olympiad set to give you recreation over the summer

months, we give the 8th Iberoamerican Mathematical Olympiad written

September 14{15, 1993 in Mexico.

8th IBEROAMERICAN MATHEMATICAL OLYMPIADSeptember 14{15, 1993 (Mexico)

First Day | 4.5 hours

1. (Argentina) Let x1 < x2 < � � � < xi < xi+1 < � � � be all the

palindromic natural numbers, and for each i, let by yi = xi+1 � xi. How

many distinct prime numbers belong to the set fy1; y2; y3; : : : g?2. (Mexico) Show that for any convex polygon of unit area, there exists

a parallelogram of area 2 which contains the polygon.

3. (Mexico) Let N� = f1; 2; 3; : : : g. Find all the functions

f : N�! N�

such that

(i) If x < y, then f(x) < f(y)

(ii) f(yf(x)) = x2 � f(xy), for all x, y belonging to N�.

Second Day | 4.5 hours

4. (Spain) Let ABC be an equilateral triangle, and � its incircle. If

D and E are points of the sides AB and AC, respectively, such that DE is

tangent to �, show thatAD

DB+AE

EC= 1:

160

5. (Mexico) Let P and Q be distinct points of the plane. We denote

m(PQ) the perpendicular bisector of the segment PQ. Let S be a �nite

subset of the plane, withmore than one element, which satis�es the following

properties:

(i) If P and Q are distinct points of S, thenm(PQ) intersects S.

(ii) IfP1Q1, P2; Q2 andP3Q3 are three distinct segments with extreme points

belonging to S, then no point of S belongs simultaneously to the three lines

m(P1Q1),m(P2Q2), m(P3Q3).

Determine the number of possible points of S.

6. (Argentina) Two non-negative integer numbers, a and b, are \cu-

ates" (friends in Mexican) if the decimal expression of a+b is formed only by

0's and 1's. Let A and B be two in�nite sets of non-negative integers, such

that B is the set of all the numbers which are \cuates" of all the elements of

A, and A is the set of all the numbers which are \cuates" of all the elements

of B. Show that in one of the sets A or B there are in�nitely many pairs of

numbers x, y such that x� y = 1.

We now turn to the readers' comments and solutions to problems given

in the December 1994 number of the Corner and the Nordic Mathematical

Contest, 1992 [1994: 277].

1. [1994: 277] Determine all real numbers x, y, z greater than 1,

satisfying the equation

x+ y + z +3

x� 1+

3

y � 1+

3

z � 1= 2

�px+ 2 +

py + 2 +

pz + 2

�:

Solutions by �Sefket Arslanagi�c, Berlin, Germany; by Cyrus C. Hsia, stu-dent, Woburn Collegiate Institute, Toronto; by Chandan Reddy, Rochester,Michigan; Michael Selby, University of Windsor, Windsor, Ontario; and byD.J. Smeenk, Zaltbommel, the Netherlands. We give Hsia's solution.

For a > 1, a 2 R we have the Arithmetic Mean{Geometric Mean

inequality

a� 1 +a+ 2

a� 1� 2

pa+ 2;

with equality if and only if a � 1 = a+2

a�1 , or a2 � 3a � 1 = 0, giving a =

(3 +p13)=2, since a > 1.

For each of a = x, y, z we have this same result and adding them gives

x+ y+ z� 3+x+ 2

x� 1+y+ 2

y� 1+z + 2

z � 1� 2(

px+ 2+

py + 2+

pz + 2)

whence

x+ y + z +3

x� 1+

3

y� 1+

3

z � 1� 2(

px+ 2+

py+ 2+

pz + 2)

161

with equality if and only ifx = y = z = (3+p13)=2. Since there is equality,

the unique solution is x = y = z = (3 +p13)=2.

2. [1994: 277] Letn be an integer greater than 1 and let a1; a2; : : : ; anbe n di�erent integers. Prove that the polynomial f(x) = (x � a1)(x �a2) : : : (x�an)�1 is not divisible by any polynomial of positive degree less

than n and with integer coe�cients and leading coe�cient 1.

Solutions by Cyrus C. Hsia, Woburn Collegiate Institute, Toronto, On-tario; by Murray S. Klamkin, University of Alberta, Edmonton, Alberta; byChandan Reddy; Rochester, Michigan; and by Michael Selby, Universityof Windsor, Windsor, Ontario. We give the solution sent in by Selby andKlamkin's comment.

Suppose p(x) divides f(x), where 1 � degp < n, and p is monic with

integral coe�cients. Therefore f(x) = p(x)q(x) where q(x) is also monic,

with integral coe�cients and 1 � deg q < n.

Now p(ai)q(ai) = �1, i = 1; 2; : : : ; n. Since p(ai), q(ai) are integers,

p(ai) = 1 and q(ai) = �1 or p(ai) = �1 and q(ai) = 1 for each i =

1; 2; : : : ; n.

In either case p(ai) + q(ai) = 0 for i = 1; 2; : : : ; n.

Consider p(x)+ q(x). This is a polynomial of positive degree less than

n, since both p(x) and q(x) are monic with degree less than n. However

p(ai) + q(ai) = 0 for n distinct values, but has positive degree less than n,

an impossibility. Therefore no such p(x) exists.

[Editor's Note.] Klamkin (whose solution was similar) points out that the

problem is well-known.

3. [1994: 277] Prove that among all triangles with given incircle, the

equilateral one has the least perimeter.

Solutionsby �Sefket Arslanagi�c, Berlin, Germany; byMurray S. Klamkin,University of Alberta, Edmonton, Alberta; by Bob Prielipp, University ofWisconsin{Oshkosh, Wisconsin, USA; and by Chandon Reddy, Rochester,Michigan. We �rst give Reddy's solution.

Let the three sides be a, b, c. Then the inradius r times the semiperime-

ter is the area A = rs, r = 1 so, using Heron's formulaps(s� a)(s� b)(s� c) = s

or p(s� a)(s� b)(s� c) =

ps

so

s = (s� a)(s� b)(s� c): (1)

Also

s = (s� a) + (s� b) + (s� c): (2)

162

We minimize the perimeter and therefore s by the AM{GM inequality

and (1) and (2)

p(s� a)(s� b)(s� c) � (s� a) + (s� b) + (s� c)

3

The RHS is minimized when we have equality, that is s� a = s� b = s� c,

and a = b = c.

Next we give Klamkin's comments and alternative approaches.

Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta. This is a well-known result. It follows immediately from the isoperi-

metric theorem for polygons, that is, of all n-gons with given perimeter, the

regular one has the maximum area and, dually, of all polygons of given area,

the regular one has the least perimeter. For triangles we have the inequality

p2

F� p20

F0= 12

p3

where p and F denote the perimeter and area of a general triangle and p0; F0correspond to the same for an equilateral triangle. Since F = rp=2 (r =

inradius), we have

p � 6p3 r

and with equality if and only if the triangle is equilateral.

For another proof in terms of the angles A, B, C of the triangle, it is

equivalent to establishing the known triangle inequality

2r

�cotA

2+

cotB

2+

cotC

2

�� 6

p3 r

It also follows that of all n-gons circumscribed to a given circle, the

regular one has the least perimeter (and area as well).

Comment: These results generalize to simplexes. In particular for the

tetrahedron, we have the following known inequalities, where E = the sum

of the 6 edgesPEi, F = the sum of the areas of the 4 faces

PFi, V = the

volume and r = the inradius,

E � 6Y

E1=6

i � k1V1=3; (1)

F � 4Y

F1=4

i � k2V2=3; (2)

and there is equality if the tetrahedron is regular so that the constants k1,

k2 are determined by taking Ei = 1. Then, Fi =p3=4 and V =

p2=12.

On multiplication of (1) and (2), we get EF � k1k2V = k1k2rF3

or that

E � k1k2r3

. This proves a 3-dimensional extension of the given result.

163

4. [1994: 277] Peter has a great number of

squares, some of them are black, some are white. Us-

ing these squares, Peter wants to construct a square,

where the edge has length n, and with the following

property: The four squares in the corners of an arbi-

trary subrectangle of the big square, must never have

the same colour. How large a square can Peter build?n = 6

Comment by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta.

That the answer is 4 � 4 is given in the solution of problem #1 of the

U.S.A. 1976 Mathematical Olympiad [1].

Reference[1]M.S.Klamkin, U.S.A.Mathematical Olympiads 1972{1986,M.A.A., Wash-

ington, D.C. 1988, pp. 93{94.

To complete this number we given two comments by Murray Klamkin

about earlier solutions.

3. [1994: 279; 1993: 131] The abscissa of a point which moves in the

positive part of the axis Ox is given by x(t) = 5(t + 1)2 + a=(t + 1)5, in

which a is a positive constant. Find the minimum a such that x(t) � 24 for

all x � 0.

Comment by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta.

A simpler non-calculus solution is given by applying the AM{GM in-

equality, that is,

5(t+ 1)2 + 2n

1

2a

(t+1)5

o[5 + 2]

��a2

4

�1=7:

Then we have

7

�a2

4

�1=7� 24;

so that

mina = 2

�24

7

�7:

5. [1994: 281; 1993: 132]

For each natural number n, let (1 +p2)2n+1 = an + bn

p2 with an and bn

integers.

(a) Show that an and bn are odd for all n.

(b) Show that bn is the hypotenuse of a right triangle with legs

an + (�1)n2

andan � (�1)n

2:

164

Comment by Murray S. Klamkin, University of Alberta, Edmonton, Al-

berta.

A more direct solution of part (a) is as follows:

an =(p2 + 1)2n+1� (

p2� 1)2n+1

2

=

�2n+ 1

1

�2n +

�2n+ 1

3

�sn�1 + � � �+ 1;

bn =(p2 + 1)2n+1+ (

p2� 1)2n+1

2p2

= 2n +

�2n+ 1

2

�2n�1 + � � �+

�2n+ 1

2n

�:

That completes the Corner for this issue. Have a good summer { spend

some time solving problems and send me your nice solutions as well as

Olympiad Contest materials.

Book wanted!

Bruce Shawyer would like to purchase a copy of the out-of-print book:

On Mathematics and Mathematicians (Memorabilia Mathematica)

by Robert Edouard Moritz.

Dover Edition published 1942.

Originally published as:

Memorabilia Mathematica or The Philomath's Quotation-Book.

Original Edition published 1914.

Anyone willing to part with a copy please send him details. Thank you.

165

THE ACADEMY CORNER

No. 3

Bruce Shawyer

All communications about this column should be sent to ProfessorBruce Shawyer, Department of Mathematics and Statistics, Memorial Uni-versity of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Spring is approaching in parts of Canada, and, as undergraduate stu-

dents in North America prepare for �nal examinations, some �nd time to

try their skills on some mathematics problems. Memorial University o�ers

some modest prizes for the best students. How do you compare? Please

send me your best solutions to these problems.

MEMORIAL UNIVERSITY OF NEWFOUNDLANDUNDERGRADUATE MATHEMATICS COMPETITION

March, 1996

1. Prove that if n is a positive integer, thenn2 + 3n+ 1

n2 + 4n+ 3is an irreducible

fraction.

2. A jar contains 7 blue balls, 9 red balls and 10 white balls. Balls are

drawn at random one by one from the jar until either four balls of the

same colour or at least two of each colour have been drawn. What is

the largest number of balls that one may have to draw?

3. Find all functions u(x) satisfying u(x) = x+

Z 1

2

0

u(t)dt.

4. Show that (p5 + 2)

1

3 � (p5� 2)

1

3 is

a rational number and �nd its value.

5. In a quadrilateral ABCD (vertices named in clockwise order), AC and

BD intersect in X. You are given that AB k DC, that AB is twice as

long as CX and that AC is equal in length to DC. Show that AB and

CD are equal in length (and hence ABCD is a parallelogram).

6. Prove that among any thirteen distinct real numbers it is possible to

choose two, x and y, such that 0 <x� y

1 + xy< 2�

p3.

166

7. A coast guard boat is hunting a bootlegger in a fog. The fog rises dis-

closing the bootlegger 4 miles distant and immediately descends. The

speed of the boat is 3 times that of the bootlegger, and it is known that

the latter will immediately depart at full speed on a straight course of

unknown direction. What course should the boat take in order to over-

take the bootlegger?

Historical Titbit

Taken from a 1950's University Scholarship Paper.

A, C, B are collinear points.

Prove that there is one and only one pointD such that (ACBD) is harmonic.

The following result is stated in a book on geometry:

if the pencils O(ACBD) and O0(ACBD) are harmonic and if

A, C, B are collinear, then D lies on the line ACB.

Give an example to show that this result is not always true.

What alteration is required in order to make it true?

Prove the result after making this alteration.

167

BOOK REVIEWS

Edited by ANDY LIU

All the Math That's Fit to Print, by Keith Devlin,

published by the Mathematical Association of America, 1994,

ISBN 0-88385-515-1, paperbound, 330+ pages, US$29.50.

Reviewed by A. Sharma, University of Alberta.

In the foreword to his famous bookMathematical Snapshots, Steinhaussays, \My purpose is neither to teach in the usual sense of the word, nor to

amuse the reader with some charades. One �ne day it happened that I was

asked the question: `You claim to be a mathematician, well what does one

do all day when one is a mathematician?' " Steinhaus conceived the idea of

his book as he tried to explain to his questioner a few geometric problems

solved and unsolved while sitting on a bench in a public park. Keith Devlin's

book is meant for a much wider audience in a completely di�erent milieu. It

is also not meant to teach or to amuse the reader with riddles, although it

succeeds eminently in doing a bit of both. As he explains in the preface, it

is meant for \anyone who regularly reads a serious newspaper and has some

interest in matters scienti�c, mathematical or is just curious".

This very interesting book, with its catching title is a collection of 143

articles written by the author for the Manchester Guardian over a period

of eight years from 1983 to 1990. From letters received by him, the author

claims that his audience was a mixed bunch: \Students at schools in their

early teens, retired people in their nineties, from prison inmates to executives

in the computer industry, from truckers to school teachers, both men and

women".

The articles in this book are chronologically arranged under fancy head-

ings some of which we sample here: Seven-up, A Farey Story, The VerticalConfusion, World's Most Wanted Number, Rabbit Pi, Biblical Fingers GetStuck Into Pi, Rabbits Do It By Numbers, and so on. Each article can be read

and enjoyed independently of the others, although cross references to allied

topics are given almost everywhere. It is a book which can be enjoyed at

one's leisure and with a pencil and a piece of paper can give hours of delight

and education.

Devlin follows the style of his mentor Martin Gardner. He is informal

and states results when he chooses and sketches a proof if it is short and

well within the reach of a layman (see Article 15). Article 19, The measureof all things, begins with the news about the discovery that a long standing

conjecture in number theory called Merten's conjecture is false. The author

slowly and carefully explains the M�obius function, and tells the reader what

Merten's conjecture is, and who proved that it is false, while giving a bird's

eye view of the result proved.

168

Article 25 is just half a page, but it tells about the Institute for Pi Re-

search at Emporia University. The Institute, we learn, is campaigning for the

value � = 3 to be given equal time with the more conventional value in state

schools.

I personally found the Article 29, Question Time, very interesting, sinceI spent hours working out some of its problems, specially questions 11, 12, 13

and 15. I was particularly pleased to read in Article 99 about D.R. Kaprekar

whom I knew while in India. There were no computers in those days, but

I have witnessed his phenomenal abilities at doing long multiplication and

other operations without pen and paper almost instantly at several meetings

of the Indian Mathematical Society.

It would be surprising if a book of almost 320 pages had no misprints.

I point out some minor ones which I came across:

� p. xi, line 2 from bottom | ready should be read;

� p. 27, column 2, line 6 from top | 33 should be 3;

� p. 46, column 1, line 15 from top | the should be to;

� p. 71, column 1, line 10 from top | 70-71 should be 72-73;

� p. 322, column 1, line 3 from bottom | savaged should be salvaged.

The range of topics covered in these articles is very impressive - Fer-

mat Primes, Carmichael numbers, Palindromic numbers, information, arti�-

cial intelligence, Hilbert's tenth problem, Bieberbach's conjecture, Banach-

Tarski paradox, cartography and many more. It is a book which should be

read piecemeal | an article or two at a time, and as Devlin says, \it is a book

for delving". I am sure this book will soon be in most libraries of schools and

colleges for the bene�t of both teachers and students.

Do you know the equation of this curve?

169

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 that 1 December 1996. They may alsobe sent by email to cruxeditor@cms.math.ca. (It would be appreciated ifemail proposals and solutions were written in LATEX, preferably in LATEX2e).Graphics �les should be in epic format, or plain postscript. Solutions re-ceived after the above date will also be considered if there is su�cient timebefore the date of publication.

2138. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.

ABC is an acute angle triangle with circumcentre O. AO meets the

circle BOC again at A0, BO meets the circle COA again at B0, and CO

meets the circle AOB again at C0.Prove that [A0B0C0] � 4[ABC], where [XY Z] denotes the area of

triangle XY Z.

2139. Proposed by Waldemar Pompe, student, University of War-saw, Poland.

Point P lies inside triangle ABC. Let D, E, F be the orthogonal pro-

jections from P onto the lines BC, CA, AB, respectively. Let O0 and R0

denote the circumcentre and circumradius of the triangleDEF , respectively.

Prove that

[ABC] � 3p3R0

qR02 � (O0P )2;

where [XY Z] denotes the area of triangle XY Z.

2140. Proposed by K.R.S. Sastry, Dodballapur, India.

Determine the quartic f(x) = x4+ ax3+ bx2+ cx� c if it shares two

distinct integral zeros with its derivative f 0(x) and abc 6= 0.

170

2141. Proposed by Toshio Seimiya, Kawasaki, Japan.A1A2A3A4 is a quadrilateral. Let B1, B2, B3 and B4 be points on the

sides A1A2, A2A3, A3A4 and A4A1 respectively,

such that

A1B1 : B1A2 = A4B3 : B3A3 and A2B2 : B2A3 = A1B4 : B4A4:

Let P1, P2, P3 and P4 be points on B4B1, B1B2, B2B3

and B3B4 respectively, such that

P1P2kA1A2; P2P3kA2A3 and P3P4kA3A4:

Prove that P4P1kA4A1.

2142. Proposed by Victor Oxman, Haifa, Israel.In the plane are given an arbitrary quadrangle and bisectors of three

of its angles. Construct, using only an unmarked ruler, the bisector of the

fourth angle.

2143. Proposed by B. M???y, Devon, Switzerland.My lucky number, 34117, is equal to 1662 + 812 and also equal to

1592 + 942, where j166� 159j = 7 and j81� 94j = 13; that is,

it can be written as the sum of two squares of positive integers intwo ways, where the �rst integers occurring in each sum di�er by7 and the second integers di�er by 13.

What is the smallest positive integer with this property?

2144. Proposed by B. M???y, Devon, Switzerland.My lucky number, 34117, has the interesting property that 34 = 2 � 17

and 341 = 3 � 117� 10, that is,

it is a 2N + 1-digit number (in base 10) for some N , such that

(i) the number formed by the �rst N digits is twice the numberformed by the last N , and

(ii) the number formed by the �rst N + 1 digits is three timesthe number formed by the last N + 1, minus 10.

Find another number with this property.

2145. 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.

171

2146. Proposed by Toshio Seimiya, Kawasaki, Japan.ABC is a triangle with AB > AC, and the bisector of \A meets BC

at D. Let P be an interior point on the segment AD, and let Q and R

be the points of intersection of BP and CP with sides AC and AB

respectively. Prove that PB � PC > RB �QC > 0.

2147. Proposed by Hoe Teck Wee, student, Hwa Chong Junior Col-lege, Singapore.

Let S be the set of all positive integers x such that there exist positive

integers y andm satisfying x2 + 2m = y2.

(a) Characterize which positive integers are in S.

(b) Find all positive integers x so that both x and x+ 1 are in S.

2148. Proposed by Aram A. Yagubyants, Rostov na Donu, Russia.Suppose thatAD, BE andCF are the altitudes of triangleABC. Sup-

pose that L, M , N are points on BC, CA, AB, respectively, such that

BL = DC, CM = EA, AF = NB.

Prove that:

1. the perpendiculars to BC, CA, AB at L, M , N , respectively are con-

current;

2. the point of concurrency lies on the Euler line of triangle ABC.

2149. Proposed by Juan-Bosco Morero M�arquez, Universidad deValladolid, Valladolid, Spain.

Let ABCD be a convex quadrilateral andO is the point of the intersec-

tion of the diagonalsAC andBD. Let A0B0C0D0 be the quadrilateral whosevertices, A0, B0, C0, D0, are the feet of the perpendiculars drawn from the

point O to the sides BC, CD, DA, AB, respectively.

Prove that ABCD is an inscribed (cyclic) quadrilateral if and only if

A0B0C0D0 is a circumscribing quadrilateral (A0B0, B0C0, C0D0, D0A0 aretangents to a circle).

2150. Proposed by �Sefket Arslanagi�c, Berlin, Germany.Find all real solutions of the equation

p1� x = 2x2 � 1 + 2x

p1� x2:

172

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

2035. [1995: 130] Proposed by V�aclav Kone �cn �y, Ferris State Univer-sity, Big Rapids, Michigan, USA.

If the locus of a point E is an ellipse with �xed foci F andG, prove that

the locus of the incentre of triangle EFG is another ellipse.

Solution by the Science Academy Problem Solvers, Austin, Texas.

Let F = (�c; 0) andG = (c; 0) so that the locus ofE = (x; y) satis�es

GE + EF = 2a (where a is the length of the semimajor axis). Let I be the

incentre and I0 be the point where the incircle is tangent to FG (so that

II0 ? FG). We have

FI0 = a+ c�GE =EF �GE

2+ c

[since s = a + c is half the perimeter of �EFG, and the tangents to the

incircle from the vertices are equal in pairs and sum to 2s]. If O is the origin

then FO = c and

OI0 = FI0 � c =EF �GE

2=EF 2� EG2

4a

=(x2 + 2cx+ c2 + y2)� (x2 � 2cx+ c2 + y2)

4a=cx

a:

Since r = II0 is the radius of the incircle [and the area of �EFG is rs]

II0 =area(EFG)

a+ c=

cy

a+ c:

Thus, I =

�c

ax;

c

a+ cy

�. The mapping (x; y) !

�c

ax;

c

a+ cy

�is there-

fore an a�ne transformation that maps each point E of the given ellipse to

the point I. A�ne transformations map ellipses to ellipses so that I traces

out an ellipse.

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; �SEFKET ARSLANAGI �C, Berlin, Germany; FRANCISCO BELLOTROSADO, I.B. Emilio Ferrari, andMARIA ASCENSI �ON L �OPEZ CHAMORRO,I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol, UK; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain;JORDI DOU, Barcelona, Spain; KEE-WAI LAU, Hong Kong; P. PENNING,Delft, the Netherlands; CRIST �OBAL S �ANCHEZ{RUBIO, I.B. Penyagolosa,Castell �on, Spain; D.J. SMEENK, Zaltbommel, the Netherlands; and theproposer.

173

2040. [1995: 130] Proposed by Frederick Stern, San Jose State Uni-versity, San Jose, California.

Let a < b be positive integers, and let

t =2a � 1

2b � 1:

What is the relative frequency of 1's (versus 0's) in the binary expansion of t?

[Ed. My interpretation of the question asked is to �nd the ratio of the num-

ber of 1's to the number of 0's; most solvers also read it this way.

The proposer was the only solver to actually mention that the ratio we are

to compute must be done asymptotically. I feel that should have been part

of the problem statement.]

Solutionby Carl Bosley, student,Washburn Rural High School, Topeka,Kansas, USA.

Let x = :000 : : : 01, where there are b � 1 zeros (the pattern below

the superbar repeats in�nitely often). Then (2b � 1)x = :111 : : : 1 = 1, so

x = 1=(2b�1). Therefore, (2a�1)=(2b�1) = :00 : : : 01 : : : 11, where there

are b�a zeros and a ones in every period, so there are a zeros for every b�aones when counted from the left.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; KEITH EKBLAW, Walla Walla, Washington, USA; TOBY GEE, student,the John of Gaunt School, Trowbridge, England; RICHARD I. HESS, RanchoPalos Verdes, California, USA; CYRUS HSIA, student, University of Toronto,Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck,Austria; P. PENNING, Delft, the Netherlands; Science Academy ProblemSolvers, Austin, Texas, USA; DAVID STONE, Georgia Southern University,Statesboro, Georgia, USA; PAUL YIU, Florida Atlantic University, Boca Ra-ton, Florida, USA; and the proposer.

Janous generalizes the problem to show that in base z + 1 we have

(z + 1)a

(z + 1)b � 1= 000 : : : 0zz : : : z

where there are b� a zeros and a copies of z when a < b.

2041. [1995: 157] Proposed by Toshio Seimiya, Kawasaki, Japan.P is an interior point of triangle ABC. AP , BP , CP meet BC, CA,

AB at D, E, F respectively. Let M and N be points on segments BF and

CE respectively so that BM : MF = EN : NC. Let MN meet BE and

CF at X and Y respectively. Prove that MX : Y N = BD : DC.

174

Solution by Jari Lappalainen, Helsinki, Finland.

First, we apply Menelaus' theorem to triangles CNY and Y FM with

line EB. Dividing the results, we get

�1�1 =

CEEN

� NXXY

� Y PPC

FBBM

� MXXY

� Y PPF

= 1

and substituting CE : EN = FB : BM (which follows directly from

BM :MF = EN : NC), we get

NX

MX=PC

PF;

or equivalently

FC

PF=MN

XM: (1)

In a similar way using Menelaus's theorem for triangles BXM and XNE

with line FC, and substitutingMF : FB = NC : CE, we get

EB

PE=NM

YN: (2)

Finally, applying Ceva's theorem to triangle ABC and Menelaus's theorem

to triangles CEP and BEA, with lines AB and CF respectively, we �nd

1

�1� (�1) =

BDDC

� CEEA

� AFFB

CAAE

� EBBP

� PFFC

� BPPE

� ECCA

� AFFB

=BD

DC� PEEB

� FCPF

= 1:

Using (1) and (2)

BD

DC� YNNM

� MN

XM= �BD

DC� Y NXM

= 1;

or MX : Y N = BD : DC.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; D.J. SMEENK,Zaltbommel, the Netherlands; HOE TECK WEE, student, Hwa Chong JuniorCollege, Singapore; and the proposer.

175

2042. [1995: 157] Proposed by Jisho Kotani, Akita, Japan, andK.R.S. Sastry, Dodballapur, India.

If A and B are three-digit positive integers, let A � B denote the six-

digit integer formed by placing them side by side. Find A and B such that

A; B; B � A; A � B andA �BB

are all integer squares.

Solution by Kathleen E. Lewis, SUNY Oswego, New York, NY, USA.

Since A, B�A andB are all integer squares, their square roots form a

Pythagorean triple. Thus, the easiest place to look for numbers meeting the

stated conditions is in the multiples of the 3; 4; 5 triple. So, letting A = 9t2

and B = 25t2 for an integer t, we see that B �A = 16t2,

A � B = 1000A+B = 9000t2 + 25t2 = 9025t2 = (95t)2

andA �BB

=(95t)2

(5t)2= 192:

Any choice of t would yield A and B satisfying all the other conditions, but

in order to make A and B three{digit numbers, t would have to be 4, 5 or 6,

yielding values of

(144; 400); (225; 625); and (324; 900)

for (A;B).

These are in fact the only possibilities. Suppose A = x2, B � A = y2

and B = z2 satisfy the given conditions. Let q = gcd(x; y; z), x0 = x=q,

y0 = y=q and z0 = z=q.

Then x02+y0

2= z0

2and z0

2divides (A�B�B)=q2 = 1000x0

2. Since

x0 and z0 are relatively prime, this means that z02must divide 1000, so z0

must be 1, 2, 5 or 10. By inspection, none of the others are possible values,

so z0 must be 5. We can also rule out the remaining case of x0 = 4, y0 = 3

and z0 = 5 [that is, A = 16t2 and B = 25t2] since (A � B)=t2 = 16025 is

not a perfect square.

Editor's note. As some other solvers point out, the case B � A =

0 should be considered. This is the case y0 = 0, x0 = z0 = 1, which is

impossible because A � B = 1001x2 is not a perfect square.

Also solved by CARL BOSLEY, student, Washburn Rural High School,Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton College, Bris-tol, UK; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; J.K. FLOYD,Newnan, Georgia, USA; TOBY GEE, student, the John of Gaunt School, Trow-bridge, England; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, On-tario; DAVID HANKIN, Hunter College Campus Schools, New York, NY,USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOHN G.

176

HEUVER, Grande Prairie Composite High School, Grande Prairie, Alberta;CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Fer-ris State University, Big Rapids, Michigan, USA; MICHAEL PARMENTER,Memorial University of Newfoundland, St. John's, Newfoundland;P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymna-sium, Graz, Austria; DAVID R. STONE, Georgia Southern University, States-boro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; HOE TECKWEE, student, Hwa Chong Junior College, Singapore; SUSAN SCHWARTZWILDSTROM, Kensington,Maryland, USA; PAUL YIU, Florida Atlantic Uni-versity, Boca Raton, Florida, USA; and the proposers.

Most solvers found all three solutions.

The two proposers actually had sent the editor this problem, or some-thing quite similar, independently and at almost the same time.

2043. [1995: 158] Proposed by Aram A. Yagubyants, Rostov naDonu, Russia.

What is the locus of a point interior to a �xed triangle that moves so

that the sum of its distances to the sides of the triangle remains constant?

Editor's Comment.

The solution to the problem as stated in the proposal was quite easy us-ing analytic/trigonometric arguments and this was the path chosen by mostof the solvers. Some contributors, anxious no doubt to uphold the impec-cable standards of CRUX, tried to do something more with the proposal -much to the delight of the Editor. One such example is to look at the locusfor di�erent triangles as portrayed in the featured solutionby Hess. Just theobservation that all points in the interior is the locus when the given triangleis equilateral can do wonders for the morale of the Editor. The second solu-tion, while restricted to the interior of the triangle, did o�er a respite fromthe analytic. Several solvers made comments of varying substance relatingto the case when the point was exterior to the triangle. The third solutionby Fritsch did the job nicely and was novel.

I. Solution by Richard I. Hess, Rancho Palos Verdes, California, USA.

From the �gure, where the largest

side is the base, we have

h1 = y;

h2 = (a� x) sin�� y cos�;

h3 = x sin� � y cos �: � �

ph1

h3h2

a

P

Note:

P = P (x; y)

177

From h1 + h2 + h3 = k (a constant), we get

y(1� cos�� cos �) = k� a sin�+ x(sin�� sin �):

There are several cases:

1. When� = � = 60�, the whole interior of the triangle hash1+h2+h3 =ap3=2.

2. When cos � + cos� = 1 but � 6= �, the locus is x =k � a sin�

sin� � sin�.

3. When � = � 6= 60�, the locus is y =k� a sin�

1� 2 cos�.

4. Otherwise, the locus is a straight line:

y =k� a sin�

1� cos � � cos�+

x(sin�� sin �)

1� cos � � cos�.

II. Solution by Toshio Seimiya, Kawasaki, Japan.

Let P be a point interior to trian-

gle ABC, so that the sum of its

distances to the sides of triangle

ABC is constant, and letD, E, F

be the feet of the perpendiculars

to BC, CA, AB respectively.

We assume that triangle ABC is

non-equilateral and that BC is

the least side.

We put PD + PE + PF = k,

where k is a constant.

B C

A

D

E

H

F

G

P

Let G, H be the points on the side AB, AC respectively, such that BG =

CH = BC. Then

[PBC] + [PCH] + [PBG] = 12a(PD+ PE + PF ) = 1

2ak;

which is constant ([XY Z] denotes the area of triangle XY Z).

Since the area of quadrilateral BCHG is constant, so also is the area of

triangle PGH. Therefore P lies on a �xed line `, parallel to GH.

Hence the locus of P is the segment of ` contained within the triangle ABC.

III. Solution by Rudolf Fritsch, Ludwig-Maximilians-Universit�at,M�unchen, Germany.

Let ABC be a triangle in the real plane; without loss of generality we

assume a � b � c. Then we may choose cartesian coordinates such that

A = (0; h) ; B = (�p; 0) ; C = (q;0)

178

with h, p, q > 0. The signed distance of a point in the plane from a side

of this triangle is taken positive if the point is in the same half plane as the

vertex opposite to this side, thus, the signed distance of P = (x; y) from the

line BC is just y.

For the other sides of triangle we choose the following equations

AB � x � sin� � y � cos� + h � cos� = 0 ;

AC � �x � sin � y � cos + h � cos = 0 :

These equations are called the Hessian normal forms of the lines under con-

sideration. The namesake is Otto Ludwig Hesse (born in K �onigsberg/East

Prussia 1811/4/22, died in M�unchen 1874/8/4) although this form has al-

ready been used by Gauss in a paper published in 1810. The general idea is

to normalize the line equation

g � ux+ vy+ w = 0

by u2 + v2 = 1, w � 0 (or w � 0), the line g being unique if it does not

pass through the origin. The advantage of this form is that for any point

P = (x; y) in the plane the expression

d(x; y) = ux + vy+ w

gives the signed distance of P from g, positive if and only if P is on the same

side of g as the origin (origin not on g).

Thus, the sum of the signed distances of P = (x; y) from the reference

triangle is

s(x; y) = x � (sin� � sin ) + y � (1� cos� � cos ) + h � (cos� + cos ) :

Since this expression is linear, the equation s(x; y) = k describes a line, for

any k 2 R. Since the slope is independent of k, all lines obtained in this

way form a pencil of parallel lines. A distinguished member of this pencil

is obtained by taking k = 0 giving the line connecting the points where the

exterior angle bisectors meet the opposite sides.

Taking absolute distances we get buckled lines for the desired locus. If

the problem is restricted to the interior points of the triangleABC, then the

result gives parallel line segments.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; NIELS BEJLEGAARD, Stavanger, Norway; DAVIDHANKIN, Hunter Col-lege Campus Schools, New York, NY, USA; WALTHER JANOUS, Ursulinen-gymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Al-berta, Edmonton, Alberta; V �ACLAV KONE �CN �Y, Ferris State University, BigRapids, Michigan, USA; P. PENNING, Delft, the Netherlands; CRIST �OBALS �ANCHEZ{RUBIO, I.B. Penyagolosa, Castell �on, Spain; ASHISH KR. SINGH,Kanpur, India; D.J. SMEENK, Zaltbommel, the Netherlands; and the pro-poser.

179

Klamkin comments: since a1r1 + a2r2 + a3r3 = 2� (where ri is thedistance to side ai, and � is the area of the triangle), the constant sum ofthe distances must be bounded by

2�

a3� r1 + r2 + r3 �

2�

a1;

where it is assumed that a1 � a2 � a3. Otherwise, there are no points inthe locus. This is also true, if, for example, a3 is smaller than the other sidesand the constant sum is 2�

a3.

2044. [1995: 158] Proposed by Murray S. Klamkin, University ofAlberta, Edmonton, Alberta.

Suppose that n � m � 1 and x � y � 0 are such that

xn+1 + yn+1 � xm � ym:

Prove that xn + yn � 1.

I. Solution by Toshio Seimiya, Kawasaki, Japan.

There is nothing to prove if x = y since in this case the given inequality

implies x = y = 0. Suppose 0 � y < x, then xm � ym > 0. From

xn+1 � xn+1 + yn+1 � xm � ym � xm and n + 1 > m, it follows that

x � 1. Using n�m � 0 andm � 1, we now obtain

(xn + yn)(xm � ym) = xn+m � yn+m � (xy)m(xn�m � yn�m)

� xn+m � xn+1

� xn+1 + yn+1 � xm � ym:

Dividing through by xm � ym gives xn + yn � 1.

II. Solution by the proposer.

It is easy to see that 1 � x � y � 0, and thus xm � x, xyn � ym and

xny � y. Therefore,

(xn + yn)(x+ y) = xn+1 + yn+1+ xny + xyn

� xm � ym + y + ym � x+ y:

Dividing through by x + y [the case when x + y = 0 being trivial { Ed.]

yields xn + yn � 1. Equality holds if and only if x = 1, y = 0.

Also solved by NIELS BEJLEGAARD, Stavanger, Norway; F.J.FLANIGAN, San Jose State University, San Jose, California, USA; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y,Ferris State University, Big Rapids, Michigan, USA; and PANOS E.TSAOUSSOGLOU, Athens, Greece.

180

Janous showed that if 1 � m � n � 1, then the condition can berelaxed to xn+1+yn+1 � xm�yn�1. Flanigan obtained the stronger resultthat if k =

�n+1m

�, then x(k�1)m + (k � 1)y(k�1)m � 1 if mjn + 1 and

xkm + kykm � 1 ifm 6 j n+ 1.

2045. [1995: 158] Proposed by V�aclav Kone �cn �y, Ferris State Uni-versity, Big Rapids, Michigan, USA.

Show that there are an in�nite number of Pythagorean triangles (right-

angled triangles with integer sides) whose hypotenuse is an integer of the

form 3333 : : : 3.

Once again, our readers have been extremely inventive! Most provided

a way to construct an in�nite sequence (or more than one sequence) of suit-

able Pythagorean triples (a; b; c), where a2 + b2 = c2. We summarize the

results below.

I. Solution by: Niels Bejlegaard, Stavanger, Norway; Carl Bosley, stu-dent, Washburn Rural High School, Topeka, Kansas, USA; Christopher J.Bradley, Clifton College, Bristol, UK;Miguel Angel Cabez �onOchoa, Logro ~no,Spain; Toby Gee, student, the John of Gaunt School, Trowbridge, England;David Hankin, Hunter College Campus Schools,New York, NY, USA; RichardI. Hess, Rancho Palos Verdes, California, USA; Friend H. Kierstead Jr., Cuya-hoga Falls, Ohio, USA; Kathleen E. Lewis, SUNY Oswego, NY, USA; MariaAscensi �on L �opez Chamorro, I.B. Leopoldo Cano, Valladolid, Spain;P. Penning, Delft, the Netherlands; Gottfried Perz, Pestalozzigymnasium,Graz, Austria; David R. Stone, Georgia SouthernUniversity, Statesboro, USA;Panos E. Tsaoussoglou, Athens, Greece; and the proposer.

First, note that neither 3 nor 33 is the hypotenuse of a Pythagorean

triangle.

Let (a; b; c) be a Pythagorean triple, where c = 33 : : : 3 is a k-digit

integer, k > 2. Then (ma;mb;mc) is also a Pythagorean triple, for all

m = 10k + 1; 102k + 10k + 1; : : : ;Pn

i=0 10ik; : : : and

mc = 33 : : : 3 has 2k; 3k; : : : ; (n+ 1)k; : : : digits.

Each of the following triples can each be used in this way to generate

an in�nite sequence of triples:

(108; 315;333)

(660; 3267;3333)

(7317; 32520;33333)

(128205;307692; 333333)

(487560;3297483; 3333333)

(25114155;21917808;33333333).

181

II. Solution by: Heinz-J �urgen Sei�ert, Berlin, Germany.

�2

310n(102n � 1);

1

3(102n � 1)2;

1

3(104n � 1)

�; where n 2 N;

gives an in�nite sequence of Pythagorean triples with hypotenuse the

4n-digit integer 33 : : : 3. Note that the �rst triple, (660; 3267; 3333), is

the same as one given above, but the rest of the sequence is di�erent.

Also solved (in a non-constructive way) by ASHISH KR. SINGH, stu-dent, Kanpur, India; MURRAY S. KLAMKIN, University of Alberta, Edmon-ton, Alberta; and CHRIS WILDHAGEN, Rotterdam, the Netherlands. Therewas one partial solution.

2047. [1995: 158] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.

ABC is a non-equilateral triangle with circumcentre O and incentre I.

D is the foot of the altitude from A to BC. Suppose that the circumradius

R equals the radius ra of the excircle to BC. Show that O, I and D are

collinear.

Solution by Waldemar Pompe, student, University of Warsaw, Poland.

If AB = AC, the I; O lie on AD; hence O; I and D are collinear.

Henceforth assume AB 6= AC.

Let the angle bisector of the angle \A intersect the circumcircle of ABC at

E and let DO and AE meet at J - see the �gure.

A

B C

O

J

D

E

OE is perpendicular to BC, so AD and OE are parallel. Therefore, since

R = ra

AJ

JE=AD

OE=ha

R=ha

ra=

2S

ara=

2ra(s� a)

ara=b+ c� a

a; (1)

182

where ha, S, s denote the altitude AD, the area, and the semiperimeter of

ABC, respectively. To complete the solution it is enough to show that J is

the incentre ofABC. Using Ptolemy's theorem on the quadrilateralACEB,

and the fact that BE = CE, we get

(AJ + JE)a = BE(b+ c): (2)

Since (1) is equivalent to (AJ + JE)a = JE(b + c), comparing it with (2)

we obtain JE(b+ c) = BE(b+ c), which gives JE = BE. Hence

\JBA+\BAJ = \BJE = \JBE = \CBJ+\CBE = \CBJ+\EAC:

Since \BAJ = \EAC, we get \JBA = \CBJ , which means that BJ

bisects \B. And since AJ bisects \A, J is the incentre of ABC, as we

wished to prove.

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; P. PENNING,Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan; ASHISH KR.SINGH, student, Kanpur, India; and the proposer.

Several solvers noted that the result was mentioned in the Editor'scomments following the solution to problem 1918 (CRUX, Vol. 21, No. 1).

2046. [1995: 158] Proposed by Stanley Rabinowitz, Westford, Mas-sachusetts, USA.

Find integers a and b so that

x3 + xy2 + y3 + 3x2 + 2xy + 4y2 + ax + by + 3

factors over the complex numbers.

Solution by the proposer.If the cubic is to factor in C [x; y], then one of the factors must be linear.

Without loss of generality, we may assume this factor is of the form x�py�qwhere p and q are complex numbers.

Substituting x = py+ q in the original cubic, we get a polynomial in y

that must be identically 0. Thus each of its coe�cients must be 0. This gives

us the four equations:

1 + p+ p3 = 0

4 + 2p+ 3p2 + q + 3p2q = 0

3 + aq+ 3q2 + q3 = 0

b+ ap+ 2q + 6pq + 3pq2 = 0:

Solving these equations simultaneously, yields a = 4 and b = 5. (The editorbeing a mere mortal needed Maple to verify this claim!)

183

As a check we note that the resulting polynomial can be written as

(x+ y+2)(y+ 1)2+ (x+ 1)3. The reducibility of this polynomial will not

change if we let x = X � 1 and y = Y � 1. This produces the polynomial

X3+XY 2+Y 3. Letting z = X=Y shows that this polynomial factors over

C [x; y] since z3 + z + 1 factors over C [z].

Also solved by RICHARD I. HESS, Rancho Palos Verdes, California,USA. Rabinowitz also remarks that,

\except for a few exceptional cases, if f(x; y) is a cubic polynomial in C [x; y],

there will be unique complex constants a and b such that f(x; y) + ax+ by

factors over C [x; y]."

2049?. [1995: 158] Proposed by Jan Ciach, Ostrowiec �Swi�etokrzyski,Poland.

Let a tetrahedron ABCD with centroidG be inscribed in a sphere of ra-

diusR. The linesAG;BG;CG;DGmeet the sphere again atA1; B1; C1; D1

respectively. The edges of the tetrahedron are denoted a; b; c; d; e; f . Prove

or disprove that

4

R� 1

GA1

+1

GB1

+1

GC1

+1

GD1

� 4p6

9

�1

a+

1

b+

1

c+

1

d+

1

e+

1

f

�:

Equality holds if ABCD is regular. (This inequality, if true, would be a

three-dimensional version of problem 5 of the 1991 Vietnamese Olympiad;

see [1994: 41].)

Solution to right hand inequality by Murray S. Klamkin, University ofAlberta, Edmonton, Alberta.

We consider a generalization of the right hand inequality:

Let G and O be the centroid and circumcentre of an n-dimensional

simplex A0A1 : : : An inscribed in a sphere of radius R. Let the lines AiG

meet the sphere again in points A0i : i = 0; 1; : : : ; n. If the edges are

denoted by ej ; j = 1; 2; : : : ; n(n+ 1)=2, then

nXi=0

1

GA0i�s8(n+ 1)

n3

n(n+1)=2Xj=0

1

ej:

By the Power-of-a Point Theorem, we have A01G � AiG = R2 � OG2.

By the power mean inequality, we have

X AiG

n+ 1��X (AiG)

2

n+ 1

�1

2

:

184

So it su�ces to prove the stronger inequality:

�X(AiG)

2(n+ 1)�1

2 �s8(n+ 1)

n3

�R2 � (OG)2

�X 1

ej: (1)

It is known that X(AiG)

2 = (n+ 1)�R2 � (OG)2

�(2)

and

�R2 � (OG)2

�=

X e2j

(n+ 1)2: (3)

Using (2) and (3), we see that (1) becomes, after raising both sides to the

power 2=3,

n(n+ 1)

2��X

e2j

�1=3 �X 1

ej

�2=3;

and the result follows immediately from H�older's inequality.

There is equality only if the simplex is regular.

Comment: Corresponding to the given left hand inequality, the analo-

gous one for the simplex (and not as yet proved) is

n+ 1

R�X 1

A0iG

or, equivalently

RX

AiG � (n+ 1)�R2 � (OG)2

�=X

(AiG)2:

Even more generally, I conjecture that, for p � 1, we have

2R (np + 1)

(n+ 1) (np�1 + 1)�P(AiG)

p+1P(AiG)p

:

Except for the case p = 1, there is no equality for a regular simplex, but for

a degenerate one with n vertices coinciding at one end of a diameter and the

remaining vertex at the other end of the diameter.

No other solutions were received.

2050. [1995: 158] Proposed by �Sefket Arslanagi�c, Berlin, Germany.Find all real numbers x and y satisfying the system of equations

2x2+y + 2x+y

2

= 128;px+

py = 2

p2:

185

Solution. Essentially identical solutionswere submittedby Carl Bosley,student, Washburn Rural High School, Topeka, Kansas, USA; ChristopherJ. Bradley, Clifton College, Bristol, UK; Toby Gee, student, the John of GauntSchool, Trowbridge, England; DavidHankin, Hunter College Campus Schools,New York, NY, USA; Cyrus Hsia, student, University of Toronto, Toronto,Ontario; Walther Janous, Ursulinengymnasium, Innsbruck, Austria; Kee-WaiLau, Hong Kong; Waldemar Pompe, student, University of Warsaw, Poland;Panos E. Tsaoussoglou, Athens, Greece; and the proposer.

Squaringpx+

py = 2

p2, we obtain that

x+ y = 8� 2pxy � 8� (x+ y);

whence

x+ y � 4: (1)

Further

x2 + y2 � 12(x+ y)2 � 8: (2)

From (1) and (2), we have

64 =2x

2+y + 2x+y2

2

� 21

2(x2+y+x+y2) � 26 = 64:

Thus x + y = 4 and x2 + y2 = 8. These are easily solved to obtain that

x = y = 2.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, andMARIA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid,Spain; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA;F.J. FLANIGAN, San Jose State University, San Jose, California, USA;MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; BEATRIZMARGOLIS, Paris, France; J.A.MCCALLUM,MedicineHat, Alberta; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College,Calgary, Alberta; DAVID R. STONE, MARTHA BELL and JIM BRASELTON;Georgia Southern University, Statesboro, Georgia, USA; STAN WAGON,Macalester College, St. Paul,Minnesota, USA; CHRISWILDHAGEN, Rotter-dam, the Netherlands; and SUSAN SCHWARTZ WILDSTROM, Kensington,Maryland, USA. One other submission was received which assumed that xand y were integers.

Janous pointed out that the result can be generalized to:

Let x1; : : : ; xn (n � 2) be non-negative real numbers such thatx1 + : : :+ xn = nw, and let b 6= 1 be a positive real number.

186

Let �1; : : : ; �n be real numbers, each greater than or equal to 1.

Then the equation

bx�11

+x�22

+:::+x�nn + bx�12

+x�23

+:::+x�nn + : : : + bx�1n +x

�21

+:::+x�nn�1

= bw�1+w�2+:::+w�n+logb(n)

has as its only solution x1 = x2 = : : : = xn = w.

This can be proved by applications of the AM{GM and power meaninequalities.

2051. [1995: 202] Proposed by Toshio Seimiya, Kawasaki, Japan.A convex quadrilateral ABCD is inscribed in a circle � with centre O.

P is an interior point ofABCD. Let O1, O2, O3, O4 be the circumcentres of

triangles PAB, PBC, PCD, PDA respectively. Prove that the midpoints

of O1O3, O2O4 and OP are collinear.

Combination of solutions by Jordi Dou, Barcelona, Spain and the pro-poser.

The result holds without restriction on the point P . The proof is in two

steps.

Step 1. The result is trivially true when P is on � (and all the cir-

cumcentres coincide with O), so let P be a point o� � and let be the

conic with foci O and P whose major axis has length R (the radius of �).

is an ellipse when P is interior to � and a hyperbola when P is exte-

rior. Let Q be the intersection of O1O2 and OB. Because O1B = O1P and

O2B = O2P , we have thatO1O2 is the perpendicular bisector ofBP . There-

foreBQ = PQ and\BQO2 = \PQO2. WhenP is interior we conclude that

OQ+ PQ = OQ+BQ = R; when exterior, OQ� PQ = OQ� BQ = R.

Thus, in either case, Q is a point on . Moreover, as O1O2 is a bisector

of \OQP; O1O2 is tangent to at Q. Similarly O2O3; O3O4, and O4O1 are

tangent to . [Editor's comment by Chris Fisher. Dou refers to � as the focalcircle of . I was unable to con�rm that terminology in any handy reference,

but I did �nd the circle mentioned as the basis of a construction of a central

conic by folding; see, for example, E.H. Lockwood, A Book of Curves: Drawa circle � on a sheet of paper and mark an arbitrary point P not on �. For

any number of positions B on � fold P onto B and crease the paper. The

creases (i.e. the perpendicular bisectors of PB) envelope a conic .]

Step 2. By Newton's theorem the midpoints of O1O3; O2O4, and the

centre of (which is the midpoint ofOP ) are collinear as desired. As a bonus,

also on that line is the midpoint of the segment joiningO5 := O1O4\O2O3

to O6 := O1O2 \ O3O4. Here is a simple projective proof of this theorem.

Consider the pencil of dual conics tangent to the sides of the complete quadri-

lateral O1O2O3O4. As a consequence of the Desargues' involution theorem

187

(see, for example, Dan Pedoe, A Course of Geometry for Colleges and Univer-sities, Theorem II, page 342), the poles of a line l with respect to the conics

of the pencil lie on a line. In particular, when l is the line at in�nity the line

of poles is the line of centres of these conics. The centre of is one such

pole. Furthermore, the line of centres passes through the midpoints of the

degenerate dual conics of the pencil (consisting of pairs of opposite vertices

of the quadrilateral), namely O1 and O3, O2 and O4, O5 and O6.

Also solved byMARIA ASCENSI �ON L �OPEZ CHAMORRO, I.B. LeopoldoCano, Valladolid, Spain.

2053. [1995: 202] Proposed by Jisho Kotani, Akita, Japan.

A �gure consisting of two equal and externally tangent circles is in-

scribed in an ellipse. Find the eccentricity of the ellipse of minimum area.

Solution by David Hankin, Hunter College Campus Schools, New York,NY, USA.

[Editor's note: By symmetry, we have:

(a) the centre of the circumscribing ellipse, E, must be the point of tangency

of the two given circles;

(b) an axis of E passes through the centres of the two given circles.

-

6

All solvers assumed this, most without stating that they had done so.]

} } } } }Let the equations of the circles be (x � r)2 + y2 = r2, and let the

equation of the ellipse bex2

a2+y2

b2= 1.

Solving the �rst and third of these equations gives

x =�ar2 � a

pa2r2 + b4 � a2b2

b2 � a2:

188

Therefore, at the two points of tangency (in the right half-plane), we have

a2r2 + b4 � a2b2 = 0. From this we get

a =b2p

b2 � r2:

Since the area of the ellipse is given by K = �ab, we have

K =�b3pb2 � r2

:

ThereforedK

db=

�b2(2b2 � 3r2)

(b2 � r2)3=2. From this, we obtain that K is minimal

when b2 =3r

2. At this point a2 = 3b2, and from

e2 =a2 � b2

a2;

we obtain that e =

r2

3.

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; CARL BOSLEY, student,WashburnRural High School, Topeka, Kansas,USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JORDI DOU,Barcelona, Spain; DAVID HANKIN, Hunter College Campus Schools, NewYork, NY, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA;MARIA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid,Spain; and the proposer. Two other submissions were received that werealmost correct: they used an incorrect formula for the eccentricity.

2054. [1995: 202] Proposed by Murray S. Klamkin, University ofAlberta, Edmonton, Alberta.

Are there any integral solutions of the Diophantine equation

(x+ y+ z)3 = 9�x2y + y2z + z2x

�other than (x; y; z) = (n;n; n)?

I. Solution by Adrian Chan, student, Upper Canada College, Toronto,Ontario.

No, there are no integral solutions other than (x; y; z) = (n; n;n).

Without loss of generality, let x � y and x � z. Let y = x + a and

z = x+b, where a and b are non-negative integers. Then the given equation

becomes

(3x+ a+ b)3 = 9�x2(x+ a) + (x+ a)2(x+ b) + (x+ b)2x

�:

189

After expanding and simplifying, this is (a+ b)3 = 9a2b, or

a3 � 6a2b+ 3ab2 + b3 = 0: (1)

Let a = kb, where k is a rational number. Then (1) becomes

b3(k3 � 6k2 + 3k + 1) = 0:

By the Rational Root Theorem, k3 � 6k2 + 3k + 1 = 0 does not have any

rational roots. So, since k is rational, k3 � 6k2 + 3k + 1 6= 0. Therefore

b = 0 and a = 0, so x = y = z.

II. Solution by the proposer.

Letting y = x+ u and z = x+ v, the equation reduces to

(u+ v)3 = 9u2v (2)

where u and v are integers. We now show that the only solution to (2) is

u = v = 0 so that (x; y; z) = (n;n; n) is the only solution of the given

equation. Letting u+ v = w, (2) becomes

w3 = 9u2(w� u): (3)

Hence w = 3w1 where w1 is an integer, and (3) is 3w31 = u2(3w1 � u). It

follows that u = 3u1 for some integer u1, and we get

w31 = 9u21(w1� u1):

Comparing this equation to (3), we see by in�nite descent that the only so-

lution to (3) is u = w = 0, which gives the negative result.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; THEODORE CHRONIS, student, Aristotle University of Thessaloniki,Greece; RICHARD I. HESS, Rancho PalosVerdes, California, USA; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, HongKong; and HOE TECK WEE, student, Hwa Chong Junior College, Singapore.There was also one incorrect solution sent in.

2055. [1995: 202] Proposed by Herbert G �ulicher, WestfalischeWilhelms-Universit�at, M�unster, Germany.

In triangle ABC let D be the point on the ray from B to C, and E

on the ray from C to A, for which BD = CE = AB, and let ` be the line

through D that is parallel to AB. If M = ` \ BE and F = CM \ AB,

prove that

(BA)3 = AE � BF � CD:

Solution by Toshio Seimiya, Kawasaki, Japan; essentially identicalsolutions were submitted by Jordi Dou, Barcelona, Spain; Mitko ChristovKunchev, Baba Tonka School of Mathematics, Rousse, Bulgaria; andGottfried Perz, Pestalozzigymnasium, Graz, Austria.

190

F A B

D `

E

M

C

By Menelaus' Theorem applied to triangle ACF and transverse line BEM ,

we haveAB

BF� FMMC

� CEEA

= �1:

Since DMkBF , we have FMMC

=BD

DC, giving

AB

BF� BDDC

� CEEA

= �1:

Hence we have

AE:BF:CD = �AB:BD:CE = BA:AB:AB = (BA)3:

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUELANGEL CABEZ �ON OCHOA, Logro ~no, Spain; P. PENNING, Delft, the Nether-lands; D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.

2057?. [1995: 203] Proposed by Jan Ciach, Ostrowiec �Swi�etokrzyski,Poland.

Let P be a point inside an equilateral triangleABC, and letRa; Rb; Rc

and ra; rb; rc denote the distances of P from the vertices and edges, respec-

tively, of the triangle. Prove or disprove that�1 +

ra

Ra

��1 +

rb

Rb

��1 +

rc

Rc

�� 27

8:

Equality holds if P is the centre of the triangle.

Solution by G.P. Henderson, Campbellcroft, Ontario. We will prove

that the inequality is true.

191

We have

sin(PAB) =rc

Ra

; sin(PAC) =rb

Ra

:

Therefore

cos(A) =1

2=

s�1� r2b

R2a

��1� r2c

R2a

�� rbrc

R2a

:

From this, we have

Ra =2p3

qr2b + rbrc + r2c ;

and similar expressions for Rb and Rc.

We can assume, without loss of generality, that P is in the section of

the triangle de�ned by ra � rb � rc. Set x = rb=rc and y = ra=rc, so that

0 � y � x � 1. The left side of the given inequality is now 1 +

p3y

2px2 + x+ 1

! 1 +

p3x

2py2 + y + 1

! 1 +

p3

2px2 + xy + y2

!: (1)

We will replace these factors by smaller quantities which do not involve

square roots. [Ed. This is a brilliant move. I recommend that the reader

draws some graphs to see how e�ective this is. In fact6� x7 + 3x

is \almost" the

�Ceby�sev-Pad �e approximant of

p3

2px2 + x+ 1

, which is6:08409� 1:33375x

7 + 2:53143x.]

We will show that, for 0 � x � 1,

p3

2px2 + x+ 1

� 6� x

7 + 3x: (2)

This is equivalent to

(1� x)2�3 + 36x� 4x2

�= (1� x)2 (4x(1� x) + 32x+ 3) � 0:

Similarly, we have, for 0 � y � 1,

p3

2py2 + y + 1

� 6� y

7 + 3y:

Replacing x by y=x(� 1) in (2), we have

p3

2px2 + xy + y2

� 6x� y

x(7x+ 3y):

Using these three expression in (1), it is su�cient to prove that�1 +

y(6� x)

7 + 3x

��1 +

x(6� y)

7 + 3y

��1 +

6x� y

x(7x+ 3y)

�� 27

8:

192

Now, this is equivalent to

f(x; y) = Py3 +Qy2 + Ry + S � 0;

where

P = P (x) = 8(x� 6)(x� 3)(3x� 1);

Q = Q(x) = 56x4 � 672x3 + 663x2 + 427x� 504;

R = R(x) = �504x4 + 35x3 + 714x2 � 273x� 392;

S = S(x) = 1008x4 + 423x3 � 3493x2 + 2352x:

We will prove this by showing that f is a decreasing function of y for 0 �y � x [for �xed x] and that f(x; x) � 0. We �nd that

f(x;x) = 10x(1� x)2�8x3 � 124x2 � 35x+ 196

� � 0:

[Note that 8 + 196 > 124 + 35.]

It remains to show that

F (x; y) =@f

@y= 3Py2 + 2Qy + R � 0: (3)

First, we note that

Q = �56x3(1� x)� 616x(1� x)2 � �569x2 � 1043x+ 504�< 0;

R = �35x(1� x2)� 238x� �504x2 � 714x+ 392�< 0:

[Also, P � 0 for 0 � x � 13and P � 0 for 1

3� x � 1.] In equation (3),

when P � 0, all three terms are negative, and so F (x; y) < 0. When P > 0,

F (x; y) � 0 provided that

F (x; 0) � 0 and F (x;x) � 0;

[since @2F@y2

> 0].

The �rst of these terms is R, which is negative. The second is

(1� x)��184x4 + 2336x3 � 537x2 � 1673x� 392

�= (1� x)

��1184x4 � 537x2(1� x)� 1673x(1� x2)

�392(1� x3)� 266x3�

� 0:

Also solved by MARCIN E. KUCZMA, Warszawa, Poland.

193

Ceva meets Pythagoras!

K.R.S. SastryDodballapur, Karnataka, India

I had assumed that nobody could beat Ptolemy in providing a shortest

proof of the Pythagorean theorem. To a rectangle you simply apply his \In

a convex cyclic quadrilateral, the product of the diagonals equals the sum

of the products of the opposite sides" theorem. In fact we can do better.

With the assumed knowledge of ratios of line segments, the area measure

of a polygonal region, the de�nitions of trigonometrical ratios and the the-

orems of Ceva and Menelaus, I am going to describe a construction to you

about erecting parallelograms on the sides of a triangle. That will enable us

to discover a new necessary and su�cient condition for the concurrency of

three cevians of a triangle. Then, by treating the altitudes of a right-angled

triangle as the degenerate case of three concurrent cevians, we deduce the

Pythagorean theorem. Intrigued?

PRELIMINARIES.

Suppose D, E, F are points respectively on the sides BC, CA, AB of tri-

angle ABC as in Figure 1. Let BD : DC = 1 : �, and CE : EA = 1 : �,

but AF : FB = � : 1. We let a, b, c denote the side lengths BC, CA, AB

respectively. You will immediately tell me that

BD = a

1+�; DC = a�

1+�;

CE = b

1+�; EA = b�

1+�;

AF = c�

1+�; FB = c

1+�:

9>>>=>>>;

(1)

r

r r

A

B CD

F E

Figure 1

We both know from Ceva that the cevians AD, BE, CF concur if and only if

� = ��: (2)

In the title of our discussion, Menelaus did not meet Pythagoras, but he has

his role to play. So let us recall that the points D, E, F will be collinear if

and only if � = ���.

194

CONSTRUCTION (?).

In addition to the points D, E, F mentioned above, let us construct paral-

lelograms outwardly on the sides BC, CA, AB having widths w1, w2, w3

respectively. These parallelograms, in general, are not similar but mutually

equiangular containing an angle �, 0� < � < 180�. Furthermore, we split

each parallelogram into two by drawing parallels through the points D, E,

F (see Figure 2). We use the notation M [DC; w1] for the area measure of

the parallelogram havingDC and w1 as side lengths.

q

q q

A

B C

D

F E

�

�

�

w1

w2

w3

Figure 2

It is now a simple matter for either of us to compute the area measures of the

three pairs of parallelograms, each pair sharing exactly one vertex of triangle

ABC:

M [DC; w1] =a�w1 sin�

1+�; M [CE;w2] =

bw2 sin�

1+�;

M [EA;w2] =b�w2 sin�

1+�; M [AF; w3] =

c�w3 sin�

1+�;

M [FB;w3] =cw3 sin�

1+�; M [BD;w1] =

aw1 sin�

1+�:

9>>>=>>>;

(3)

WHEN DO CEVIANS AD, BE, CF CONCUR?

I will provide an answer to that question by relating the area measures of

the parallelograms listed in (3).

Theorem 1. The cevians AD, BE, CF concur if and only if there exist

mutually equiangular parallelograms of appropriate widths w1, w2, w3, de-

pending on the values a, b, c, �, �, so that the construction (?) yields paral-

lelogram pairs of equal area measure in (3).

195

Proof: Allow me to assume that AD, BE, CF are concurrent cevians.

This leaves me with the task of providing consistent expressions for w1, w2,

w3 in terms of a, b, c, �, �.

From my assumption and (2), � = �� follows. So I will substitute � = ��

and equate parallelogram area measures of each pair in (3). Observe that the

factor sin� disappears from our equations:

a�w1

1 + �=

bw2

1 + �;

b�w2

1 + �=

c��w3

1 + ��;

cw3

1 + ��=

aw1

1 + �; �� 6= 0; �; � 6= �1:

Indeed we have the consistent equations

a�w1

1 + �=

bw2

1 + �=

c�w3

1 + ��= k say:

These yield, for an appropriate constant k,

w1 =k(1 + �)

a�; w2 =

k(1 + �)

b; w3 =

k(1 + ��)

c�: (4)

True, we get an in�nity of dissimilar parallelograms of varying widths as we

vary the value of k.

To establish the converse, presently I am obliged to assume that there

are parallelograms of appropriate widths w1, w2, w3 so that the construc-

tion (?) leads to parallelogram pairs of equal area measure in (3). You will

therefore allow me to assume the equations

a�w1

1 + �=

bw2

1 + �;

b�w2

1 + �=

c�w3

1 + �;

cw3

1 + �=

aw1

1 + �:

But wait! The above equations imply that

a��w1

1 + �=

b�w2

1 + �=

c�w3

1 + �=a�w1

1 + �:

Hence � = �� and we have established the concurrency of the cevians AD,

BE, CF by virtue of (2).

You will agree with me that, although the proof thus far has proceeded

as if �, �, � are all positive, the same goes through should two of them be

negative. So, to complete the proof of Theorem 1, I must now answer the

question: What if � (or �) is 0?

As you can see from Figure 3, when� = 0, both pointsD andF coincide

with the vertex B. We therefore determine w1, w2, w3 from the equations

aw1 =bw2

1 + �;

b�w2

1 + �= cw3:

196

This yields

w1 =k

a�; w2 =

k(1 + �)

b�; w3 =

k

c

for some appropriate constant k. And this springs on us the following de-

lightful surprise!

M [AB; w3] +M [BC; w1] = M [CA; w2]: (5)

r

A

B C

E

(D)

(F )

Figure 3

As this equation (5) is an important result, rivalling Pythagoras, let us give

it the status of a theorem.

Theorem 2. Let E be a point on side AC of triangle ABC distinct from

the vertex A or C. Suppose CE : EA = 1 : �. Then there exist mutually

equiangular parallelograms of appropriate widths w1, w2, w3, depending on

the values a, b, c, �, so that the sum of the area measures of the parallel-

ograms on the sides AB, BC equals the area measure of the parallelogram

on side AC.

THE PYTHAGOREAN THEOREM.

I know your impatience is growing at an alarming rate to meet and greet

Pythagoras. But I am honour bound to show you, in some speci�c instance,

how I actually compute the w's. For this purpose let us assume that AD,

BE, CF are the altitudes of triangle ABC. Then you know BD = c cosB,

DC = b cosC, � � � , BD : DC = (c cosB) : (b cosC) = 1 : �, � � � . Hence

� =b cosC

c cosB; � =

c cosA

a cosC:

I am very glad to learn that on using (4) you already deduced the nice con-

clusion

w1 = ka; w2 = kb; w3 = kc: (6)

For the proof of the Pythagorean theorem, look at Figure 4. If you have

the triangle right-angled at B, the altitudes concur at the vertex B. So from

Theorem 2, (5) and (6) we have

M [AB; kc] +M [BC; ka] = M [CA; kb]:

197

B

A

C

E

Figure 4

That is

kc2 sin�+ ka2 sin� = kb2 sin�;

which is more than Pythagoras is said to have said. Speci�cally, k = 1,

� = 90� yields the Pythagorean theorem. You may wish to establish the

more general result and its converse: In triangleABC the cosine law follows

by treating its altitudes as three concurrent cevians.

CONCLUSION.

For the sake of completeness, I will state Theorem 3. It can be proved in the

manner of Theorem 1.

Theorem 3. The pointsD, E, F on the sides of a triangleABC are collinear

if and only if there exist mutually equiangular parallelograms of appropriate

widths w1, w2, w3, depending on the values a, b, c, �, �, so that the con-

struction (?) yields parallelogram pairs of equal area measure in (3).

ACKNOWLEDGEMENT.

The author thanks the referee for his suggestions.

198

THE SKOLIAD CORNERNo. 15

R.E. Woodrow

First this issuewe pause to correct an error which crept into the solution

given for problem 10 of the Eleventh W.J. Blundon Contest, February 23,

1994. Thanks go to Bob Prielipp for noting that we used 6 in place of 5 in the

last line on page 155, thus producing a wrong answer! Here is his correction.

10. [1996: 102, 1996: 155] The Eleventh W.J. Blundon Contest.Two numbers are such that the sum of their cubes is 5 and the sum of

their squares is 3. Find the sum of the two numbers.

Solution by Bob Prielipp, University of Wisconsin{Oshkosh.Let the two numbers be denoted by x and y. Then x3 + y3 = 5 and

x2 + y2 = 3. Since x2 + y2 = 3, (x2 + y2)(x+ y) = 3(x+ y). Thus (x3 +

y3)+xy(x+y) = 3(x+y). Because x3+y3 = 5, xy(x+y) = 3(x+y)�5.

Hence 5 = x3+y3 = (x+y)3�3xy(x+ y) = (x+ y)3� 9(x+y)+15, so

(x+ y)3� 9(x+ y) + 10 = 0:

Let x+ y = s. Then

s3 � 9s+ 10 = 0

(s� 2)(s2 + 2s� 5) = 0

(s� 2)(s2 + 2s+ 1� 6) = 0

s = 2 or s+ 1 = �p6:

Thus s = 2 or s = �1�p6.

In this issue, we give the problems of the 14th annual American Invi-

tational Mathematics Examination written March 28, 1996. These problems

are copyrighted by the Committee on the American Mathematical Competi-

tions of the Mathematical Association of America andmay not be reproduced

without permission. Solutions, and additional copies of the problems, may

be obtained for a nominal fee from Professor Walter E. Mientka, C.A.M.C.

Executive Director, 917 Oldfather Hall, University of Nebraska, Lincoln, NE,

USA 68588{0322. As always we welcome your original \nice" solutions and

generalizations which di�er from the published o�cial solutions.

199

14th ANNUAL AMERICAN INVITATIONAL

MATHEMATICS EXAMINATION (AIME)

Thursday, March 28, 1996

1. In a magic square, the sum of the three entries in any row, column,

or diagonal is the same value. The �gure shows four of the entries of a magic

square. Find x.

1

x 19 96

2. For each real number x, let bxc denote the greatest integer that doesnot exceed x. For how many positive integers n is it true that n < 1000 and

that blog2 nc is a positive even integer?

3. Find the smallest positive integern for which the expansion of (xy�3x+7y�21)n, after like terms have been collected, has at least 1996 terms.

4. A wooden cube, whose edges are one centimetre long, rests on a

horizontal surface. Illuminated by a point source of light that isx centimetres

directly above an upper vertex, the cube casts a shadow on the horizontal

surface. The area of the shadow, which does not include the area beneath

the cube, is 48 square centimetres. Find the greatest integer that does not

exceed 1000x.

5. Suppose that the roots of x3 + 3x2 + 4x� 11 = 0 are a, b, and c,

and that the roots of x3 + rx2 + sx + t = 0 are a + b, b + c, and c + a.

Find t.

6. In a �ve-team tournament, each team plays one game with every

other team. Each team has a 50% chance of winning any game it plays. (There

are no ties.) Let m=n be the probability that the tournament will produce

neither an undefeated team nor a winless team, wherem andn are relatively

prime positive integers. Findm+ n.

7. Two of the squares of a 7� 7 checkerboard are painted yellow, and

the rest are painted green. Two color schemes are equivalent if one can be

obtained from the other by applying a rotation in the plane of the board.

How many inequivalent color schemes are possible?

8. The harmonic mean of two positive numbers is the reciprocal of the

arithmetic mean of their reciprocals. For how many ordered pairs of positive

integers (x; y) with x < y is the harmonic mean of x and y equal to 620?

200

9. A bored student walks down a hall that contains a row of closed

lockers, numbered 1 to 1024. He opens the locker numbered 1, and then

alternates between skippingand opening each closed locker thereafter. When

he reaches the end of the hall, the student turns around and starts back.

He opens the �rst closed locker he encounters, and then alternates between

skipping and opening each closed locker thereafter. The student continues

wandering back and forth in this manner until every locker is open. What is

the number of the last locker he opens?

10. Find the smallest positive integer solution to

tan19x� =cos 96� + sin96�

cos 96� � sin96�:

11. Let P be the product of those roots of z6 + z4 + z3 + z2 +1 = 0

that have positive imaginary part, and suppose that P = r(cos ��+i sin ��),

where 0 < r and 0 � � < 360. Find �.

12. For each permutation a1; a2; a3; : : : ; a10 of the integers 1, 2, 3,

: : : ; 10, form the sum

ja1 � a2j+ ja3 � a4j+ ja5 � a6j+ ja7 � a8j+ ja9 � a10j:

The average value of all such sums can be written in the form p=q, where p

and q are relatively prime positive integers. Find p+ q.

13. In triangleABC, AB =p30, AC =

p6, andBC =

p15. There

is a pointD for which AD bisects BC and \ADB is a right angle. The ratio

Area (4ADB)

Area (4ABC)

can be written in the formm=n, wherem and n are relatively prime positive

integers. Findm+ n.

14. A 150 � 324 � 375 rectangular solid is made by gluing together

1�1�1 cubes. An internal diagonal of this solid passes through the interiors

of how many of the 1� 1� 1 cubes?

15. In parallelogram ABCD, let O be the intersection of diagonals

AC andBD. AnglesCAB andDBC are each twice as large as angleDBA,

and angle ACB is r times as large as angle AOB. Find the greatest integer

that does not exceed 1000r.

201

In the last number, we gave the problems of the European \Kangaroo"

Mathematical Challenge written 23 March 1995. Here are solutions.

1. B 2. E 3. C 4. E 5. C

6. C 7. C 8. C 9. B 10. D

11. B 12. 27 13. C 14. C 15. D

16. E 17. A 18. E 19. E 20. C

21. D 22. D 23. C 24. E 25. E

That completes the Skoliad Corner for this issue. Thanks go to readers

who have sent in contest materials and solutions. Please send me more!

Congratulations!

Professor Andy Liu

We are delighted to congratulate Andy Liu, Department of Mathematical Sci-

ences, University of Alberta, Edmonton, on being awarded, this year, the

prestigiousDavid Hilbert International Award at the World Federation of Na-

tional Mathematics Competitions meeting at ICME-8, in Seville.

Andy has a long and distinguished record in Mathematics. He has coached

both the American and Canadian teams for the IMO, and also participated in

the training of teams from Hong Kong and China. He is a well respected prob-

lemist. His mathematical interests span discrete mathematics, hypergraph

theory, combinatorial geometry, foundations of mathematics, mathematical

education and recreational mathematics. He says that the common charac-

teristics of the research problems in which he works are that they are easy to

understand, but not so easy to solve.

Andy has long been associated with CRUX, and is at present the Editor of the

Book Reviews section.

Past recipients of the award, who are associated with CRUX, include Ed. Bar-

beau, Murray Klamkin and Marcin Kuczma.

202

THE OLYMPIAD CORNERNo. 175

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 with the 1996 CanadianMathematical Olympiad which we re-

produce with the permission of the Canadian Mathematical Olympiad Com-

mittee of the Canadian Mathematical Society. My thanks go to Daryl Tin-

gley, Chair of the Committee, for forwarding the questions and promising to

provide \o�cial" and interesting contestant solutions.

1996 CANADIANMATHEMATICAL OLYMPIAD

1. If �, �, are the roots of x3 � x� 1 = 0, compute

1 + �

1� �+

1 + �

1� �+

1+

1� :

2. Find all real solutions to the following system of equations. Care-

fully justify your answer. 8>>>>>>>>>><>>>>>>>>>>:

4x2

1 + 4x2= y;

4y2

1 + 4y2= z;

4z2

1 + 4z2= x:

3. We denote an arbitrary permutation of the integers 1; : : : ; n by

a1; : : : ; an. Let f(n) be the number of these permutations such that

(i) a1 = 1;

(ii) jai � ai+1j � 2, i = 1; : : : ; n� 1.

Determine whether f(1996) is divisible by 3.

4. Let 4ABC be an isosceles triangle with AB = AC. Suppose

that the angle bisector of \B meets AC at D and that BC = BD + AD.

Determine \A.

203

5. Let r1; r2; : : : ; rm be a given set of m positive rational numbers

such thatPm

k=1 rk = 1. De�ne the function f by f(n) = n �Pm

k=1[rkn]

for each positive integer n. Determine the minimum and maximum values

of f(n). Here [x] denotes the greatest integer less than or equal to x.

The next set of problems are from the twenty-�fth annual United States

of America Mathematical Olympiadwritten May 2, 1996. These problems are

copyrighted by the Committee on the American Mathematical Competitions

of theMathematical Association of America andmay not be reproduced with-

out permission. Solutions, and additional copies of the problems, may be

obtained for a nominal fee from Professor Walter E. Mientka, C.A.M.C. Ex-

ecutive Director, 917 Oldfather Hall, University of Nebraska, Lincoln, N.E.,

USA 68588{0322. As always we welcome your original, \nice" solutions and

generalizations which di�er from the published o�cial solutions.

25th UNITED STATES OF AMERICA

MATHEMATICAL OLYMPIAD

Part I 9 a.m. - 12 noon

May 2, 1996

1. Prove that the average of the numbers n sinn� (n = 2, 4, 6, : : : ;

180) is cot 1�.

2. For any nonempty set S of real numbers, let �(S) denote the sum

of the elements of S. Given a set A of n positive integers, consider the

collection of all distinct sums �(S) as S ranges over the nonempty subsets

of A. Prove that this collection of sums can be partitioned into n classes so

that in each class, the ratio of the largest sum to the smallest sum does not

exceed 2.

3. Let ABC be a triangle. Prove that there is a line ` (in the plane of

triangleABC) such that the intersection of the interior of triangleABC and

the interior of its re ection A0B0C0 in ` has area more than 2=3 the area of

triangle ABC.

Part II 1 p.m. - 4 p.m.

May 2, 1996

4. An n-term sequence (x1; x2; : : : ; xn) in which each term is either 0

or 1 is called a binary sequence of length n. Let an be the number of binary

sequences of length n containing no three consecutive terms equal to 0, 1,

0 in that order. Let bn be the number of binary sequences of length n that

contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that

order. Prove that bn+1 = 2an for all positive integers n.

204

5. Triangle ABC has the following property: there is an interior point

P such that \PAB = 10�, \PBA = 20�, \PCA = 30�, and \PAC =

40�. Prove that triangle ABC is isosceles.

6. Determine (with proof) whether there is a subset X of the integers

with the following property: for any integer n there is exactly one solution

of a+ 2b = n with a; b 2 X.

As a third Olympiad for this issue we give the problems of the Ital-

ian Mathematical Olympiad written May 6, 1994. My thanks go to Richard

Nowakowski who collected this (and many others) when he was Canadian

Team Leader to the International Mathematical Olympiad in Hong Kong.

ITALIANMATHEMATICAL OLYMPIAD

May 6, 1994

1. Show that there exists an integerN such that for all n � N a square

can be represented as a union of n pairwise non-overlapping squares.

2. Find all integer solutions of the equation

y2 = x3 + 16:

3. A journalist wants to report on the island of scoundrels and knights,

where all inhabitants are either scoundrels (and they always lie) or knights

(and they always tell the truth). The journalist interviews each inhabitant

exactly once, and he gets the following answers:A1: on this island there is at least one scoundrel;

A2: on this island there are at least two scoundrels;

: : :

An�1: on this island there are at least n� 1 scoundrels;

An: on this island everybody is a scoundrel.

Can the journalist decide whether there are more scoundrels or more knights?

4. Let r be a line in the plane and let ABC be a triangle contained in

one of the halfplanes determined by r. Let A0, B0, C0 be the points symmet-

ric to A, B, C with respect to r; draw the line through A0 parallel to BC,

the line through B0 parallel to AC and the line through C0 parallel to AB.

Show that these three lines have a common point.

5. Let OP be a diagonal of the unit cube. Find the minimum and

the maximum value of the area of the intersection of the cube with a plane

through OP .

6. On a 10 � 10 chessboard the squares are labelled with the num-

bers 1; 2; : : : ; 100 in the following way: the �rst row contains the numbers

1; 2; : : : ; 10 in increasing order from left to right, the second row contains the

numbers 11; 12; : : : 20 in increasing order from left to right, ... , the last row

205

contains the numbers 91, 92, : : : , 100 in increasing order from left to right.

Now change the signs of �fty numbers in such a way that each row and each

column contains �ve positive and �ve negative numbers. Show that after this

change the sum of all numbers on the chessboard is zero.

Before turning to reader's solutions to problems from the 1995 numbers

of Crux, I want to give some reactions to material from the Corner. First,

comments on the 1995 Canadian Mathematical Olympiad [1995: 190; 1995:

223{226] by Murray S. Klamkin, The University of Alberta.

2. Let a, b, and c be positive real numbers. Prove that

aabbcc � (abc)(a+b+c)=3:

Comment. This same exact problem appeared in the 1974U.S.A.Math-

ematical Olympiad [1]. In the solution given, it was also noted that since

lnxx is convex for x > 0, it follows more generally and immediately by

Jensens's inequality that

aa11 aa22 � � � aann � (a1a2 � � � an)(a1+a2+���+an)=n:

3. Let n be a �xed positive integer. Show that for any nonnegative

integer k, the Diophantine equation

x31 + x3

2 + � � �+ x3n = y3k+2

has in�nitely many solutions in positive integers xi and y.

Comment. The following problem, which implies the solution, ap-

peared in the 1985 U.S.A. Mathematical Olympiad [1]:

Determine whether or not there are any positive integral solutions of

the simultaneous equations

x21 + x22 + � � �+ x21985 = y3;

x31 + x32 + � � �+ x31985 = z2;

with distinct integers x1; x2; : : : ; x1985.

It was shown that there are in�nitely many solutions. The solution

given easily extends to take care of the case if z2 is replaced by z3k+2. For a

single equation, we have more generally that

xa11 + xa22 + � � �xann = ya

has in�nitely many positive solutions if a is relatively prime to all the ai's.

We just let

xi = uifX

uaii gkp=ai where p =Y

ai so thatX

xaii = fX

uaii gkp+1:

206

Then we can take y = fPuaii gl. Finally, since p and a are relatively prime,

intgers k and l exist such that kp+ 1 = al.

Much more general results are to appear in the proceedings of the Bel-

gian Royal Academy.

In view of the above comments on previously published problems, the

Canadian Olympiad Examination Committee should be more careful about

the problems they set in their competitions.

Reference[1] M.S. Klamkin, U.S.A. Mathematical Olympiads 1972{1986, Mathemat-

ical Association of America, Washington, D.C., 1988, pp. 81, 33{34.

Next we give a comment by Murray S. Klamkin of the University of

Alberta about one of the solutions published in the Corner.

2. [1994: 39; 1995: 193{194] 1992 Czechoslovak MathematicalOlympiad, Final Round.

Let a, b, c, d, e, f be lengths of edges of a given tetrahedron and S its

surface area. Prove that

S � (p3=6)(a2 + b2 + c2 + d2 + e2 + f2):

Comment by Murray S. Klamkin, University of Alberta. It is just as

easy as the previous solution to obtain the stronger inequality

S � (p3=4)f(def)2=3 + (abd)2=3 + (bce)2=3 + (caf)2=3g:

This follows from the known inequality

F � (p3=4)(uvw)2=3

for the area F of a triangle of sides u, v, w, and which is equivalent to

the equilateral triangle being the triangle of maximum area which can be

inscribed in a given circle. The rest follows from the AM{GM inequality

(uvw)2=3 � 3(u2 + v2 + w2):

207

The next two solutions are in response to challenges to the readership

\to �ll the gaps" in solutions to problem sets discussed in 1995 and 1996

numbers of the Corner. My thanks to members of the Singapore team to the

35th IMO.

4. [1994: 64{65; 1995: 274] 9th Balkan Mathematical Olympiad(Romania).

For every integer n > 3 �nd the minimum positive integer f(n) such

that every subset of the set A = f1;2; 3; : : : ; ng which contains f(n) ele-

ments contains elements x; y; z 2 A which are pairwise relatively prime.

Solution by the joint e�orts of Siu Taur Pang and Hoe Teck Wee, silvermedallists on the Singapore team to the 35th IMO.

Let Tn = ft j t � n; 2 j t or 3 j tg. Therefore, among any three

elements of Tn, two have a common factor (either 2 or 3). Hence, f(n) �jTnj+ 1.

For 4 � n � 24 < 52, there are jTnj � 2 composite numbers in the set

An = f1; 2; 3; : : : ; ng since every composite number in An is a multiple of

2 or 3. Hence, there exist three irreducible elements (either 1 or a prime) in

any (jTnj+ 1)-element subset of An, which are in pairs relatively prime.

Therefore, we have, for 4 � n � 24, f(n) = jTnj+1. By the Principle

of Inclusion and Exclusion,

jTnj =�n

2

�+

�n

3

���n

6

�;

f(n) =

�n

2

�+

�n

3

���n

6

�+ 1: (�)

Assume that (�) also holds for some integer k > 3. Consider any (jTk+6j+1)-

element subsetB ofAk+6. If at least f(k) elements ofB lie inAk, then there

exist three elements in B which are pairwise relatively prime. Otherwise,

since jTk+6j = Tk + 4, at least �ve elements of B are from the �ve-element

set C = fk + 1; k + 2; : : : ; k + 6g. Note that the di�erence between any

two elements of C is at most 5, so the greatest common divisor between

any two elements of C is 1; 2; 3; 4 or 5. If any three elements of B \ C are

odd, then there are three elements in B which are pairwise relatively prime,

namely these three odd elements since the di�erence between any two of

these three elements is either 2 or 4 which have no odd factor larger than 1.

Otherwise, there are two odd numbers and three even numbers in B \ C.

The di�erence between any two of these even numbers is either 2 or 4, so at

most one of these three numbers is divisible by 3, and at most one is divisible

by 5. Hence, there exists an even number inB\C which is neither divisible

by 3 nor 5. Hence, this even number and the two odd numbers are pairwise

relatively prime. Thus, f(k+ 6) = jTk+6j+ 1, so (�) holds for k+ 6.

Therefore, we conclude by the Principle ofMathematical Induction that

(�) holds for all integers n > 3.

208

5. [1994: 129{130; 1996; 24{27] Final Round of the 43rdMathematicalOlympiad in Poland.

The regular 2n-gon A1; A2; : : : ; A2n is the base of a regular pyramid

with vertex S. A sphere passing through S cuts the lateral edges SAi in the

respective points Bi (i = 1; 2; : : : ; 2n). Show that

nXi=1

SB2i�1 =

nXi=1

SB2i:

Solution by Hoe Teck Wee, silver medallist on the Singapore team atthe 35th IMO.

It can be proved easily, using vectors that for any regular polygon

A1A2 : : : A2n with centre C (where n > 1 is a positive integer) and any

point P in space, that

nXj=1

PA22j�1 =

nXj=1

PA22j;

by writing PAi �PAi = CP �CP+CAi �CAi�2CAi �CP (i = 1; 2; : : : ; 2n),

and using the identity

nXj=1

CA2j�1 =

nXj=1

CA2j:

Let O denote the center of the sphere passing through S and R denote its

radius. Note that we may extend each of the lateral edges so that OAi > R

for i = 1; 2; : : : 2n. Therefore, by considering the power of Ai, we have

OA2i � R2 = SAi � AiBi = SAi � (SAi � SBi) = SA2

i � SAi � SBi:

nXj=1

(SA22j�1 � SA2j�1 � SB2j�1) =

nXj=1

(SA22j � SA2j � SB2j):

Taking P = S, we have

nXj=1

SA22j�1 =

nXj=1

SA22j )

nXj=1

SB2j�1 =

nXj=1

SB2j :

209

We now turn to reader's solutions to problems from the 1995 numbers

of the corner. We begin with the First Stage Exam of the 10th Iranian Math-

ematical Olympiad [1995: 8{9].

1. Find all integer solutions of

1

m+

1

n� 1

mn2=

3

4:

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; Christopher J. Bradley, Clifton College, Bristol, UK; Hans Engelhaupt,Franz{Ludwig{Gymnasium, Bamberg, Germany; Cyrus Hsia, student, Uni-versity of Toronto, Toronto, Ontario; Stewart Metchette, Gardena, Cali-fornia, USA; Michael Selby, University of Windsor, Windsor, Ontario; andEdward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We giveHsia's write-up, although most solutions were similar.

1

m+ 1

n� 1

mn2= 3

4. Note m;n 6= 0. Then

n2 +mn� 1

mn2=

3

4

giving

m =4(n2� 1)

n(3n� 4)=

4(n+ 1)(n� 1)

n(3n� 4):

Now (n+ 1; n) = 1 = (n� 1; n), i.e. n is relatively prime to both n + 1

and n � 1. Since m is an integer, n j 4(n+ 1)(n� 1), giving n j 4. Thus

n = �1;�2;�4.

Forn = �1; m = 0 which is impossible

n = �2; m =4(�1)(�3)

(�2)(�10); not an integer

n = 2; m = 3

n = �4; m =4(�3)(�5)

(�4)(�16); not an integer

n = 4; m =4(5)(3)

4 � 8; not an integer.

Thus the only solution is (n;m) = (2;3).

210

2. Let X be a set with n elements. Show that the number of pairs

(A;B) such that A, B are subsets of X, A is a subset of B, and A 6= B is

equal to:

3n � 2n:

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;Mansur Boase, student, St. Paul's School, London, England; ChristopherJ. Bradley, Clifton College, Bristol, UK; Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany; CyrusHsia, student, University of Toronto,Toronto, Ontario; Joseph Ling, University of Calgary, Calgary, Alberta; BobPrielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; Michael Selby,University of Windsor, Windsor, Ontario; and Edward T.H. Wang, WilfridLaurier University, Waterloo, Ontario. We give Boase's argument.

The number of subsets of an n element set is 2n. Thus the number of

pairs (A;B) withA = B is 2n. As a result we need to show that the number

of pairs (A;B) with A a subset of B, (possibly equal to B), is 3n. We do so

by induction on n.

If n = 1, let X = fa1g. The pairs (A;B) are (;; ;), (;; X), (X;X).

Assume, by induction, that for k elements the number of pairs is 3k.

Let X = fa1; : : : ; ak; ak+1g. By hypothesis there are 3k pairs (A;B) with

A � B � fa1; : : : ; akg. Each of these can be extended in three ways to

yield a pair (A0; B0) with A0 � B0 � fa1; : : : ; ak; ak+1g as follows:

(i) A0 = A, B0 = B

(ii) A0 = A [ fakg, B0 = B [ fakg(iii) A0 = A, B0 = B [ fakg.

Obviously no pairs are repeated, so the number of pairs (A0; B0) is

3 � 3k = 3k+1. This completes the induction, as required.

4. Let a, b, c, be rational and one of the roots of ax3 + bx+ c = 0 be

equal to the product of the other two roots. Prove that this root is rational.

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;Mansur Boase, St. Paul's School, London, England; Christopher J. Bradley,Clifton College, Bristol, UK; Paul Colucci, student, University of Illinois;Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany; CyrusHsia, student, University of Toronto, Toronto, Ontario; Michael Selby, Uni-versity ofWindsor,Windsor, Ontario; and Edward T.H. Wang,Wilfrid LaurierUniversity, Waterloo, Ontario. We give Amengual Covas's solution (whichwas similar to the others).

Let r1, r2, r3, be the roots of the given cubic equation and also let

r1 = r2r3. Then

r1 + r2 + r3 = 0; (1)

211

r1(r2 + r3) + r1 = r1r2 + r1r3 + r2r3 =b

a; (2)

r1r2r3 = r21 = � c

a: (3)

(Note that a 6= 0 for the existence of three solutions.) From (1) r2+r3 = �r1and substituting for r2 + r3 in (2) we �nd that

�r21 + r1 =b

a;

or equivalently

r1 =b

a+ r2

1 :

Finally, substituting �ca

for r21 from (3), we obtain r1 = b�ca

, which is

rational.

5. Find all primes p such that (2p�1 � 1)=p is a square.

Solutions by Christopher J. Bradley, Clifton College, Bristol, UK;Stewart Metchette, Gardena, California, USA; Michael Selby, University ofWindsor, Windsor, Ontario; and Edward T.H. Wang, Wilfrid Laurier Univer-sity, Waterloo, Ontario. We give Bradley's solution.

Lemma 1. The only solution in natural numbers of 2n = x2 � 1 is

n = 3, x = 3.

Proof. 2n = x2 � 1 = (x� 1)(x+ 1). The factors of 2n that di�er by

2 are 22 and 21 with n = 3 and no others.

Lemma 2. The only solutions in natural numbers of 2n = x2 + 1 are

x = 0, n = 0, n = 1, x = 1.

Proof. The cases mentioned apart, n � 2 and the lefthand side is

0 mod 4. Now x2 � 0; 1 mod 4, so x2 + 1 � 1; 2 mod 4. Contradiction

establishes the result.

Let p = 2. Then (2p�1�1)=2 = 1

2, so pmust be odd. Put p = 2k+1.

Then 2p�1�1p

= 22k�12k+1

which is an integer by Fermat's Little Theorem. Now

22k�12k+1

=(2k+1)(2k�1)

2k+1, and since 2k + 1 is prime it must be the case that

2k+1 j 2k +1 or 2k+1 j 2k� 1. Indeed 2k +1 and 2k� 1 are coprime, so

the only ways in which 2p�1�1p

can be a perfect square is for either 2k+1

2k+1and

2k � 1 to be perfect squares or 2k + 1 and 2k�12k+1

to be both perfect squares.

Case 1. 2k � 1 a perfect square. By Lemma 2 we have k = 0 or k = 1.

Then 2k+1

2k+1= 2 (k = 0) or 1 (k = 1). So k = 1, p = 3 gives a solution.

Case 2. 2k + 1 a perfect square. By Lemma 1 we have k = 3. Then2k�12k+1

= 7

7= 1, so p = 7 gives a solution, and there can be no others.

212

6. Let O be the intersection of diagonals of the convex quadrilateral

ABCD. If P and Q are the centres of the circumcircles of AOB and COD

show that

PQ � AB + CD

4:

Solution by D.J. Smeenk, Zaltbommel, the Netherlands.

A

B

C

D

Q

RO

Q1

Q2

R1

R2

Let P1, Q1, P2, Q2 be the midpoints of OB, OD, OA and OC respec-

tively.

ThenPQ � P1Q1 ) PQ � 1

2BD;

PQ � P2Q2 ) PQ � 1

2AC:

From this, we get

PQ � 1

4(BD+ AC)

� 1

4(OB + OD +OA+ OC)

� 1

4[(OB + OA) + (OC + OD)]:

Now OB + OA > AB and OC + OD > CD. Thus PQ � 1

4(AB + CD).

That completes the Corner for this issue. Please send me your nice

solutions as well as Olympiad contests for use in the future.

213

THE ACADEMY CORNERNo. 4

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

In the February 1996 issue, we gave the �rst set of problems in the

Academy Corner. Here we present solutions to the �rst three questions, as

sent in by �Sefket Arslanagi�c, Berlin, Germany.

Memorial University Undergraduate

Mathematics Competition 1995

1. Find all integer solutions of the equation x4 = y2 + 71.

Solution. From the equation x4 = y2 + 71, we get

(x2 � y)(x2 + y) = 71 � 1:

Since 71 is prime, we have

I

(x2 � y = 1

x2 + y = 71or II

(x2 � y = 71

x2 + y = 1;

so that

2x2 = 72, or 2x2 = 72,

giving

x1;2 = �6, y1;2 = 35, x3;4 = �6 y3;4 = �35.

That is,

(x; y) 2 f(6; 35); (�6; 35); (6;�35); (�6;�35)g:

The other possibility is (x2 � y)(x2+ y) = (�71) � (�1), which gives:

x = �6i =2Z.Thus we have found all solutions.

2. (a) Show that x2 + y2 � 2xy for all real numbers x, y.

(b) Show that a2 + b2 + c2 � ab+ bc+ ca for all real numbers a, b, c.

214

Solution.

(a) x2 + y2 � 2xy, (x� y)2 � 0 for all x; y 2 R.(b) From (a), we get

a2 + b2 � 2ab; a2 + c2 � 2ac; b2 + c2 � 2bc;

that is

2(a2 + b2 + c2) � 2(ab+ ac+ bc)

or

a2 + b2 + c2 � ab+ ac+ bc; for all a; b 2 R:

Remark: Let a, b, c 2 C . We observe that the equality

x3 � 1 = 0; that is, (x� 1)(x2 + x+ 1) = 0

has roots x1 = 1, x2;3 = 1

2(�1 � i

p3), so that x1 6= x2 6= x3. Let

a = x1, b = x2, c = x3, giving a + b + c = 0, ab + ac + bc =1

2(�1 + i

p3) + 1

2(�1 � i

p3) + 1

4(1 + 3) = 0 and a2 + b2 + c2 =

(a+ b+ c)2 � 2(ab+ ac+ bc) = 0.

(i) Further, in (b) the equality holds because

a2+b2+c2 = ab+ac+bc, 1

2[(a�b)2+(a�c)2+(b�c)2] = 0

if and only if a = b = c 2 R.(ii) Also, (b) holds as well, since a2 + b2 + c2 = ab + ac + bc for

a; b; c 2 C and a 6= b 6= c.

3. Find the sum of the series

1

2!+

2

3!+

3

4!+

4

5!+ : : :+

99

100!:

Solution. We have

1

k!� 1

(k+ 1)!=

1

k!� 1

(k+ 1)=

k

(k+ 1)!:

Thus

1

2!+

2

3!+

3

4!+

4

5!+ : : :+

n

(n+ 1)!= 1� 1

(n+ 1)!:

For n = 99, we get

1

2!+

2

3!+

3

4!+

4

5!+ : : :+

99

100!= 1� 1

100!:

215

BOOK REVIEWS

Edited by ANDY LIU

She Does Math! real-life problems from women on the job, edited by

Marla Parker.

Published by the Mathematical Association of America, 1995, paper-

bound, 272+ pages, ISBN# 0-88385-702-2, list price US$24.00.

Reviewed by Katherine Heinrich, Simon Fraser University.

Thirty eight women working in a variety of careers (engineering, math-

ematics, computing, biology, management) describe their lives, their feel-

ings about mathematics and the work they currently do. In all cases, com-

petence, con�dence and skill in mathematics are essential to their work.

Shelley J. Smith avoided math and chose archaeology because it required

no math. When she reached graduate school she knew it was needed (to

compute shapes of pottery from broken pieces and to do statistical analysis).

She overcame her fear and discovered not only that she could do math but

that she also enjoyed it.

These are \regular" women who, on discovering what they wanted to

do, pursued it. They met the usual challenges: of changing goals (Donna

McConnaha Sheehy started in commercial art but after a work-study expe-

rience in the Forest Service, moved to civil engineering); of having children

(Beth MacConnell, with a new baby, still had to rise at 4 a.m. to track grizzly

bears); and of taking care of parents (Nancy G. Roman retired from her job

as astronomer at NASA to take care of her mother) to mention a few.

Their stories are written for young women of all ages: telling them it

is possible to be successful and to have interesting and challenging jobs; and

that the �rst step is to develop math skills and to have con�dence in one's

abilities.

With each story the author presents problems based on either her work

or her other interests. All are mathematical. The problems range from the

very simple (formatted stock prices) to the far too obscure and di�cult (de-

termining prism diopters, determining a star's perigalactic distance). Many

were confusing with important aspects unde�ned { they could be answered

only after one �rst studied the solution. Many were simply substitutions

into an unexplained formula - these same problems could often be solved by

careful thought without the formula.

Some required formulas and ideas from physics which were presented

only in the solutions, and then often without explanation. Some were simple

arithmetic and others required calculus. On the other hand, there were some

well-written problems that encouraged the reader to explore and search for

understanding: such problems were regrettably in the minority.

216

Perhaps the greatest di�culty I had was in trying to understand who

the book was written for. I fear that in trying to reach all young women,

regardless of math background or feelings about math, it will fail them all.

I hope that any future edition will begin by targeting the audience and then

ensuring that the problems and their solutions are written speci�cally for

them.

Mathematical Literacy

Here are the answers to the questions posed in the February 1996 issue.

1. Who thought that the binary system would convince the Emperor of

China to abandon Buddhism in favour of Christianity?

Leibnitz

2. Who asked which king for one grain of wheat for the �rst square of a

chess board, two grains for the second square, four grains for the third

square, and so on?

Grand Vizier Sissa Ben Dahir asked King Shirham

3. In which well-known painting, by whom, does a Magic Square appear?

Melancholia by D�urer

4. Where was bread cut into \Cones, Cylinders, Parallelograms, and sev-

eral other Mathematical Figures"?

Laputa | (Gulliver's Travels)

5. Which mathematician said: \Philosophers count about two-hundred

and eighty eight views of the sovereign good"?

Pascal

217

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 that 1 March 1997. They may also besent by email to cruxeditor@cms.math.ca. (It would be appreciated if emailproposals and solutionswere written in LATEX, preferably in LATEX2e). Graph-ics �les should be in epic format, or plain postscript. Solutions received afterthe above date will also be considered if there is su�cient time before thedate of publication.

2151. Proposed by Toshio Seimiya, Kawasaki, Japan.4ABC is a triangle with \B = 2\C. Let H be the foot of the per-

pendicular from A to BC, and let D be the point on the side BC where the

excircle touches BC. Prove that AC = 2(HD).

2152. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,Austria.

Let n � 2 and 0 � x1 � : : : � xn � �

2be such that

nXk=1

sinxk = 1.

Consider the set Sn of all sums x1 + : : :+ xn.

1. Show Sn is an interval.

2. Let ln be the length of Sn. What is limn!1

ln?

2153. Proposed by �Sefket Arslanagi�c, Berlin, Germany.Suppose that a, b, c 2 R. If, for all x 2 [�1; 1], jax2 + bx + cj � 1,

prove that

jcx2 + bx+ aj � 2:

2154. Proposed by K.R.S. Sastry, Dodballapur, India.

218

In a convex pentagon, the medians are concurrent. If the concurrence

point sections each median in the same ratio, �nd its numerical value. (A

median of a pentagon is the line segment between a vertex and the midpoint

of the third side from the vertex.)

2155. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.

Prove there is no solution of the equation

1

x2+

1

y8=

1

z2

in which y is odd and x; y; z are positive integers with highest common fac-

tor 1.

Find a solution in which y = 15, and x and z are also positive integers.

2156. Proposed by Hoe Teck Wee, student, Hwa Chong Junior Col-lege, Singapore.

ABCD is a convex quadrilateral with perpendicular diagonalsAC and

BD. X and Y are points in the interior of sides BC and AD respectively

such thatBX

CX=BD

AC=DY

AY:

EvaluateBC �XYBX � AC :

2157. Proposed by �Sefket Arslanagi�c, Berlin, Germany.Prove that 21997�1996� 1 is exactly divisible by 19972.

2158. Proposed by P. Penning, Delft, the Netherlands.Find the smallest integer in base eight for which the square root (also

in base eight) has 10 immediately following the `decimal' point.

In base ten, the answer would be 199, with sqrt (199) = 14:10673 : : : .

2159. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,Austria.

Let the sequence fxn; n � 1g be given by

xn =1

n(1 + t+ : : :+ tn�1)

where t > 0 is an arbitrary real number.

Show that for all k, l � 1, there exists an indexm = m(k; l) such that

xk � xl � xm.

2160. Proposed by Toshio Seimiya, Kawasaki, Japan.4ABC is a triangle with\A < 90�. LetP be an interior point ofABC

such that \BAP = \ACP and \CAP = \ABP . Let M and N be the

219

incentres of4ABP and4ACP respectively, and letR1 be the circumradius

of 4AMN . Prove that

1

R1

=1

AB+

1

AC+

1

AP:

2161. Proposed by Juan-Bosco Romero M�arquez, Universidad deValladolid, Valladolid, Spain.

Evaluate1Xn=1

1

(2n� 1)(3n� 1):

2162. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.In4ABC, the Cevian lines AD, BE, and CF concur at P . [XY Z] is

the area of4XY Z. Show that

[DEF ]

2[ABC]=PD

PA� PEPB

� PFPC

2163. Proposed by Theodore Chronis, student, Aristotle Universityof Thessaloniki, Greece.

Prove that if n;m 2 N and n � m2 � 16, then 2n � nm.

Corrections.

2139. Proposed by Waldemar Pompe, student, University of War-saw, Poland.

Point P lies inside triangle ABC. Let D, E, F be the orthogonal pro-

jections from P onto the lines BC, CA, AB, respectively. Let O0 and R0

denote the circumcentre and circumradius of the triangleDEF , respectively.

Prove that

[ABC] � 3p3R0

qR02 � (O0P )2;

where [XY Z] denotes the area of triangle XY Z.

Note: 3p3R0 instead of 3

p3R0, as was printed in the May 1996 issue.

Also, problem 2149 was ascribed to Juan-Bosco Morero M�arquez in-

stead of Juan-Bosco Romero M�arquez.

220

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

Some solvers are reminded about the instructions given in the Problems

section above:

: : : please send your proposals and solutions on signed andseparate : : : sheets of paper.

The emphasis here is on \signed". The secretarial sta� may not notice

if a submitted solution is not signed, and when it comes time for an editor

to read the �le, there is no way of knowing who the anonymous person is!

1823. [1993: 77; 1994: 54] Proposed by G. P. Henderson, Campbell-croft, Ontario.

A rectangular box is to be decorated with a ribbon that goes across the

faces and makes various angles with the edges. If possible, the points where

the ribbon crosses the edges are chosen so that the length of the closed path

is a local minimum. This will ensure that the ribbon can be tightened and

tied without slipping o�. Is there always a minimal path that crosses all six

faces just once?

Correction and editor's comment by Bill Sands.The previous editor's remark on [1994: 55] that Jordi Dou's second

example of a minimal path, namely

""""""""""""""""""

a

c a

b

b a

c

does not result in any further boxes that can be decorated, is false. As Gerd

Baron has since pointed out, and Dou and the proposer knew all along, there

are boxes that can be decorated in the above way but whose sides a; b; c do

not satisfy the triangle inequality. For example, a = 4, b = 2, c = 1 yield

221

with the ribbon as shown.

The following is derived from Baron's and the proposer's correct solu-

tions for this problem. From the �rst �gure above, one can see that such a

ribbon exists in this case if and only if

b

2a+ c<

b+ c

2a+ b+ c<b

c;

where the middle fraction is the slope of the slanted line. This simpli�es to

c2 � b2 + 2ac > 0 and b2 � c2 + 2ab > 0:

By permuting a; b; c we obtain two other su�cient conditions for an a by b

by c box to allow a ribbon:

a2 � c2 + 2ab > 0 and c2 � a2 + 2bc > 0;

and

b2 � a2 + 2bc > 0 and a2 � b2 + 2ac > 0:

Moreover, these (and the previous solution) exhaust all possible ways for a

ribbon to cross all six faces of the box. If we assume that a � b � c, then

an a by b by c box will not possess a ribbon satisfying the condition of the

problem exactly if the following three inequalities all hold:

b2 � a2 + 2bc � 0; c2 � b2 + 2ac � 0; b+ c � a:

For example, the 4 by 3 by 1 box does not have a ribbon.

So in retrospect, only Baron, Dou and the proposer sent in complete

solutions of this problem. The previous editor (that is, me, Bill Sands) thanks

Baron and the proposer for calling attention to my too{hasty reading of the

solutions the �rst time around.

2044. [1995: 158] Proposed by Murray S. Klamkin, University ofAlberta, Edmonton, Alberta.

Suppose that n � m � 1 and x � y � 0 are such that

xn+1 + yn+1 � xm � ym:

Prove that xn + yn � 1.

A typographical error in the LATEX codes used in typesetting resulted in

Toshio Seimiya, Kawasaki, Japan being listed as the solver. The solver was

indeed Heinz-J �urgen Sei�ert, Berlin, Germany. The editor o�ers his apolo-

gies to Heinz-J �urgen Sei�ert for any embarrassment caused by this error.

222

2048. [1995: 158]ProposedbyMarcin E. Kuczma, Warszawa, Poland.Find the least integer n so that, for every string of length n composed

of the letters a; b; c; d; e; f; g; h; i; j; k (repetitions allowed), one can �nd a

non-empty block of (consecutive) letters in which no letter appears an odd

number of times.

Solution by Douglas E. Jackson, Eastern New Mexico University, Por-tales, New Mexico, USA.

Let S be a string of length 211 = 2048 of the 11 letters, a, b, : : : , k. For

i = 1; ;, : : : , 2048, let na(i), nb(i), : : : , nk(i) be the number of occurrences

mod 2 of the letters a, b, : : : , k respectively in the �rst i positions of S.

Then, for each position i in S, we have an associated binary string, n(i) =

na(i)nb(i) : : : nk(i), for length 11. If, for some i, we have n(i) = 00 : : : 0,

then in the �rst i positions of S, each letter appears an even number of times.

If no position is associated with the zero bit string, then, since there are only

2047 non-zero bit strings of length 11, there must exists i and j such that

1 � i � j � 2048 and n(i) = n(j). [The Pigeon-hole Principle one more

time! | Ed.]

Then each letter appears an even number of times in positions i+1 through

j of S. Therefore, any string on 11 letters with length at least 2048 will have

a block in which no letter appears an odd number of times.

In the following sequence of strings, each string is constructed by taking two

copies of the previous string and inserting a new letter between these copies.

This construction allows a simple inductive argument, that in these strings,

every non-empty block contains some letter an odd number of times. Let B

be a block in one of the strings of this sequence, other than the �rst. If B

contains the central (new) letter, then obviously it contains that letter an odd

number of times (once). Otherwise, B must be a block of the previous string,

and hence is inductively assumed to contain some letter an odd number of

times.

a

a b a

a b a c a b a

a b a c a b a d a b a c a b a

: : :

Clearly string number 11 in this sequence has 11 letters and length 2047. So

the minimum value of n which forces the required property is 2048.

Also solved by CARL BOSLEY, student, Washburn Rural High School,Topeka, Kansas, USA; ASHSIH KR. SINGH, student, Kanpur, India; HOETECK WEE, student, Hwa Chong Junior College, Singapore (all with virtuallyidentical arguments to the one given above); and the proposer.

Wee actually considered the general case when the number of letters is kinstead of 11. Using the same argument, he showed that the answer is 2k.

223

2052. [1995: 202] Proposed by K.R.S. Sastry, Dodballapur, India.The in�nite arithmetic progression 1+ 3+ 5+ 7+ : : : of odd positive

integers has the property that all of its partial sums

1; 1 + 3; 1 + 3 + 5; 1 + 3 + 5 + 7; : : :

are perfect squares. Are there any other in�nite arithmetic progressions, all

terms positive integers with no common factor, having this same property?

Solution by Toby Gee, student, the John of Gaunt School, Trowbridge,England Let the nth term of the arithmetic progression be a+(n� 1)d, and

let the sum of the �rst n terms be

Sn = an+n(n� 1)d

2:

Let p be any prime and consider n = 2p. Then, since S2p is a perfect square,

we have

S2p = 2ap+ p(2p� 1)d = y2

for some integer y. Thus p j y. Let y = px. Then

2a+ 2pd� d = px2 � 0 (mod p)

Therefore, p j (2a�d). For all su�ciently large primes pwe have p > 2a�d,implying 2a = d, whence Sn = an+an(n�1) = an2. Since these numbers

have no common factor, we must have a = 1, so there are no other such

progressions.

Also solved by CARL BOSLEY, student, Washburn Rural High School,Topeka, Kansas, USA; JEFFREY K. FLOYD,Newnan, Georgia, USA; RICHARDI. HESS, Rancho Palos Verdes, California, USA; MICHAEL PARMENTER,Memorial University of Newfoundland, St. John's, Newfoundland; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College,Calgary; and the proposer. There were two incorrect solutions submitted.

Remark by Christopher J. Bradley, Clifton College, Bristol, UK.The correct generalization of the

1 1 + 3 1 + 3 + 5 1 + 3 + 5 + 7 : : :

phenomenon is to be found not in the problem stated, but rather by produc-

ing the partial sums of an arithmetic progression such as

15 15 + 33 15 + 33 + 51 15 + 33 + 51 + 69 : : :

(a = 15, d = 18) and then adding 1 providing

16 49 100 169 : : :

and then tacking on 1 at the beginning giving

12; 42; 72; 102; 132; : : :

224

The integers whose squares appear are always in arithmetic progression them-

selves.

It always works. For example, if we want to do the reverse for

12; 52; 92; 132; 172; : : :

we look at 24, 80, 168, 288, etc. and perceive this to be

24; 24 + 56; 24 + 56 + 88; 24 + 56 + 88 + 120; : : :

which are the partial sums for a = 24, d = 32. Proofs covering these remarks

are straightforward. In the above, d is twice a perfect square and a = 1

2d+p

2d. It also works when squares other than 1 (or 12) are tacked on provided

a and d are carefully chosen.

BRADLEY mentions that he has seen this in some magazine before,but does not recall when or where. Perhaps a reader can supply us with theproper reference.

2056. [1995: 203] Proposed by Stanley Rabinowitz, Westford, Mas-sachusetts.

Find a polynomial of degree �ve whose roots are the tenth powers of

the roots of the polynomial x5 � x� 1.

I Solution by Christopher J. Bradley, Clifton College, Bristol, UK.If x5 � 1 = x has roots �i (i = 1; 2; : : : ; 5), then the substitution

y = x5 will provide an equation with roots �5i . Now, (x

5�1)5 = x5, so the

equation required is (y�1)5 = y; that is, y5�5y4+10y3�10y2+4y�1 = 0.

If we now put u = y2, the resulting equation in u will have roots �10i .

Since y(y4 + 10y2 + 4) = 5y4 + 10y2 + 1, we have, by squaring, that

u(u2 + 10u+ 4)2 = (5u2 + 10u+ 1)2. Multiplying out and collecting like

terms, we get u5 � 5u4 + 8y3� 30u2 � 4u� 1 = 0, and the polynomial on

the left of this equation is the answer.

II Solution by Carl Bosley, student, Washburn Rural High School, Top-eka, Kansas, USA.

Let ri (1 � i � 5) be the roots of the given polynomial. Since x5 =

x + 1, we have r10i = (ri + 1)2. It is easy to construct a polynomial with

roots ri + 1; it is

f(x) = (x� 1)5� (x� 1)� 1 = x5 � 5x4 + 10x3 � 10x2 + 4x� 1:

Now, the polynomial

�f(x)f(�x) = (x5 � 5x4 + 10x3 � 10x2 + 4x� 1)

�(x5 + 5x4 + 10x3 + 10x2 + 4x+ 1)

= x10 � 5x8 + 8x6 � 30x4 � 4x2 � 1

225

is even, and has roots ri + 1, �(ri + 1) (1 � i � 5). Replacing each x2k

term with an xk term, we get a polynomial x5� 5x4+8x3�30x2�4x�1,

with roots (ri + 1)2, that is, r10i .

Also solved by HAYO AHLBURG, Benidorm, Spain; MIGUEL ANGELCABEZ �ON OCHOA, Logro ~no, Spain; THEODORE CHRONIS, student, Aristo-tle University of Thessaloniki, Greece; RONALD HAYNES, student, Memo-rial University of Newfoundland, St. John's, Newfoundland; RICHARD I.HESS, Rancho Palos Verdes, California, USA; PETER HURTHIG, ColumbiaCollege, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck,Austria; KEE-WAI LAU, Hong Kong; MARIA ASCENSI �ON L �OPEZCHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; VEDULA N. MURTY,Andhra University, Visakhapatnam, India; DIGBY SMITH,MountRoyal Col-lege, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; MITKOCHRISTOV VINCHEO, Rousse, Bulgaria; and the proposer. One reader sub-mitted a one-line answer only, and another submitted a partially incorrectanswer. The solutions given by Hess, Murty and the proposer are very sim-ilar to Solution I above.

2058. [1995: 203] Proposed by Christopher J. Bradley, Clifton Col-lege, Bristol, UK.

Let a; b; c be integers such that

a

b+b

c+c

a= 3:

Prove that abc is the cube of an integer.

Solution by Michael Parmenter, Memorial University of Newfound-land, St. John's, Newfoundland, modi�ed by the editor.

Without loss of generality, we may assume that gcd(a; b; c) = 1, since,

if d j a; b; c and a0 = a

d, b0 = b

d, c0 = c

d, then a

b+ b

c+ c

a= 3 if and only if

a0

b0+ b0

c0+ c0

a0= 3, and abc is a perfect cube if and only if so is a0b0c0. Rewrite

the given equation as

a2c+ b2a+ c2b = 3abc: (1)

If abc = �1, then we are done. Otherwise, let p be any prime divisor of

abc. Then from (1), it is clear that p divides exactly two of a, b and c. By

symmetry, we may assume that p j a, p j b and p6 j c.Suppose that pm k a (that is, pm j a, but pm+1 6 j a) and pn k b. Then

pm+n j 3abc. We claim that n = 2m.

If n < 2m, then n + 1 < 2m, and thus pn+1 j a2c. Since pn+1 j b2a, butpn+1 6 j c2b, we have pn+1 6 j a2c + b2a + c2b. This is a contradiction since

n+ 1 � m+ n implies pn+1 j 3abc.

226

If n > 2m, then n � 2m + 1, and thus p2m+1 j c2b. Since p2m+1 j b2a,but p2m+1 6 j a2c, we have that p2m+1 6 j a2c + b2a + c2b. This is again a

contradiction since 2m+ 1 < m+ n implies that p2m+1 j 3abc.Hence n = 2m, which implies that pm k abc. It follows immediately that

abc is a perfect cube.

Also solved by CARL BOSLEY, student, Washburn Rural High School,Topeka, Kansas, USA; ADRIAN CHAN, student, Upper Canada College,Toronto, Ontario; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain;THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece;F.J. FLANIGAN, San Jose State University, San Jose, California, USA;PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; PANOS E. TSAOUSSOGLOU,Athens, Greece; and the proposer.

Six solvers submitted incorrect solutions. All of them assumed inadvertentlythat a, b, c are positive. (One other reader sent in a partial solutionbased onthis assumption.) In this case, the problem becomes trivial since, from theArithmetic-Geometric Mean inequality, one can deduce immediately thata = b = c. Several other incomplete, or partially incomplete, solutionswere also received.

Using an argument similar to the one given above, Bosley showed that allsolutions are given by a = b = c or a = kr(r + s)2, b = ksr2, c =

�k(r+s)s2 for some integers k, r, s, provided that abc 6= 0. Setting k = 1,r = n, s = n+1 (n 6= 0; �1) yields the in�nite family of solutions (a; b; c),where a = n(2n+1)2, b = n2(n+1), c = �(n+1)2(2n+1) [together withall integer multiples of them, and triples obtained by cyclically permuting a,b and c. | Ed.].

Setting k = 1, r = n � 1, s = n + 1, or k = 1, r = n + 1, s = n � 1

(n 6= 0; �1) yields the in�nite families of solutions (a; b; c), where a =

4n2(n� d), b = (n2 � 1)(n� d), c = �2n(n+ d)2, where d = �1. Allthese families were also found by Penning. Setting k = 1, r = 2, s = 3, oneobtains the solution (50; 12;�45), also found by Hurthig. The special case(4; 1;�2), obtained by letting k = r = s = 1 was also found by Chronis,Flannigan, Tsoussaglou, and the proposer.

Janous asked whether the \n{analogue" of this problem is true (the casen = 2 is trivial); for example, if a

b+ b

c+ c

d+ d

a= 4, where a, b, c, d are

integers, must abcd be a perfect 4th power? The answer is clearly negative,as shown by the example (a; b; c; d) = (�2; 4; 1; 1), found by this editor.

227

2059. [1995: 203] Proposed by �Sefket Arslanagi�c, Berlin, Germany.

Let A1A2 : : : An be an n-gon with centroid G inscribed in a circle.

The linesA1G;A2G; : : : ; AnG intersect the circle again at B1; B2; : : : ; Bn.

Prove thatA1G

GB1

+A2G

GB2

+ � � �+ AnG

GBn

= n:

Solution by Toshio Seimiya, Kawasaki, Japan.

Let O and R be the centre and the radius of the circumscribed circle

A1A2 : : : An.

O

R

R

An

A1

A2

Ai�1

Ai

Ai+1

SinceAiG

GBi

=AiG

2

AiG �GBi

=AiG

2

OA2i � OG2

=AiG

2

R2 � OG2;

for i = 1; 2; : : : ; n, we get

nXi=1

AiG

GBi

=

nXi=1

AiG2

R2 � OG2

: (1)

Since G is the centroid of A1A2 : : : An, we have

nXi=1

AiG = 0:

Thus we have

nXi=1

AiG �GO =

nXi=1

AiG

!�GO = 0 �GO = 0:

228

Hence we have

nXi=1

AiO2

=

nXi=1

�AiG+GO

�2

=

nXi=1

AiG2

+ 2

nXi=1

AiG �GO +

nXi=1

GO2

=

nXi=1

AiG2

+ n GO2

; that is

nR2 =

nXi=1

AiG2 + nGO2:

Therefore

nXi=1

AiG2 = n(R2 �GO2): (2)

From (1) and (2), we obtain that

nXi=1

AiG

GBi

= n.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK;MIGUELANGEL CABEZ �ONOCHOA, Logro ~no, Spain; JORDI DOU,Barce-lona, Spain; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;MARIA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid,Spain; P. PENNING, Delft, the Netherlands; HOE TECK WEE, student, HwaChong Junior College, Singapore; and the proposer.

2060. [1995: 203] Proposed by Neven Juri�c, Zagreb, Croatia.Show that for any positive integers m and n, the integerj�

m+pm2 � 1

�nkis odd (bxc denotes the greatest integer less than or equal to x).

I Solution by David Doster, Choate Rosemary Hall, Wallingford, Con-necticut, USA.

Let xn =�m+

pm2 � 1

�n+�m�

pm2 � 1

�n. Then, using the

Binomial Theorem, we get

xn = 2

bn=2cXk=0

�n

2k

�mn�2k

�m2 � 1

�k;

and so xn is an even integer.

Since 0 <�m�

pm2 � 1

�n=

1

m+pm2 � 1

� 1, we have that

229

xn � 1 ��m+

pm2 � 1

�n< xn, and thus thatj�

m+pm2 � 1

�nk= xn � 1, which is odd.

II Solution by P. Penning, Delft, the Netherlands. (modi�ed slightlyby the editor).

Let u = m+pm2 � 1 and

Fn = un + u�n =�m+

pm2 � 1

�n+�m�

pm2 � 1

�n:

Then F1 = 2m. Since

2mFn = F1Fn =�u+ u�1

� �un + u�n

�= Fn+1 + Fn�1;

we obtain the recurrence relation

Fn+1 = 2mFn � Fn�1:

Since F2 = 2�2m2 � 1

�is even, we conclude that Fn is even for all n. Since

un � 1, we have 0 < u�n � 1, and it follows that bunc =�Fn � u�n

�=

Fn � 1, which is odd.

Also solved by HAYO AHLBURG, Benidorm, Spain; �SEFKETARSLANAGI �C, Berlin, Germany; SEUNG-JIN BANG, Seoul, Korea;FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CARLBOSLEY, student, Washburn Rural High School, Topeka, Kansas, USA;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN,student, Upper Canada College, Toronto, Ontario; TOBY GEE, student, theJohn of Gaunt School, Trowbridge, England; DAVID HANKIN, Hunter Col-lege Campus Schools, New York, NY, USA; RICHARD I. HESS, Rancho PalosVerdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; KEE-WAI LAU, Hong Kong; VEDULA N. MURTY, AndhraUniversity, Visakhapatnam, India; GOTTFRIED PERZ, Pestalozzigymnasium,Graz, Austria; CORY PYE; student, Memorial University, St. John's, New-foundland; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; PANOS E.TSAOUSSOGLOU, Athens, Greece; CHRIS WILDHAGEN, Rotterdam, theNetherlands; PAUL YIU, Florida Atlantic University, Boca Raton, Florida,USA; and the proposer.

Most submitted solutions are similar to one of the two solutions above.Bellot Rosado pointed out that the problem is a very old one. He locatedit as Part (a) of Problem 363, Americal Mathematical Monthly, 1912, p. 51.

He also gave four similar or related problems.

230

2061. [1995: 234] Proposed by Toshio Seimiya, Kawasaki, Japan.ABC is a triangle with centroid G, and P is a variable interior point of

ABC. Let D, E, F be points on sides BC, CA, AB respectively such that

PD k AG, PE k BG, and PF k CG. Prove that [PAF ]+ [PBD]+ [PCE]

is constant, where [XY Z] denotes the area of triangle XY Z.

I. Solution by Jaosia Jaszunska, student, Warsaw, Poland.In the �gure, the line through P :

parallel to AB intersects the sidesBC and CA at I and L respectively;

parallel to BC intersects the sides CA and AB at K and N respec-

tively;

parallel to CA intersects the sides AB and BC at M and J respec-

tively.

B I D J C

K

E

L

A

N

F

M

P

Since PD and AG are parallel, the dilatation that takes triangle PIJ

to ABC, takes D to the mid-point of BC. Thus D is the mid-point of IJ.

Similarly, E and F are the mid-points of KL and MN respectively. It is

easily seen that the shaded area is equal to the unshaded area. [The vertically

shaded pieces such as PID form half of a triangle, while the horizontally

shaded pieces such as PBI form half of a parallelogram.] Thus

[PAF ] + [PBD] + [PCE] = 1

2[ABC];

so that the sum is constant as desired.

[Editor's comment: Several solvers mentioned that since the result belongs

to a�ne geometry, one can assume without loss of generality that the given

triangle is equilateral. We have therefore used an equilateral triangle in the

�gure.]

231

II. Solution by Christopher J. Bradley, Clifton College, Bristol, UK.The result holds for any point P in the plane { there is no need to

restrict P to the interior of the given triangle. In areal coordinates, AG =

(�2

3; 13; 13), and since PDkAG, the point D must have coordinates (� �

2�

3; �+ �

3; �+ �

3), where P is (�; ��), normalised so that �+�+� = 1, and

� is chosen so thatD lies onBC. Thus � = 3�=2 andD is (0; �+ �

2; �+ �

2).

From the formula for ratio of areas,

[PBD]

[ABC]=

������� 0 0

� 1 �+ �

2

� 0 � + �

2

������ = 1

2�2 + ��:

Similar formulae hold for the other areas, from which we deduce that

[PAF ] + [PBD] + [PCE]

= [ABC]�1

2�2 + 1

2�2 + 1

2�2 + ��+ ��+ ��

�= 1

2(�+ �+ �)2[ABC]

= 1

2[ABC];

which is constant for a given triangle, as P varies.

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; JORDI DOU, Barce-lona, Spain; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg,Germany; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; WALDEMARPOMPE, student, University of Warsaw, Poland; and the proposer.

2062. [1995: 234] Proposed by K.R.S. Sastry, Dodballapur, India.Find a positive integer n so that both the continued rootss

1995 +

rn+

q1995 +

pn+ � � �

and sn+

r1995 +

qn+

p1995 + � � �

converge to positive integers.

Solution by Miguel Angel Cabez �on Ochoa, Logro ~no, Spain (loosely

translated from the Spanish).

232

Let a0 = 0, a1 =p1995, and ap+2 =

q1995 +

pn+ ap for p � 0.

Similarly let b0 = 0, b1 =pn, and bp+2 =

qn+

p1995 + bp for p � 0.

We will now show by induction that ap < K for all p � 0, where K =

maxf1995; ng. This is clearly true for p = 0 and p = 1. Suppose that

ap < K for all p, 0 � p < t with t � 2. Then

at =p1995 +

pn+ at�2 <

pK +

pK +K =

pK +

p2K

= K

s1

K+

p2K

K2= K

s1

K+

r2

K3< K

whence ap < K for all p � 0.

Next we show that fapg is an increasing sequence by induction. It is clear

that a0 < a1 < a2. Suppose that ap < ap+1 for some p � 0. Then

ap+2 =

q1995 +

pn+ ap <

q1995 +

pn+ ap+1 = ap+3

which establishes that the sequence is increasing.

Similar proofs exist for the sequence fbpg. Thus both fapg and fbpg con-

verge to limits, say x and y, respectively. This leads to the equations x =p1995 + y and y =

pn+ x. Let y = (p+ 44)2 � 1995; then x = p+ 44

and n = y2 � x. (Editor's comment: this is clearly a solution for all integer

values of p � 1.) The �rst two values for p give the following solutions:

(p; n; x; y) = (1;855; 45; 30) and (p; n; x; y) = (2; 14595; 46; 121).

Also solved by SABIN CAUTIS, student, Earl Haig Secondary School,North York, Ontario; ADRIAN CHAN, student, Upper Canada College,Toronto, Ontario; TIM CROSS, King Edward's School, Birmingham, Eng-land; KEITH EKBLAW,WallaWalla, Washington,USA; HANS ENGELHAUPT,Franz{Ludwig{Gymnasium, Bamberg, Germany; ROBERT GERETSCHL �AGER,Bundesrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph ScollardHall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, Cali-fornia, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;FRIENDH. KIERSTEAD JR., Cuyahoga Falls, Ohio,USA; JAMSHIDKHOLDI,New York, NY, USA; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; KEE-WAI LAU, Hong Kong; BEATRIZ MARGOLIS, Paris,France; CORY PYE, student, Memorial University, St. John's, Newfound-land; CRIST �OBAL S �ANCHEZ{RUBIO, I.B. Penyagolosa, Castell �on, Spain;HEINZ-J�URGEN SEIFFERT, Berlin, Germany; ASHSIH KR. SINGH, student,Kanpur, India; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRISWILDHAGEN, Rotterdam, the Netherlands; KENNETH M. WILKE, Topeka,Kansas, USA; and the proposer. There was one incomplete and one incorrectsolution submitted.

233

Most solvers found all of the solutions above, but some only foundone. Very few solvers directly examined the question of convergence of thecontinued roots.

2063. [1995: 234] Proposed by Aram A. Yagubyants, Rostov naDonu, Russia.

Triangle ABC has a right angle at C.

(a) Prove that the three ellipses having foci at two vertices of the given

triangle, while passing through the third, all share a common point.

(b) Prove that the principal vertices of the ellipses of part (a) (that is the

points where an ellipse meets the axis through its foci) form two pairs

of collinear triples.

Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria.(a) Let P be the fourth vertex of the rectangle APBC. Then

PA = BC; PB = AC; PC = AB;

whence

PA+PB = CB+CA; PB+PC = AC+AB; PC+PA = BA+BC;

which means that each of the three ellipses passes through P .

(b) Let M1, M2, M3 be the centres of the three ellipses, and there-

fore the mid-points of AB, BC and CA respectively. Denote the principal

vertices of the ellipses by Xi and Yi, with B lying between A and X1 and

between C and X2, and with C lying between A and X3. Since

M1X1 =AC +BC

2; M2X2 =

AB + AC

2; M3X3 =

AB + BC

2;

we get with AB = c, BC = a, AC = b (and recalling that a2 + b2 = c2),

AX1

BX1

� BX2

CX2

� CX3

AX3

=

a+ b+ c

2a+ b� c

2

�a+ b+ c

2a+ b+ c

2

a� b+ c

2a+ b+ c

2

=�a2 + 2ab� b2 + c2

a2 + 2ab+ b2 � c2

=2ab

2ab= 1:

This means, according to Menelaus' Theorem, thatX1,X2, X3 are collinear.

234

Analogously, Y1, Y2, Y3 are collinear.

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MARIAASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGELCABEZ �ON OCHOA, Logro ~no, Spain; ROBYNM. CARLEY (student) and ANAWITT, AustinPeay State University, Clarksville, Tennessee, USA; JORDIDOU,Barcelona, Spain; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bam-berg, Germany; CAMILLA FOX, Toronto, Ontario; SHAWNGODIN, St. JosephScollardHall, North Bay, Ontario; WALTHER JANOUS, Ursulinengymnasium,Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan,USA; WALDEMAR POMPE,student,University ofWarsaw, Poland;CRIST �OBAL S �ANCHEZ{RUBIO, I.B. Penyagolosa, Castell �on, Spain; TOSHIOSEIMIYA, Kawasaki, Japan; and the proposer. One anonymous solutionwasreceived - see note at the start of this section.

Bellot Rosado and L �opez Chamorro, Cabez �on Ochoa, Dou, Pompe andSeimiya all sent in solutions that are essentially the same as Perz's, whosesolution was chosen by the time-honoured method of selecting the supe-rior paper after having thrown the lot down the stairs. The other solversused coordinates; teachers may wish to note that this problem provides anon-standard exercise that illustrates e�ectively several aspects of analyticgeometry.

2064. [1995: 234] Proposed by Murray S. Klamkin, University ofAlberta, Edmonton, Alberta.

Show that

3max

�a

b+b

c+c

a;b

a+c

b+a

c

�� (a+ b+ c)

�1

a+

1

b+

1

c

�

for arbitrary positive real numbers a; b; c.

I. Solutionby Sabin Cautis, student, Earl Haig Secondary School,NorthYork, Ontario.

Assume without loss of generality that

a

b+b

c+c

a� b

a+c

b+a

c:

Using the AM{GM inequality,

a

b+b

c+c

a� 3

3

rabc

bca= 3;

thus

3max

�a

b+b

c+c

a;

b

a+c

b+a

c

�= 3

�a

b+b

c+c

a

�

235

��a

b+b

c+c

a

�+

�b

a+c

b+a

c

�+

�a

b+b

c+c

a

�

��a

b+b

c+c

a

�+

�b

a+c

b+a

c

�+ 3

= (a+ b+ c)

�1

a+

1

b+

1

c

�:

II. Solution by Chris Wildhagen, Rotterdam, the Netherlands.It is evident that

x+ y � 2z =) maxfx; yg � z

for all real numbers x, y and z. Hence it is su�cient to show that

3

�a

b+b

c+c

a+

b

a+c

b+a

c

�� 2(a+ b+ c)

�1

a+

1

b+

1

c

�;

or �a

b+b

a

�+

�b

c+c

b

�+

�c

a+a

c

�� 6:

This is obviously true, since if p; q > 0 then

p

q+q

p=p2 + q2

pq� 2:

III. Editorial comments.Note that Solution II says that \max" in the problem statement can

be replaced by \average". This stronger result was noticed by several other

solvers as well. Here are some other remarks and generalizations made by

readers.

If a; b; c are the sides of a triangle, then the inequality is true if \max"

is replaced by \min", that is, the simpler inequality

3

�a

b+b

c+c

a

�� (a+ b+ c)

�1

a+

1

b+

1

c

�(1)

holds whenever a; b; c are the sides of a triangle. The proposer included

this problem as #2 in his list of six Quickies published in the February 1996

Olympiad Corner (for a solution, see [1996: 59{60]). The proposer also notes

that this inequality need not hold otherwise. For example, it fails when

a = 8, b = 3 and c = 1.

Bradley proves that in fact (1) holds wheneverpa;pb;pc are the sides

of a triangle. (This happens whenever a; b; c are sides of a triangle and in

other cases besides.) He �rst proves that if x; y; z are the sides of a triangle

and l; m; n are any real numbers, then

x2l2 + y2m2 + z2n2 � (y2 + z2 � x2)mn

�(z2 + x2 � y2)nl� (x2 + y2 � z2)lm � 0; (2)

236

by completing the square. (If we multiply the left side of (2) by 4x2, then it

can be rewritten in the form�2x2l� (z2 + x2 � y2)n� (x2 + y2 � z2)m

�2+(x+ y+ z)(x+ y� z)(y+ z � x)(z + x� y)(m� n)2; (3)

with the help of the identity

(x+ y + z)(x+ y � z)(y+ z � x)(z + x� y)

= 4x2z2 � (z2 + x2 � y2)2; (4)

and (3) is clearly � 0 if x; y; z are sides of a triangle.) Bradley then puts

a = x2 = n, b = y2 = l, c = z2 = m in (2); the result is equivalent to

(1) and holds ifpa;pb;pc are the sides of a triangle. Consultation with

helpful expert (and current proposer) Murray Klamkin yielded the following

alternate way to derive (2). It is known (for example, see Theorem 3, page

306 of Gantmacher'sMatrix Theory (The Theory ofMatrices), Vol. I, Chelsea,1959, or any other advanced book on matrix theory) that (2) holds if and only

if the matrix266666664

x2z2 � x2 � y2

2

y2 � z2 � x2

2

z2 � x2 � y2

2y2

x2 � y2� z2

2

y2 � z2 � x2

2

x2 � y2 � z2

2z2

377777775

is non-negative de�nite, which is true if and only if the principal minors are

non-negative. Since the minor x2 is non-negative, and the determinant of

the entire matrix is zero (just add all the rows), all we have to prove is that�������x2

z2 � x2 � y2

2z2 � x2 � y2

2y2

������� � 0:

But this is true when x; y; z are the sides of a triangle because

4

�������x2

z2 � x2 � y2

2z2 � x2 � y2

2y2

������� = 4x2y2� (z2 � x2 � y2)2

which is � 0 by a version of (4).

Janous shows that for any n � 3 and for all x1; : : : ; xn > 0,

(x1 + � � �+ xn)

�1

x1+ � � �+ 1

xn

�� n

n� 1

X1�k<l�n

�xk

xl+

xl

xk

�;

237

which generalizes the result of Solution II and can be proved the same way.

This can be written as

M1(x1; : : : ; xn)

M�1(x1; : : : ; xn)�M1

��xk

xl: k 6= l

��;

where Mt denotes the tth power mean:

Mt(x1; : : : ; xn) =

�xt1 + � � �+ xtn

n

�1=t:

He asks for the minimum t > 0 such that

M1(x1; : : : ; xn)

M�1(x1; : : : ; xn)�Mt

��xk

xl: k 6= l

��

for all x1; : : : ; xn > 0. (His conjecture: t = 1.) Janous also writes the

inequality of Solution II as

M1(a; b; c)M1

�1

a;1

b;1

c

�� 1

2

�M1

�a

b;b

c;c

a

�+M1

�b

a;c

b;a

c

��;

and then wonders what the minimum t > 0 is so that

M1(a; b; c)M1

�1

a;1

b;1

c

��Mt

�M1

�a

b;b

c;c

a

�;M1

�b

a;c

b;a

c

��:

For an integer n > 3 and positive real numbers a1; a2; : : : ; an, it is notalways true that

n �max

�a1

a2+a2

a3+ � � �+ an

a1;a2

a1+a3

a2+ � � �+ a1

an

�

� (a1 + a2 + � � �+ an)

�1

a1+

1

a2+ � � �+ 1

an

�:

A counterexample when n = 4 is (a1; a2; a3; a4) = (4;2; 1; 2), for which the

left side of the inequality is 20 and the right side is 81=4.

For readers who still need more problems to solve, here are two more,

inspired by the above contributions.

(i) Find the smallest t > 0 so that (1) holds whenever at; bt; ct are the

sides of a triangle. Could the answer be t = 1=2? (Klamkin's quickie says

t � 1, and Bradley proved that t � 1=2.)

(ii) Find the smallest t > 0 so that

1

3(a+ b+ c)

�1

a+

1

b+

1

c

�� t

�a

b+b

c+c

a

�+ (1� t)

�b

a+c

b+a

c

�

whenever a; b; c > 0 satisfy

a

b+b

c+c

a� b

a+c

b+a

c:

238

(The original problem says t � 1, and Solution II shows that t � 1=2.)

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; CARL BOSLEY,student,Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHERJ. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN, student, Up-per Canada College, Toronto, Ontario; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; TOBY GEE, student, the John of GauntSchool, Trowbridge, England; ROBERT GERETSCHL �AGER, Bundesrealgym-nasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall, North Bay,Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOEHOWARD, New MexicoHighlands University, Las Vegas, New Mexico, USA;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; DAGJONSSON, Uppsala, Sweden; V �ACLAV KONE �CN �Y, Ferris State University,Big Rapids, Michigan, USA; SAI C. KWOK, San Diego, California, USA; KEE-WAI LAU, Hong Kong; THOMAS LEONG, Staten Island, New York, USA;VEDULA N.MURTY, Andhra University, Visakhapatnam, India; GOTTFRIEDPERZ, Pestalozzigymnasium, Graz, Austria; WALDEMAR POMPE, student,University of Warsaw, Poland; BOB PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; CRIST �OBAL S �ANCHEZ{RUBIO, I.B. Penyagolosa,Castell �on, Spain; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; ASHSIH KR.SINGH, student, Kanpur, India; PANOS E. TSAOUSSOGLOU,Athens, Greece;JOHANNES WALDMANN, Friedrich-Schiller-Universit�at, Jena, Germany(two solutions); EDWARD T.H. WANG, Wilfrid Laurier University, Water-loo, Ontario; and the proposer. There was also one anonymous solution.

Many solvers gave solutions similar to I or II. Kone �cn �y included a gen-eralization to n numbers which is weaker than Janous's generalization givenabove.

2065. [1995: 235] Proposed by Stanley Rabinowitz, Westford,Massachusetts.

Find a monic polynomial f(x) of lowest degree and with integer coef-

�cients such that f(n) is divisible by 1995 for all integers n.

Solutionby Carl Bosley, student,Washburn Rural High School, Topeka,Kansas, USA.

We have 1995 = 3 � 5 � 7 � 19. It can be shown that the congruence

f(x) � 0 (mod p) of degree n with integer coe�cients has at most n solu-

tions for prime p (see, for example, Niven, Zuckerman, and Montgomery's

An Introduction to the Theory of Numbers, 5th edition, p. 93). Hence if

f(n) � 0 (mod 19) for all n and f(x) is monic, f(x) must be of degree 19

or higher. On the other hand,

f(x) = x(x+ 1)(x+ 2) � � � (x+ 18);

of degree 19, is easily seen to be divisible by 1995 for all integers x.

239

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany;F.J. FLANIGAN, San Jose State University, San Jose, California, USA; TOBYGEE, student, the John of Gaunt School, Trowbridge, England; SHAWNGODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS,Rancho Palos Verdes, California, USA; KEE-WAI LAU, Hong Kong; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; PAUL YIU, Florida Atlantic University,Boca Raton, Florida, USA; an anonymous solver; and the proposer. Therewere �ve incorrect solutions submitted.

Pompe comments that \this problem has been discussed earlier in theliterature; see Problem 17 in Donald J. Newmann's A Problem Seminar".

2066. [1995: 235] Proposed by John Magill, Brighton, England.

The inhabitants of Rigel III use, in their arithmetic, the same operations

of addition, subtraction, multiplication and division, with the same rules of

manipulation, as are used by Earth. However, instead of working with base

ten, as is common on Earth, the people of Rigel III use a di�erent base,

greater than two and less than ten.

BC

AB)CBC

AB

BDC

BDC

- - - -

This is the solution to one of their long division problems, which I have

copied from a school book. I have substituted letters for the notation orig-

inally used. Each of the letters represents a di�erent digit, the same digit

wherever it appears.

As the answer to this puzzle, substitute the correct numbers for the

letters and state the base of the arithmetic of Rigel III.

Solutionby Shawn Godin, St. Joseph ScollardHall, North Bay, Ontario.

From the long division, note that B � AB = AB so B must be 1;

CB � AB = BD so D must be zero; since there was no \borrowing" from

the C in the previous subtraction, C = A+ 1. From C � AB = BDC and

the �rst three conditions, we get (A+ 1)� A = 10; thus the base must be

the product of consecutive numbers. Since the base is greater than two and

less than ten, it must be six; thus A = 2 and C = 3.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; SABIN CAUTIS, student, Earl Haig Secondary School, North York, On-tario; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;

240

KEITH EKBLAW,WallaWalla,Washington,USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; ROBERT GERETSCHL �AGER, Bun-desrealgymnasium, Graz, Austria; RICHARD I. HESS, Rancho Palos Verdes,California, USA; FRIEND H. KIERSTEAD JR., Cuyahoga Falls, Ohio, USA;V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;DAVID E. MANES, State University of New York, Oneonta, NY, USA; J.A.MCCALLUM,MedicineHat, Alberta; JOHN GRANT MCLOUGHLIN, Okana-ganUniversity College, Kelowna, BritishColumbia; GOTTFRIED PERZ, Pesta-lozzigymnasium, Graz, Austria; CORY PYE, student, Memorial Universityof Newfoundland, St. John's, Newfoundland; CHRIS WILDHAGEN, Rot-terdam, the Netherlands; SUSAN SCHWARTZ WILDSTROM, Kensington,Maryland, USA; KENNETH M. WILKE, Topeka, Kansas, USA; and the pro-poser.

Congratulations!

The editors would like to congratulate some regular contributors for making

their country's team for the 37th International Mathematical Olympiad, held

in July 1996, in Mumbai, India, and on their performances at the IMO. Any

student who makes a national team is a champion is his/her own right.

Name Country Award (if any)

Carl Bosley United States Gold Medal

Sabin Cautis Canada Bronze Medal

Adrian Chan Canada Bronze Medal

Toby Gee Great Britain Silver Medal

Ashish Kr. Singh India

241

Dissecting Squares into SimilarRectangles

Byung-Kyu Chun (student)

Harry Ainlay Composite High School

Andy Liu

University of Alberta

and

Daniel van Vliet (student)University of Alberta

1 Introductory Remarks.

Karl Scherer andMartin Gardner [1] have proposed the following three-part

problem. Cut a square into three similar pieces, where

(1) all three are congruent;

(2) exactly two are congruent;

(3) no two are congruent.

Note that (2) really consists of two sub-problems, where the congruent

pieces are

(2a) smaller than the third;

(2b) larger than the third.

All their solutions to (2) are to (2a). It appears that, alas, (2b) is not to be.

Also considered in [1] are related dissections of an equilateral triangle, but

they do not concern us here.

We generalize the problem of Scherer and Gardner for the square as

follows. Given any integer m > 1 and any of its 2m�1 compositions, or

ordered partitions, m = a1 + a2 + � � �+ an, dissect a square into m similar

pieces so that there are a1 congruent pieces of the largest size, a2 congruent

pieces of the next largest size, and so on. In the original problem, m = 3

and the compositions are: (1) 3; (2a) 1 + 2; (2b) 2 + 1; (3) 1 + 1 + 1.

Our main result is that the dissection problem always admits a solution

using rectangular pieces if and only if the composition is not of the form k+1,

where k is any positive integer. These solvable cases are covered by two

constructions which are only slightly di�erent.

242

2 Construction for the case m = a1 + a2 + � � � +

an; an > 1.

Suppose the composition ism = a1 + a2 + � � �+ an; an > 1. We start with

a rectangle R and divide it into n rectangles R1; R2; : : : ; Rn as follows. R1

is the right half of R;R2 is the bottom half of R � R1; R3 is the right half

of R � R1 � R2; R4 is the bottom half of R � R1 � R2 � R3, and so on,

except that Rn = R � R1 � R2 � � � � � Rn�1. The dimensions of these

rectangles will be adjusted later. Divide Ri into ai rectangular pieces, using

vertical lines if i is odd and horizontal lines if i is even. Let the horizontal and

vertical dimensions of each piece in Ri be xi and yi respectively, as shown

in Figure 1 for the case 6 = 1 + 3 + 2.

x1x2

x3y1

y2

y3

Figure 1

Next, we make all the pieces similar to one another by setting yi = xit

for all i, where t is a given positive number. It follows from the assumption

an > 1 that a piece in Ri is larger than a piece in Rj if i > j. We choose

xn = 1, so that yn = t. If n is even, yn�1 = anyn = ant and xn�1 = an.

If n > 1 is odd, xn�1 = anxn = an and yn�1 = ant.

Going backwards step by step on the basis of the recursive formulae

yi�1 = aiyi+yi+1 for i even and xi�1 = aixi+xi+1 for i odd, we can com-

pute the dimensions of the pieces in Rn�2; Rn�3; : : : ; R1 and R. Figure 2

is obtained by applying this process to Figure 1. Note that the xi are always

integers while the yi are always integral multiples of t.

Finally, we want to �nd a t so that R is a square. In other words, t is

a solution of y1 = a1x1 + x2. Since y1 is linear in t while a1x1 + x2 is a

constant, y1 > a1x1+x2 for su�ciently large t. On the other hand, if t = 1,

then the pieces will be square pieces, so that y1 = x1 � a1x1 � a1x1 + x2.

Since m > 1, either a1 > 1 or x2 > 0, so that y1 < a1x1 + x2. It follows

that there exists a real number t > 1 for which y1 = a1x1 + x2.

243

2 7

1

7t

2t

t

Figure 2

From Figure 2, we have 7t = 9 or t = 9

7. In general, since t is the

solution of a linear equation with integral coe�cients, it is a rational number.

Hence we can apply a uniform magni�cation to make all pieces have integral

sides. Figure 3 is a magni�cation of Figure 2 by a factor of 7.

3 Construction for the case m = a1 + a2 + � � � +

an; an = 1 and n � 3.

Suppose the composition ism = a1+a2+ � � �+an; n � 3 and an = 1. The

�rst step is identical to the construction in the last Section. We make all the

pieces similar to one another by setting yi = xit, except that xi = yit for

the last odd i. We shall prove that the equation y1 = a1x1 + x2 has a real

solution. Note that, as before, we have y1 < a1x1 + x2 if t = 1.

Suppose n � 4 is even. Let xn = 1 and yn = t. Then yn�1 =

t; xn�1 = t2; xn�2 = t2 + 1 and yn�2 = t3 + t. By induction, xi is of the

form at2 + b while yi is of the form ct3 + dt for all i, where a, b, c and d are

integers dependent only on i. Thus y1 > a1x1 + x2 for su�ciently large t.

Supposen � 3 is odd. Let xn = t and yn = 1. Then xn�1 = t; yn�1 =

t2; yn�2 = t2 + 1 and xn�2 = t+ 1

t. By induction, xi is of the form at+ b

t

while yi is of the form ct2 + d for all i, where a; b; c and d are integers

dependent only on i. Thus y1 > a1x1 + x2 for su�ciently large t.

244

14 49

18

9

7

63

Figure 3

Figure 4 illustrates the cases 6 = 1 + 2 + 2 + 1 and 6 = 2 + 3 + 1. In

many cases covered by this construction, we can also use pieces with integral

sides in the same orientation. Figure 5 illustrates one such example, namely,

11 = 4 + 6 + 1.

4 Impossibility proof for the case m = k + 1.

We now consider the remaining cases, compositions of the form m = k+ 1,

where k is any positive integer. We shall prove that the dissection problem

is not solvable with rectangular pieces. Our approach is indirect. We assume

that a solution exists, with a \1 by t" piece and k \s by st" pieces, where

t � 1 and s > 1 are real numbers.

Let the side length of the square be `. Consider a horizontal or vertical

segment from one side to the other, not running along any side of a piece.

Then ` = as+ bst for some integers a and b if the segment does not cut the

small piece. If it does, we have either ` = cs+ dst+ 1 or ` = es+ fst+ t

for some integers c, d, e and f .

From as+ bst = cs+ dst = 1, it follows that s is rational if and only

if t is. We �rst dispose of the case where s and t are rational. Let s =g

h

and t = ij, where g and h are relatively prime integers, as are i and j. Then

agj + bgi = cgj + dgi + hj = egj + fgi+ hi. Hence g divides hi and

hj. Since g and h are relatively prime, g divides i and j. Since i and j are

relatively prime, we must have g = 1. However, s = 1

h� 1, which is a

contradiction.

245

2t2 + 1 4t2 + 3

2t3 + t

t

1 t2

4t3 + 3t

t 3t+ 1

t

t2

1

3t2 + 1

Figure 4

We now consider the case where s and t are irrational, so that t > 1.

Consider a horizontal or vertical segment from one side of the square to the

other, not running along any side of a piece and not cutting the small piece.

As before, we have ` = as+ bst for some integers a and b.

We make the important observation that a and b are independent of

the choice of such a segment. If we have ` = a0s + b0st for some integers

a0 6= a and b0 6= b, then t = a�a0

b0�bwill be rational. Moreover, since the

segment can be horizontal or vertical, we must have a > 0 and b > 0.

We now use this to prove that the dissected square does not have a

fault-line. This is de�ned as a segment which divides the square into two

rectangles, each containing only complete pieces. Suppose to the contrary

that there is a horizontal fault-line. Then one of the rectangles is dissected

into only large pieces. Divide it into horizontal strips of width s from the top.

Each strip contains b \st by s" pieces and a or 2a squares of side s, each of

which contains parts of one or two s by st pieces. It follows that the height

of the rectangle is equal to ps for some positive integer p.

We can rearrange the pieces within this rectangle so that all the hor-

izontal ones are to the left and all the vertical ones are to the right. We

then have ps = qst for some positive integer q. However, t will then be

rational, which is a contradiction. It follows that the dissected square has no

fault-lines.

We now combine the large pieces into rectangles until the union of any

two of them is not a rectangle. The small piece is not part of any rectangle.

If this combination is not unique, we choose the one for which the number r

of rectangles is minimum. If r � 3, there will be a fault-line. Hence r � 4.

246

21 28

51

68

28 21

68

51

Figure 5

Figure 6

Suppose r = 4. The only possible con�guration has the four rectangles

at the corners and the small piece in the middle, as shown in Figure 6. Now

each of the rectangles is dissected into large pieces all of which are in the

same orientation. Since ` = as + bst with constants a > 0 and b > 0,

two opposite rectangles must be dissected into a columns and b rows of s by

st pieces, with the other two into a rows and b columns of st by s pieces.

However, the small piece will then be a square, and t = 1 is a contradiction.

Suppose r � 5. Then there is a rectangle at a corner of the square

which is not adjacent to the small piece. At least one of its two sides within

the square is the union of the sides of at least two other rectangles. If the

vertical side is not, as illustrated in Figure 7, then the horizontal side must

be.

247

Figure 7

Now this corner rectangle and those immediately below it are dissected

into large pieces in the same orientation within each. In fact, it must be the

same in all of them, as otherwise we can show that t is rational. Now we

can expand the corner rectangle until it swallows up a rectangle below it.

However, this reduction in r contradicts its minimality assumption.

5 Concluding Remarks.

We conjecture that for the compositions m = k+ 1, the dissection problem

is not solvable even if non-rectangular polygonal pieces are permitted. This

is true for the simplest case, namely, 2 = 1 + 1. We give a proof using an

indirect argument.

Suppose that a solution exists. Clearly, the smaller polygon P cannot

contain two opposite corners of the square S. Hence the larger polygon P 0

contains a side of S. We claim that this is the longest sides of P 0. Otherwise,

the longest side will be inside S, and it separates P 0 from P . Hence it will

also be the longest side of P , and the two polygons are in fact congruent.

This contradiction justi�es the claim. It follows that the longest side of P

is shorter than a side of S. Now P and P 0 have the same number of sides

inside S, but P 0 has more sides than P on the boundary of S. Hence P and

P 0 cannot be similar.

On the other hand, Figure 8 shows a \solution" using fractal-like pieces.

For the cases covered by our two constructions, we have many solutions

using polygonal pieces that are not rectangles. Figure 9 illustrates one such

example, namely, 7 = 1 + 1 + 1 + 1 + 1 + 1 + 1, based on a solution to a

problem in [2].

248

P 0

P

!

P 0

P

! � � �

Figure 8

������������

��

����

@@

@@

@@

Figure 9

6 References:

[1]Martin Gardner, Six Challenging Dissection Tasks, Quantum, Volume 4,

Issue 5 (1994) 26-27.

[2]Peter Taylor, \InternationalMathematics Tournament of the Towns (1980-

1984)", Australian Mathematics Trust, Canberra (1993) 106.

249

THE SKOLIAD CORNERNo. 16

R.E. Woodrow

As a contest this issue, we give the Saskatchewan Senior Mathemat-

ics Contest. This contest was written Wednesday, February 22, 1995. My

thanks go to Garreth Gri�th, Mathematics Department, The University of

Saskatchewan, long time organizer of the contests in Saskatchewan, for per-

mission to use the contest and for the solutions we shall give next issue.

1995 SASKATCHEWAN SENIOR MATHEMATICS

CONTESTFebruary 22, 1995 | Time: 1.5 hours

1. They sell regular and jumbo sized orders of �sh at Jerry's Fish &

Chips Emporium. The jumbo order costs�4

3

�times as much as a regular one

and an order of chips costs $1.30.

Last Thursday, the Emporium was quite busy over the lunch break

(11:30 am - 1:30 pm). Exactly thirteen jumbo sized orders of �sh along with

a quantity of regular orders of �sh and chips were sold. $702.52 had been

placed in the till.

During the period 1:30 - 4:30, business slackened o�. 26 regular orders

of �sh were sold during this time. Four times as many regular orders had

been sold during the lunch break. No jumbo portions were sold and the

number of orders of chips declined to one �fth the number that had been

sold during the lunch break. At 4:30 pm there was $850.46 in Jerry's till.

What is the price of a regular sized order of �sh at Jerry's? [5 marks]

2.ABCD is a square with side of length s. A circle, centre A and

radius r is drawn so that the arc of this circle which lies within the square

divides the square into two regions of equal area. Write r as a function of s.

[6 marks]

3.(a) Solve the equation 3y = 10y. [3 marks]

(b) Solve the equation 3y = 10. [3 marks]

(c) Write t log8px� 2 log8 y as a single logarithm. [4 marks]

4. Establish the identity 2 cotA = cot A2� tan A

2: [5 marks]

250

5. ABC is a triangle, right angled at C. Let a, b, c denote the lengths

of the sides opposite angles A, B, C respectively. Given that a = 1, \B =

75� and that tan75� = 2 +p3, express b and c in the form p + q

p3 such

that p = 2 orp2. [8 marks]

6. Determine the function f(x) which satis�es all of the following

conditions: [8 marks]

(i) f(x) is a quadratic function.

(ii) f(x+ 2) = f(x) + x+ 2.

(iii) f(2) = 2.

7.Prove that if n is a positive integer (written in base 10) and that if 9

is a factor of n, then 9 is also a factor of the sum of the digits of n.[8 marks]

Last number we gave the problems of the American Invitational Math-

ematics Examination. These problems and their solutions are copyrighted by

the Committee on the American Mathematical Competitions of the Mathe-

matical Association of America and may not be reproduced without permis-

sion. Full solutions, and additional copies of the problems, may be obtained

for a nominal fee from Professor Walter E. Mientka, C.A.M.C. Executive

Director, 917 Oldfather Hall, University of Nebraska, Lincoln, NE, U.S.A.

68588{0322.

Solutions to the 1996 A.I.M.E.

1. 200 2. 340 3. 044 4. 166 5. 023

6. 049 7. 300 8. 799 9. 342 10. 159

11. 276 12. 058 13. 065 14. 768 15. 777

That completes the Skoliad Corner for this issue. Please send me suit-

able contest materials, your students' nice solutions, and comments or advice

for the future of the Corner.

251

THE OLYMPIAD CORNERNo. 176

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 of the 10th Iberoamerican

Mathematical Olympiad, written September 26{27, 1995 at Valparaiso (Chile).

My thanks go to Professor Francisco Bellot Rosado of Valladolid, Spain for

sending me an English translation of the problems.

10th IBEROAMERICAN MATHEMATICALOLYMPIAD

September 26{27, 1995 (Valparaiso, Chile)FIRST DAY | Time: 4.5 hours

1. (Brazil). Determine all the possible values of the sum of the digits

of all the perfect squares.

2. (Spain). Let n be an integer bigger than 1. Determine the real num-

bers x1; x2; : : : xn � 1, and xn+1 > 0, such that the following conditions are

simultaneously ful�lled:

(a)px1 + 3

px2 + � � �+ n+1

pxn = n � pxn+1

(b)x1 + x2 + � � �+ xn

n= xn+1.

3. (Brazil). Let r and s be two orthogonal straight lines, not belonging

to the same plane. Let AB be their common perpendicular, with A 2 r andB 2 s. (Note that the plane which contains B and r is perpendicular to s).

Consider the sphere with diameter AB. The points M 2 r, and N 2 s, arevariable, with the condition thatMN is tangent to the sphere at some point

T . Find the locus of T .

SECOND DAY | Time: 4.5 hours

4. (Argentina). Coins are situated on an m � m board. Each coin

situated on the board \dominates" all the cells of the row ( -� ), the column

( 6?) and the diagonal (@R@I) to which the coin belongs. Note that the coin does

not \dominate" the diagonal (���). Determine the smallest number of coins

which must be placed in order that all the cells of the board be dominated.

252

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 this

circumference meets AD again at its mid-point X, that is, AX = XD. The

lines XB and XC meet the inscribed circumference again at Y and Z, re-

spectively. Show that EY = FZ.

6. (Chile{Brazil). Let N = f1; 2; 3; : : : g. A function f : N ! N is

called circular if for each p 2 N there exists n 2 N with n � p such that

fn(p) = f(f(: : : f| {z }n times

(p)) : : : ) = p:

The function f has repulse degree k, 0 < k < 1, if for each p 2 N, f i(p) 6= p

for all i � [k � p], in which [x] is the integer part of x. Determine the biggest

repulse degree that can be reached by a circular function.

The next contest we present in its o�cial language. Here is your oppor-

tunity to brush up your French. (Of course we will accept your nice solutions

in either English or French!) The questions are to the mini and maxi �nals of

the 19th Belgian Mathematical Olympiad, written in 1994. My thanks go to

Richard Nowakowski, Canadian Team Leader at the 35th IMO in Hong Kong,

for collecting this contest (and others) and forwarding them to the Olympiad

Corner for its use.

DIX-NEUVIEME OLYMPIADE MATHEMATIQUEBELGE

Mini Finale 1994

1. Combien de nombres enteiers naturels de trois chi�res non nuls

distincts (en base 10) sont primiers avec 10?

2. En prenant comme sommets les points d'intersection des cot �es pro-

long �es d'un hexagone r �egulier, Jean obtient un nouvel hexagone r �egulier.

Il applique ensuite la meme construction �a ce nouvel hexagone, et recom-

mence de meme : : : Combien de fois Jean doit-il e�ectuer cette construction

pour que l'aire du dernier hexagone construit d �epasse 1994 fois l'aire de

l'hexagone initial?

3. Trente-huit lampes num�erot �ees de l �a 38 sont dispos �ees en cercle

autour d'une lampe centrale num�erot �ee 0. Ces lampes forment des groupes

de quatre:

f0; 1; 2; 3g; f0; 3; 4; 5g; f0; 5; 6; 7g;f0; 7; 8; 9g; : : : ; f0; 35; 36; 37g; f0; 37; 38; 1g

deux op �erations seulement sont r �ealisables:

253

(�) �eteindre les quatre lampes d'un meme groupe;

(�) changer l' �etat de chacune des lampes d'un meme groupe (c'est- �a-

dire, une lampe allum �ee est �eteinte, une �eteinte est allum �ee).

Tout �etat initial des 39 lampes est-il transformable par une suite de telles

op �eratious en

(a) l' �etat o �u toutes les lampes sont allum �ees?

(b) l' �etat o �u seule la lampe num�ero 0 est allum �ee?

4. Sur un terrain plat et carr �e de 32 ares (ou 3; 200m�etres carr �es) dont

les cot �es sont orient �es NO-SE et NE-SO se trouve une villa rectangulaire de

16 m�etres sur 20 m�etres, dont les quatre fa�cades font face aux quatre points

cardinaux. Le centre de la villa co��ncide avec le centre du terrain. Le reste du

terrain est am �enag �e en pelouse. Quelle fraction de la pelouse est constitu �ee

de points d'o �u sont visibles deux fa�cades de la villa?

Maxi Finale 1994

1. Un pentagone plan covexe a deux angles droits non adjacents. Les

deux cot �es adjacents au premier angle droit out des longueurs �egales. Les

deux 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 un

seul cot �e de ces angles droits, nous formons un quadrilat �ere. Ce quadrilat �ere

admet-il n �ecessairement un angle droit?

2. Des lampes en nombre 2n (avec n � 2) et num�erot �ees de 1 �a

2n sont dispos �ees en cercle autour d'une lampe centrale num�erot �ee 0. Ces

lampes forment des groupes de quatre:

f0; 1; 2; 3g; f0; 3; 4; 5g; : : : f0; 2k�3; 2k�2; 2k�1g; f0; 2k�1; 2k; 2k+1g;f0; 2k+ 1; 2k+ 2; 2k+ 3g; : : : ; f0; 2n� 1; 2n; 1g

et deux op �erations seulement sont r �ealisables:

(�) �eteindre les quatre lampes d'un meme groupe;

(�) changer l' �etat de chacune des lampes d'un meme groupe (c'est- �a-

dire, une lampe allum �ee est �eteinte, une �eteinte est allum �ee).

Pour quelles valeurs de n tout �etat initial des 2n + 1 lampes est-il trans-

formable par une suite de telles op �erations en

(a) l' �etat o �u toutes les lampes sont allum �ees?

(b) l' �etat o �u seule la lampe nuim �ero 0 est allum �ee?

3. Existe-t-il une num�erotation des aretes d'un cube par douze nom-

bres naturels cons �ecutifs telle que

(a) la somme des nombres attribu �es aux aretes aboutissant en un sommet

soit toujours la meme?

(b) la somme des nombres attribu �es aux aretes d'une face soit toujours la

meme?

254

4. Le plan contient il 1994 points (distincts) non tous align �es tels que

la distance entre deux quelconques d'entre eux soit un nombre entier?

As promised last issue, we now give the \o�cial" solutions to prob-

lems of the 1996 Canadian Mathematical Olympiad. My thanks go to Daryl

Tingley, Chair of the Canadian Mathematical Olympiad Committee of the

Canadian Mathematical Society, for forwarding the problems and solutions.

Problems with communications during summer break did not allow us to in-

corporate the novel solutions of some contest participants. Hopefully next

year we will be on track earlier and have time to solicit permission to use the

submitted solutions.

1996 CANADIANMATHEMATICAL OLYMPIAD

1. If �, �, are the roots of x3 � x� 1 = 0, compute

1 + �

1� �+

1 + �

1� �+

1+

1� :

Solution. If f(x) = x3 � x � 1 = (x � �)(x � �)(x � ) has roots

�; �; standard results about roots of polynomials give � + � + = 0,

�� + � + � = �1, and �� = 1.

Then

S =1+ �

1� �+

1+ �

1� �+

1 +

1� =

N

(1� �)(1� �)(1� )

where the numerator simpli�es to

N = 3� (�+ � + )� (�� + � + � ) + 3��

= 3� (0)� (�1) + 3(1)

= 7:

The denominator is f(1) = �1 so the required sum is �7.2. Find all real solutions to the following system of equations. Care-

fully justify your answer. 8>>>>>>>>>><>>>>>>>>>>:

4x2

1 + 4x2= y

4y2

1 + 4y2= z

4z2

1 + 4z2= x

255

Solution. For any t; 0 � 4t2 < 1 + 4t2, so 0 � 4t2

1 + 4t2< 1. Thus

x; y and z must be non-negative and less than 1.

Observe that if one of x y or z is 0, then x = y = z = 0:

If two of the variables are equal, say x = y, then the �rst equation

becomes4x2

1 + 4x2= x:

This has the solution x = 0, which gives x = y = z = 0 and x =1

2

which gives x = y = z =1

2.

Finally, assume that x; y and z are non-zero and distinct. Without

loss of generality we may assume that either 0 < x < y < z < 1 or

0 < x < z < y < 1. The two proofs are similar, so we do only the �rst case.

We will need the fact that f(t) =4t2

1 + 4t2is increasing on the interval

(0; 1):

To prove this, if 0 < s < t < 1 then

f(t)� f(s) =4t2

1 + 4t2� 4s2

1 + 4s2

=4t2 � 4s2

(1 + 4s2)(1 + 4t2)

> 0:

So 0 < x < y < z ) f(x) = y < f(y) = z < f(z) = x, a contradic-

tion.

Hence x = y = z = 0 and x = y = z =1

2are the only real solutions.

Alternate Solution. Notice that x; y and z are non-negative. Adding

the three equations gives

x+ y+ z =4z2

1 + 4z2+

4x2

1 + 4x2+

4y2

1 + 4y2:

This can be rearranged to give

x(2x� 1)2

1 + 4x2+y(2y� 1)2

1 + 4y2+z(2z � 1)2

1 + 4z2= 0:

Since each term is non-negative, each term must be 0, and hence each vari-

able is either 0 or 1

2. The original equations then show that x = y = z = 0

and x = y = z = 1

2are the only two solutions.

Alternate Solution. Notice that x, y, and z are non-negative. Multiply

both sides of the inequality

y

1 + 4y2� 0

256

by (2y� 1)2, and rearrange to obtain

y � 4y2

1 + 4y2� 0;

and hence that y � z. Similarly, z � x, and x � y. Hence, x = y = z and,

as in Solution 1, the two solutions follow.

Alternate Solution. As for solution 1, note that x = y = z = 0 is a

solution and any other solution will have each of x; y and z positive.

The arithmetic-geometric mean inequality (or direct computation)

shows that1 + 4x2

2�p1 � 4x2 = 2x and hence x � 4x2

1 + 4x2= y, with

equality if and only if 1 = 4x2, that is, x =1

2. Similarly, y � z with equality

if and only if y =1

2and z � x with equality if and only if z =

1

2. Adding

x � y; y � z and z � x gives x+ y + x � x+ y+ z. Thus equality must

occur in each inequality, so x = y = z =1

2.

3. We denote an arbitrary permutation of the integers 1; : : : ; n by

a1; : : : ; an. Let f(n) be the number of these permutations such that

(i) a1 = 1;

(ii) jai � ai+1j � 2, i = 1; : : : ; n� 1.

Determine whether f(1996) is divisible by 3.

Solution. Let a1; a2; : : : ; an be a permutation of 1; 2; : : : ; nwith prop-

erties (i) and (ii).

A crucial observation, needed in Case II (b) is the following: If ak and

ak+1 are consecutive integers (i.e. ak+1 = ak � 1), then the terms to the

right of ak+1 (also to the left of ak) are either all less than both ak and ak+1or all greater than both ak and ak+1.

Since a1 = 1, by (ii) a2 is either 2 or 3.

CASE I: Suppose a2 = 2. Then a3; a4; : : : ; an is a permutation of

3; 4; : : : ; n. Thus a2; a3; : : : ; an is a permutation of 2; 3; : : : ; n with a2 = 2

and property (ii). Clearly there are f(n� 1) such permutations.

CASE II: Suppose a2 = 3.

(a) Suppose a3 = 2. Then a4; a5; : : : ; an is a permutation of 4; 5; : : : ; nwith

a4 = 4 and property (ii). There are f(n� 3) such permutations.

(b) Suppose a3 � 4. If ak+1 is the �rst even number in the permutation then,

because of (ii), a1; a2; : : : ; ak must be 1; 3; 5; : : : ; 2k�1 (in that order). Thenak+1 is either 2k or 2k � 2, so that ak and ak+1 are consecutive integers.

Applying the crucial observation made above, we deduce that ak+2; : : : ; anare all either greater than or smaller than ak and ak+1. But 2must be to the

right of ak+1. Hence ak+2; : : : ; an are the even integers less than ak+1. The

only possibility then, is

1; 3; 5; : : : ; ak�1; ak; : : : ; 6; 4; 2:

257

Cases I and II show that

f(n) = f(n� 1) + f(n� 3) + 1; n � 4: (?)

Calculating the �rst few values of f(n) directly gives

f(1) = 1; f(2) = 1; f(3) = 2; f(4) = 4; f(5) = 6:

Calculating a few more f(n)'s using (?) and mod 3 arithmetic, f(1) = 1,

f(2) = 1, f(3) = 2, f(4) = 1, f(5) = 0, f(6) = 0, f(7) = 2, f(8) = 0,

f(9) = 1, f(10) = 1, f(11) = 2. Since f(1) = f(9), f(2) = f(10) and

f(3) = f(11)mod 3, (?) shows that

f(a) = f(a mod 8); mod 3; a � 1:

Hence f(1996)� f(4) � 1 (mod 3) so 3 does not divide f(1996).

4. Let 4ABC be an isosceles triangle with AB = AC. Suppose

that the angle bisector of \B meets AC at D and that BC = BD + AD.

Determine \A.

Solution. Let BE = BD with E on BC, so that AD = EC:A

B CE

D

x

x 2x

2x

4x

4x

By a standard theorem,AB

CB=AD

DC; so in 4 CED and 4 CAB

we have a common angle and

CE

CD=

AD

CD=

AB

CB=

CA

CB:

Thus,4CED � 4CAB, so that \CDE = \DCE = \ABC = 2x:

Hence \BDE = \BED = 4x, whence 9x = 180� so x = 20� :

Thus \A = 180� � 4x = 100� :

Alternate Solution. Apply the law of sines to 4ABD and 4BDC to

getAD

BD=

sinx

sin 4xand 1 +

AD

BD=

BC

BD=

sin3x

sin2x:

Now massage the resulting trigonometric equation with standard identities

to get

sin2x (sin4x+ sinx) = sin2x (sin5x+ sinx) :

258

Since 0 < 2x < 90�, we get

5x� 90� = 90� � 4x ;

so that \A = 100�.

5. Let r1; r2; : : : ; rm be a given set of m positive rational numbers

such thatPm

k=1 rk = 1. De�ne the function f by f(n) = n �Pm

k=1[rkn]

for each positive integer n. Determine the minimum and maximum values

of f(n). Here [x] denotes the greatest integer less than or equal to x.

Solution. Let

f(n) = n�mXk=1

[rkn]

= n

mXk=1

rk �mXk=1

[rkn]

=

mXk=1

frkn� [rkn]g:

Now 0 � x� [x] < 1, and if c is an integer, (c+ x)� [c+ x] = x� [x].

Hence 0 � f(n)<

mXk=1

1 = m. Because f(n) is an integer, 0 � f(n) �m� 1.

To show that f(n) can achieve these bounds for n > 0, we assume that

rk =ak

bkwhere ak; bk are integers; ak < bk.

Then, if n = b1b2 : : : bm; (rkn) � [rkn] = 0; k = 1; 2; : : : ; m and thus

f(n) = 0.

Letting n = b1b2 : : : bn � 1, then

rkn = rk(b1b2 : : : bm � 1)

= rkf(b1b2 : : : bm � bk) + bk � 1)g= integer+ rk(bk � 1):

259

This gives

rkn� [rkn] = rk(bk � 1)� [rk(bk � 1)]

=ak

bk(bk � 1)�

�ak

bk(bk � 1)

�

=

�ak �

ak

bk

���ak �

ak

bk

�

=

�ak �

ak

bk

�� (ak � 1)

= 1� ak

bk= 1� rk:

Hence

f(n) =

mXk=1

(1� rk) = m� 1:

Next we give reader solutions to problems of the Second Stage Exam of

the 10th Iranian Mathematical Olympiad [1995: 9{10].

1. In the right triangleABC (A = 90�), let the internal bisectors ofB

and C intersect each other at I and the opposite sidesD andE respectively.

Prove that the area of quadrilateral BCDE is twice the area of the triangle

BIC.

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;

Christopher J. Bradley, CliftonCollege, Bristol, UK; Hans Engelhaupt, Franz{

Ludwig{Gymnasium, Bamberg, Germany; Cyrus Hsia, student, University

of Toronto, Toronto, Ontario; Joseph Ling, University of Calgary, Calgary,

Alberta; and by Dieter Ruo�, Department of Mathematics and Statistics,

The University of Regina, Regina, Saskatchewan. We give the solution of

Covas.

If b and c are the legs, a the hypotenuse, s the semiperimeter and r

the inradius of the given right triangle then it is known that r = s� a.The area of such a triangle is bc=2. On the other hand, the area of any

triangle is sr. Setting the two expressions equal we have

bc = 2sr = 2s(s� a) (1)

260

We see that (see �gure 1.1)

A E B

C

D

Figure 1.1

area of quadrilateral BCDE = area (4ABC)� area (4AED): (2)

Since AE = bc=(a + b) and AD = bd=(a + c) we have Area (4AED) =1

2

bca+b� bca+c

. Substituting, using (2), we have

area of quadrilateral BCDE

=1

2bc� 1

2

bc

a+ b� bc

a+ c

=1

2bc

�1� bc

(a+ b)(a+ c)

�

=1

2bc

a(a+ b+ c)

(a+ b)(a+ c)=

abcs

(a+ b)(a+ c): (3)

Since (a+ b)(a+ c) = a(a+ b+ c) + bc = 2as+ 2s(s� a) = 2s2 we can

write (3) in the form

Area of quadrilateral BCDE =abc

2s:

Finally, we substitute 2rs for bc from (1), simplify, and obtain Area of

quadrilateral BCDE = ar = 2(area of4BIC), which was to be proved.

Editor's Note. Both Ling and Ruo� generalized the result proving more. Here

is Ruo�'s generalization.

Let 4ABC be a triangle with angles 2�, 2�, 2 , D the intersecting

point of the angle bisector at B and AC, E the intersecting point of the

angle bisector at C and AB, and I the incentre of 4ABC. Then

j4BCIj >=<

1

2jBCDEj i� �

>=<

�

4: (1)

261

AE

B

C

D

Ic

Ib

Ia

I

��

�

�r

r

r

0

�0

Figure 1.1a

Proof. Let Ia, Ib, Ic be the projections of I to BC, CA, AB respec-

tively and note the LHS of (1) is equivalent to

j4BIIcj+ j4CIIcj >=<jBICDEBj;

jBICABj � jIIbAIcj >=<jBICABj � j4ADEj

and

j4ADEj >=<jIIbAIcj: (2)

The essence and heuristic departure point of the following is the proof of

j4BCIj = 1

2jBCDEj (1�)

respectively

j4ADEj = jIIbAIcj (2�)for triangles with right angle at A (� = �

4). In this case IIcAIc is obviously

a square, 4ADE a right triangle, \IcID = �0 = �, \IcIE = 0 = ,

consequently �0 = �4� 0, and (2*) becomes for IIb = IIc = r.

AIc E

B

Ib

D

C

I

��

�

�

r

r

�0

0

Figure 1.1b

262

r2

2(1 + tan 0)(1 + tan(

�

4� 0)) = r2; (3*)

which holds because of the trigonometric identity

(1 + tanx)(1 + tan(�

4� x)) = 2:

Returning to the general case, we �rst assume that Ic lies between A

and E, and Ib between A and D, in which case (2) amounts to

r2

2(cot�+ tan 0)(cot�+ tan�0) � sin2 � >

=<r2 cot� (3)

respectively to

(1 + tan� tan 0)(1 + tan� tan�0) � cos2 � >=<

1:

Multiplying this equation by1

cos2 �= 1+ tan2 �, we obtain

tan 0 + tan�0

1� tan 0 � tan�0 = tan( 0 + �0)>=<

tan�: (4)

The angle sum in the quadrilateral AEID is

2�+ (2� + ) + ( 0 + � � 2�+ �0) + (2 + �) = 2� (5)

and hence

0 + �0 = � � 3(� + ) = � � 3

��

2� �

�

= 3�� �

2

>=<� for

8<:

� > �

4

� = �4

� < �4

: (6)

The conditions on the RHS of (6) determine the signs of (4) and one by one

those of (3), (2) and (1), q.e.d.

A B

C

Ic E

I

D

Ib

��

�

�

0

�0#

Figure 1.1c

263

If Ic lies between A and E, but D between A and Ib then, in (3) and

in the following, �0 has to be multiplied by �1, and (4), (5), (6) turn into

tan( 0 � �0)>=<

tan�; (4*)

2�+ (2� + ) + ( 0 + � � 2�� �0) + (2 + �) = 2� (5*)

and

0 � �0 = 3�� �

2

>=<� for

8<:

� > �4

� = �4

� < �

4

; (6*)

A B

C

E Ic

D

Ib

I

2� �

Figure 1.1d

from which, however, the same conclusions result as before. Since 2�+� <�2, the only de facto applicable condition is � < �

4. Also, if E lies between

A and Ic and D between A and Ib, � < �4is the only applicable condition;

that the <-sign holds in (2) follows directly from the �gure.

2. Given the sequence a0 = 1, a1 = 2, an+1 = an+an�1

1+(an�1)2, n > 1,

show that 52 < a1371 < 65.

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Ontario.

Note �rst that \n > 1" in the statement should have been \n � 1" for

the problem to be correct. We show that in general,

p2n+ 1 � an �

p3n+ 2 for all n � 0: (1)

Since when n = 1371,p2n+ 1 =

p2743 � 52:374 and

p3n+ 2 =p

4115 � 64:148, 52 < a1371 < 65 would follow. To establish (1), we

�rst show by induction that

an = an�1 +1

an�1for all n � 1: (2)

This is clearly true for n = 1 since a1 = 2 = a0 +1

a0. Suppose (2) holds for

some n � 1. Then

an =(an�1)

2 + 1

an�1) 1

an=

an�1

1 + (an�1)2

264

and thus, from the given recurrence relation, we get an+1 = an + 1

an, com-

pleting the induction. Since clearly an > 0 for all n, we see from (2) that the

sequence fang is strictly increasing. In particular, 1

a2n�1

� 1 for all n � 1

and so from a2n = a2n�1 + 2+ 1

a2n�1

we get

a2n�1 + 2 < a2n � a2n�1 + 3 for all n � 1: (3)

Now we use (3) and induction to establish (1). The case when n = 0 is trivial

since a0 = 1 <p2. Suppose (1) holds for some n � 0. Then by (3),

an+1 �qa2n + 3 � p3n+ 2+ 3 =

p3(n+ 1) + 2

and

an+1 >

qa2n + 2 � p2n+ 1+ 2 =

p2(n+ 1) + 1

and our proof is complete.

3. There is a river with cities on both of its sides. Some boat lines

connect these cities in such a way that each line connects a city of one side

to a city on the other side, and each city is joined exactly to k cities on the

other side. One can travel between every two cities. Prove if one of the boat

lines is cancelled, one can travel between every two cities.

Solution by Cyrus Hsia, student, University of Toronto, Toronto, On-

tario.

Originally, one can travel between every two cities. If we consider cities

as vertices and boat lines as edges on a graph, then this graph is connected.

We must show that if an edge is removed, then the graph is still connected.

We interpret cities on the two sides of the river as a bipartite graph

since each line connects a city of one side to a city on the other side. As well,

each vertex has k edges. If we count the number of edges for each vertex on

one side we are counting all the vertices because each edge has exactly one

end on that side. Thus, originally we have ak edges (where a is the number

of vertices on one of the sides). This shows that we must have a vertices on

both sides.

Now suppose, on the contrary, that by removing an edge we have two

disjoint graphs. Suppose it is divided into parts U , V , X, Y where U [ Vand X [ Y are the vertices of the two sides and the remaining edges are

between U and X, and V and Y respectively. Further, since one edge was

removed, let U and V have kjU j � 1 and kjV j incident edges respectively.IfX has kjXj � 1 incident edges, that would mean that originally the whole

graph was not connected. Therefore X has kjXj edges remaining and Y has

kjY j�1 edges remaining. This is impossible! kjU j�1 6= kjV j unless k = 1.

If k = 1, a = 1 or else we would have disjoint lines!

Now there are cities on both sides, so a > 1, so k > 1, and it is

impossible to have two disjoint bipartite graphs by removing one edge.

265

4. Prove that for each natural number t, 18 divides

A = 1t + 2t + � � �+ 9t � (1 + 6t + 8t):

Remarks by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg,

Germany; and by Stewart Metchette, Gardena, California, USA.

There must be a misprint.

For t = 1, A = 1+ 2 + 3 + � � �+ 9� (1 + 6 + 8) = 30,

and t = 2,A = 1+22+32+� � �+92�(1+62+82) = 285�101 = 184.

Both are not divisible by 18.

5. In the triangle ABC we have A � 90� and B = 2C. Let the

internal bisector of C intersect the median AM (M is the mid-point of BC)

at D. Prove that \MDC � 45�. What is the condition for \MDC = 45�?

Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain.

It su�ces to show that tan(\MDC) � 1.

If 4ABC has sides a, b, c, in the usual order, then the condition

B = 2C is equivalent to the condition b2 = c(c+ a) (see this journal [1976:

74] and [1984: 287]). (1)

We introduce a Cartesian frame with origin at B and x-axis along BC:

-

6

B M C

A

D

Figure 5

The coordinates of A are (c �cosB; c � sinB), the coordinates ofC are (a; 0),

and those of M are (a=2; 0).

The internal angle bisector of C has slope

m1 = tan(180� � C=2) = � tan(C=2)

and the slope of the median AM is

m2 =c � sinB

c � cosB � a=2: (2)

The law of cosines gives

b2 = c2 + a2 � 2ca � cosB:

266

Substituting for b2 from (1), we obtain

c(c+ a) = c2 + a2 � 2ca � cosB;

and hence

c � cosB = (a� c)=2:Substituting this into (2), we obtain

m2 = �2 � sinB:

Since B = 2C and sin 2C = 2 sinC cosC, this equation may be rewritten

as

m2 = �4 sinC cosC:

Using the formula for the tangent of the angle between two lines, we get

tan(\MDC) =m1 �m2

1 +m1m2

=� tan(C=2) + 4 sinC cosC

1 + 4(tan(C=2)) � sinC cosC: (3)

We express sinC and cosC in terms of tan(C=2). Putting t = tan(C=2),

we get

sinC =2t

(1 + t2); cosC =

(1� t2)

(1 + t2):

When these are substituted into (3), it becomes

tan(\MDC) =�t+ 8t(1�t)2

(1+t2)2

1 +8t2(1�t2)

(1+t2)2

=�t(1 + t2)2 + 8t(1� t2)

(1 + t2)2 + 8t2(1� t2):

We now prove that tan(\MDC) � 1. This holds if and only if

8t(1� t2)� t(1 + t2)2 � 8t2(1� t2) + (1 + t2)2;

or, equivalently,

8t(1� t2)� 8t2(1� t2) � (1 + t2)2 + t(1 + t2)2;

or

8t(1� t2)(1� t) � (1 + t2)2(1 + t):

Dividing both sides by the positive number 1 + t, we get

8t(1� t)2 � (1 + t2)2;

equivalent to

8t� 16t2 + 8t3 � 1 + 2t2 + t4;

which is indeed true since

t4 � 8t3 + 18t2 � 8t+ 1 = (t2 � 4t+ 1)2 � 0:

267

Equality \MDC = 45� occurs if and only if t2 � 4t + 1 = 0, where

t = tan(C=2). This is satis�ed when t = 2�p3, i.e. C = 30� + 360�k; or

t = 2 +p3, i.e. C = 150� + 360�k (k = : : : ;�2;�1; 0; 1; 2; : : : ).

The only acceptable value for C is 30�.

We conclude that \MDC = 45� i� A = 90�, B = 60�, C = 30�.

6. Let X be a non-empty �nite set and f : X ! X a function such

that for all x in X, fp(x) = x, where p is a constant prime. If Y = fx 2X : f(x) 6= xg, prove that the number of elements of Y is divisible by p.

Solution by Cyrus Hsia, student, University of Toronto, Toronto, On-

tario; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, On-

tario. We give Wang's solution and remarks.

For each x 2 Y consider that orbit, B(x), of x de�ned by

B(x) = fx; f(x); f2(x); : : : ; fp�1(x)g:We claim that all the elements of B(x) are distinct.

Suppose not. Then let j be the least positive integer such that f i(x) = f j(x)

for some integer i with 0 � i < j � p� 1. (We de�ne f i(x) = x if i = 0.)

Then

x = fp(x) = fp�j(f j(x)) = fp�j(f i(x)) = fp�j+i(x)

) f j�i(x) = f j�i(fp�j+i(x)) = fp(x) = x:

Since 0 < j � i � j, we must have i = 0 and thus f j(x) = x. Now let

p = qj + r, where q, r are integers with q > 0 and 0 � r < j. Clearly

f j(x) = x implies fqj(x) = x and hence

fr(x) = fr(fqj(x)) = fp(x) = x:

Since r < j, we must have r = 0 and thus p = qj. Since f(x) 6= x,

j > 1. On the other hand, since j < p, q > 1. Hence p is a composite, a

contradiction. Therefore, f i(x) 6= f j(x) for all i = 0; 1; 2; : : : ; p�1, we see

that B(x) � Y .

Next we show that the orbits of two elements of Y are either disjoint or

identical. Let x; y 2 Y and suppose B(x)\B(y) 6= ;. Then f l(x) = fk(y)

for some integers l and k, with 0 � l � k � p� 1. Hence

y = fk(y) = fp�k(fk(y)) = fp�k(f l(x)) = fp�k+l(x);

which show that y 2 B(x). It then follows that B(x) = B(y). Therefore

Y can be partitioned into disjoint orbits each having cardinality p and the

result follows.

Remarks. (1) Actually, the result still holds even when p = 1 since in this

case Y = ; and thus jY j = 0. (2) The result need not hold if p is composite.

A counterexample is given by X = f1;2; 3; 4; 5; 6g, f(1) = 2, f(2) = 1,

f(3) = 4, f(4) = 5, f(5) = 6, and f(6) = 3. In this case, p = 4 and

Y = X, jY j = 6.

That completes the Olympiad Corner for this issue. Send me your con-

tests and nice solutions.

268

THE ACADEMY CORNERNo. 5

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

In the February 1996 issue, we gave the �rst set of problems in the

Academy Corner. Here we present solutions to the last three questions, as

sent in by �Sefket Arslanagi�c, Berlin, Germany.

Memorial University Undergraduate

Mathematics Competition 1995

4. If a, b, c, d are positive integers such that ad = bc, prove that a2 +

b2 + c2 + d2 is never a prime number.

We will give a generalization: If a, b, c, d, r, s are positive integers such

that rad = sbc, prove that r(a2 + d2) + s(b2 + c2) is never a prime

number.

Solution. All positive integers greater than or equal to 2 can be written

as products of �nitely many prime numbers. Therefore rad = sbc =

p1 : : : p` (pi; pj not necessarily di�erent) are the prime factors of r, a,

d and similarly of factors s, b, c. Also, there exist positive integers

x11; x12; : : : ; x33 such that

r = x11x12x13 = y1; a = x21x22x23 = y2; d = x31x32x33 = y3;

s = x11x21x31 = z1; b = x12x22x32 = z2; c = x13x23x33 = z3:

We write this as

x11 x12 x13 ! Q= r = y1

x21 x22 x23 ! Q= a = y2

x31 x32 x33 ! Q= d = y3

# # #Q= s

Q= b

Q= c

= z1 = z2 = z3

269

Let

xiji =Y

px

px 2 yipx 2 zj

k = 1; : : : ; `

Now, we get

r(a2 + s2) + s(b2 + c2)

= x11x12x13(x221x

222x

223 + x231x

232x

233)

+x11x21x31(x212x

222x

232 + x2

13x223x

233

= x11x12x13x221x

222x

223 + x11x12x13x

231x

232x

233

+x11x21x31x212x

222x

232 + x11x21x31x

212x

223x

233

= x11x12x21x222(x13x21x

223 + x12x31x

232)

+x11x13x31x233(x12x31x

232 + x12x21x

223)

= (x11x12x21x222 + x11x13x31x

233)

�(x13x21x223 + x12x31x232):

Thus, r(a2 + d2) + s(b2 + c2) is the product of two factors of positive

integers, and is greater than or equal to 2, because xij � 1 for all i; j.

Therefore, r(a2 + d2) + s(b2 + c2) is never a prime number.

For r = s + 1, that is, ad = bc, the sum a2 + b2 + c2 + d2 is never a

prime number.

5. Determine all functions f : R! Rwhich satisfy

(x� y)f(x+ y)� (x+ y)f(x� y) = 4xy(x2 � y2)

for all real numbers x, y.

Solution. This has already appeared in CRUX [1993: 41{42]. The

problem was used in the XLI Mathematics Olympiad in Poland.

6. Assume that when a snooker ball strikes a cushion, the angle of inci-

dence equals the angle of re ection.

For any position of a ball A, a point P on the cushion is determined as

shown.

Prove that if the ball A is shot at point P , it will go into the pocket B.

270

i i i

i i i

w

r

r

r

r

r

P

A

B

Solution. We will use analytic geometry. We let the axes beOx � OP ,

Oy = OB, Ox ? Oy; and give coordinates: O(0;0), A(a; b), B(0; c),

A0(a; 0); we have

6

-O x

B

B0

y

P

S

A0

A

OA : y = bax

BA0 : y = � cax+ c;

fSg = OA \BA0; i,e,

S�

ac

b+c; bc

b+c

�and P

�ac

b+c; 0�

AP : y = b+c

ax� c;

that is, B0(0;�c):Therefore jOBj = jOB0j = c and

�OBP ' �OB0P;

that is, \BPO = \B0PO:

Because \B0PO = \APA0,

we obtain \BPO = \APA0:

271

BOOK REVIEWS

Edited by ANDY LIU

Experience in Problem Solving - a W. J. Blundon Commemorative,edited by R. H. Eddy and M. M. Parmenter.

Published by the Atlantic Provinces Council on the Sciences, 1994,

ISBN # 0-9698965-0-6, 82+ pages, hardbound, $25.00.

Reviewed byMurray S. Klamkin, University of Alberta.

This book is the culmination of e�orts of its editors, whose idea it was

to honour the late W. J. Blundon, by collecting his problem solving contri-

butions to the mathematical literature. The preface contains a short bibli-

ography of Blundon. CRUX readers before 1990 may remember some of his

contributions.

The book consists of all (as far as its editors are aware) the published

problems and/or solutions of Blundon and which have appeared in Crux,

American Mathematical Monthly, Mathematics Magazine, College Mathe-

matics Journal, Elemente der Mathematik, SIAMReview, and Nieuw Archief

voor Wiskunde. The problems and solutions are given under four categories:

Geometry, Geometric Inequalitites, Number Theory, and Miscellaneous.

Jack Blundon, whom I knew personally from my days on the Canadian

Mathematical Olympiad Committee when he was the chairman, was a very

personable man as well as mathematician, and was particularly fond of ge-

ometric inequalities (as can be seen from the corresponding section of the

book). Further evidence of this is his several papers on this topic. In partic-

ular, his seminal paper Inequalities associated with the triangle, Canadian

Mathematics Bulletin 8 (1963) 615-627, was a catalyst for a subsequent se-

ries of papers from around the world dealing with the theory for triangle

inequalities, on which very little had been done before. He was also, as

the editors note and I agree, an ardent disciple of elegance in mathematical

exposition, often rewriting a proposal or solution several times in order to

meet his exacting standards. As a problem section editor, this is something

I would hope my contributors would also do.

As a very small sampling of his nice, easily understood proposals from

each of the four sections and which did not appear in Crux, we have:

� If, in triangle ABC, we have

sinA+ sinB + sinC

cosA+ cosB + cosC=p3;

prove that at least one angles of the triangle is 60�.

272

� For any triangle (other than equilateral), with circumcentre O, incentre

I and orthocentre H, let the angles have measures � � � � . Prove

that

(1) 1 < OHIO

< 3;

(2) 0 < IHOH

< 2

3;

(3) 0 < IHIO

< 1 if � < 60�, IH = IO if � = 60�;

1 < IHIO

< 2 if � > 60�.

� Find all solutions in integers of the equation

y2 + y = x4 + x3 + x2 + x:

� Find necessary and su�cient conditions on a, b and c in order that the

system of equations

x+1

x= a; y +

1

y= b; xy +

1

xy= c

has at least one solution.

Finally, it is to be noted that the editors, who were younger colleagues

of Professor Blundon at Memorial University, have done a good job in com-

piling this nice collection of eighty-one problems and solutions. Not only

did it remind me of results that I had forgotten and was looking for, it also

suggested to me a number of new problems which I will be submitting to

Crux!

[Ed.: The book is available, only from:

Atlantic Provinces Council on the SciencesMemorial University of NewfoundlandP.O. Box 4200, St. John's, NewfoundlandCanada A1C 5S7

for CDN $25 plus CDN $3 for shipping and handling.]

273

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 812"�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 that 1 April 1997. They may also be

sent by email to cruxeditor@cms.math.ca. (It would be appreciated if email

proposals and solutionswere written in LATEX, preferably in LATEX2e). Graph-

ics �les should be in epic format, or plain postscript. Solutions received after

the above date will also be considered if there is su�cient time before the

date of publication.

2164. Proposed by Toshio Seimiya, Kawasaki, Japan.

Let D be a point on the side BC of triangle ABC, and let E and F be

the incentres of triangles ABD and ACD respectively. Suppose that B, C,

E, F are concyclic. Prove that

AD +BD

AD+ CD=AB

AC:

2165. Proposed by Hoe Teck Wee, student, Hwa Chong Junior Col-

lege, Singapore.

Given a triangle ABC, prove that there exists a unique pair of points

P and Q such that the triangles ABC, PQC and PBQ are directly similar;

that is, \ABC = \PQC = \PBQ and \BAC = \QPC = \BPQ,

and the three similar triangles have the same orientation. Find a Euclidean

construction for the points P and Q.

2166. Proposed by K.R.S. Sastry, Dodballapur, India.

In a right-angled triangle, establish the existence of a unique interior

point with the property that the line through the point perpendicular to any

side cuts o� a triangle of the same area.

274

2167. Proposed by �Sefket Arslanagi�c, Berlin, Germany.

Prove, without the aid the di�erential calculus, the inequality, that in a

right triangle

a2(b+ c) + b2(a+ c)

abc� 2 +

p2;

where a and b are the legs and c the hypotenuse of the triangle.

2168. Proposed by Jan Ciach, Ostrowiec �Swi�etokrzyski, Poland.

Let P be a point inside a regular tetrahedron ABCD, with circumra-

dius R and let R1; R2; R3; R4 denote the distances of P from vertices of the

tetrahedron. Prove or disprove that

R1R2R3R4 �4

3R4;

and that the maximum value of R1R2R3R4 is attained.

2169. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

AB is a �xed diameter of circle �1(0;R). P is an arbitrary point of its

circumference. Q is the projection ontoAB ofP . Circle �2(P1PQ) intersects

�1 at C and D. CD intersects PQ at E. F is the midpoint of AQ. FG ?CD, where G 2 CD. Show that:

1. EP = EQ = EG,

2. A, G and P are collinear.

2170. Proposed by Tim Cross, King Edward's School, Birmingham,

England.

Find, with justi�cation, the positive integer which comes next in the

sequence 1411; 4463;4464; 1412; 4466; 4467;1413; 4469; : : : .

[Ed.: the answer is NOT 4470.]

2171. Proposed by Juan-Bosco Romero M�arquez, Universidad de

Valladolid, Valladolid, Spain.

Let P be an arbitrary point taken on an ellipse with foci F1 and F2,

and directrices d1, d2, respectively. Draw the straight line through P which

is parallel to the major axis of the ellipse. This line intersects d1 and d2 at

points M and N , respectively. Let P 0 be the point where MF1 intersects

NF2.

Prove that the quadrilateral PF1P0F2 is cyclic.

Does the result also hold in the case of a hyperbola?

2172. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,

Austria.

Let x; y; z � 0 with x + y + z = 1. For �xed real numbers a and b,

determine the maximum c = c(a; b) such that a+ bxyz � c(yz+ zx+xy).

275

2173. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,

Austria.

Let n � 2 and x1; : : : ; xn > 0 with x1 + : : :+ xn = 1.

Consider the terms

ln =

nXk=1

(1 + xk)

s1� xk

xk

and

rn = Cn

nYk=1

1 + xkp1� xk

where

Cn = (px� 1)n+1(

pn)n=(n+ 1)n�1:

1. Show l2 � r2.

2. Prove or disprove: ln � rn for n � 3.

2174. Proposed by Theodore Chronis, student, Aristotle University

of Thessaloniki, Greece.

Let A be an n� n matrix. Prove that if An+1 = 0 then An = 0.

2175. Proposed by Christopher J. Bradley, Clifton College, Bristol,

UK.

The fraction1

6can be represented as a di�erence in the following ways:

1

6=

1

2� 1

3;

1

6=

1

3� 1

6;

1

6=

1

4� 1

2;

1

6=

1

5� 1

30:

In how many ways can the fraction1

2175be expressed in the form

1

2175=

1

x� 1

y;

where x and y are positive integers?

2176. Proposed by �Sefket Arslanagi�c, Berlin, Germany.

Prove that

n

vuut nYk=1

(ak + bk) � n

vuut nYk=1

ak + n

vuut nYk=1

bk

where a1; a2; : : : ; an > 0 and n 2 N.

276

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

1987. [1994: 250; 1995: 283-285] Proposed by Herbert G �ulicher,

Westfalische Wilhelms-Universit�at, M�unster, Germany.

In the �gure, B2C1kA1A2, B3C2kA2A3

and B1C3kA3A1. Prove that B2C1, B3C2

and B1C3 are concurrent if and only if

A1C3

C3B3

� A2C1

C1B1

� A3C2

C2B2

= 1:

A2 A3

A1

B3

C3

C1

C2

B2

B1

II. Solutionby anonymous (spotted on the internet byWaldemar Pompe,

student, University of Warsaw, Poland).

Just apply Ceva's theorem to 4B1B2B3.

Editor's comments by Chris Fisher.

1. Note that if we permit Bi to lie on the side of the triangle extended be-

yond Ai�1 or Ai+1 (as Bradley did in solution I) then Ceva's theorem

fails to apply exactly when B1; B2; B3 are collinear. Thus the alterna-

tive condition in the middle of page 284 (namely u = pq=(p+ q� 1) in

Bradley's notation) applies if and only ifB1 2 B2B3. As a consequence,

Bradley's extended version of the problem can be restated as

A1A3

C3B3

� A2C1

C1B1

� A3C2

C2B2

= 1 if and only if either

B2C1; B3C2; and B1C3 are concurrent or B1 2 B2B3:

2. It makes an amusing exercise to prove directly that B1 2 B2B3 if and

only if C1 2 C2C3 (in the notation of G �ulicher's problem).

277

2067. [1995: 235] Proposed by Moshe Stupel and Victor Oxman,

Pedagogical Religious Girls' College \Shanan", Haifa, Israel.

Triangle ABC is inscribed in a circle �. Let AA1; BB1 and CC1 be the

bisectors of angles A;B and C, with A1; B1 and C1 on �. Prove that the

perimeter of the triangle is equal to

AA1 cos�A2

�+ BB1 cos

�B2

�+ CC1 cos

�C2

�:

Solution by Miguel Angel Cabez �on Ochoa, Logro ~no, Spain.

Let the internal bisector of \BAC meet the circumcircle of 4ABCagain in A1. Let R be the radius of the circumcircle.

A

B

CA1

A2

A2

From 4BAA1, we haveAA1

sin�B + A

2

� = 2R, so that

AA1 = 2R sin�B + A

2

�:

Thus

AA1 cos�A2

�= 2R sin

�B + A

2

�cos

�A2

�= R

�sin(B + A) + sinB

�= R sinC +R sinB = b+ c

2:

Similarly

BB1 cos�B

2

�= c+ a

2; CC1 cos

�C

2

�= a+ b

2:

Therefore

AA1 cos�A2

�+BB1 cos

�B2

�+ CC1 cos

�C2

�= b+ c

2+ c+ a

2+ a+ b

2:

278

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,

Spain; �SEFKET ARSLANAGI �C, Berlin, Germany; FRANCISCO BELLOT

ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CARL BOSLEY, student,

Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER J.

BRADLEY, Clifton College, Bristol, UK; SABIN CAUTIS, student, Earl Haig

Secondary School,North York, Ontario; HANS ENGELHAUPT, Franz{Ludwig{

Gymnasium, Bamberg, Germany; HIDETOSI FUKAGAWA, Gifu, Japan; TOBY

GEE, student, the John of Gaunt School, Trowbridge, England; RICHARD

I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursul-

inengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State Uni-

versity, Big Rapids, Michigan, USA; MITKO CHRISTOV KUNCHEV, Rousse,

Bulgaria; VEDULA N. MURTY, Andhra University, Visakhapatnam, India;

GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; WALDEMAR

POMPE, student, University of Warsaw, Poland; CRIST �OBAL S �ANCHEZ{

RUBIO, I.B. Penyagolosa, Castell �on, Spain; HEINZ-J�URGEN SEIFFERT,

Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; ASHSIH KR. SINGH,

student, Kanpur, India; PANOS E. TSAOUSSOGLOU, Athens, Greece; PAUL

YIU, Florida Atlantic University, Boca Raton, Florida, USA; and the pro-

posers. There was one anonymous solution.

Amengual Covas points out that this problem has already appeared

in print, in A Treatise on Plane Trigonometry by E.W. Hobson, Cambridge

University Press, 2nd Edition, 1897, Example 2 on page 194.

2069. [1995: 235] Proposed by D.J. Smeenk, Zaltbommel, the Neth-

erlands.

M is a variable point of side BC of triangle ABC. A line through M

intersects the lines AB in K and AC in L so that M is the mid-point of

segment KL. Point K0 is such that ALKK0 is a parallelogram. Determine

the locus of K0 asM moves on segment BC.

I. Essentially the same solution by Gottfried Perz, Pestalozzigymna-

sium, Graz, Austria; Waldemar Pompe, student, University of Warsaw, Pol-

and; and Toshio Seimiya, Kawasaki, Japan.

Suppose that the dilatation with centre A and ratio two transforms the

points B, C andM to B0, C0 and N respectively. SinceM is the mid-point

of bothKL andAN , we have that AKNL is a parallelogram. By de�nition,

ALKK0 is a parallelogram as well, so thatK is the mid-point ofK0N ; also,

K0N and AC0 are parallel. De�ne C00 to be the point such that A is the

mid-point of C0C00. We conclude that, as M varies on BC, N varies on

B0C0, so that the position of K0 must vary on the segment B0C00.

Editor's comments:

1. Note that this solution provides an explicit construction of the points

K and L.

279

2. Smeenk points out that K0 helps in �nding the position ofM for which

the corresponding segmentKL hasminimum length. (KL = AK;, and

AK0 has its minimum length whenK0 is the foot of the perpendicular

from A to the line B0C00.)

II. Essentially the same solution by Tim Cross, King Edward's School,

Birmingham, England; Hidetosi Fukagawa, Gifu, Japan; Walther Janous,

Ursulinengymnasium, Innsbruck, Austria; Mitko Christov Kunchev, Rousse,

Bulgaria; Ashsih Kr. Singh, student, Kanpur, India.

Take A to be the origin and set the vectors AB = B, etc. Then M =

tB+(1� t)C, where t varies from 0 to 1 asM moves from C to B. Suppose

that K = �B and that L = �C. Because M is the mid-point of KL, we

have

�B + �C

2= tB + (1� t)C;

so that � = 2t and � = 2(1� t) (since B and C are linearly independent).

Since ALKK0 is a parallelogram, it follows that

K = K � L= t(2B + (1� t)(�2C):

Thus the locus ofK0 is the segment joining�2C (where t = 0) to 2B where

t = 1).

Editor's comment: The algebra makes clear that it is not necessary to

restrict M to the segment BC; as t ranges over the real numbers,K0 moves

along its line, whileM moves along the line BC.

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; CARL BOSLEY,

student, Washburn Rural High School, Topeka, Kansas, USA; JORDI DOU,

Barcelona, Spain; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bam-

berg, Germany; CRIST �OBAL S �ANCHEZ{RUBIO, I.B. Penyagolosa, Castell �on,

Spain; TOSHIO SEIMIYA, Kawasaki, Japan; and the proposer. One anony-

mous solution was received - see note at the start of this section. There was

one incorrect solution received.

2070. [1995: 236] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,

Alcorc �on, Madrid, Spain.

For which positive integers n is the Catalan number

1

n+ 1

�2n

n

�

odd?

280

I Virtually identical solutions by Toby Gee, student, the John of Gaunt

School, Trowbridge, England; Douglas E. Jackson, Eastern New Mexico Uni-

versity, Portales, New Mexico, USA; Thomas Leong, Staten Island, NY, USA;

Andy Liu, University of Alberta, Edmonton, Alberta; Waldemar Pompe, stu-

dent, University of Warsaw, Poland; and the proposer.

It is well-known that the nth Catalan number, Cn = 1

n+1

�2n

n

�, satis�es

the recurrence relation

Cn+1 = C0Cn + C1Cn�1 + : : :+CnC0 n � 0:

We show that Cn is odd if and only if n = 2k � 1 for some integer k � 0.

We use induction on n.

Since C0 = 1, the assertion is true for n = 0. Assume that among the

numbers C0, C1, : : : , Cn, only those with n of the form 2k � 1 are odd. If

n+ 1 is even, then

Cn+1 = 2

(n�1)=2Xk=0

CkCn�k;

which is even.

On the other hand, if n+ 1 is odd, then

Cn+1 = 2

(n�1)=2Xk=0

CkCn�k + C2n=2;

showing that Cn+1 is odd if and only if Cn=2 is odd. But Cn=2 is odd if and

only if n2= 2k � 1 for some integer k � 0. Thus Cn+1 is odd if and only if

n+ 1 = 2(2k � 1) + 1 = 2k+1 � 1, which completes the proof.

II Solution by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg,

Germany.

Let Cn = 1

n+1

�2n

n

�denote the nth Catalan number, n = 1, 2, 3, : : : ,

and let F (n) denote the highest power of 2 that divides n. Then it is well

known that

F (n!) =

�n

2

�+

�n

4

�+

�n

8

�+ : : : :

From this, it is easy to see that F (n!) � n� 1, with equality if and only if

n = 2k for some integer k � 0. [More generally, it is known, and easy to

show, that if p is prime andn = arar�1 : : : a1a0 is the base-p representation

of n, then

h =

n�rX

i=0

ai

p� 1;

where phkn! [See, for example, Theorem 2.30 in Elementary Introduction to

Number Theory, 3rd edition, by Calvin T. Long | Ed.]

281

Since

Cn =(2n)!

(n+ 1)!n!=

2n � 1 � 3 � 5 � � � (2n� 1)

(n+ 1)!;

Cn is odd if and only ifF ((n+1)!) = n, which is true if and only ifn+1 = 2k

or n = 2k�1 for some integer k � 1. We can also allow k = 0 since C0 = 1

is odd.

Solutions similar or equivalent to II above were submitted by �SEFKET

ARSLANAGI �C, Berlin, Germany; CARL BOSLEY, student, Washburn Rural

High School, Topeka, Kansas, USA; CHRISTOPHER J. BRADLEY, Clifton

College, Bristol, UK; SABIN CAUTIS, student, Earl Haig Secondary School,

North York, Ontario; EMERIC DEUTSCH, Brooklyn, NY, USA; SHAWN

GODIN, St. Joseph Scollard Hall, North Bay, Ontario; SOLOMON W.

GOLOMB, Univ of Southern California, Los Angeles, CA, USA; RICHARD

I. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, New Mex-

ico Highlands University, Las Vegas, NM, USA; WALTHER JANOUS, Ursul-

inengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State Uni-

versity, Big Rapids, Michigan, USA; HARRY SEDINGER, St. Bonaventure

University, St. Bonaventure, NY, USA; KEE-WAI LAU, Hong Kong; HEINZ-

J�URGEN SEIFFERT, Berlin, Germany; SKIDMORE COLLEGE PROBLEM

GROUP Saratoga Springs, NY, USA; CHRIS WILDHAGEN, Rotterdam, the

Netherlands; KENNETH M. WILKE, Topeka, Kansas, USA; PAUL YIU,

Florida Atlantic University, Boca Raton, Florida, USA (two solutions).

Several solvers pointed out the well-known fact that the highest power of 2

that divides�2n

n

�is equal to the number of ones in the binary representation

of n. This is, of course, an immediate consequence of the more general fact

mentioned in the editor's comment in solution II above.

This problem is certainly not new. Deutsch supplied the reference: \The Par-

ity of the Catalan Numbers via Lattice Paths" by �Omer E�gecio�glu, Fibonacci

Quarterly, (21), 1983, 65{66. Godin gave the reference: \Time Travel and

other Mathematical Bewilderments" byMartin Gardner, W.H. Freeman and

Co. In fact, the result of this problem was also mentioned byMartin Gardner

in his article \Mathematical Games; Catalan number: an integer sequence

that materializes in unexpected places", Scienti�c American, 234(6), 1976,120{125, but he did not indicate any proof.

2071. [1995: 277] Proposed by Toshio Seimiya, Kawasaki, Japan.

P is an interior point of an equilateral triangleABC so that PB 6= PC,

and BP and CP meet AC and AB at D and E respectively. Suppose that

PB : PC = AD : AE. Find angle BPC.

Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.

We claim that \BPC = 120�. Let the side of the triangle be 1. Fur-

ther, let x = AE; y = AD; � = \BCE and ' = \CBD. By the law of

282

cosines (for 4ABD and4AEC respectively) we get (since cos 60� = 1=2):

BD =p1 + y2 � y and CE =

p1 + x2 � x. Via the law of sines applied

to

4BCD : sin' =1� y

BDsin60� =

1� yp1 + y2 � y

�p3

2

4EBC : sin� =1� x

CEsin 60� =

1� xp1 + x2 � x

�p3

2

4BCP :sin'

sin�=PC

PB=x

y[by assumption for the problem!]

Hence, y�y2p1+y2�y

= x�x2p1+x2�x

, that is (with X = x� x2 and Y = y� y2)

Xp1�X

=Yp1� Y

: (1)

Here, since 0 < x; y < 1; 0 < X;Y � 1=4:

Now, (1)() X2(1� Y ) = Y 2(1�X)() (X � Y )(XY �X � Y ) = 0:

Thus, two cases are to be considered.

(a) XY �X � Y = 0() (X � 1)(Y � 1) = 1; which is impossible.

(b) X = Y () x� x2 = y� y2() (y� x)(x+ y � 1) = 0()y = 1� x (since x 6= y).

Hence 4ABD is congruent to 4BCE ) \ABD = � ) � + ' = 60� )\BPC = 180� � (� + ') = 120�:

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and

MARIA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid,

Spain; CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol, UK; JORDIDOU,

Barcelona, Spain; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bam-

berg, Germany; RICHARD I. HESS, Rancho Palos Verdes, California, USA;

V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;

P. PENNING, Delft, the Netherlands; D.J. SMEENK, Zaltbommel, the Neth-

erlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; HOE TECK WEE, stu-

dent, Hwa Chong Junior College, Singapore; one anonymous solver and the

proposer. There were two incorrect solutions.

2073?. [1995: 277] Proposed by Jan Ciach, Ostrowiec �Swi�etokrzyski,

Poland.

Let P be an interior point of an equilateral triangle A1A2A3 with cir-

cumradius R, and let R1 = PA1, R2 = PA2, R3 = PA3. Prove or disprove

that

R1R2R3 �9

8R3:

283

Equality holds if P is the mid-point of a side. [Compare this problem with

Crux 1895 [1995: 204].]

I Solution by Kee-Wai Lau, Hong Kong.

Without loss of generality we assume that R = 2, so that the length of

each side of the triangle is 2p3. Let \A1PA2 = �;\A2PA3 = �; so that

\A1PA3 = 2� � � � �. Applying the cosine law to �A1PA2;�A2PA3

and �A1PA3, respectively, we obtain

cos � = (R12 + R2

2 � 12)=(2R1R2);

cos� = (R22 +R3

2 � 12)=(2R2R3) and

cos(� + �) = (R12 + R3

2 � 12)=(2R1R3):

Since [cos � cos�� cos(�+�)]2 = (1� cos2 �)(1� cos2 �), we obtain after

some simpli�cation that

R14 + R2

4 + R34 �R1

2R22 � R2

2R32 � R3

2R12�

12(R12 + R2

2 + R32) + 144 = 0:

Thus

R32 =

1

2(z + 24� (12y� 3z2)1=2); (1)

where y = R12R2

2; z = R12 + R2

2 � 12 and z2 � 4y:

By considering R1 = R2 = R3 = 2, we see that we should choose in (1) the

� sign from �. It follows that

f(y;z) := R12R2

2R32 =

1

2y[z + 24� (12y� 3z2)1=2]:

From df=dz = 0, we obtain z = �py and f(y;�py) = y(12�2py) � 64:

Now at the boundary z2 = 4y, we have jR1�R2j =p12 orR1+R2 =

p12,

so that�6 � z � 0. Hence f = z2(z+24)=8 � 81 and equality holds when

z = �6 or P is the mid-point of a side. This proves the required inequality.

II Solution by Manuel Benito Munoz and Emilio Fern�andez Moral, I.B.

Sagasta, Logro ~no, Spain

The proposed inequality holds (equality only on the mid-point of a

side), as a particular case (n = 3) of the following:

R1R2 : : : Rn � (1 + cosn�

n) � Rn

when P is any point from a closed regular n-gon,A1A2 : : : An; with circum-

radius R and Ri = PAi. (Equality holds only on the mid-point of a side.)

Without loss of generality, let us suppose that R = 1 and that the vertices of

the n-gon are the points �1; �2; : : : ; �n of the complex plane, where �k =

e2�in

(k�1). Let z = �ei� represent the point P .

284

By the maximum modulus principle applied to the function �nk=1(z � �k);

the modulus of that function, i.e., PA1 � PA2 : : : PAn = �nk=1jz � �kj,

assumes its greatest value on the boundary of the n-gon.

Now, geometrical considerations permit us to avoid much analytical treat-

ment and we conclude that the maximum value is attained on the mid-point

of any side AiAi+1: Let P be a point on the side A1A2 of the n-gon; the

vertices A3; : : : ; An of the n-gon are pairwise arranged symmetrically with

respect to the mediatrix of the segment A1A2 (except for the single point

An+12

+1when n is odd).

Let Ak; Al be one of such pairs of vertices; if M is the mid-point of A1A2

we have, by the GA-means inequality and \shortest path principle", that

PAk � PAl ��PAk + PAl

2

�2��MAk +MAl

2

�2= MAk �MAl

(the latter because MAk = MAl).

When n is even, as PA1 � PA2 �MA1 �MA2, we are done.

When n is odd, let Am be the unpaired vertex and put A1A2 = 2l, PM = x

and MAm = a. Therefore,

PA1 � PA2 � PAm = (l2 � x2) �pa2 + x2;

and the maximum value of that function for x 2 [0; l] is attained at x = 0;

that is, when P = M: (There are no other critical points because a > l:) So

that, for every n, we have shown

PA1 � PA2 : : : PAn �MA1 �MA2 : : :MAn;

where M is the mid-point of a side.

Now �nally

(PA1 � PA2 : : : PAn)2

= �nk=1jz � �kj2 = �n

k=1(z � �k) ��nk=1(z � �k)

= (zn � 1) � (zn � 1) = �2n � 2�n cosn� + 1;

and if P is the mid-point of A1A2 (� = cos��n

�; � = �

n); that value is�

cos��n

��2n � 2 cosn��n

� � cos� + 1 =�cosn

��n

�+ 1

�2, as required.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK; WOLFGANG GMEINER, Bundesgymnasium Spittal/Drau, Spittal, Aus-

tria (also found the generalization of �); WALTHER JANOUS, Ursulinen-

gymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN, University of Al-

berta, Edmonton, Alberta(with a generalization from triangles to tetrahedra);

V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;

HEINZ-J�URGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki,

285

Japan; JOHANNES WALDMANN, Friedrich-Schiller-Universit�at, Jena,

Germany; and the proposer.

2074. [1995: 277] Proposed by Stanley Rabinowitz, Westford,

Massachusetts.

The number 3774 is divisible by 37, 34 and 74 but not by 77. Find

another four-digit integer abcd that is divisible by the two-digit numbers

ab, ac, ad, bd and cd but is not divisible by bc.

Solution by Paul Yiu, Florida Atlantic University, Boca Raton, Florida,

USA.

The two possible solutions are

1995 and 2184:

LetN = abcd = 100 �(ab)+cd be divisible by ab. Then cdmust be divisible

by ab as well; say, cd = k � (ab) where k < 10. SinceN is divisible by cd, so

is 100 � (ab). It follows that k is a divisor of 100. As such, it must be one of

1, 2, 4, 5. Now, a direct computer search reveals that only 1995, 2184 and

3774 satisfy the requirement.

Note: apart from numbers of the form aaaa, 1155 is the only four-digit

number divisible by all two-digit numbers \contained in it".

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK; KEITH EKBLAW, Walla Walla, Washington, USA; HANS ENGELHAUPT,

Franz{Ludwig{Gymnasium, Bamberg, Germany; JEFFREY K. FLOYD, New-

nan, Georgia, USA; SHAWN GODIN, St. Joseph Scollard Hall, North Bay,

Ontario; 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, Ferris State

University, Big Rapids, Michigan, USA; KEE-WAI LAU, Hong Kong; J.A.

MCCALLUM, Medicine Hat, Alberta; STEWART METCHETTE, Gardena,

California, USA; P. PENNING, Delft, the Netherlands; ROBERT P.

SEALY, Mount Allison University, Sackville, New Brunswick; HEINZ-

J�URGEN SEIFFERT, Berlin, Germany; DAVID R. STONE, Georgia Southern

University, Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens,

Greece; CHRIS WILDHAGEN, Rotterdam, the Netherlands; KENNETH M.

WILKE, Topeka, Kansas, USA; and the proposer.

A bit over half of the solvers found both solutions, often by computer,

as there does not appear to be a short way to do the problem by hand. The

proposer also used a computer, but it seems to have let him down, as he

thought that 1995 was the only answer!

Metchette also gave the \solution" 0315, which works, but is not a

true four-digit number. He wonders why all the solutions are multiples of

3; can any reader see an easy explanation?

286

2075. [1995: 278] Proposed by Christopher J. Bradley, Clifton Col-

lege, Bristol, UK

ABC is a triangle with \A < \B < \C, and I is its incentre. BCL,

ACM , ABN are the sides of the triangle with L on BC produced, etc., and

the points L, M , N chosen so that

\CLI =1

2(\C�\B); \AMI =

1

2(\C�\A); \BNI =

1

2(\B�\A):

Prove that L, M , N are collinear.

Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria.

We have

\AIN = 180� � (\BAI + \BNI)

= 180� ��\A

2+\B � \A

2

�

= 180� � \B

2= \NBI;

which implies that triangles4ANI and4INB are similar, or

AN : NI : AI = NI : BN : BI

Hence it follows (via AN �BN = NI2) that

AN

BN=

NI2

BN2=AI2

BI2:

Analogously we getBL

CL=BI2

CI2;

CM

AM=CI2

AI2

and �nallyAN

BN� BLCL� CMAM

=AI2

BI2� BI

2

CI2� CI

2

AI2= 1

whence (by the converse of Menelaus' Theorem) M , N and L are collinear.

Perz also remarks that, more generally, L, M , N are collinear, if P is

a point inside or outside4ABC and L;M;N are such that

\CLP = \BCP � \PBC; \AMP

= \CAP � \PCA; \BNI

= \ABP � \PAB;

where all angles are oriented angles.

287

Also solved by TOBY GEE, student, the John of Gaunt School, Trow-

bridge, England; CYRUS HSIA, student, University of Toronto, Toronto, On-

tario; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,

USA; KEE-WAI LAU, Hong Kong; P. PENNING, Delft, the Netherlands;

TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Neth-

erlands; and the proposer.

2077. [1995: 278] Proposed by Joseph Zaks, University of Haifa,

Israel.

The determinant ��������z1z1 z1 z1 1

z2z2 z2 z2 1

z3z3 z3 z3 1

z4z4 z4 z4 1

��������equals 0 if and only if the four complex numbers z1, z2, z3, z4 satisfy what

simple geometric property?

Solutionby V�aclav Kone �cn �y, Ferris State University, Big Rapids, Michi-

gan, USA. [modi�ed slightly by the editor.]

Let D denote the given determinant. If any two (or more) of the four

complex numbers are equal, z1 = z2, say, thenD = 0, and the points z2, z3,

z4 must be collinear (or concyclic). We thus assume that the four numbers

are all distinct.

It is well-known that an equation of the circle passing through three

non-collinear points (x2; y2), (x3; y3, (x4; y4), can be written as��������x2 + y2 x y 1

x22 + y22 x2 y2 1

x23 + y23 x3 y3 1

x24 + y24 x4 y4 1

��������= 0:

Letting z1 = x+ {y and zk = xk + {yk for k = 2; 3, 4, we have

D =

��������z1z1 x z1 1

z2z2 x2 z2 1

z3z3 x3 z3 1

z4z4 x4 z4 1

��������= �2{

��������x2 + y2 x y 1

x22 + y22 x2 y2 1

x23 + y23 x3 y3 1

x24 + y24 x4 y4 1

��������:

Thus D = 0 if and only if z1 lies on the circle passing through the three

points z2, z3 and z4. There the sought property is that z1, z2, z3, z4 are

either concyclic or collinear (which can be viewed as a degenerate case).

[Editor's note: This is certainly a known result. The references supplied by

several readers are listed below. The number, if there is one, after a solver's

name corresponds to the reference given by that solver.

288

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,

Spain; �SEFKET ARSLANAGI �C, Berlin, Germany; FRANCISCO BELLOT

ROSADO, I.B. Emilio Ferrari, Valladolid, Spain [3]; CHRISTOPHER J.

BRADLEY, Clifton College, Bristol, UK; F.J. FLANIGAN, San Jose State Uni-

versity, San Jose, California, USA; WOLFGANG GMEINER, Millstatt,

Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA [2];

WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI

LAU, Hong Kong; HEINZ-J�URGEN SEIFFERT, Berlin, Germany [1]; CHRIS

WILDHAGEN, Rotterdam, the Netherlands; and the proposer. One anony-

mous solution [4] and one incomplete solution were also received.

References:

[1] Conway, J.B., Functionsof One Complex Variable, Springer-Verlag, 2nd

ed., 1978, p. 49, Proposition 3.10.

[2] Dodge, Clayton W., Complex Numbers, 1993, p. 135 and p. 199.

[3] Krzyz, Problems in Complex Variable Theory, Elsevier, PWN, 1971, P. 5

and p. 142.

[4] Schwerdtfeger, H., Geometry for Complex Numbers, Dover, NY, 1962.

In the May 1996 issue of CRUX [1996: 168], we asked \Do you know

the equation of this curve?"

Here is a hint { it is known as the \butter y"!

289

Letter from the Editor

Changes to CRUX

Readers will be interested to learn about some of the changes that arein the works for CRUX.

First, CRUX is now available as an on-line supplement for individualsubscribers at

http://camel.math.ca/CRUX

This new service was instituted at the beginning of September, and isproving to be very popular. A survey of the tra�c accessing CRUX on-line

shows that many people are sur�ng to take a look, and as a result, we haveseveral new subscriptions.

Information about each issue will be available to anyone who accessesthe CRUX on-line site. As well as seeing the table of contents, a two pagesynopsis is also available. This gives a short summary of what is in eachpublished article, as well as some highlights of what is in the various corners.One of the problems is given as a \free sample". This is to try to encouragemore problem solvers to become involved.

When individual subscribers renew their subscription for 1997, they willreceive information from the CMS Executive O�ce on how they can activatetheir account on CAMEL and gain access to CRUX on-line. In the future,CRUX on-line will be an extra free service to individuals with a normal (hardcopy) subscription. As well, an on-line only service (no hard copy) will beavailable. The rates will be published in the next issue of CRUX.

Since CRUX is copyrighted by the CanadianMathematical Society, thereare restrictions on the use of the on-line version (as there is for the printedversion). Here is the statement that appears on-line:

All rights reserved. For private and personal use only. No fur-ther distribution of these materials is permitted without the ex-press written permission of the Canadian Mathematical Society(managing-editor@cms.math.ca).

Teachers wishing to use these materials are particularly encour-aged to contact the CMS for permission.

Also, in 1997, CRUX is proposing to increase the quantity of high schoollevel materials in each issue. This will not involve any decrease in the quan-tity of each of the present sections. We are proposing to increase the issuesize to 64 pages (at no increase in the subscription price over that planned for1997 based on 48 pages per issue). Thus, subscribers, especially those whohave connections with high schools, would get a lot more for their money.Watch for a further announcement in the next issue.

Bruce ShawyerEditor-in-Chief

290

A Note on the Mean Value Theorem

Finbarr HollandDepartment of Mathematics,

University College, Cork, Ireland

Every student of Calculus knows that given two distinct points A;Bon the graph of a smooth function there is a point C on the arc of the curvejoiningA andB where the tangent to the graph is parallel to the chord joiningA and B. Moreover, it is easy to construct examples where the numberof such points C is either �nite, countably in�nite or uncountable; see theexercises.

The purpose of this note is to show that if the function is convex, thenthere are only two possibilities: namely, either there is a unique point oruncountably many such points.

We recall the de�nition of convexity. Geometrically it means that theportion of the graph joining any two points on it must lie below the chordjoining them. Analytically, f is convex on an interval [a; b], with a < b, if

f(x) � f(s) +f(t)� f(s)

t� s(x� s);whenever a � s � x � t � b.

The result we wish to establish depends on a lemma which we will dealwith �rst.

Lemma 1 Suppose f is convex on [a; b] and di�erentiable at c 2 (a; b). Then

f(c) + f 0(c)(x� c) � f(x); 8 x 2 [a; b]:

Proof. To see this, suppose c < t < x � b. Then

f(t)� f(c) +f(x)� f(c)

x� c(t� c):

Hencef(t)� f(c)

t� c� f(x)� f(c)

x� c

and so, letting t! c+, we deduce that

f 0(c) � f(x)� f(c)

x� c;

291

that is, (x�c)f 0(c) � f(x)�f(c)which proves the desired result for x > c.Next, if a � x < t < c, then

f(t) � f(x) +f(c)� f(x)

c� x(t� x)

= f(x) +f(c)� f(x)

c� x(t� c+ c� x)

= f(x) +f(c)� f(x)

c� x(t� c) + f(c)� f(x)

= f(c) +f(x)� f(c)

x� c(t� c):

Hencef(t)� f(c)

t� c� f(x)� f(c)

x� c

and so, letting t! c�, we deduce that

f 0(c) � f(x)� f(c)

x� c;

that is, (x � c)f 0(c) � f(x) � f(c) which proves the desired result forx < c. This completes the proof of the lemma since its conclusion certainlyholds when x = c.

This result means that the graph of y = f(x) lies above the tangentline y = f(c) + f 0(c)(x� c).

Theorem 1 Let �1 < a < b < 1. Suppose f is continuous and convex on[a; b], and di�erentiable on (a; b). Then the equation

f 0(c) =f(b)� f(a)

b� a

has either a unique solution in (a; b) or in�nitely many solutions.

Proof. The Mean Value Theorem guarantees that the solution set is non-empty. Suppose c1; c2 are two distinct solutions in (a; b), so that

f 0(c1) =f(b)� f(a)

b� a= f 0(c2):

We can and do suppose that c1 < c2. It follows from the lemma that

f(c1) + f 0(c1)(c2 � c1) � f(c2)

and

f(c2) + f 0(c2)(c1 � c2) � f(c1):

292

Hence

f 0(c1) = f 0(c2) =f(c2)� f(c1)

c2 � c1:

But, by convexity, the line joining the two points (c1; f(c1)); (c2; f(c2)), liesabove the graph of f that is, we have

f(x) � f(c1) + f 0(c1)(x� c1); 8x 2 [c1; c2]:

Applying the lemma once more, we deduce that

f(x) = f(c1) + f 0(c1)(x� c1); 8 x 2 [c1; c2]:

In other words, f 0(x) = (f(b)� f(a))=(b�a) for all x 2 [c1; c2]. It followsthat there are in�nitely many points with this property as claimed.

Exercises

1. Concoct examples to show that the number of points on the arc of asmooth curve where the tangent to the curve is parallel to the chordjoining the ends of the arc can be (i) �nite; (ii) uncountable.

2. Show that, for every non-negative integer n, the equation

tanx =x+ n�

3

has a unique solution on the interval [0; �=2).

3. Consider the function g de�ned on [�1; 1] as follows:

g(x) =

�x3 sin(1=x); if 0 < jxj � 1,0; if x = 0.

Show that g is continuous on [�1; 1] and has a continuous derivative on(�1; 1). Using the previous exercise, or otherwise, show that there isa countably in�nite number of points on the graph of y = g(x) wherethe tangent is parallel to the chord joining the points (�1; g(�1)) and(1; g(1)).

4. Deduce, and/or prove directly, that the function g is not convex on[�1; 1].

5. Show that f is convex on [a; b] if and only if the line segment joiningany two points in the set

S = f(x; y) : f(x) � y; a � x � bg

is a subset of S.

293

6. Suppose f is continuous on [a; b], di�erentiable on (a; b) and the tan-gent at any point on the graph of y = f(x) lies below the graph. Showthat f is convex on [a; b]. (This is a converse of the lemma.)

7. Suppose f is convex on [a; b] and di�erentiable on (a; b). Show that f 0

is increasing on (a; b).

8. Suppose f is twice continuously di�erentiable on (a; b) and f 00 � 0 on(a; b). Show that f is convex on [a; b]. (This provides a simple | andwell-known | test for convexity.) [Hint: if x; c 2 (a; b), then

f(x) = f(c) + f 0(c)(x� c) +

Z x

c

(x� t)f 00(t)dt:]

9. Suppose f is twice di�erentiable on (a; b) with f 00 > 0 on (a; b). Showthat the function

f�(x) = supa<s<b

fxs� f(s)g

is convex on the range of f 0. (The function f� is called the Fenchel-Legendre transform of f .)

10. Determine the Fenchel-Legendre transforms of each of the followingfunctions de�ned on (0;1):

ex, x logx� x, 1=x, �2px, xp=p, xq=q (p > 1, q = p=(p� 1)):

DID YOU KNOW� � �

| that the following numbers, containing only the digits 1, 4, 9, are allperfect squares:

1; 4; 9; 49; 144; 441; 1444; 11449; 44944;

991494144; 4914991449; 149991994944 ?

Are there in�nitely many such numbers?

| Neven Juri�c

294

THE SKOLIAD CORNER

No. 17

R.E. Woodrow

As a problem set in this issue, we give the grade 12 level contest fromNova Scotia, written April 26, 1996. My thanks go to Professor Michael Nuttof Acadia University, Wolfville, Nova Scotia, for sending the contest to us forour use.

CANADIANMATHEMATICAL SOCIETY PRIZE EXAMFriday, April 26, 1996 | Time: 2.5 hours

1. (a) Solvepx+ 20�px+ 1 = 1.

(b) Try to solve 3px+ 20� 3

px+ 1 = 1.

2. Suppose a function is de�ned so that f(xy) = f(x) + f(y).(a) Show that if f is de�ned at 1, f(1) = 0.

(b) Similarly if f is de�ned at 0, its value at any x will be 0, so it is a\trivial" function.

(c) Show that if f is de�ned only for f1; 2; 3; : : : g there are many non-trivial ways to de�ne such an f (give an example).

3. Three circles of equal radii r all touch each other to enclose a threecornered concave area A. How big is the area of A?

4. Show that for all real numbers x,

(a) x4 � 4x� 3.

(b) x4 is not greater than 3x� 2, even though it appears to be true.

x �1 0 14

13

12

1 2 3

x4 1 0 :004 0:12 :063 1 16 81

4x� 3 �7 �3 �2 �1:67 �1 1 5 9

3x� 2 �5 �2 �1:25 �1 �:5 1 4 7

5. Two integers are called equivalent, written x � y, if they are di-visible by the same prime numbers (primes are 2; 3; 5; 7; : : : ) so 2 � 2 � 4,3 � 27 but 2 6� 3.

(a) Show that 10 � 80 but 10 6� 90.

(b) Prove that if x � y, then x2 � y2.

6. We can describe certain fractions in terms of others all with biggerdenominators (always in lowest terms). For instance 1

3= 1

4+ 1

12and 2

3=

1

4+ 1

4+ 1

6but 2

3= 2

6+ 2

6does not work since 2

6= 1

3and 2

3= 1

2+ 1

6does

not since 2 < 3.

295

(a) Can you write 12as a sum 1

a+ 1

bfor integers 2 < a < b?

(b) Try to write 11996

as 1a+ 1

bfor integers 1996 < a < b.

Last issue we gave the problems of the Saskatchewan Senior Math-ematics Competition. Here are the o�cial solutions. Many thanks go toProfessor Garreth Gri�th, of the Department of Mathematics of the Univer-sity of Saskatchewan, and long time contest organizer in Saskatchewan, forproviding the problems and solutions.

SASKATCHEWAN SENIOR MATHEMATICS CONTESTWednesday, February 22, 1995

Time: 1.5 hours

1. They sell regular and jumbo sized orders of �sh at Jerry's Fish &Chips Emporium. The jumbo order costs

�43

�times as much as a regular one

and an order of chips costs $1.30.Last Thursday, the Emporium was quite busy over the lunch break

(11:30 am - 1:30 pm). Exactly thirteen jumbo sized orders of �sh along witha quantity of regular orders of �sh and chips were sold. $702.52 had beenplaced in the till.

During the period 1:30 - 4:30, business slackened o�. 26 regular ordersof �sh were sold during this time. Four times as many regular orders hadbeen sold during the lunch break. No jumbo portions were sold and thenumber of orders of chips declined to one �fth the number that had beensold during the lunch break. At 4:30 pm there was $850.46 in Jerry's till.

What is the price of a regular sized order of �sh at Jerry's Emporium?[5 marks]

Solution. Let x be the price of a regular order of �sh at Jerry's. Then, thecost of a jumbo order is 4x

3. Let y be the number of orders of chips sold

at lunchtime. Since 26 regular orders of �sh were sold during the afternoonperiod, 104 were sold at lunch time. Also, y

5orders of chips were sold during

the afternoon. The revenue at lunchtime was

(1:30)y+ 104x+ 13

�4x

3

�= 702:52 (L)

and the revenue during the afternoon was

(1:30)y

5+ 26x = 850:46� 702:52

= 147:94: (PM)

To solve these two equations, consider

5(PM)� (L) = 130x� 104x� 13

�4x

3

�= 37:18

296

which is equivalent to

2x

3= 2:86 or x = 4:29:

2. ABCD is a square with side of length s. A circle, centre A andradius r is drawn so that the arc of this circle which lies within the squaredivides the square into two regions of equal area. Write r as a function of s.[6 marks]

Solution. The area of the square is s2. The area of the (whole) circle is �r2.One quarter of this area lies within the square. This area is �r2=4. Therefore

�r2

4=

s2

2or r2 =

2s2

�

so that

r =

r2

�s:

(The � sign is unnecessary since r and s are distances.)

3.(a) Solve the equation 3y = 10y. [3 marks]

(b) Solve the equation 3y = 10. [3 marks]

(c) Write t log8px� 2 log8 y as a single logarithm. [4 marks]

Solution. (a) Take (common) logs of both sides:

log 3y = log 10y

y log 3 = y log 10 = y:

0 = y(1� log 3) so that the only solution is y = 0.(b) Similarly, y log 3 = 1 so that

y =1

log 3:

If we use base 3 instead, then

y log3 3 = log3 10 or y = log3 10:

(The two answers are equal.)(c)

6 log8px� 2 log8 y = 6 log8 x

1=2 � 2 log8 y

= log8(x1=2)6 � log8 y

2

= log8

�x3

y2

�:

297

4. Establish the identity 2 cotA = cot A2� tan A

2: [5 marks]

Solution.

The right side =cosA=2

sinA=2� sinA=2

cosA=2

=cos2 A

2� sin2 A

2

sin A2cos A

2

=cosA12sinA

=2 cosA

sinA

= 2 cotA = the left side:

5. ABC is a triangle, right angled at C. Let a, b, c denote the lengthsof the sides opposite angles A, B, C respectively. Given that a = 1, \B =

75� and that tan75� = 2 +p3, express b and c in the form p + q

p3 such

that p = 2 orp2. [8 marks]

Solution. Since ABC is a right angled triangle, tanB = ba. Since a = 1,

b = 2+p3. By the theorem of Pythagoras,

c2 = a2 + b2 = 1 + b2

= 1 + (2 +p3)2 = 8 + 4

p3 = r(2 +

p3):

Therefore c = 2p2 +

p3. (As in problem 2, we dismiss the � sign.) The

problem now is to write 2p2 +

p3 as p+ q

p3 where p = 2 or

p2. Square

both sides4(2 +

p3) = p2 + 3q2 + 2pq

p3:

If p = 2, then

8 = 4 + 3q2 (1)

and 4p3 = 4q

p3: (2)

From (2), q = 1. This is not consistent with (1).If p =

p2, then

8 = 2 + 3q2 (3)

and 4p3 = 2

p2qp3: (4)

From (4), q =p2 which is consistent with (3) so that

p = q =p2 and c =

p2 +

p2p3 or

p2(1 +

p3):

Check!

c2 = 2(1 +p3)2

= 2(4 + 2p3)

= 4(2 +p3):

298

6. Determine the function f(x) which satis�es all of the followingconditions: [8 marks]

(i) f(x) is a quadratic function.

(ii) f(x+ 2) = f(x) + x+ 2.

(iii) f(2) = 2.

Solution. (i) Let f(x) = ax2 + bx+ c.(ii) f(x+ 2) = a(x+ 2)2 + b(x+ 2) + c so that

a(x+ 2)2 + b(x+ 2) + c = ax2 + bx+ c+ x+ 2

orax2 + 4ax+ 4a+ bx+ 2b+ c = ax2 + bx+ c+ x+ 2

so that 4ax+ 4a+ 2b = x+ 2.Since this is an identity,

4a = 1 and 4a+ 2b = 2 so that a =1

4and b =

1

2:

(iii) f(2) = 4a + 2b + c = 2. Therefore 2 + c = 2 so that c = 0. Itfollows that

f(x) =x2

4+x

2:

7.Prove that if n is a positive integer (written in base 10) and that if 9is a factor of n, then 9 is also a factor of the sum of the digits of n.[8 marks]

Solution. Let a be the units digit of n; let b be the tens digit of n; let c bethe hundreds digit of n and so on. Then

n = a+ 10b+ 100c+ � � �= a+ (1 + 9)b+ (1 + 99)c+ � � �= (a+ b+ c+ � � � ) + 9b+ 99c+ � � �

Since 9 divides 9b + 99c + � � � it follows that 9 divides n if and only if 9divides (a+ b+ c+ � � � ) which is the sum of the digits of n.

That completes the Skoliad Corner for this issue. Send me your con-tests, your suggestions, and your recommendations to improve this feature.

299

THE OLYMPIAD CORNER

No. 177

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 some of the problems proposed to the jury,but not used at the 36th International Olympiad held at Toronto, Ontario. Iwelcome your novel, nice solutions that di�er from the \o�cial" publishedsolutions.

36th INTERNATIONALMATHEMATICAL OLYMPIADCanadian Problems for Consideration

by the International Jury

Algebra

1. Let a and b be non-negative integers such that ab � c2, where cis an integer. Prove that there is a number n and integers x1; x2; : : : ; xn,y1; y2; : : : yn such that

nXi=1

x2i = a;

nXi=1

y2i = b; and

nXi=1

xiyi = c:

2. Let n be an integer, n � 3. Let a1; a2; : : : ; an be real numbers,where 2 � ai � 3 for i = 1; 2; : : : ; n. If s = a1 + a2 + � � �+ an, prove that

a21 + a22 � a23a1 + a2 � a3

+a22 + a23 � a24a2 + a3 � a4

+ � � �+ a2n + a21 � a22an + a1 � a2

� 2s� 2n:

3. Let a, b and c be givenpositive real numbers. Determine all positivereal numbers x, y and z such that

x+ y + z = a+ b+ c

and4xyz � (a2x+ b2y+ c2z) = abc:

300

Geometry

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.

5. The incircle of ABC touches BC, CA and AB at D, E and Frespectively. 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.

6. An acute triangle ABC is given. Points A1 and A2 are taken onthe side BC (with A2 between A1 and C), B1 and B2 on the side AC (withB2 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.

Number Theory and Combinatorics

7. Let k be a positive integer. Prove that there are in�nitely manyperfect squares of the form n2k � 7, where n is a positive integer.

8. Let Z denote the set of all integers. Prove that, for any integers Aand B, one can �nd an integer C for which M1 = fx2 + Ax + B : x 2 Zgand M2 = f2x2 + 2x+C : x 2Zg do not intersect.

9. Find all positive integers x and y such that x + y2 + z3 = xyz,where z is the greatest common divisor of x and y.

Sequences

10. Does there exist a sequence F (1);F (2);F (3); : : : of non-negativeintegers which simultaneously satis�es the following three conditions?

(a) Each of the integers 0; 1; 2; : : : occurs in the sequence.

(b) Each positive integer occurs in the sequence in�nitely often.

(c) For any n � 2,

F (F (n163)) = F (F (n))+ F (F (361)):

11. For an integer x � 1, let p(x) be the least prime that does notdivide x, and de�ne q(x) to be the product of all primes less than p(x). Inparticular, p(1) = 2. For x having p(x) = 2, de�ne q(x) = 1. Consider thesequence x0; x1; x2; : : : de�ned by x0 = 1 and

xn+1 =xnp(xn)

q(xn)

for n � 0. Find all n such that xn = 1995.

301

12. Suppose that x1; x2; x3; : : : are positive real numbers for which

xnn =

n�1Xj=0

xjn

for n = 1; 2; 3; : : : . Prove that for all n,

2� 1

2n�1� xn < 2� 1

2n:

Next we turn to the \o�cial" results of the 37th IMOwhichwas writtenin Mumbai, India, July 10{11, 1996. My source this year was Ravi Vakil,former star Olympian and this year's Canadian Team Leader. I hope that Ihave made no serious errors in compiling the results and transcribing names.

This year a total of 426 students from 75 countries took part. Thisis somewhat up from last year. Sixty-�ve countries sent teams of six (thenumber invited to participate in recent years). But there were ten teams ofsmaller size, two of �ve members; �ve of four, and one of each of sizes three,two and one.

The contest is o�cially an individual competition and the six problemswere assigned equal weights of seven marks each (the same as the last �fteenIMOs for a maximum possible individual score of 42 and a total possible of252 for a national team of six students). For comparison see the last �fteenIMO reports in [1981: 220], [1982: 223], [1983: 205], [1984: 249], [1985:202], [1986: 169], [1987: 207], [1988: 193], [1989: 193], [1990: 193],[1991: 257], [1992: 263], [1993: 256], [1994: 243], and [1995: 267].

There was only one perfect score. The jury awarded �rst prize (Gold) tothe thirty-�ve students who scored 28 or more. Second (Silver) prizes wentto the sixty-six students with scores from 20 to 27, and third (Bronze) prizeswent to the ninety-nine students with scores from 12 to 19. Any studentwho did not receive a medal, but who scored full marks on at least one prob-lem, was awarded honourable mention. This year there were twenty-onehonourable mentions awarded. The median score on the examination was11.

Congratulations to the Gold Medalists.

Name Country Score

Ciprian Manolescu Romania 42

Yu Seek Kong Korea 39

Sug Woo Shin Korea 38

Nikolai Dourov Russia 37

Alexander Harry Saltman U.S.A. 37

Ngo Dac Tuan Vietnam 37

302

Name Country Score

Peter Burcsi Hungary 36

Sachiko Nakajima Japan 36

Serguei Norine Russia 36

Christopher C. Chang U.S.A. 36

Arend Boyer Germany 35

Peter Frenkel Hungary 35

Juliy Sannikov Ukraine 34

Ivan Ivanov Bulgaria 33

David Chkhaidze Georgia 33

Constantin Chiscanu Romania 33

David William Bibby United Kingdom 33

Ngo Duc Duy Vietnam 33

Serguei Chikh Belarus 32

Chen Huayi China 32

Gunther Vogel Germany 31

Carl Bosley U.S.A. 31

Michael Korn U.S.A. 31

Nguyen Thai Ha Vietnam 31

He Xuhua China 30

Yan Jun China 30

Bertram Felgenhauer Germany 30

Gyula Pap Hungary 30

Ajay C. Ramdoss India 30

Lev Buhovski Israel 30

Adrian Dumitru Corouneanu Romania 30

Senkodan Thevendran Singapore 29

Eaman Eftekhary Iran 28

Stefan Radu Niculescu Romania 28

Michael Comyn Ching United Kingdom 28

Next we give the problems from this year's IMO Competition. Solu-tions to these problems, along with those of the 1995 USA MathematicalOlympiad will appear in a booklet entitled Mathematical Olympiads 1996which may be obtained for a small charge from: Dr. W.E. Mientka, ExecutiveDirector, MAA Committee on H.S. Contests, 917 Oldfather Hall, Universityof Nebraska, Lincoln, Nebraska, 68588, USA.

37th INTERNATIONALMATHEMATICAL OLYMPIADJuly 10{11, 1996 (Mumbai, India)

First Day | Time: 4.5 hours

1. Let ABCD be a rectangular board with jABj = 20, jBCj = 12.The board is divided into 20 � 12 unit squares. Let r be a given positiveinteger. A coin can be moved from one square to another if and only if thedistance between the centres of the two squares is

pr. The task is to �nd a

303

sequence of moves taking the coin from the square which has A as a vertexto the square which has B as a vertex.

(a) Show that the task cannot be done if r is divisible by 2 or 3.

(b) Prove that the task can be done if r = 73.

(c) Can the task be done when r = 97?

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.

3. Let S = f0; 1; 2; 3; : : : g be the set of non-negative integers. Findall functions f de�ned on S and taking their values in S such that

f(m+ f(n)) = f(f(m))+ f(n) for all m;n 2 S:

Second Day | Time: 4.5 hours

4. The positive integers a and b are such that the numbers 15a+ 16band 16a� 15b are both squares of positive integers. Find the least possiblevalue that can be taken by the minimum of these two squares.

5. LetABCDEF be a convex hexagon such thatAB is parallel toED,BC is parallel to FE andCD is parallel to AF . Let RA, RC , RE denote thecircumradii of triangles FAB, BCD, DEF respectively, and let p denotethe perimeter of the hexagon. Prove that

RA +RC + RE �p

2:

6. Let n, p, q be positive integers with n > p+ q. Let x0; x1; : : : ; xnbe integers satisfying the following conditions:

(a) x0 = xn = 0;

(b) for each integer i with 1 � i � n,

either xi � xi�1 = p or xi � xi�1 = �q:

Show that there exists a pair (i; j) of indices with i < j and (i; j) 6= (0; n)such that xi = xj .

Although the IMO is o�cially an individual event, the compilation ofteam scores is uno�cial, if inevitable. These totals and the prize awards aregiven in the following table.

304

Rank Country Score Gold Silver Bronze Total

1. Romania 187 4 2 | 6

2. U.S.A. 185 4 2 | 6

3. Hungary 167 3 2 1 6

4. Russia 162 2 3 1 6

5. United Kingdom 161 2 4 | 6

6. China 160 3 2 1 6

7. Vietnam 155 3 1 1 5

8. Korea 151 2 3 | 5

9. Iran 143 1 4 1 6

10. Germany 137 3 1 1 5

11.{12. Bulgaria 136 1 4 1 6

11.{12. Japan 136 1 4 | 5

13. Poland 122 | 3 3 6

14. India 118 1 3 1 5

15. Israel 114 1 2 2 5

16. Canada 111 | 3 3 6

17. Slovakia 108 | 2 4 6

18. Ukraine 105 1 | 5 6

19. Turkey 104 | 2 3 5

20. Taipei 100 | 2 3 5

21. Belarus 99 1 1 2 4

22. Greece 95 | 1 5 6

23. Australia 93 | 2 3 5

24. Yugoslavia 87 | 1 2 3

25.{26. Italy 86 | 2 2 4

25.{26. Singapore 86 1 | 3 4

27. Hong Kong 84 | 1 4 5

28. Czech Republic 83 | 2 1 3

29. Argentina 80 | 1 3 4

30. Georgia 78 1 | 2 3

31. Belgium 75 | | 4 4

32. Lithuania 68 | 1 2 3

33. Latvia 66 | | 3 3

34.{35. Armenia 63 | | 1 1

34.{35. Croatia 63 | 1 1 2

36. France 61 | 2 | 2

37.{38. New Zealand 60 | | 3 3

37.{38. Norway 60 | | 3 3

39.{40. Colombia 58 | 1 | 1

39.{40. Finland 58 | | 2 2

41. Sweden 57 | 1 1 2

42. Moldava (Team of 5) 55 | | 2 2

43. Austria 54 | 1 | 1

44. Republic of South Africa 50 | | 2 2

45.{46. Mongolia 49 | | 2 2

45.{46. Slovenia 49 | | 2 2

47. Thailand 47 | | 1 1

48.{51. Denmark 44 | | 2 2

48.{51. Macao 44 | | 1 1

48.{51. Former Yugoslav 44 | | 2 2

Republic of Macedonia

48.{51. Spain 44 | | | |

52. Brazil 36 | | | |

53.{54. Mexico 34 | | | |

53.{54. Sri Lanka 34 | | 1 1

305

Rank Country Score Gold Silver Bronze Total

55. Estonia 33 | | | |

56. Iceland 31 | | 1 1

57. Bosnia-Herzegovina (Team of 4) 30 | | 1 1

58. Azerbaijan 27 | | | |

59. The Netherlands 26 | | | |

60. Trinidad & Tobago 25 | | | |

61. Ireland 24 | | | |

62. Switzerland (Team of 4) 23 | | 1 1

63. Portugal 21 | | | |

64. Kazakhstan 20 | | | |

65. Morocco 19 | | 1 1

66. Cuba (Team of 1) 16 | | 1 1

67.{68. Albania (Team of 4) 15 | | | |

67.{68. Kyrgyztan 15 | | | |

69. Cyprus (Team of 5) 14 | | | |

70. Indonesia 11 | | | |

71. Chile (Team of 2) 10 | | | |

72.{73. Malaysia (Team of 4) 9 | | | |

72.{73. Turkmenistan (Team of 4) 9 | | | |

74. Philippines 8 | | | |

75. Kuwait (Team of 3) 1 | | | |

This year the Canadian Team rose to 16th place from 19th last year and24th the previous year. The Team members were:

Richard Hoshino 22 Silver

Derek Kisman 22 Silver

Saroosh Yazdani 22 Silver

Byung Kyu Chun 18 Bronze

Adrian Chan 14 Bronze

Sabin Cautis 13 Bronze

This is the �rst time every member of the Canadian Team was awardeda medal.

The Team Leader was Ravi Vakil, a former Canadian Team Memberand now a graduate student; and P.J. Grossman was Deputy Leader, also aformer Olympian and current graduate student.

The Romanian Team placed �rst this year. Its members were:

Ciprian Manolescu 42 Gold

Constantin Chiscanu 33 Gold

Adrian Dumitru Corouneanu 30 Gold

Stefan Rodu Niculescu 28 Gold

Dragos Ghioca 27 Silver

Nicolae Dragos Oprea 27 Silver

Congratulations to the Romanian Team!!

306

To �nish this number of the Corner we discuss solutions sent in bythe readers to problems of the 6th Korean Mathematical Olympiad, whichwe gave in the February 1995 number (6th Korean Mathematical Olympiad,Final Round, April 17{18, 1993, [1995: 45{46]).

1. Let there by a 9�9 array of white squares. Find the largest positiveinteger n satisfying the following property: There always remains either a1 � 4 or a 4� 1 array of white squares no matter how you choose n out of81 white squares and colour them black.

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany;and by Solomon Golomb, University of Southern California. We giveGolomb's solution and comments.

The largest positive integer n such that, no matter which n squares ofa 9� 9 board are blocked, it will still be possible to place a 1� 4 rectangle[\straight tetromino"] on the board, either horizontally or vertically, is 19.We prove the following clearly equivalent proposition: \The smallest numberof squares whichmust be blocked on a 9�9 board so that a straight tetrominowill no longer �t is 20."

First here is a con�guration with 20 blocking squares such that no straighttetromino will �t:

X

X

X

X

X

X

X

X

X

X

X

X

X

XX

X

X

X

X

X

Clearly, a blocking con�guration must involve at least two excludedsquares in each row and in each column. (One excluded square cannot blocka row or column of length nine.) This already shows that 2 � 9 = 18 ex-cluded squares are required, to block all rows. But consider the two (ormore) excluded squares in the bottom row. Their columns each require twoadditional excluded squares to block those columns. So all the columns haveat least two excluded squares, but at least two columns must have at leastthree excluded squares, for a total of at least 7 � 2 + 2 � 3 = 20 excludedsquares.

307

Comment. Methods for this type of problem are described in Chap-ter 3: \Where PentominoesWill Not Fit," in my book Polyominoes. (Originaledition, Charles Scribner's Sons, 1965; expanded, revised edition, PrincetonUniversity Press, 1994.)

2. Let ABC be a triangle with BC = a, CA = b, AB = c. Find thepoint P for which

a � AP 2+ b � BP 2

+ c � CP 2

is minimal, and �nd the minimum.

Solutions by �Sefket Arslanagi�c, Berlin, Germany; and by Panos E.Tsaoussoglou, Athens, Greece. We give the solution of Arslanagi�c.

We have (e.g. pages 278 and 280 of [1]):Let P be a point in the plane of a triangle ABC andM be an arbitrary

point in space. Then�!

MP= (Pxi

�!

MA)=(Pxi), where x1, x2, x3 are real

numbers, and summations are taken cyclically. The following generalizationof the well-known Leibnitz identity is valid:

�Xxi

�2MP

2=�X

xi

�XxiMA

2 �X

a2x2x3:

For P = I, (where I is the incentre of�ABC), we get (because we can takePxi = 2s,

Pa2xixj = abc � 2s)

XaMA

2= 2sMI

2+ abc;

and as a consequence we have the following inequality,XaMA

2 � abc;

and equality holds only for M = I.From this take M = P ,

aAP2+ bBP

2+ cCP

2 � abc

andmin(aAP

2+ bBP

2+ cCP

2) = abc

with P = I.

Reference

[1] D.S. Mitrinovi�c, Y.E. Pe�cari�c and V. Volenec, Recent Advances in Geo-metric Inequalities.

3. Find the smallest positive integer x for which

7x25 � 10

83

is an integer.

308

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany; andby Stewart Metchette, Gardena, California, USA. We give the solution ofBoase.

We want 7x25 � 10 mod 83, or 7x25 = 83a+ 10.So 83a+ 10 � 0 mod 7 whence a � 3 mod 7 and x25 � 37 mod 83.Now 83 is prime and congruent to 3 mod 5, so there is only one solution tothis congruence mod83.

(x5)5 � 37 mod 83

x5 � 16 mod 83

x � 69 mod 83:

The smallest such x is 69.

4. An integer is called a Pythagorean number if it is the area of aright triangle whose sides are of integral lengths, say x; y; z 2 N such thatx2 + y2 = z2. Prove that for each positive integer n (n > 12), there existsa Pythagorean number between n and 2n.

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; and by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Ger-many. We give the solution of Engelhaupt.

The smallest Pythagorean numbers are

x 3 6 5 9 8y 4 8 12 12 15

z 5 10 13 15 17Pythagorean Number 6 24 30 54 60

Thus for n = 13 to 23 we have 24 in the interval [n; 2n],for n = 24 to 29 we have 30 in the interval [n; 2n], andfor n = 30 to 53 we have 54 in the interval [n; 2n].Now the numbers 6k2, with k 2 N are Pythagorean numbers (x = 3k,

y = 4k). Observe that6(k+ 1)2

6k2< 2 if 2k+ 1 < k2, or k > 2.

Henceforth, k � 3.For n = 6k2 to 6(k+ 1)2 � 1 we have 6(k+ 1)2 lies in the interval [n; 2n],for example

for n = 54 to 95, 96 lies in [n; 2n].for n = 96 to 149, 150 lies in [n; 2n];for n = 150 to 215, 216 lies in [n; 2n];

and so on.

5. Let n be a given natural number. Find all the continuous functionsf(x) satisfying: n

0

!f(x)+

n

1

!f�x2�+

n

2

!f�x22�+ � � �+

n

n� 1

!f�x2n�1

�+

n

n

!f�x2n��

309

Solution by the editor.We prove by induction that f(x) � 0 if f is a continuous function satis-

fying the condition (*). With n = 0 there is nothing to prove as the conditionbecomes f(x) = 0 for all x. For n = 1, assume that f(x) + f(x2) = 0

for all x with f continuous. Now f(�x) = �f(x2) = f(x) so f is an evenfunction and it su�ces to prove the result for x � 0. From f(0) = �f(0)and f(1) = �f(1) we get f(0) = 0 = f(1). So consider a �xed valueof x > 0. Now limk!1 x1=k = 1 from which we obtain limm!1 x1=2

m

= 0. It is easy to see that f(x1=2m

) = (�1)mf(x) for m � 0. However,limm!1 f(x1=2

m

) = f(1) = 0. From this it follows that f(x) = 0, estab-lishing the result for n = 1.

Now assume f is a continuous function satisfying

n+1Xj=0

�n+ 1

j

�f(x2

j

) = 0 for all x:

Set

g(x) =

nXl=0

�n

l

�f(x2

l

):

Now

g(x) + g(x2) =

nXl=0

�n

l

�f(x2

l

) +

nXl=0

�n

l

�f((x2)2

l

)

=

nXl=0

�n

l

�f(x2

l

) +

nXl=0

�n

l

�f(x2

l+1

)

=

n+1Xj=0

�n+ 1

j

�f(x2

j

) = 0

since�n+1

j

�=�n

j

�+�n

j�1

�.

But then g(x) is a continuous function satisfying g(x) + g(x2) = 0 forall x. By the case n = 1, g(x) = 0 for all x, i.e.

g(x) =

nXl=0

�n

l

�f(x2

l

) = 0 for all x:

By the induction hypothesis f(x) = 0 for all x and the induction is complete.

6. Let ABC be a triangle with BC = a, CA = b and AB = c. Let Dbe the mid-point of the sideBC, and letE be the point on BC for which theline segment AE is the bisector of angleA. Let the circle passing throughA,D, E intersect with the sides CA, AB at F , G respectively. Finally let Hbe the point on AB for which BG = GH, i.e. BH = 2BG. Prove that thetriangles EBH and ABC are similar and then �nd the ratio �EBH

�ABCof these

areas.

310

Solution by Mansur Boase, St. Paul's School, London, England.

q

q

q

q q

B C

A

D E

FG

H

Now \EAF = \EGF = \FDE and \GAD = \GFD = \GED.Thus �GED � �ABD since two angles are the same. Thus

BE

AB=

BG

BD=

GE

ADand

BE

AB=

BH

BC=HE

AC:

Thus �BHE � �ABC.The ratio �BHE

�ABC=�BEC

�2. By the sine rule

BE

sin A2

=AE

sinB

a� BE

sin A2

=AE

sinC:

Therefore BE sinB = (a� BE) sinC, and so

BE =a sinC

sinB + sinC=

a c2R

b2R

+ c2R

=ac

b+ c;

and �BE

c

�2: 1 =

�ac

c(b+ c)

�2: 1 =

a2

(b+ c)2:

That completes the solutions we have and the Olympiad Corner for thisissue. Send me your contests and nice solutions!

311

THE ACADEMY CORNER

No. 5

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 Mathematics Archives on the World Wide Web.Everyone knows that there is a plethora of information on the World

Wide Web. I was directed recently to the Mathematics Archives: Lessons,

Tutorials and Lecture Notes at

http://archives.math.utk.edu/tutorials.html

This contains over 140 links to mathematical assistance. The whole is toolarge to mention here, but I mention a few of the links that could be of useto CRUX readers.

1. http://www.lance.colostate.edu/auth/lessons/1/

Aristotle University of Thessaloniki

Lesson on the Newton-Raphson method

2. http://www.math.uga.edu/�andrew/Binomial/index.html

Arithmetic Properties of Binomial Coe�cients

Binomial Coe�cients is an on-line dynamic survey, available on theWorld Wide Web, and accessible from various mathematical sites ande-journals. It is an edited document, in HTML, and can and will beedited as new developments arise. Content is at the discretion of theeditors. Editors are Andrew Granville and Richard Witt.

3. http://www2.ncsu.edu/math/Projects/MA141Manual/Contents.html

CALCULUS I WITH MAPLE V, North Carolina State University

4. http://www.math.ilstu.edu/�day/305S96.html

Combinatorics: Topics for K-8 Teachers

Information and materials from a course taught by Roger Day at IllinoisState University.

5. ftp://ftp.utirc.utoronto.ca/pub/ednet/maths/model_html/ExpMath.html

Experiments in Mathematics Using Maple

A Computer Lab Book For High Schools and Home Computers by C. T.J. Dodson and E. A. Gonzalez, University of Toronto.

312

6. http://millbrook.lib.rmit.edu.au/fractals/exploring.html

Exploring Chaos and Fractals

Exploring Chaos and Fractals is an electronic textbook which includesfull text, work sheets, sound, video and animation. Parts of the mate-rial have been placed on a Web server as an experiment in electronicpublishing of hypertext based material.

7. http://math.holycross.edu/�davids/fibonacci/fibonacci.html

The Fibonacci Numbers

8. http://william-king.www.drexel.edu/top/class/game-toc.html

Game Theory: An Introductory Sketch

9. http://www.geom.umn.edu/docs/education/institute91/

Geometry and the Imagination

Notes and handouts for an innovative geometry course developed atPrinceton and the Geometry Center by John Conway, Peter Doyle, JaneGilman and Bill Thurston.

10. http://www.utm.edu:80/departments/math/graph/

Graph Theory : Tutorials

This is the home page for a series of short interactive tutorials by ChrisK. Caldwell introducing the basic concepts of graph theory. They aredesigned with the needs of future high school teachers in mind.

11. http://www.intergalact.com/threedoor/threedoor.html

The Three Door Puzzle

12. http://www.wam.umd.edu/�krc/numtheory/prime.html

Prime Numbers

These pages were written by Kevin Coombes as an experiment to aidstudents to better understand proofs of theorems. From an email mes-sage: \It seems to me that one of the main di�culties that students of(higher) mathematics encounter stems from the insistence on getting allthe logical prerequisites set up before trying to explain anything inter-esting. (Of course, this style of presentation goes back at least as far asEuclid.)"

13. http://www.math.ilstu.edu/�day/326S96.html

Technology Tools for Secondary School Mathematics

Materials developed by Roger Day at Illinois State University.

Acknowledgment: Many of the descriptions were taken from and/or modi�edfrom the documents from the listed site.

313

BOOK REVIEWS

Edited by ANDY LIU

Five Hundred Mathematical Challenges

by Edward J. Barbeau, Murray S. Klamkin and William O.J. Moser.Published by The Mathematical Association of America.softcover, 227+ pages, ISBN # 0-88385-619-4, US $ 29.50.Reviewed byMarcin E. Kuczma, University of Warsaw.

The authors' names alone are a su�cient recommendation of this book,which is a must for your book-shelf | unless you have already the previouslypublished collection of booklets, some 20 years ago, by the same authors,containing the same problems. Actually, even in that case, it is still worthhaving the new book, since it is a revised and expanded version. Solutionsare presented with greater care and elegance. Multiple solutions are oftensupplied, with extensions and remarks providing profound insight into thenature of the problems. There are also new bibliographical references.

I could have stopped the review right here. However, I should perhapssay some more about the book than just express my enchantment with it.

The title is a little bit misleading. Why so? The word \Challenges"calls for re ection. The readers may be relieved to learn that some of theproblems presented are no more di�cult than very simple \puzzlers" oftenfound in non-mathematical journals. Here is an example.

Problem 471Ma and Pa and brother and me, The sum of our ages is eighty-three, Six times Pa's is seven times Ma's, And Ma's is three times me|nofractions please.

Another frequent type is:

Without using a calculator tell which one of the two numbers (expressed interms of horrifying root combinations) is greater.

I am sure the readers will welcome with much pleasure such exercises,which provide healthy entertainment. They bring a refreshing e�ect as theyappear, from time to time, amidst heavier artillery: the more or less typicalcompetition problems.

As mentioned, the book arose from a series of earlier publications. Itmust be kept in mind that in those years, publications of that type were rare,at least in America. Problem solving competitions had not yet expanded intoa \branch of industry" as they are today. Students were not acquainted withany o�-curricular techniques. That explains the presence of many problemswhich are immediately reducible to direct applications of, say, Jensen's In-equality, Menelaus' Theorem, the Factor Theorem, the Pigeonhole Princi-ple, congruence considerations, and so on. Nowadays, those techniques aretaught at every training course. Any student trying to tackle the olympiad

314

or other serious mathematics competitions must be familiar with them, andthis takes the challenge somewhat away.

A few problems can be quite \challenging" to a high school student,while being standard at the �rst-year college level. Examples are one-sidedor two-sided estimates of the partial sums of some typical series or of ex-pressions involving factorials and binomial coe�cients; rate of growth: poly-nomial versus logarithmic or exponential; and maximization of functions bymultivariable calculus methods. I would hesitate to call any one of thosequestions a \challenge".

What has been said so far should by nomeans be construed as criticism.Even if we agree that some trick has become \standard", this fact is notnecessarily re ected in the standard problem literature. The book underreview is a treasury of such training material.

About a quarter of the book consists of problems at the olympiad level|true challenges. I must confess to having real di�culties with many of them,such as the following. Problem 432 Show that �ve or more great circles ona sphere, no three of which are concurrent, determine at least one sphericalpolygon having 5 or more sides.

With some of the problems I had no trouble only because they werewell-known to me. Again, this is no criticism! Problems do circulate in theliterature. I am quite sure that several problems were actually born and �rstpublished many years ago, in one of those booklets by the authors. Theymay have since been repeatedly reused in other publications and contests, toreappear in the present edition.

Several problems have been reprinted, with reference given, from otherjournals. Special mention must be made about those taken from the issuesofMath. Gazette in the previous century(!)|a precious gift to every problemcollectioner.

The authors write in the Preface: \ : : : we make no claim for the origi-nality of most of the problems; we acknowledge our debt to the unsung cre-ators : : : ". Toomodest! Many of the problems have without any doubt beencreated by the authors of the book. The majority of those truly challengingand beautiful problems come from these domains: algebra of polynomials, inone or more variables; sophisticated inequalities, often involving symmetricforms; and smart combinatorial reasonings. Each one of the authors has lefthis own imprint.

The correlation between \challenging" and \beautiful" is well-knownto every problemist; nice problems are usually the di�cult ones. Now, it isa pleasant feature of this book that it also contains relatively many problemsof medium or even less thanmedium di�culty, which nevertheless have morethan average charm, sometimes revealed only after arriving at the solution.Here are some examples.

315

Problem 28

A boy lives in each of n houses on a straight line. At what point should then boys meet so that the sum of the distances that they walk from their housesis as small as possible?(Two solutions are provided, contrasting a common sense argument againstan analytic one.)

Problem 292

Let f(x) be a non-decreasing function of a real variable. Let c be any realnumber. Solve the equation x = c� f(x+ f(c)).

Problem 327

Let three concentric circles be given such that the radius of the largest is lessthan the sum of the radii of the two smaller. Construct an equilateral trianglewhose vertices lie one on each circle.

Problem 445

Prove that if the top 26 cards of an ordinary shu�ed deck contain more redcards than there are black cards in the bottom 26, then there are in the deckat least three consecutive cards of the same colour.

This sample re ects my taste; another reader may point out other prob-lems she/he considers particularly nice. While it is impossible that all 500problems should be equally appealing to everybody, it is rather certain thateverybody shall �nd in this book enough items matching her/his taste, needsand \sense of pleasure" accurately.

With a certain dose of pride, I can say that I was able to solve mostof the problems I tried. And with a certain dose of shame I must say that,in most cases, I failed to �nd solutions as elegant as those included in thebook. For many problems, di�erent approaches are shown. The neatnessof presentation must be emphasised. The solutions are clear and concise.Occasionally, a rigorous argument is preceded by a heuristic one.

The solution to Problem 327 (see above) ends with a \rider": Givenequilateral triangle ABC such that PA = 3, PB = 4, PC = 5 for an inte-rior point P , �nd AB. Anyone who solved the problem (or has read andunderstood the solution) is invited to tackle this new one, employing thesimilar lines of reasoning, and yet revealing some new and unexpected fea-tures. Such \riders" follow the solutions to many problems. Here is anotherexample.

Problem 380

Prove that the function f(x; y) = 12(x+ y)(x+ y+ 1) + x is a one-to-one

map from the set of lattice points (x; y)with x; y � 0, other than (0; 0), ontothe set N of positive integers.

Too easy? or well-known? Try the \rider", then: For each n 2 N es-tablish the existence of a polynomialFn(x1; : : : ; xn) such thatFn : N

n ! Nis a bijection.

316

Problem 447

Ifm and n are positive integers, show that m�1=n + n�1=m > 1:

Known? Perhaps, but consider the intriguing \rider" which I very muchrecommend to every reader: Can you give a non-calculus (!) proof thatxy + yx > 1

for x; y > 0 ?

These \riders" show extensions (sometimes far-reaching), teach analo-gies, and provide further training to learn the techniques from the given so-lutions. Their value cannot be over-stated.

Five hundred is a huge number (in fact, there are altogether some sixhundred problems, since the \riders" might be counted separately). In sucha voluminous text, small mistakes are unavoidable. The solution to Problem

177 (a functional equation) misses certain discontinuous functions. Accord-ing to the statement of Problem 365, the father's age has to be a perfectsquare and a prime number, at the same time. A mistake in sign occurs inthe statement of the Inclusion-Exclusion Principle. Minor typos are not en-tirely absent, but they are rare enough not to a�ect the impression that thebook has been prepared and edited with great care.

The book ends with a Tool Chest section. It is a handy help for thereaders. For authors of problem collections, this is a model example of howsuch a section ought to be written.

In conclusion, I repeat that this book is a must.

Rider:

To buy or not to buy?

(Answer on page ??.)

317

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 that 1 May 1997. They may also besent by email to cruxeditor@cms.math.ca. (It would be appreciated if emailproposals and solutions were written in LATEX). Graphics �les should be inepic format, or plain postscript. Solutions received after the above date willalso be considered if there is su�cient time before the date of publication.

Correction to 2137

2137. [1996: 124] Proposed by Aram A. Yagubyants, Rostov naDonu, Russia.

Three circles of (equal) radius t pass through a point T , and are eachinside triangle ABC and tangent to two of its sides. Prove that:

(i) t =rR

R+ 2,

[NB: r instead of 2]

(ii) T lies on the line segment joining the cen-tres of the circumcircle and the incircle of�ABC.

2177. Proposed by Toshio Seimiya, Kawasaki, Japan.ABCD is a convex quadrilateral, with P the intersection of its diag-

onals and M the mid-point of AD. MP meets BC at E. Suppose thatBE : EC = (AB)2 : (CD)2. Characterize quadrilateral ABCD.

318

2178. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.

If A;B;C are the angles of a triangle, prove that

sinA sinB sinC � 8�sin3A cosB cosC + sin3B cosC cosA

+sin3C cosA cosB�

� 3p3�cos2A+ cos2B + cos2C

�:

2179. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,Austria.

For real numbers a � �1, we consider the sequenceF (a) := f(1 + 1

n)pn(n+a); n � 1g.

Determine the sets D, respectively I, of all a, such that F (a) strictly de-creases, respectively increases.

2180. Proposed by Juan-Bosco Romero M�arquez, Universidad deValladolid, Valladolid, Spain.

Prove that if a > 0; x > y > z > 0; n � 0 (natural), then

1. ax(yz)n(y� z) + ay(xz)n(z � x) + az(xy)n(x� y) � 0;

2. ax coshx(y � z) + ay cosh y(z� x) + az cosh z(x� y) � 0.

2181. Proposed by �Sefket Arslanagi�c, Berlin, Germany.Prove that the product of eight consecutive positive integers cannot be

the fourth power of any positive integer.

2182. Proposed by Robert Geretschl�ager, Bundesrealgymnasium,Graz, Austria.

Many CRUX readers are familiar with the card game \Crazy Eights", of whichthere are many variations. We de�ne the game of \Solo Crazy Eights" in thefollowing manner:

We are given a standard deck of 52 cards, and are dealt k of these atrandom, 1 � k � 52. We then attempt to arrange these k cards according tothree rules:

1. Any card can be chosen as the �rst card of a sequence;

2. A card can be succeeded by any card of the same suit, or the same num-ber, or by any eight;

3. Anytime in the sequence that an eight appears, any suit can be \called",and the succeeding card must be either of the called suit, or anothereight. (This means that, in e�ect, any card can follow an eight).

The game is won if all dealt cards can be ordered into a sequence accordingto rules 1{3. If no such sequence is possible, the game is lost.

What is the largest value of k for which it is possible to lose the game?

319

2183. Proposed by V�aclav Kone �cn �y, Ferris State University, Big Rapids,Michigan, USA.

Suppose that A, B, C are the angles of a triangle and that k, l, m � 1.Show that:

0 < sinkA sinlB sinm B

� kkllmmSS2

�(Sk2 + P )�

k2

��(Sl2 + P )�

l2

��(Sm2 + P )�

m2

�;

where S = k + l +m and P = klm.

2184. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Madrid, Spain.

Let n be a positive integer and let an denote the sum

bn=2cXk=0

(�1)k�n� k

k

�:

Prove that the sequence fan : n � 0g is periodic.

2185. Proposed by Bill Sands, University of Calgary, Calgary, Al-berta.

Notice that

22 + 42 + 62 + 82 + 102 = 4 � 5 + 5 � 6 + 6 � 7 + 7 � 8 + 8 � 9;

that is, the sum of the �rst n (in this case 5) even positive squares is equalto the sum of some n consecutive products of consecutive pairs of positiveintegers.

Find another value of n for which this happens.

(NOTE: this problem was suggested by a �nal exam that I marked recently.)

2186. Proposed by Vedula N. Murty, Andhra University, Visakhap-atnam, India.

Let a, b, c respectively denote the lengths of the sides BC, CA, AB oftriangle ABC. Let G denote the centroid, let I denote the incentre, let Rdenote the circumradius, r denote the inradius, and let s denote the semi-perimeter.

Prove that

GI2 =1

9(a+ b+ c)

�(a� b)(a� c)(b+ c� a)

+ (b� c)(b� a)(c+ a� b) + (c� a)(c� b)(a+ b� c)�:

320

Deduce the (known) result

GI2 =1

9

�s2 + 5r2 � 16Rr

�:

2187. Proposed by Syd Bulman-Fleming and Edward T.H. Wang,Wil-frid Laurier University, Waterloo, Ontario.

It is easy to show that the maximum number of bishops that can be placedon an 8� 8 chessboard, so that no two of them attack each other, is 14.

(a) Prove or disprove that in any con�guration of 14 non-attacking bishops,all the bishops must be on the boundary of the board.

(b) Describe all of the con�gurations with 14 non-attacking bishops.

2188. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

Suppose that a, b, c are the sides of a triangle with semi-perimeter s andarea �. Prove that

1

a+

1

b+

1

c<

s

�:

DID YOU KNOW� � �

| that the nine consecutive integers

2; 3; 4; 5; 6; 7; 8; 9; 10;

partitioned into three groups of three as shown, can be permuted inside eachgroup to form three squares

324 = 182; 576 = 242; 1089 = 332 ?

Is there another sequence of nine consecutive positive integers with thisproperty?

| K. R. S. Sastry

321

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

1940. [1994: 108; 1995: 107; 1995: 206]Proposedby Ji Chen, NingboUniversity, China.

Show that if x; y; z > 0,

(xy + yz+ zx)

�1

(x+ y)2+

1

(y+ z)2+

1

(z + x)2

�� 9

4:

In CRUX [1995: 206], the editor asked `if anyone �nds a \nice" so-lution to this problem [CRUX [1994:108; 1995: 107]], the editor would beinterested to see it.'

Comment by Vedula N. Murty, Andhra University, Visakhapatnam,India. Without loss of generality, assume that 0 < x � y � z, and leta = x+y

2, b = x+z

2, and c = y+z

2. Then we have

0 < a � b � c; (1)

and the inequality proposed in CRUX [1994: 108] is equivalent to

(2bc+ 2ca+ 2ab� a2 � b2 � c2)�1a2

+ 1b2

+ 1c2

�� 9: (2)

We now proceed to establish (2). For this, we note that a, b and c representthe side lengths of a triangle. The left side of (2) is identical to

9 + (b� c)2�2bc� 1

a2

�+ (c� a)2

�2ca� 1

b2

�+ (a� b)2

�2ab� 1

c2

�� 0:

Hence, (2) is established by proving

bc(b� c)2(2a2 � bc) + ca(c� a)2(2b2� ca)

+ ab(a� b)2(2c2 � ab) � 0: (3)

To prove (3), we note that (1) implies that either

0 � 2a2 � bc � 2b2 � ca � 2c2 � ab (4)

or

2a2 � bc < 0 < 2b2 � ca � 2c2 � ab: (5)

If (4) holds, then (3) is true, and we are done.

If (5) holds, we note that the sum of the �rst two terms of (3) is non-negative,and hence that (3) is true.

322

2068. [1995: 235] Proposed by �Sefket Arslanagi�c, Berlin, Germany.Find all real solutions of the equationp

17 + 8x� 2x2 +p4 + 12x� 3x2 = x2 � 4x+ 13:

Solution by Toby Gee, student, the John of Gaunt School, Trowbridge,England.

We have p17 + 8x� 2x2 +

p4 + 12x� 3x2

=p25� 2(x� 2)2 +

p16� 3(x� 2)2

�p25 +

p16 = 9 � 9 + (x� 2)2

= x2 � 4x+ 13

with equality throughout if and only if x� 2 = 0. Thus x = 2.

Also solved by SEUNG-JIN BANG, Seoul, Korea; FRANCISCO BELLOTROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CARL BOSLEY, student,Washburn Rural High School, Topeka, Kansas, USA; PAUL BRACKEN, Uni-versity of Waterloo; CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; SABIN CAUTIS, student, Earl Haig Secondary School, North York, On-tario; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;TIM CROSS, King Edward's School, Birmingham, England; HANSENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; F.J.FLANIGAN, San Jose State University, San Jose, California, USA; JEFFREYK. FLOYD, Newnan, Georgia, USA; ROBERT GERETSCHL �AGER, Bundes-realgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall,North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California,USA; JOEHOWARD, NewMexicoHighlandsUniversity; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; FRIENDH. KIERSTEAD JR., Cuya-hoga Falls, Ohio,USA; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; MITKO CHRISTOV KUNCHEV, Baba Tonka School of Math-ematics, Rousse, Bulgaria; SAI CHONG KWOK, San Diego, CA, USA; KEE-WAI LAU, Hong Kong; THOMAS LEONG, Staten Island, NY, USA; DAVIDE. MANES, State University of New York, Oneonta, NY, USA; BEATRIZMARGOLIS, Paris, France; J.A. MCCALLUM, Medicine Hat, Alberta;VEDULA N.MURTY, Andhra University, Visakhapatnam, India; GOTTFRIEDPERZ, Pestalozzigymnasium, Graz, Austria; BOB PRIELIPP, University ofWisconsin{Oshkosh,Wisconsin,USA; CORY PYE, student,Memorial Univer-sity, St. John's, Newfoundland; JUAN-BOSCOROMEROM�ARQUEZ, Univer-sidad de Valladolid, Valladolid, Spain; CRIST �OBAL S �ANCHEZ{RUBIO, I.B.Penyagolosa, Castell �on, Spain; K.R.S. SASTRY, Dodballapur, India; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; ASHSIH KR. SINGH, student, Kan-pur, India; DIGBY SMITH, Mount Royal College, Calgary, Alberta; PANOSE. TSAOUSSOGLOU, Athens, Greece; EDWARD T.H. WANG, Wilfrid LaurierUniversity, Waterloo, Ontario; CHRIS WILDHAGEN, Rotterdam, the Neth-erlands; SUSAN SCHWARTZ WILDSTROM, Kensington, Maryland, USA;

323

KENNETH M. WILKE, Topeka, Kansas, USA; PAUL YIU, Florida AtlanticUniversity, Boca Raton, Florida, USA; an anonymous solver; and the pro-poser.

2072. [1995: 277] Proposed by K.R.S. Sastry, Dodballapur, India.Find positive integers x, y, u, v such that

x2 + y2 = u2 and x2 � xy + y2 = v2:

(Equivalently, �nd a right-angled triangle with integral sidesx, y surroundingthe right angle and a triangle with sides x, y surrounding a 60� angle, andwith the third side an integer in both cases.)

Solution by Paul Yiu, Florida Atlantic University, Boca Raton, Florida,USA. We need only consider x and y relatively prime, in which case x2 + y2

and x2 � xy + y2 are also relatively prime. These are squares if and onlyif their product is a square. It is convenient to consider the possibility thatthe second equation admit negative integral values of x as well (in that casejxj and y surround a 120� angle instead). Writing s = x

y, we consider the

rational points of the quartic curve

t2 = (s2 + 1)(s2 � s+ 1) = s4 � s3 + 2s2 � s+ 1:

This corresponds to the elliptic curve

v2 = u3 + 2u2 � 3u� 6:

The quartic curve has an obvious rational point (0;1), which corresponds toa rational point P = (2; 2) on the elliptic curve. By standard procedures,we determine the multiples nP , 2 � n � 8. These are rational points onelliptic curves with corresponding rational points on the quartic curve. Forn = 2; 3; : : : we have rational points on the quartic curve with

s =15

8;�1768

2415; : : :

with corresponding values of x, y, u, and v given below. We separate thosewith di�erent signs of x. The two solutions corresponding to n = 2; 3 can befound in Dickson's history of the Theory of Numbers, vol.2, p. 481.

n 2 4 6 7

x 15 8109409 101477031226926255 4676030077060796052820312

y 8 10130640 422390893185635192 1382348542917116969367345

u 17 12976609 434409546986238833 4876078831979879983187537

v 13 9286489 381901401745295077 4160798170065530232858973

Corresponding to n = 3; 5; 8 we have solutions with negative x (corre-sponding to triangles with a 120� angle):

324

n 3 5 8

�x 1768 498993199440 227124445985970945806894399956799

y 2415 136318711969 654056791401866496244333771257120

u 2993 517278459169 692369699180530962038852086430401

v 3637 579309170089 792419135769606228834850391429041

Remark. It is, however, not possible to �nd two triangles with integralsides, one with sides x, y surrounding a 60� angle, another with sides x, ysurrounding a 120� angle, and with the third side an integer in both cases.This is because the quadratic forms x2 + xy + y2 and x2 � xy + y2 cannotbe simultaneously made squares. See Dickson, ibid.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany;JEFFREY FLOYD, Newnan, Georgia; ROBERT GERETSCHL �AGER, Bundes-realgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall,North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California,USA; CYRUS HSIA, student, University of Toronto, Toronto, Ontario;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; KEE-WAILAU, Hong Kong; P. PENNING, Delft, the Netherlands; HEINZ-J�URGENSEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary;DAVID R. STONE, Georgia SouthernUniversity, Statesboro, Georgia; PANOSE. TSAOUSSOGLOU, Athens, Greece; CHRIS WILDHAGEN, Rotterdam, theNetherlands; KENNETHM.WILKE, Topeka, Kansas, USA; and the proposer.There was one incorrect solution. Most solvers simply gave the solutioncorresponding to n = 2 in YIU's solution above; several found su�cientconditions �rst.

2078?. [1995: 278] Proposed by �Sefket Arslanagi�c, Berlin, Germany.Prove or disprove that

pa� 1 +

pb� 1 +

pc� 1 �

pc(ab+ 1)

for a; b; c � 1.

Solution by Theodore Chronis, student, Aristotle University of Thessa-loniki, Greece.

Let a � 1 = x2, b� 1 = y2, c� 1 = z2, where x; y; z � 0. Then theinequality to be proved becomes

x+ y+ z �p(z2 + 1)[(x2 + 1)(y2+ 1) + 1]:

From the Cauchy{Schwarz inequality,

x+ y = x � 1 + 1 � y �p(x2 + 1)(y2 + 1)

325

and

z +p(x2 + 1)(y2 + 1) �

pz2 + 1

p(x2 + 1)(y2+ 1) + 1;

and we are done.

Also solved by MANUEL BENITO MU ~NOZ and EMILIO FERN �ANDEZMORAL, Logro ~no, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bris-tol, UK; WOLFGANG GMEINER, Millstatt, Austria; RICHARD I. HESS, Ran-cho Palos Verdes, California, USA; JOE HOWARD, New Mexico HighlandsUniversity, Las Vegas, New Mexico, USA; CYRUS HSIA, student, Universityof Toronto, Toronto, Ontario; PETER HURTHIG, Columbia College, Burn-aby, BC;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; DAGJONSSON, Uppsala, Sweden; MURRAY S. KLAMKIN, University of Alberta,Edmonton, Alberta; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; KEE-WAI LAU, Hong Kong; JUAN-BOSCO ROMEROM�ARQUEZ, Universidad de Valladolid, Valladolid, Spain; HEINZ-J�URGENSEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary,Alberta; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, On-tario; and the proposer.

As several solvers pointed out, and as can be seen from the above proof,equality holds in the original inequality if and only if

a = 1 + t; b = 1 +1

t; c = 1+

1

ab

for some positive real number t.

2079. [1995: 278]Proposedby Crist �obal S �anchez{Rubio, I. B. Penya-golosa, Castell �on, Spain.

An ellipse is inscribed in a rectangle. Prove that the contact pointsof the ellipse with the sides of the rectangle lie on the rectangular hyperbolawhich passes through the foci of the ellipse andwhose asymptotes are parallelto the sides of the rectangle.

Solution by P. Penning, Delft, the Netherlands.

Let the ellipse be�xa

�2+�y

b

�2= 1 with a > b. Suppose thatm andm0

are the slopes of the sides of the rectangle, so thatmm0 = �1. By symmetry,the centre of the hyperbola will coincide with the origin. The square of onehalf of the distance between the foci is a2 � b2. Thus, the equation of thehyperbola is

(mx� y)(m0x� y) = a2 � b2:

Let x = a = cos(�), y = b sin(�) be any of the four points of contact of therectangle with the ellipse. The tangent to the ellipse at this point is

a cos(�)

a+y sin(�)

b= 1:

326

Thus we have

m =�b

a tan(�); m0 =

a tan(�)

b;

mx� y =�b cos2(�)sin(�)

� b sin(�) =�b

sin(�);

m0x� y =�(a2 � b2) sin(�)

b:

Substitution into the equation of the hyperbola shows that this point of con-tact lies on the hyperbola.

[Ed: This argument fails if the rectangle has sides parallel to the axes. OnlyKone �cn �y pointed out that in this case, the hyperbola is degenerate and is, infact, given by the axes.]

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; JORDI DOU, Barcelona, Spain; ROBERT GERETSCHL �AGER, Bundes-realgymnasium, Graz, Austria; WALTHER JANOUS, Ursulinengymnasium,Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; KEE-WAI LAU, Hong Kong; D.J. SMEENK, Zaltbommel, theNetherlands; and the proposer.

2080. [1995: 278]ProposedbyMarcin E. Kuczma, Warszawa, Poland.

Let S = f1; 2; 3; 4; 5; 6; 7g. Find the number of maps f from S to Ssuch that f2080(x) = x for every x 2 S. (Here the superscript denotesiteration: f1(x) = f(x) and fn(x) = f(fn�1(X)) for all n > 1.)

Solution by Robert Geretschl�ager, Bundesrealgymnasium, Graz, Aus-tria.

Perhaps not too surprisingly, knowing Marcin, there are 2080 suchmaps. We can see this in the following manner.

First of all, each f must be one-to-one in S, since there would other-wise exist an x 2 S such that there is no y 2 S with f(y) = x, which wouldcontradict f(f2079(x)) = f2080(x) = x. f must therefore be a permutationof the elements of S.

We know that every permutation of a �nite set can be expressed uniquelyas the product of cyclic permutations with no common elements. The prob-lem is therefore equivalent to �nding the number of such products of cycles,such that the 2080-th iteration is the identity. In order for this to be thecase, the length of each cycle must be a divisor of 2080. The lengths of suchcycles can therefore be either 1, 2, 4 or 5. We now consider three cases.

(a) There exists a cycle of length �ve.

There are�7

5

�ways to select the �ve numbers in the cycle, and 4!ways to

build a cycle of �ve given numbers. Since the remaining numbers can either

327

build a cycle of length two, or two of length one, there are�7

5

�� 4! � 2 = 1008

such maps.

(b) There exists a cycle of length four.

There are�7

4

�ways to select the four numbers in the cycle, and 3! ways

to build a cycle of four given numbers. The remaining three numbers caneither each build a cycle of length one, or there are three ways in which tochoose two to build a cycle of length two, leaving the other to build a cycleof length one. There are therefore

�7

4

�� 3! � 4 = 840 such maps.

(c) The only other cycles that exist are of lengths one and two.

There is one such map with no cycle of length two (the identity). Thereare

�7

2

�= 21 such maps with precisely one cycle of length two. There are

12��7

2

���5

2

�= 105 such maps with precisely two cycles of length two (choosing

2 from 7, then 2 from the remaining 5, and then dividing by the number ofrepetitions of the chosen cycles). Finally there are 1

3!��7

2

���5

2

���3

2

�= 105

such maps with precisely three cycles of length two (for the analogous reasonas for two such cycles).

Adding up, we have

1008 + 840 + 1 + 21 + 105 + 105 = 2080

maps with the desired property.

Also solved by HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bam-berg, Germany; F.J. FLANIGAN, San Jose State University, San Jose, Cal-ifornia, USA; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, On-tario; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, HongKong; ROBERT P. SEALY, Mount Allison University, Sackville, New Bruns-wick; HOE TECKWEE, student, Hwa Chong Junior College, Singapore; CHRISWILDHAGEN, Rotterdam, the Netherlands; and the proposer. There werealso six incorrect solutions sent in.

Sealy �nds all positive integers n such that there are exactly n mapsf : S �! S (with S = f1; : : : ; 7g) satisfying fn(x) = x for all x 2 S. Hegets:

n = 1; 351; 505;721; 1072; 1225; 1575;2080; 2800; 3312;4320; 5040:

As Sealy remarks, \You now have proposed problems 2800, 3312, 4320 and5040."! This list contains two obvious members, 1 and 5040 = 7!, and theanalogous numbers would appear in the corresponding list if we were to useS = f1; : : : ; Ng for an arbitrary positive integerN . Howmany numbers arein the list forN? What are the second{smallest and second{largest numbersin the list?

328

2081. [1995: 306] Proposed by K.R.S. Sastry, Dodballapur, India.

In base ten, if we write down the �rst double-digit integer (10) followedby the last two single-digit integers (8 and 9) we form a four-digit number(1089) which is a perfect square (332). What other bases exhibit this sameproperty?

Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.

In base b � 2, Nb has the four digits 10(b� 2)(b� 1), so

Nb = b3 + (b� 2)b+ (b� 1) = (b� 1)(b+ 1)2;

which shows thatNb is a perfect square if and only if b�1 is a perfect square.

(Editor's note: a number of solvers ignored base b = 2, where 10012 =

910 = (310)2 = (112)

2:) Benito Mu~noz, Fern~andez Moral, and Geretschl�agerall noted that if b� 1 = m2, then

pNb is the base b two-digit number with

both digits equal to m; Vella remarked that in base 10 thispNb is just the

product ofm and b+1. For example, if b = 5 = 22+1,pN5 = 225 = 1210.)

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; MANUEL BENITO MU ~NOZ and EMILIO FERN �ANDEZ MORAL, I.B.Sagasta, Logro ~no, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bris-tol, UK; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;TIM CROSS, King Edward's School, Birmingham, England; LUIS V.DIEULEFAIT, IMPA, Rio de Janeiro, Brazil; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; JEFFREY K. FLOYD, Newnan,Georgia, USA; TOBY GEE, student, the John of Gaunt School, Trowbridge,England; ROBERT GERETSCHL �AGER, Bundesrealgymnasium, Graz, Austria;SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARDI. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursul-inengymnasium, Innsbruck, Austria; KATHLEEN E. LEWIS, SUNY Oswego.Oswego, NY, USA; DAVID LINDSEY, Austin Peay State University, TN, USA;DAVID E. MANES, State University of New York, Oneonta, NY, USA; J.A.MCCALLUM, Medicine Hat, Alberta; P. PENNING, Delft, the Netherlands;GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; CORY PYE, student,Memorial University of Newfoundland, St. John's, Newfoundland; CRIST �

OBAL S �ANCHEZ{RUBIO, I.B. Penyagolosa, Castell �on, Spain; DAVID R. STONE,Georgia Southern University, Statesboro, Georgia, USA; PANOS E. TSAOUS-SOGLOU, Athens, Greece; DAVID C. VELLA, Skidmore College, SaratogaSprings, NY, USA; CHRIS WILDHAGEN, Rotterdam, the Netherlands; KEN-NETH M. WILKE, Topeka, Kansas, USA; and the proposer. There was oneincorrect solution.

2082. [1995: 306] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle with \A > 90�, and AD, BE and CF are its alti-tudes (with D on BC, etc.). Let E0 and F 0 be the feet of the perpendicularsfrom E and F to BC. Suppose that 2E0F 0 = 2AD + BC. Find \A.

329

Solution by Kee-Wai Lau, Hong Kong.As usual, let a = BC, b = CA and c = AB. We have

AD = b sinC; and E0F 0 = a(1� sin2B � sin2C);

where the latter equation comes from E0F 0 = a�CF 0�BE0, with CF 0 =BC � BF 0 = a � BF cosB = a � a cos2B = a sin2B, and, similarly,BE0 = a sin2C.

Since we are given 2E0F 0 = 2AD +BC, by the sine law, we have

2 sinA(1� sin2B � sin2C) = 2 sinB sinC + sinA:

Hence,sinA(1� 2 sin2B � 2 sin2C) = 2 sinB sinC;

and so

sinA�2 cos(B � C) cos(B + C)� 1

�= cos(B �C)� cos(B + C);

or (since cosA = � cos(B + C))

� cos(B � C)(1 + 2 sinA cosA) = cosA+ sinA;

or(cosA+ sinA)

�1 + (cosA+ sinA) cos(B � C)

�= 0:

Since \A is obtuse, we have

1 + (cosA+ sinA) cos(B � C) = 1 +p2 cos(A� 45�) cos(B � C)

> 1 +p2

��1p2

�= 0:

Thus, cosA+ sinA = 0, and therefore \A = 135�.

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; MANUELBENITO MU ~NOZ and EMILIO FERN �ANDEZ MORAL, I.B. Sagasta, Logro ~no,Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; RICHARDI. HESS, Rancho Palos Verdes, California, USA; V �ACLAV KONE �CN �Y,Ferris State University, Big Rapids, Michigan, USA; P. PENNING,Delft, the Netherlands; D.J. SMEENK, Zaltbommel, the Netherlands;PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.

2083. [1995: 306] Proposed by Stanley Rabinowitz, Westford, Mas-sachusetts.

The numerical identity cos2 14� � cos 7� cos 21� = sin2 7� is a specialcase of the more general identity cos2 2x�cosx cos 3x = sin2 x. In a similarmanner, �nd a generalization for each of the following numerical identities:

330

(a) tan55� � tan35� = 2 tan20�;

(b) tan70� = tan20� + 2 tan40� + 4tan 10�;

(c)? csc 10� � 4 sin 70� = 2.

[Ed: it was not easy to decide which submitted solution to highlight here,since the answers to the problem are not unique. We decided to give theproposer's solution to (a) and (b), and one of several submitted for (c).]

Solutions: (a) and (b) by the proposer, (c)? by several solvers.

(a) tanx� tan(90� � x) = 2 tan(2x� 90�),

(b) tan(x+ 60�) = tan(30� � x) + 2 tan(60� � 2x) + 4 tan(4x� 30�),

(c)? csc(2x)� 4 sin(15� � x) sin(75� � x) csc(2x) = 2.

[Ed: all solvers listed below solved parts (a) and (b). A ? before a solver'sname indicates that part (c) was also solved, while a y before a solver's nameindicates that a restricted case of part (c) was also solved.]

Solved by ? �SEFKET ARSLANAGI �C, Berlin, Germany; ?Manuel BenitoMu~noz and EMILIO FERN �ANDEZ MORAL, I.B. Sagasta, Logro ~no, Spain;?CARL BOSLEY, student,WashburnRural High School, Topeka, Kansas, USA;?CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; ADRIAN CHAN,student, Upper Canada College, Toronto, Ontario; ?THEODORECHRONIS, student, Aristotle University of Thessaloniki, Greece; CURTISCOOPER, Central Missouri State University, Warrensburg, Missouri, USA;?TIM CROSS,KingEdward's School,Birmingham, England; ?DAVIDDOSTER,Choate Rosemary Hall, Wallingford, Connecticut,USA; HANS ENGELHAUPT,Franz{Ludwig{Gymnasium, Bamberg, Germany; SHAWN GODIN, St. JosephScollard Hall, North Bay, Ontario; ?RICHARD I. HESS, Rancho PalosVerdes,California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-tria; yV �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,USA; yDAVID E. MANES, State University of New York, Oneonta, NY, USA;yP. PENNING, Delft, the Netherlands; ?BOB PRIELIPP, University of Wis-consin{Oshkosh, Wisconsin, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Ger-many; ?TOSHIO SEIMIYA, Kawasaki, Japan; yDIGBY SMITH, Mount RoyalCollege, Calgary, Alberta; PANOS E. TSAOUSSOGLOU, Athens, Greece; andthe proposer.

2084. [1995: 306] Proposed by Murray S. Klamkin, University ofAlberta.

Prove that

cos B2cos C

2+ cos C

2cos A

2+ cos A

2cos B

2� 1� 2 cos A

2cos B

2cos C

2;

where A, B, C are the angles of a triangle.

331

Solution by the proposer.

Let x = cos A2, y = cos B

2, z = cos C

2. The inequality is now equivalent

to

1 � 1� x

1 + x+

1� y

1 + y+

1� z

1 + z;

or

1 � tan2 A4+ tan2 B

4+ tan2 C

4:

Since tan2 x4is convex for x 2 [0; �], the latter inequality follows by the

majorization inequality. Note that

(�; 0; 0) � (A;B; C) ���3; �3; �3

�:

So there is equality only from the degenerate triangle of angles �, 0, 0. Italso follows that

tan2 A4+ tan2 B

4+ tan2 C

4� 3 tan2 �

12= 21� 12

p3:

and with equality if and only if the triangle is equilateral.

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; MANUELBENITO MU ~NOZ and EMILIO FERN �ANDEZ MORAL, I.B. Sagasta, Logro ~no,Spain; CARL BOSLEY, student,WashburnRural High School, Topeka, Kansas,USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; RICHARDI. HESS, Rancho Palos Verdes, California, USA; JOE HOWARD, New MexicoHighlands University, Las Vegas, NM, USA; WALTHER JANOUS, Ursulinen-gymnasium, Innsbruck, Austria; DAVID E. MANES, State University of NewYork, Oneonta, NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; andPANOS E. TSAOUSSOGLOU, Athens, Greece.

2085. [1995: 307] Proposed by Iliya Bluskov, student, Simon FraserUniversity, Burnaby, BC, and Gary MacGillivray, University of Victoria, B. C.

Find a closed-form expression for the n by n determinant������������������

n �1 �1 �1 � � � �1 �1 �1�1 3 �1 0 � � � 0 0 0

�1 �1 3 �1 � � � 0 0 0

�1 0 �1 3 � � � 0 0 0

.

.

....

.

.

....

. . ....

.

.

....

�1 0 0 0 � � � 3 �1 0

�1 0 0 0 � � � �1 3 �1�1 0 0 0 � � � 0 �1 3

������������������

:

Combination of solutions by Theodore Chronis, student, Aristotle Uni-versity of Thessaloniki, Greece, and Heinz-J �urgen Sei�ert, Berlin, Germany.

332

Let Dn denote the determinant under consideration. We shall provethat

Dn = F2n�1 + F2n+1 � 2 for all positive integers n;

where (Fk) is the sequence of the Fibonacci numbers de�ned by F1 = F2 = 1

and Fk = Fk�1 + Fk�2 for k � 2.Adding all rows to the �rst row and then adding all columns to the �rst

column, we get that Dn equals the n by n determinant������������������

3 1 0 0 � � � 0 0 1

1 3 �1 0 � � � 0 0 0

0 �1 3 �1 � � � 0 0 0

0 0 �1 3 � � � 0 0 0

..

....

..

....

. . ....

..

....

0 0 0 0 � � � 3 �1 0

0 0 0 0 � � � �1 3 �11 0 0 0 � � � 0 �1 3

������������������

:

By expanding this determinant in minors along the �rst row, we thereforehave

Dn = 3An�1 ��An�2 + (�1)n(�1)n�2

�+(�1)n+1

�(�1)n�2 + (�1)nAn�2

�= 3An�1 � 2An�2 � 2

for n � 3, where An is the n by n determinant����������������

3 �1 0 � � � 0 0 0

�1 3 �1 � � � 0 0 0

0 �1 3 � � � 0 0 0

..

....

..

.. . .

..

....

..

.0 0 0 � � � 3 �1 0

0 0 0 � � � �1 3 �10 0 0 � � � 0 �1 3

����������������

:

By expanding this along the �rst row, we �nd the recurrence relation

An = 3An�1 � An�2; n � 3:

Since A1 = 3 = F4 and A2 = 8 = F6, from the known identity

Fn = 3Fn�2 � Fn�4

it follows that An = F2n+2 for all n � 1. So the original determinant isequal to

Dn = 3F2n � 2F2n�2 � 2

= 2F2n + (F2n+1 � F2n�1)� 2(F2n � F2n�1)� 2

= F2n+1 + F2n�1 � 2

333

for n � 3. The cases n = 1 and n = 2 are done by direct computation[D1 = 1 = F3 + F1 � 2, D2 = 5 = F5 + F3 � 2].

Also solved by MANUEL BENITO MU ~NOZ and EMILIO FERN �ANDEZMORAL, I.B. Sagasta, Logro ~no, Spain; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; and the proposers.

Benito and Fern �andez give the answer in the form Dn = L2n � 2,where Ln is the nth Lucas number, de�ned by L1 = 1, L2 = 3, Ln =

Ln�1 + Ln�2. They also point out that the answer could be written in yetanother interesting form:

Dn =

(5F 2

n = L2n � 4; n even;

L2n = 5F 2

n � 4; n odd.

In his article \The sequence 1 5 16 45 121 320 : : : in combinatorics",in the Fibonacci Quarterly Vol. 13 (1975) pp. 51{55, Kenneth Rebman consid-ers the same sequence of integers de�ned by the determinant in this prob-lem, and gives examples involving graphs and matrices where this sequencearises. He also evaluates a determinant which can easily be shown to beequal to the �rst determinant in the above solution.

2086. Proposed by Aram A. Yagubyants, Rostov na Donu, Russia.If the side AC of the spherical triangle ABC has length 120� (that is,

it subtends an angle of 120� at the centre), prove that the median from B(that is, the arc of the great circle from B to the midpoint of AC) is bisectedby the other two medians.

Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta.

This is a special case of a known result:

If G is the concurrency point of the three medians and A0 is themidpoint of BC then sinAG

sinGA0= 2 cos a

2, etc.

So, if a = 120�, then AG = GA0. (The other possibility AG+GA0 =180� is ruled out since A and A0 would then be antipodal, forcing triangleABC to be degenerate.)

Proof. Let ~A, ~B, ~C denote unit vectors from the centre of the sphereto the vertices A, B, C, respectively. It now follows easily that

G =~A+ ~B + ~C��� ~A+ ~B + ~C

���lies on each of the medians and hence is the point of concurrency. Also,

~A0 =~B + ~C��� ~B + ~C

��� : Then

334

sinAG =

��� ~A� �~A+ ~B + ~C

������� ~A+ ~B + ~C��� =

��� ~A� �~B + ~C

������� ~A+ ~B + ~C��� ;

sinGA0 =

���� ~A+ ~B + ~C���~B + ~C

������� ~A+ ~B + ~C��� ���~B + ~C

��� =

��� ~A� �~B + ~C

������� ~A+ ~B + ~C��� ��� ~B + ~C

��� ;so that

sinAG

sinGA0=

��� ~B + ~C��� = 2 cos

a

2:

Also solved by MANUEL BENITO MU ~NOZ and EMILIO FERN �ANDEZMORAL, Logro ~no, Spain; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; P. PENNING, Delft, the Netherlands; and the proposer.

Benito and Ferna ~ndez found the general result as an exercise in a nine-teenth century text by J.A. Serret; they show it to be an easy application ofspherical trigonometry.

2088. [1995: 307] Proposed by �Sefket Arslanagi�c, Berlin, Germany.Determine all real numbers x satisfying the equation�

2x+ 1

3

�+

�4x+ 5

6

�=

3x� 1

2;

where bxc denotes the greatest integer � x.

Solution by Luis Dieulefait, IMPA, Rio de Janeiro, Brazil.LetZdenote the integers. Then, since b(2x+1)=3c+ b(4x+5)=6c =

(3x� 1)=2 must be an integer,

x =2n+ 1

3for some n 2Z: (1)

We will also use the fact that

2x+ 1

3+

1

2=

4x+ 5

6: (2)

For every real number � we have:

�� b�c <1

2()

��+

1

2

����+

1

2

�� 1

2

[that is, the fractional part of � is < 1=2 if and only if the fractional part of�+ 1=2 is � 1=2], so that

b�c > �� 1

2()

��+

1

2

�� �:

335

From this, using byc � y we obtain

b�c+��+

1

2

�� 2�; (3)

and using byc > y � 1 we obtain

b�c+��+

1

2

�> 2�� 1; (4)

both for every real number �. [Editor's remark: for example, to get (3) notethat for any � either�

�+1

2

�� � and b�c � �;

or

b�c � �� 1

2and

��+

1

2

�� �+

1

2;

and adding gives (3) in either case.] Putting inequalities (3) and (4) togetherand using (2) gives, for � = (2x+ 1)=3:

2(2x+ 1)

3� 1 <

�2x+ 1

3

�+

�4x+ 5

6

�� 2(2x+ 1)

3:

Using the original equation, this is

4x� 1

3<

3x� 1

2� 4x+ 2

3: (5)

From this we obtain 1 < x � 7. Using (1) the only possibilities are

x =5

3;7

3;9

3; : : : ;

21

3:

So this is exactly the set of all solutions.

Editorial comment. Alternatively, equations (3) and (4) imply that

b�c+��+

1

2

�= b2�c

for all �, so the original equation becomes�4x+ 2

3

�=

3x� 1

2;

which is equivalent to (5) and (1) together. Benito and Fern �andez, Wildhagenand the proposer also gave solutions along these lines.

Also solved by CLAUDIO ARCONCHER, Jundia��, Brazil; MANUELBENITO MU ~NOZ and EMILIO FERN �ANDEZ MORAL, I.B. Sagasta, Logro ~no,

336

Spain; CARL BOSLEY, student,WashburnRural High School, Topeka, Kansas,USA; ADRIAN CHAN, student, Upper Canada College, Toronto, Ontario;THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece;TIM CROSS, King Edward's School, Birmingham, England; DAVID DOSTER,Choate Rosemary Hall, Wallingford, Connecticut,USA; HANS ENGELHAUPT,Franz{Ludwig{Gymnasium, Bamberg, Germany; JEFFREY K. FLOYD, New-nan, Georgia, USA; TOBY GEE, student, the John of Gaunt School, Trow-bridge, England; ROBERT GERETSCHL �AGER, Bundesrealgymnasium, Graz,Austria; RICHARD I.HESS, Rancho PalosVerdes, California,USA; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Fer-ris State University, Big Rapids, Michigan, USA; KEE-WAI LAU, Hong Kong;DAVID E. MANES, State University of New York, Oneonta, NY, USA;P. PENNING, Delft, the Netherlands; CORY PYE, student, Memorial Uni-versity of Newfoundland, St. John's; CRIST �OBAL S �ANCHEZ{RUBIO, I.B.Penyagolosa, Castell �on, Spain; HRISTOS SARAGHIOTES, student, AristotleUniversity of Thessaloniki, Greece; ROBERT P. SEALY, Mount Allison Uni-versity, Sackville, New Brunswick; HEINZ-J�URGEN SEIFFERT, Berlin, Ger-many; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer.As well, there were ten incorrect solutions sent in, most of which assumedthat x had to be an integer.

In the May 1996 issue of CRUX [1996: 168], we asked: \Do you knowthe equation of this curve?"

In the September 1996 issue of CRUX [1996: 288], we gave the hint:\it is known as the \butter y"!"

The answer is r = ecos(�) � 2 cos(4�) + sin5(�=12).

If you have any other \nice" curves, please send them to the editor.

(Answer to Rider on page ??:)

To buy or not to buy, that is not the question. Piglet.

337

Letter from the Editors

As announced in the November issue, 1997 will see an increase in thequantity of material devoted to high school mathematics, without a decreasein any of the present composition of CRUX. We are pleased to announce thatthis will be e�ected through an amalgamation of CRUX MATHEMATICO-

RUM with Mathematical Mayhem. For those of you not familiar with it,Mayhem was started by a group of Canadian IMO Alumni, and aimed to bea \journal of high school and college level mathematics, written by and forstudents". It has run continuously for eight years, and has a circulation ofabout 130. Its previous editors have included the distinguished problemistsRavi Vakil (co-founder with Patrick Surry) and J.P. Grossman, who were theDeputy Leader and Leader of the Canadian delegation of the 1996 IMO. Thepresent editors are Naoki Sato and Cyrus Hsia, who will be joining the CRUXwith MAYHEM Editorial Board in January. Mayhem has published itemssuch as expository articles, hints of problem solving, IMO problems and so-lutions, and three levels of problems | high school, advanced and challengeboard | (set by the editors or contributors). By keeping communication aliveamongst the Canadian IMO alumni, team members and \wanabees", May-

hem has played an important role in keeping the IMO spirit alive and wellin Canada.

This amalgamation will give CRUX readers additional high school levelmaterial (if you have forgotten CRUX's mandate, please re-read the back in-side cover), as well as giving Mayhem a wider and international exposure.CRUX with MAYHEM will have 64 pages per issue and will have problemsto challenge all readers from interested high school students to senior un-dergraduate students. We hope that high school readers who will be new toCRUX withMAYHEM will �nd the range of problems useful in the classroomand stimulating.

If the issue that you habitually read is a library copy, please ensure thatyour librarian is aware of this change. Since CRUX is the older journal, andtaken by more libraries, the Mayhem editors have agreed that CRUX with

MAYHEM will continue the volume numbering of CRUX (next year will seevolume 23) as well as maintain the general external appearance. However,the table of contents will be moved to the outside back cover since it will beenlarged, and also to make space for the words Mathematical Mayhem onthe front cover.

With all best wishes,

Bruce Shawyer Graham WrightEditor-in-Chief Managing Editor

338

Lettre des r �edacteurs

Comme nous l'avons annonc �e dans le num�ero de novembre, le CRUXsera davantage ax �e sur lesmath �ematiques de niveau secondaire en 1997, sanspour autant que l'on touche �a son contenu actuel. Il nous fait plaisir de vousannoncer que nous r �eussirons ce tour de force en fusionnant le CRUXMATH-

EMATICORUM et le Mathematical Mayhem. Pour ceux d'entre vous qui neconnaitraient pas leMayhem, sachez qu'il a �et �e fond �e par d'anciens membresde l' �equipe canadienne �a l'OIM s' �etant donn �e pour mission de produire unp �eriodique sur les math �ematiques de niveau secondaire et coll �egial, con�cupar et pour des �etudiants. Ce magazine, qui parait r �eguli �erement depuishuit ans, est tir �e �a environ 130 exemplaires. Parmi ses anciens r �edacteurs,mentionnons les excellents probl �emistes Ravi Vakil (cofondateur avec PatrickSurry) et J.P. Grossman, respectivement chef d' �equipeadjoint et chef d' �equipede la d �el �egation canadienne �a l'OIM1996. Les r �edacteurs actuels, Naoki Satoet Cyrus Hsia, se joindront �a l' �equipede r �edaction du CRUX withMayhem enjanvier. LeMayhem a entre autres publi �e des articles de fond, des trucs pourfaciliter la r �esolution de probl �emes et certains probl �emes pos �es aux derni �eresOIM accompagn �es. Les probl �emes pr �esent �es sont class �es selon trois niveauxde di�cult �e, d �etermin �es par les r �edacteurs ou les collaborateurs : secondaire,avanc �e et �challenge board�. En permettant les �echanges entre les anciensparticipants canadiens �a l'OIM, les membres de l' �equipe canadienne actuelleet les aspirants �a ce titre, le Mayhem a grandement contribu �e �a pr �eserverl'esprit de l'OIM au Canada.

La fusionde ces deuxmagazines permettra aux lecteurs duCRUX d'avoiracc �es �a une plus grande quantit �e d'articles sur les math �ematiques de niveausecondaire (pour un rappel du mandat du CRUX, relire la troisi �eme de cou-verture) et procurera au Mayhem une visibilit �e accrue et internationale.Chaque num�ero du CRUX with Mayhem comptera 64 pages et pr �esenterades probl �emes qui sauront susciter l'int �eret de tous les lecteurs, des �el �eves dusecondaire aux �etudiants en �n de bac. Nous esp �erons que les lecteurs du sec-ondaire qui d �ecouvriront le CRUX withMayhem y trouveront des probl �emesstimulants et utiles pour leurs �etudes.

Si vous lisez habituellement un exemplaire provenant d'unebiblioth �eque,v �eri�ez si votre biblioth �ecaire est au courant de cette fusion. Puisque leCRUX est le plus ancien des deux magazines et le plus commun dans les bib-lioth �eques, les r �edacteurs du Mayhem ont accept �e que le CRUX with May-

hem suive la num�erotation par volumes du CRUX (l'an prochain : volumeno 23) et conserve en gros sa pr �esentation ext �erieure. On d �eplacera toutefoisla table des mati �eres en quatri �eme de couverture, puisqu'elle sera agrandieet qu'il faudra faire de la place aux mots �Mathematical Mayhem� sur lapage-titre.

Meilleurs voeux �a tous!Bruce Shawyer Graham WrightR �edacteur en chef Directeur de r �edaction

339

In memoriam | P�al Erd }os

P�al Erd }os | 1913-03-26 to 1996-09-20

In appreciation of the \Prince of Problem Posers", P �al Erd }os, who,sadly for those of us who had the privilege of knowing him, died while hewas in Warsaw, attending a mathematics meeting.

It always seemed that Erd }os was eternal; it's hard to realize that wewon't hear his delightful language again. He in uenced hundreds, probablythousands, of mathematicians, and each in an individual way. Anything onecan say must be personal.

A major part of Erd }os's genius was asking innumerable questions, and,most importantly, asking them of the right person. He seemed to know,better than you yourself, what problems you could solve. He gave the con-�dence that many of us needed to embark on research.

Erd }os was the problem proposer par excellence. Almost anyone canask questions that are impossibly di�cult or are trivially easy. To achievethe balance between these extremes is given to few of us. Erd }os's questionswere always just right. Many have remained as outstanding, but impor-tant problems, but most have been attacked and partially, if not completelysolved.

But he didn't only pose problems; he wrote more than 1500 papers.Those who didn't know him well thought that he just threw out ideas andgot others to write for him. Certainly he had a phenomenal number of co-authors, but a good fraction of his papers were solo e�orts, and, far fromdepending on his co-authors, he often wrote the paper himself and addedon the other names, even on occasions when their contributions were com-paratively minor. He opened up new areas of research in number theory,combinatorics and geometry. Notable are probabilistic number theory withR �enyi & Tur �an and the partition calculus with Rado & Hajnal.

340

Perhaps `bosses', `slaves', `captured', `epsilons', `noise', `poison', `theSF', `The Book', `the Sam & Joe show', `vot vos this ven it vos alive?', `isyour brain open?' will remain in the vocabulary of a few of his many friends,but the words won't sound the same now that it's not Erd }os saying them.

R.K. Guy

For the information of those readers who may not know who P �al Erd }oswas, here is some further information.

Erd }os was highly regarded in the mathematical community as one ofthe most brilliant and eccentric mathematicians of the 20th century. Hissole interest was in mathematics, resulting in him being only known to themathematical community. He lived a very simple life, eschewing personalcomfort and possessions. He considered money to be an encumbrance, andgave away, as prizes for solvingmathematical problems, whatever he earnedthat was over and above his basic simple needs.

Erd }os published over 1,500 papers, most of them jointly with other mathe-maticians. He gave generously of his ideas to everyone with whom he cameinto contact. He would ascertain your mathematical interests, and then pro-ceed with \I have this problem, : : : ". Over 4,000 mathematicians have pub-lished jointly with Erd }os.

P �al was born in Budapest into a Hungarian-Jewish family, the son of twomathematics teachers. While his father was a prisoner of war in Siberia forsix years, his mother taught him at home. He took his doctorate in 1934 atthe University of Budapest, and then went for further study in Manchester,England. The conditions in Europe made it unsafe for him to return to Hun-gary, and he moved to the United States. In the 1950's, the United Statesdenied him re-entry, as he was suspected of being a Soviet sympathiser. Hethen spentmuch time in Israel. He was allowed re-entry to the United Statesin the 1960's, and his mother, then in her eighties, began to travel with him.

Erd }os never saw the need to restrict himself to one institution, so he criss-crossed the world, inspiring mathematicians wherever he went. After hismother died in 1971, he immersed himself in his work with great enthusi-asm, often spending 18 hours daily on mathematical problems. He was stillfollowing this active lifestyle when the great reaper called on him inWarsaw.

341

Algebraic Integers and TensorProducts of Matrices.

Shaun M. FallatDept. of Mathematics and Statistics

University of VictoriaVictoria B.C., Canada

A complex number is called an algebraic integer if it is a zero of a monic(i.e., the coe�cient of the term with the highest exponent is one) polynomialwith integer coe�cients. Let C denote the complex numbers. A set S � C iscalled a subring if 1 2 S, and for every a; b 2 S, a+b 2 S and a �b 2 S. It iswell-known that the set of algebraic integers forms a subring of the complexnumbers (see [3]). The aim of the present note is to show that this resultfollows easily from basic facts about tensor products of matrices.

First we will need some notation. The set of n � n matrices over asubring R is denoted by Mn(R). The notation A = [aij ] means the (i; j)th

entry of A is aij. The set of polynomials with integer coe�cients is denotedby Z[x], where x is an indeterminate. Let detA denote the determinantof an n � n matrix A. If A 2 Mn(R), then pA(x) = det(xIn � A) isthe characteristic polynomial of A, where In denotes the n � n identitymatrix. Since the operations involved in computing the determinant of Aare multiplication and addition, it is readily veri�ed that if A 2 Mn(Z),then pA(x) is a monic polynomial in Z[x]. Recall that if � is a zero of pA(x),then � is called an eigenvalue of A.

We begin with an example for completeness.

Example 1. It is easy to check thatp2,p3 are algebraic integers. For exam-

ple,p2 is a zero of x2� 2. Therefore there must exist a polynomial in Z[x],

for whichp2 �p3 is a root. Since

p2 �p3 =

p6, it is easy to verify that

x2 � 6 is an example of a polynomial in Z[x] withp6 as a root. Similarly,

there must exist a polynomial in Z[x], for whichp2+p3 is a root. Consider

f(z) = (x� (p2 +p3))(x� (

p2�p3))

�(x� (�p2 +p3))(x� (�

p2�p3))

= x4 � 10x2 + 1:

Hence x4 � 10x2 + 1 is an example of a polynomial in Z[x] withp2 +p3

as a root.

The following result will establish the relationship between algebraicintegers and matrices.

342

Lemma 2. Let � be a complex number. Then � is an algebraic integer if andonly if � is an eigenvalue of a square matrix with integer coe�cients.

Proof. If� is an eigenvalue ofA 2Mn(Z), then pA(x) is a monic polynomialin Z[x], and pA(�) = 0. Thus � is an algebraic integer. To prove the con-verse, assume that f(�) = 0, where f(x) is a monic polynomial in Z[x], sayf(x) = xn + c1x

n�1 + � � �+ cn�1x+ cn. Let A be the companion matrixof f(x), i.e.

A =

26666664

0 1 0 � � � 0

0 0 1 � � � 0

..

....

. . .. . .

..

.0 0 � � � 0 1

�cn �cn�1 � � � �c2 �c1

37777775:

Expanding pA(x) by the �rst column and using a simple induction argument,it follows that pA(x) = f(x), and this fact can be found in [1; pp 230-231].Therefore � is an eigenvalue of A and A 2Mn(Z).

De�nition 3. If A = [aij ] is anm� n matrix and B = [bij] is a p� q matrixthen the Kronecker or tensor product of A and B, denoted A B, is themp� nq matrix de�ned as follows:

A B =

2664a11B � � � a1nB

.

... . .

.

..am1B � � � amnB

3775 :

Note that if A and B have integer entries, then so does A B. Fromthe above de�nition it is easy to verify that if A;B;C; D 2Mn(R), then

(AB)(C D) = AC BD:

Vectors in Cn may be identi�ed with n � 1 matrices over C, so the abovede�nition may be used to de�ne tensor products x y with x 2 Cn andy 2 Cm. The following proposition establishes a part of the well-knownresults that identify the eigenvalues of tensor products see, e.g. [2; pp 242-245]. We provide an easy proof for completeness.

Proposition 4. Let A 2 Mn(Z), B 2 Mm(Z), let � be an eigenvalue of Aand let � be an eigenvalue of B. Then

(a) �� is an eigenvalue of A B,

(b) �+ � is an eigenvalue of (A Im) + (In B).

Proof. There exist non-zero vectors x 2 Cn and y 2 Cm such that Ax = �x

and By = �y. It follows immediately that

343

(A B)(x y) = ��(x y);

and((A Im) + (In B))(x y) = (�+ �)(x y):

This proves the propostion.

Notice that if A 2 Mn(Z), B 2 Mm(Z), then A B 2 Mmn(Z).Hence, if two algebraic integers �;� are given, the tensor product is a usefultool for determining the existence of an integral matrix, for which � � � or�+ � is an eigenvalue. We summarize this in the following theorem.

Theorem 5. The set of algebraic integers forms a subring of the ring of com-plex numbers.

The proof follows directly from Lemma 2 and Proposition 4.

References

[1] K. Ho�man and R. Kunze, Linear Algebra, Prentice-Hall, New Jersey,1971.

[2] R.A. Horn and C.R. Johnson, Topics in Matrix Analysis, Cambridge Uni-versity Press, New York, 1991.

[3] H.L. Montgomery, I. Niven and H.S. Zuckerman, An Introduction to theTheory of Numbers, John Wiley and Sons, Inc. New York, 1991.

PHOTO

PROBLEM

Can you identifythis regular CRUXcontributor?

344

THE SKOLIAD CORNERNo. 18

R.E. Woodrow

Last issue we gave the problems of the Canadian Mathematical SocietyPrize Exam for Nova Scotia. Here are solutions.

CANADIANMATHEMATICAL SOCIETY PRIZE EXAMFriday, April 26, 1996 | Time: 2.5 hours

1. (a) Solvepx+ 20�px+ 1 = 1.

(b) Try to solve 3px+ 20� 3

px+ 1 = 1.

Solution. (a) Set A =px+ 20 and B =

px+ 1. We get A2 � B2 =

A + B as A � B = 1. So x + 20 � (x + 1) = 19 = A + B. Now2A = 20, A = 10 and x = 80.

(b) Let A = 3px+ 20 and B = 3

px+ 1 so A�B = 1. Multiplying by

A2+AB+B2 givesA3�B3 = A2+AB+B2 orA2+AB+B2 = 19. AsA2� 2AB+B2 = 12 = 1 we obtain 3AB = 19� 1 = 18 or AB = 6.So (x + 20)(x + 1) = A3B3 = 63 = 216. This gives the quadraticequation x2+21x� 196 = 0, or (x+28)(x� 7) = 0 and x = �28 orx = 7.

2. Suppose a function is de�ned so that f(xy) = f(x) + f(y).

(a) Show that if f is de�ned at 1, f(1) = 0.

(b) Similarly if f is de�ned at 0, its value at any x will be 0, so it is a\trivial" function.

(c) Show that if f is de�ned only for f1; 2; 3; : : : g there are many non-trivial ways to de�ne such an f (give an example).

Solution. (a) With x = 1 = y we obtain f(1 � 1) = f(1) + f(1), sof(1) = 2f(1) giving f(1) = 0.

(b) Suppose that f is de�ned at 0. Then with y = 0 and x arbitrary wehave f(0) = f(x � 0) = f(x) + f(0) and f(x) = 0.

(c) To see there are many such f with domain the natural numbers con-sider the natural logarithm, logarithms to base 10, and the followingfamily of examples: Let S be a set of prime numbers. A given naturalnumber can be written uniquely in the form

QPmi

i where P1; P2; : : : ;lists the prime numbers and mi is a non-negative integer. (Of courseonly �nitely many of themi are non-zero. De�ne fs(n) =

Pmi.

345

3. Three circles of equal radii r all touch each other to enclose a threecornered concave area A. How big is the area of A?

Solution.

D E

F

q

q q

Let the centres of the circles be labelled D, E, F . Then DEF is anequilateral triangle with side length 2r and area

p3 r2. The area A is

the area of the triangle less the three sectors, each subtending an angleof 60� in a circle of radius r, so the area of A is

p3 r2 � 3 � 1

6�r2 =

�p3� �

2

�r2:

4. Show that for all real numbers x,

(a) x4 � 4x� 3.

(b) x4 is not greater than 3x� 2, even though it appears to be true.

x �1 0 14

13

12

1 2 3

x4 1 0 :004 0:12 :063 1 16 81

4x� 3 �7 �3 �2 �1:67 �1 1 5 9

3x� 2 �5 �2 �1:25 �1 �:5 1 4 7

Solution. (a) x4 � 4x� 3 is the same as x4 � 4x+ 3 � 0.

Set f(x) = x4 � 4x+ 3. Now

f(x) = f((x� 1) + 1)

= [(x� 1) + 1]4 � 4[(x� 1) + 1] + 3

= (x� 1)4 + 4(x� 1)3 + 6(x� 1)2

+4(x� 1) + 1� 4(x� 1)� 4 + 3

= (x� 1)4 + 4(x� 1)3 + 6(x� 1)2

= (x� 1)2[(x� 1)2 + 4(x� 1) + 6]:

346

Now (x�1)2 � 0 and the discriminant ofX2+4X+6 is 42�4�1�6< 0

so both factors of f(x) are non-negative. The minimum value is 0 atx = 1, as required.

(b) Consider x = 0:9. Now x4 = 0:6561 and 3x � 2 = 0:7, so thatx4 6� 3x� 2.

5. Two integers are called equivalent, written x � y if they are divisibleby the same prime numbers (primes are 2; 3; 5; 7; : : : ) so 2 � 2 � 4,3 � 27 but 2 6� 3.

(a) Show that 10 � 80 but 10 6� 90.

(b) Prove that if x � y, then x2 � y2.

Solution. (a) 10 = 2� 5, 80 = 24� 5 so 10 and 80 are divisible by thesame primes, namely 2 and 5.

(b) Suppose x � y. Consider any �xed prime p. If p divides x2 thenp must also divide p. But then if p divides x2 it divides x and y, andthus y2. Similarly if p divides y2 it also divides x2. Thus x2 � y2, asrequired.

6. We can describe certain fractions in terms of others all with bigger de-nominators (always in lowest terms). For instance 1

3= 1

4+ 1

12and

23= 1

4+ 1

4+ 1

6but 2

3= 2

6+ 2

6doesn't work since 2

6= 1

3and 2

3= 1

2+ 1

6

doesn't since 2 < 3.

(a) Can you write 1

2as a sum 1

a+ 1

bfor integers 2 < a < b?

(b) Try to write 11996

as 1a+ 1

bfor integers 1996 < a < b.

Solution. (a)1

2=

1

3+

1

6and 2 < 3 < 6.

(b)1

1996=

1

998� 1

2=

1

998��1

3+

1

6

�=

1

2994+

1

5998:

That completes the Skoliad Corner for this issue. Send me your con-tests, suggestions, and recommendations to improve this feature.

347

THE OLYMPIAD CORNERNo. 178

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 the remaining problems proposed to the jury,but not used at the 36th International Olympiad held at Toronto, Ontario inJuly 1995. As always, I welcome your novel, nice solutions that di�er fromthe \o�cial" published solutions.

36th INTERNATIONALMATHEMATICAL OLYMPIADProblems proposed to the jury but not used

Algebra

1. Let R be the set of real numbers. Does there exist a function f :

R! Rwhich simultaneously satis�es the following three conditions?

(a) There is a positive numberM such that �M � f(x) �M for all x.

(b) f(1) = 1.

(c) If x 6= 0, then

f

�x+

1

x2

�= f(x) +

�f

�1

x

��2:

2. Let n be an integer, n � 3. Let x1; x2; : : : ; xn be real numberssuch that xi < xi+1 for 1 � i � n� 1. Prove that

n(n� 1)

2

Xi<j

xixj >

n�1Xi=1

(n� i)xi

!0@ nXj=2

(j � 1)xj

1A :

Geometry

3. Let A1A2A3A4 be a tetrahedron, G its centroid, and A01, A02, A

03

and A04 the points where the circumsphere of A1A2A3A4 intersects GA1,GA2, GA3 and GA4 respectively. Prove that

GA1 �GA2 �GA3 �GA4 � GA01 �GA02 �GA03 �GA04and

1

GA01+

1

GA02+

1

GA03+

1

GA04� 1

GA1

+1

GA2

+1

GA3

+1

GA4

:

348

4. O is a point inside a convex quadrilateral ABCD of area S. K, L,M andN are interior points of the sidesAB,BC,CD andDA respectively.If OKBL and OMDN are parallelograms, prove that

pS � pS1 +

pS2,

where S1 and S2 are the areas of ONAK and OLCM respectively.

5. Let ABC be a triangle. A circle passing through B and C intersectsthe sidesAB andAC again atC0 andB0, respectively. Prove thatBB0, CC0

and HH0 are concurrent, where H andH0 are the orthocentres of trianglesABC and AB0C0 respectively.

Number Theory and Combinatorics

6. At a meeting of 12k people, each person exchanges greetings withexactly 3k+6 others. For any two people, the number who exchange greet-ings with both is the same. How many people are at the meeting?

7. Does there exist an integer n > 1 which satis�es the followingcondition? The set of positive integers can be partitioned into n non-emptysubsets, such that an arbitrary sum of n�1 integers, one taken from each ofany n� 1 of the subsets, lies in the remaining subset.

8. Let p be an odd prime. Determine positive integers x and y forwhich x � y and

p2p�px�py is non-negative and as small as possible.

Sequences

9. For positive integers n, the numbers f(n) are de�ned inductively asfollows: f(1) = 1, and for every positive integer n, f(n+ 1) is the greatestinteger m such that there is an arithmetic progression of positive integersa1 < a2 < � � � < am = n and

f(a1) = f(a2) = � � � = f(am):

Prove that there are positive integers a and b such that f(an+ b) = n+ 2

for every positive integer n.

10. Let N denote the set of all positive integers. Prove that thereexists a unique function f : N! N satisfying

f(m+ f(n)) = n+ f(m+ 95)

for allm and n in N. What is the value ofP19

k=1 f(k)?

We now turn to readers' solutions to problems of the contest given inthe Corner as the TurkishMathematical Olympiad Committee Final SelectionTest of April 4, 1993. This problem set was given in the April 1995 numberof the corner [1995: 117-118].

349

TURKISH MATHEMATICAL OLYMPIAD

COMMITTEE

FINAL SELECTION TEST

April 4, 1993 | Part I

(Time: 3 hours)

1. Show that there is an in�nite sequence of positive integers suchthat the �rst term is 16, the number of distinct positive divisors of each termis divisible by 5, and the terms of the sequence form an arithmetic progres-sion. Of all such sequences, �nd the one with the smallest possible commondi�erence between consecutive terms.

Solutions by Cyrus Hsia, student, University of Toronto, Toronto, On-tario; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, On-tario. We give Wang's solution.

Clearly the question should have been reworded as \of all..., �nd theone with the smallest non-zero common di�erence between consecutiveterms." For otherwise the constant sequence 16; 16; 16; : : : is clearly theanswer. Let d denote the common di�erence. If d < 0 then the terms of thesequence will eventually become negative. Hence assume that d > 0.

For natural numbers n, let � (n) denote the number of positive divisorsof n. Then it is well-known that if n = P�1

1 P�22 � � �P�k

k denotes the prime

power decomposition of n, then � (n) =Qk

i=1(�i + 1). Furthermore, � is amultiplicative function, i.e., if gcd (n;m) = 1 then � (n �m) = � (n) � � (m).

Therefore if 16+d = 24(2q+1), where q = 0; 1; 2; : : : , then 5j� (16+d). To minimize d, or 16+d, we take q = 1 since q = 0 implies d = 0. Thend = 3� 24 � 24 = 25. To see that this value of d would yield an arithmeticprogression with the desired property, note �rst that for allm = 0; 1; 2; : : : ,� (16+md) = � (24+25m) = � (24(2m+1)) � 0 mod 5. Finally, straight-forward checking reveals that 56 j � (16 + t) for t = 1; 2; : : : ; 31. Therefore,the required sequence is indeed f16 + 32mg,m = 0; 1; 2; : : : with d = 32.

2. Let M be the circumcentre of an acute-angled triangle ABC, andassume the circle (BMA) intersects the segment [BC] at P , and the segment[AC] at Q. Show that the line CM is perpendicular to the line PQ.

Solution by Cyrus Hsia, student, University of Toronto, Toronto, On-tario.

Let triangle ABC have angles A, B, and C. \AMC = 2\B in circle(ABC). Thus

\MCA =180� � 2B

2(in isosceles triangle AMC)

= 90� � B;

\PQC = \ABP (in circle (AMB))

= B:

350

q

q

q

q

q

q

q

B

A

P

Q

MN

C

Thus, extending CM to meet PQ in N we have \QCN = \MCA

= 90� � B and \NQC = \PQC = B.It follows that

\QNC = 180� � (\QCN + \NQC)

= 180� � (90� � B + B)

= 90�:

That is, CM ? PQ.

3. Let fbng be sequence of positive real numbers such that

for each n � 1; b2n+1 �b21

13+b22

23+ � � �+ b2n

n3:

Show that there is a natural numberK such that

KXn=1

bn+1

b1 + b2 + � � �+ bn>

1993

1000:

Solution by Cyrus Hsia, student, University of Toronto, Toronto, On-tario.

(13 + 23 + � � �+ n3)(b2n+1)

� (13 + 23 + � � �+ n3)

�b21

13+b22

23+ � � �+ b2n

n3

�� (b1 + b2 + � � �+ bn)

2

by the Cauchy{Schwartz inequality. Thus

b2n+1

(b1 + b2 + � � �+ bn)2� 1

13 + 23 + � � �+ n3=

�2

n(n+ 1)

�2:

It follows thatbn+1

b1 + b2 + � � �+ bn� 2

n(n+ 1)

351

since the sequence fbng has only positive real terms. Thus

KXn=1

bn+1

b1 + b2 + � � �+ bn�

KXn=1

2

n(n+ 1)

= 2

KXn=1

1

n(n+ 1)

= 2

KXn=1

�1

n� 1

n+ 1

�

= 2

�1� 1

K + 1

�

=2K

K + 1:

By setting K = 999 we have

999Xn=1

bn+1

b1 + � � �+ bn� 2(999)

999 + 1=

1998

1000>

1993

1000;

as required.

April 4, 1993 | Part II

(Time: 3 hours)

1. Some towns are connected to each other by some roads with at mostone road between any pair of towns. Let v denote the number of towns, ande denote the number of roads. Show that

(a) if e < v� 1, then there are at least two towns such that it is impos-sible to travel from one to the other,

(b) if 2e > (v� 1)(v� 2), then travelling between any pair of towns ispossible.

Solution by Cyrus Hsia, student, University of Toronto, Toronto, On-tario.

(a) The minimum value that e could be such that the graph with v townsis connected is e = v � 1.

We prove this by induction. For v = 3, we must have e at least 2 sincee = 1 would only connect two towns.

(And e = 2 is possible, as shown r

r

r�@).Suppose that the minimum for v towns is e = v � 1. Consider a con-

nected situation with v+1 towns. We can assume there is at least one townwith at most one road connecting it to other towns otherwise 2e � 2(v+1)

and e � v + 1. Removing that town (and its road to another town) must

352

leave a connected situation with v towns. So e � 1 � v � 1 and e � v asrequired. It is easy to see that one can construct a connected example withv + 1 towns and v + 1� 1 = e edges, completing the induction.

(b) The statement is proved by induction. For v = 3, we have

3 >(v�1)(v�2)

2= 1, so that e � 2.

With e = 2 we have r

r

r�@ and e = 3 r

r

r�@ as the essential possibili-ties.

Suppose travelling between any pair of towns is possible for v towns if2e > (v � 1)(v� 2). Now consider v + 1 towns joined by 2e > (v)(v� 1)

roads. Consider a �xed town. If it is connected to the v remaining towns, weare done. Otherwise, let it be joined to x � v� 1 towns. If the town and itsroads are removed we are left with v towns and

e� x >v(v� 1)

2� x � v(v� 1)

2� (v� 1)

=(v� 1)(v� 2)

2

roads, that is, the remaining v towns have more than (v�1)(v�2)2

roads andare connected. If x > 0, the (v + 1)st town is joined to some one of theremaining towns, and hence connected to all of them. If x = 0 then the v

towns can have a maximum number of�v

2

�=

v(v�1)2

roads. But e > v(v�1)2

,so x 6= 0.

Thus for v towns, e > (v�1)(v�2)2

implies all towns are connected andone can travel between any pair of towns.

2. On a semicircle with diameter AB and centre O pointsE andC aremarked in such a way that OE is perpendicular to AB, and the chord ACintersects the segment OE at a point D which is interior to the semicircle.Find all values of the angle \CAB such that a circle can be inscribed into thequadrilateral OBCD.

Solutions by Cyrus Hsia, student, University of Toronto, Toronto, On-tario; and by D.J. Smeenk, Zaltbommel, the Netherlands. We give Smeenk'ssolution.

����

����

����

�AAAAAAA�

A BO

D

C

E

q q

q

q

q

q

353

We set OA = OB = 1 and denote \BAC = �. Then OD = tan�,OB = 1 and BC = 2 sin�, CD = 2cos�� 1

cos�.

For a circle to be inscribed in quadrilateral OBCD, we have

OB + CD = OD +BC

1 + 2cos�� 1

cos�= tan�+ 2 sin�

cos�+ 2 cos2 �� 1 = sin�+ 2 sin� cos�

cos�+ cos 2� = sin�+ sin2�

2 cos3

2� cos

�

2= 2 sin

3

2� cos

�

2:

Since cos�

26= 0, we have

tan3�

2= 1

3�

2=

�

4

� =�

6:

3. Let Q+ denote the set of all positive rational numbers. Find allfunctions f : Q+! Q+ such that

for every x; y 2 Q+; f

�x+

y

x

�= f(x) +

f(y)

f(x)+ 2y:

Solutions by Cyrus Hsia, student, University of Toronto, Toronto, On-tario; and by Beatriz Margolis, Paris, France. We give the solution by Mar-golis.

We show that the only solution is f(x) = x2, x 2 Q+.Take x = 1, this gives

f(1 + y) = f(1) +f(y)

f(1)+ 2y: (1)

Now taking x = y we get

f(y+ 1) = f(y)+ 1 + 2y: (2)

From (1), and (2) we obtain

(f(1)� 1)

�1� f(y)

f(1)

�= 0:

Now f can not be a constant function, so we have that

f(1) = 1: (3)

354

By induction, from (2) and (3) we get

f(k) = k2 k 2 N+: (4)

Therefore, if n, k 2 N+, using (4), we obtain that

f

�k+

n

k

�= f(k) +

f(n)

f(k)+ 2n

= k2 +n2

k2+ 2n =

�k+

n

k

�2: (5)

Let x = mk2 Q+, and N 2 N+. By (2), we have

f(x+N)� f(x) =

N�1Xj=0

[f(x+ j + 1)� f(x+ j)]

=

N�1Xj=0

[1 + 2(x+ j)]

= N(2x+ 1) + (N � 1)N

= (x+N)2 � x2:

Therefore f(x+N)� (x+N)2 = f(x)� x2, N 2 N+, x 2 Q+.In particular, using (5), we obtain that

0 = f

�n

k+ k

���n

k+ k

�2= f

�n

k

���n

k

�2= f(x)� x2;

where x = n

k, with n, k 2 N+.

Thus f(x) = x2 for x 2 Q+.

That completes the solutions we have on �le and the Olympiad Cornerfor this issue. Send me your nice problem sets and solutions!

355

THE ACADEMY CORNERNo. 7

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

In the May 1996 issue of CRUX [1996: 165], we printed the questionsof the 1996 Memorial University Undergraduate Mathematics Competition,and invited readers to send in solutions. Here is a complete solution set.

MEMORIAL UNIVERSITY OF NEWFOUNDLAND

UNDERGRADUATE MATHEMATICS COMPETITION

March, 1996

Solutions

1. Prove that if n is a positive integer, thenn2 + 3n+ 1

n2 + 4n+ 3is an irreducible

fraction.

Solution by Panos E. Tsaoussoglou, Athens, Greece [Shortened by theeditors.]

Assume that the fraction is reducible. Then there is a positive integerk such that kj(n2 + 3n+ 1) and kj(n2 + 4n+ 3).

Then kj(n2 + 3n + 1)� (n2 + 4n+ 3) or kjn + 2, so that there is apositive integer l such that n+ 2 = kl.

Now,n2 + 3n+ 1

k= nl+

n+ 1

k. The latter term is an integer only if

kjn+ 1. Since kjn+ 2, this implies that k = 1.

This is a contradiction, proving the result.

2. A jar contains 7 blue balls, 9 red balls and 10 white balls. Balls aredrawn at random one by one from the jar until either four balls of thesame colour or at least two of each colour have been drawn. What isthe largest number of balls that one may have to draw?

Solution by Panos E. Tsaoussoglou, Athens, Greece.

Assume that three balls of one colour are drawn, then three balls of another colour are drawn, and then one ball of the third colour. The nextball (of any colour) satis�es the conditions of the problem.

The answer is 8 balls.

356

3. Find all functions u(x) satisfying u(x) = x+

Z 1

2

0

u(t)dt.

\O�cial Solution" from Maurice Oleson.

Since

Z 1

2

0

u(t) dt is a constant, we have that u(x) = x + c for some

constant c.

Then

Z 1

2

0

(t+ c) dt = c, which gives c = 14.

Thus u(x) = x+ 14.

4. Show that�p

5 + 2� 1

3 ��p

5� 2� 1

3

is a rational number and �nd its

value.

Solution by Panos E. Tsaoussoglou, Athens, Greece.

Let k =�p

5 + 2�1

3 ��p

5� 2� 1

3

. Then

k3 =�p

5 + 2���p

5� 2�� 3

��p5 + 2

� �p5� 2

�� 1

3

���p

5 + 2�1

3 ��p

5� 2� 1

3

�;

ork3 = 4� 3k or (k3 � 1) + 3(k� 1) = 0:

Thus (k� 1)(k2 + 4k+ 4) = 0, so that k = 1 (since k > 0).

5. In a quadrilateral ABCD (vertices named in clockwise order), AC andBD intersect in X. You are given that AB k DC, that AB is twice aslong as CX and that AC is equal in length to DC. Show that AB andCD are equal in length (and hence ABCD is a parallelogram).

Solution by Panos E. Tsaoussoglou, Athens, Greece.

��������

��������

����

����

����

ZZ

ZZ

ZZ

ZZ

Z

D C

A B

X

We have AB = 2CX and AC = DC.

Triangles4AXB and4DXC are similar, so that

AX

CX=AB

DC

357

so thatAX + CX

CX=AB +DC

DC:

Thus, we have

2DC

AB=

AB +DC

DC

2DC2 = AB2 + AB �DC0 = DC2 � AB2 +DC2 � AB �DC

= (DC � AB)(AB + 2CD);

so that AB = DC.

6. Prove that among any thirteen distinct real numbers it is possible to

choose two, x and y, such that 0 <x� y1 + xy

< 2�p3.

\O�cial Solution" from Maurice Oleson.

We recognize the similarity ofx� y

1 + xywith tan(���) = tan�� tan�

1 + tan� tan�.

Since twelve �fteen degree angles make a straight angle, the pigeon-

hole principle gives that 0 <x� y1 + xy

< tan(15�).

We note that

tan2(15�) =sin2(15�)

cos2(15�)

=1� cos(30�)

1 + cos(30�)

=1�

p3

2

1 +p32

=2�p3

2 +p3

= (2�p3)2;

and the result follows.

7. A coastguard boat is hunting a bootlegger in a fog. The fog rises dis-closing the bootlegger 4 miles distant and immediately descends. Thespeed of the boat is 3 times that of the bootlegger, and it is known thatthe latter will immediately depart at full speed on a straight course ofunknown direction. What course should the boat take in order to over-take the bootlegger?

358

\O�cial Solution" from Maurice Oleson. We look at two diagrams:

-

V

�q q

A B

-q q

A B

ds

dr&

d� &

The distance from A to B is 1 mile.

From ds2 = dr2 + (r d�)2, dsdt

= 3v and drdt

= v, we get

9v2 = v2 + r2�d�

dt

�2:

Thus p8v = r

d�

dt= r

d�

dr� drdt

= rd�

drv;

yieldingdr

r=

1p8d�. Hence ln(r) =

�p8+ c.

Since � = 0 when r = 1, we have c = 0, giving r = e�=p8.

This is the spiral path that the coastguard must follow if she wishes tointercept the bootlegger.

PHOTO

PROBLEM

Can you identifythis regular CRUXcontributor?

359

BOOK REVIEWS

Edited by ANDY LIU

The Universe in a Handkerchief by Martin Gardner,published by Copernicus, an imprint of Springer-Verlag New York Inc., 1994,ISBN 0-387-94673-X, x+158 pages, US$19.00.Reviewed by Richard Guy, University of Calgary.

A book by Martin Gardner is always welcome, and most mathemati-cians and many other people are interested in Lewis Carroll. Although it isalmost a century since he died, there is still much to learn about him. MortonCohen's biography [2] only appeared last year and Edward Wakeling's edi-tion of Lewis Carroll's diaries [4] is still in process of appearing. The presentbook contains a bibliography of more than thirty items: and there is littleoverlap with the nearly �fty items which relate to Carroll in John Fisher'sbook [3].

Many mathematicians are interested in word play of all sorts, and thismakes up the major part of the book: Carroll's `Doublets' or `Word Links' arestill a popular pastime: can you get from `Bread' to `Toast' with less than 21links? But if you're looking for speci�cally mathematical items, then they'llneed picking out for you. The gravity-operated train from Sylvie and Brunoconcluded. Carroll's attachment to, if not obsession with, the number 42.The Butcher's piece of algebra in Fit 5 of The Hunting of the Snark. Thequadratic equation in the poem in Rhyme? and Reason? The puzzle of themonkey and weight over a pulley. John Conway's \Doomsday Rule" for �nd-ing what day of the week any date falls on has its original inspiration fromLewis Carroll. Rhymes for remembering logarithms to seven decimal places.The Telegraph Cipher. The game of Arithmetical Croquet. Conjuring tricksdepending on `casting out the nines'. Probability paradoxes which anticipatethe `motor car and goats' debate of recent years. And mathematical recre-ations of all the traditional kinds, such as are listed in Rouse Ball [1], the �rstthree editions of which appeared in Carroll's lifetime. It was probably in thefourth edition that Rouse Ball added a footnote to his geometrical fallacies,saying that `they particularly interested Mr. C. L. Dodgson; see the LewisCarroll Picture Book, London, 1899, pp. 264, 266, where they appear in theform in which I originally gave them.'

The following problem, from a letter by Carroll to Enid Stevens, wasthought by Morton Cohen to be too ambiguously stated to have a preciseanswer:

Three men, A, B and C, are to run a race of a quarter-of-a-mile. Whenever A runs against B, he loses 10 yards in everyhundred; whenever B runs against C, he gains 10 yards in everyhundred. How should they be handicapped?

360

Isn't the following a reasonable interpretation? For distances run in equaltimes, A : B = 90 : 100 and B : C = 110 : 100, so that

A : B : C = 99 : 110 : 100 = 396 : 440 : 400;

and B should start from scratch, with C at the 40 yard mark and A at44 yards.

Something I learned while reviewing this book, though not while read-ing it, andwhich Canadiansmight be particularly interested in, is that Carroll,with such de�nitions as `OBTUSE ANGER is that which is greater than RightAnger', anticipated the ideas in Stephen Leacock's Boarding House Geome-try.

And where does the title come from? In Sylvie and Bruno concludedMein Herr gives instructions, not in fact practicable in three dimensions, toLady Muriel for sewing three handkerchiefs into the Purse of Fortunatus,which is a projective plane. \Whatever is inside that Purse, is outside it; andwhatever is outside it, is inside it. So you have all the wealth of the world inthat leetle Purse!"

References

[1] W. W. Rouse Ball & H. S. M. Coxeter, Mathematical Recreations & Es-says, 12th edition, Univ. of Toronto Press, 1974.

[2] Morton Cohen, Lewis Carroll: A Biography, Knopf, New York, 1995.

[3] John Fisher (ed.), The Magic of Lewis Carroll, Penguin Books, Har-mondsworth, 1975.

[4] Edward Wakeling (ed.), Lewis Carroll's Diaries, Vols. 1, 2 & 3., LewisCarroll Society, Luton, England, 1993, 1994, 1995.

361

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 that 1 June 1997. They may also besent by email to cruxeditor@cms.math.ca. (It would be appreciated if emailproposals and solutions were written in LATEX). Graphics �les should be inepic format, or encapsulated postscript. Solutions received after the abovedate will also be considered if there is su�cient time before the date of pub-lication.

2189. Proposed by Toshio Seimiya, Kawasaki, Japan.The incircle of a triangle ABC touches BC at D. Let P and Q be

variable points on sidesAB and AC respectively such that PQ is tangent tothe incircle. Prove that the area of triangle DPQ is a constant multiple ofBP �CQ.

2190. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,Austria.

Determine the range of

sin2A

A+

sin2B

B+

sin2C

C

where A;B;C are the angles of a triangle.

2191. Proposed by �Sefket Arslanagi�c, Berlin, Germany.Find all positive integers n, that satisfy the inequality

1

3< sin

��

n

�<

1

2p2:

362

2192. Proposed by Theodore Chronis, student, Aristotle Universityof Thessaloniki, Greece.

Let fang be a sequence de�ned as follows:

an+1 + an�1 =

�a2

a1

�an; n � 1:

Show that if

����a2a1���� � 2, then

����ana1���� � n.

2193. Proposed by Luis V. Dieulefait, IMPA, Rio de Janeiro, Brazil.

(a) Prove that every positive integer is the di�erence of two relatively primecomposite positive integers.

(b) Prove that there exists a positive integer n0 such that every positive in-teger greater than n0 is the sum of two relatively prime composite positiveintegers.

2194. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.

Prove or disprove that it is possible to �nd a triangle ABC and a transversalNML withN lying betweenA andB,M lying betweenA andC, andL lyingon BC produced, such that BC, CA, AB, NB, MC, NM , ML, and CLare all of integer length, and NMCB is a cyclic inscriptable quadrilateral.

2195. Proposed by Bill Sands, University of Calgary, Calgary, Al-berta.

A barrel contains 2n balls, numbered 1 to 2n. Choose three balls at random,one after the other, and with the balls replaced after each draw.

What is the probability that the three-element sequence obtained has theproperties that the smallest element is odd and that only the smallest ele-ment, if any, is repeated?

For example, the sequences 453 and 383 are acceptable, while the sequences327 and 388 are not.

(NOTE: this problem was suggested by a �nal exam that I marked recently.)

2196. Proposed by Juan-Bosco Romero M�arquez, Universidad deValladolid, Valladolid, Spain.

Find all solutions of the diophantine equation

2(x+ y) + xy = x2 + y2;

with x > 0, y > 0.

363

2197. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Madrid, Spain.

Let n be a positive integer. Evaluate the sum:

1Xk=n

�2k

k

�(k+ 1)22k+1

:

2198. Proposed by Vedula N. Murty, Andhra University, Visakhap-atnam, 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.

2199. Proposed byDavidDoster, Choate Rosemary Hall, Wallingford,Connecticut, USA.

Find the maximum value of c for which (x+ y + z)2 > cxz for all 0 � x <

y < z.

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:

Bonus Problem for 1996

220A?. 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.

364

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

2030. [1995: 91, 1996: 132] Proposed by Jan Ciach, Ostrowiec�Swi�etokrzyski, Poland.

For which complex numbers s does the polynomial z3 � sz2 + sz � 1

possess exactly three distinct zeros having modulus 1?

Editor's comment.Readers who found #2030 interesting might want to look at \What Is

the Shape of a Triangle" by Dana N. Mackenzie [NOTE DI MATEMATICA13:2 (1993) 237-250; MATH REVIEWS 96c: 51029].

The author notes that an arbitrary triangleABC is similar to a trianglewhose vertices are represented by three complex numbers zi with jzij = 1

and z1 � z2 � z3 = 1.

He de�nes the SHAPE of the triangle to be � = z1 + z2 + z3.

It follows that the shape appears as the coe�cient of z2 in the cubicz3 � (�)z2 + (��)z � 1, whose zeros all lie on the unit circle.

His solution to the question, \what are the possible values of �?" isessentially the solution to CRUX 2030 (appearing more than two years beforeit appeared in CRUX). The author goes on to discuss properties of his \shapeinvariant"; quite surprisingly, during that discussion the Morley triangle oftriangle ABC plays a key role.

2076. [1995: 278] Proposed by John Magill, Brighton, England.

C

AC

24

4

This is a multiplicative magic square, where the product of each row,column and diagonal has the same value, ABCD. Each letter represents adigit, the same digit wherever it appears, and each cell contains an integer.Complete the square by entering the correct numbers in each of the ninecells.

Solution by Toby Gee, student, the John of Gaunt School, Trowbridge,England.

Letting AC = 10A + C and letting x be the value in the bottom lefthand corner we have 24Cx = 4(10A + C)x, implying C = 2A. If we

365

further let z be the value of the upper left hand corner and y be that of themiddle top row box, we also have 4yz = 24(10A+C)z = 24 � 12Az, whichyields y = 72A. The product of the entries in the middle column is then72A �12A �2A = 1728A3, which must be a four digit integer, whence A = 1

and we can complete the square as below:

6 72 4

8 12 18

36 2 24

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; CARL BOSLEY,student,WashburnRural High School, Topeka, Kansas, USA; ADRIAN CHAN,student, Upper Canada College, Toronto, Ontario; THEODORE CHRONIS,student, Aristotle University of Thessaloniki, Greece; HANS ENGELHAUPT,Franz{Ludwig{Gymnasium, Bamberg, Germany; ROBERT GERETSCHL �AGER,Bundesrealgymnasium, Graz, Austria; RICHARD I. HESS, Rancho PalosVerdes, California, USA; CYRUSHSIA, student, University of Toronto, Toron-to, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;POLLY KONTOPOULOU, student, Aristotle University of Thessaloniki,Greece; KEE-WAI LAU, Hong Kong; J.A. MCCALLUM, Medicine Hat, Al-berta; WOLFGANG GMEINER, Millstatt, Austria; P. PENNING, Delft, theNetherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria;HRISTOS SARAGHIOTES, student, Aristotle University, Thessaloniki, Greece;HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount RoyalCollege, Calgary; DAVID R. STONE, Georgia Southern University, States-boro, Georgia; DAVID TASCIONE, student, St. Bonaventure University, NewYork; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRIS WILDHAGEN,Rotterdam, the Netherlands; KENNETH M. WILKE, Topeka, Kansas, USA;and the proposer.

2087. [1995: 307] Proposed by Toby Gee, student, the John of GauntSchool, Trowbridge, England.

Find all polynomials f such that f(p) is a power of two for every prime p.

Solution by Luis V. Dieulefait, IMPA, Rio de Janeiro, Brazil.The only solutions are constant polynomials. Suppose instead that a poly-nomial f with deg(f) > 0 were a solution. Since f(x) takes positive valuesfor every prime value of x, it is clear that limx!1 f(x) =1. Also there isa real number x0 such that f(x) is increasing for x > x0. If we enumeratethe sequence of primes: p1, p2, p3, : : : in increasing order, then

f(pi+1) > f(pi) for pi > x0: (�)

Lemma. Given a real number r > 1, every su�ciently large pi satis�esr � pi � pi+1.

366

Proof. Let us suppose, on the contrary, that there exist in�nitely manyprimes pj satisfying pj+1 > r � pj. Then �(r � pj)� �(pj) = 0 where �(x)

is the number of primes less than or equal to x. This means that�(r�pj)�(pj)

= 1

for in�nitelymany primes. Applying the prime number theorem, we see that

limi!1

�(r � pi)�(pi)

= limi!1

r � ln(pi)ln(r � pi)

= r > 1:

This contradiction proves the lemma.

Let d = deg(f) > 0. Choose r > 1 such that rd < 2. From (*) abovewe see that f(pi+1 � 2 �f(pi) for every i large enough. Applying the lemmawe have f(r �pi) > 2 �f(pi) for every i large enough (because f is increasing

for such i). This implies limi!1f(r�pi)f(pi)

� 2. But f is a polynomial of degree

d, so we know that the limit above equals rd and rd < 2. This contradictionproves that no polynomial f with deg(f) > 0 can solve our problem.

Then the only solutions are constant, and obviously we must havef(x) = 2n, for n a non-negative integer.

Also solved by MANUAL BENITO MU ~NOZ and EMILIO FERN �ANDEZMORAL, I.B. Sagasta, Logro ~no, Spain; RICHARD I. HESS, Rancho PalosVerdes, California, USA; KEE-WAI LAU, Hong Kong;MURRAY S. KLAMKIN,University of Alberta, Edmonton, Alberta; CHRIS WILDHAGEN, Rotterdam,the Netherlands; and the proposer.

2089. [1995: 307]ProposedbyMiguel Amengual Covas, Cala Figuera,Mallorca, Spain.

Let ABCD be a trapezoid with AB k CD, and let X be a point onsegment AB. Put P = CB \ AD, Y = CD \ PX, R = AY \ BD andT = PR \AB. Prove that

1

AT=

1

AX+

1

AB:

Solution by Waldemar Pompe, student, University of Warsaw, Poland.Using Menelaus's Theorem on the triangle ABD, we get

AT

TB� BRRD� DPPA

= 1:

Since BR : RD = AB : DY and DP : PA = DY : AX, we obtain

TB

AT=AB

DY� DPPA

=AB

DY� DYAX

=AB

AX;

whence

AB

AT=TB

AT+ 1 =

AB

AX+ 1 or

1

AT=

1

AX+

1

AB;

367

as we wished.

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany; MANUELBENITO MU ~NOZ and EMILIO FERN �ANDEZ MORAL, I.B. Sagasta, Logro ~no,Spain; CARL BOSLEY, student,WashburnRural High School, Topeka, Kansas,USA; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; TIM CROSS,King Edward's School, Birmingham, England; DAVID DOSTER, Choate Rose-mary Hall, Wallingford, Connecticut,USA; HANS ENGELHAUPT, Franz{Lud-wig{Gymnasium, Bamberg, Germany; RICHARD I. HESS, Rancho PalosVerdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; KEE-WAI LAU, Hong Kong; P. PENNING, Delft, the Nether-lands; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, theNetherlands; and the proposer.

2091. [1995: 343] Proposed by Toshio Seimiya, Kawasaki, Japan.Four points A, B, C, D are on a line in this order. We put AB =

a, BC = b, CD = c. Equilateral triangles ABP , BCQ and CDR areconstructed on the same side of the line. Suppose that \PQR = 120�. Findthe relation between a, b and c.

I. Solution by Jordi Dou, Barcelona, Spain.If the middle triangle is largest then b is twice the arithmetic mean of

a and c; if smallest, b is half the harmonic mean of a and c.

Proof: We suppose that BCQ is �xed. Let S be the image of Q underre ection inBC. P andR will lie on the prolongations of SB and SC, whichwe will call p and r respectively.

As in the �gure, K, L are the points where the line through Q parallelto BC intersects p, r. P0 is the foot of the perpendicular to p from Q, andR00 is the point where that line meets r; analogously, R0 is the foot of theperpendicular to r from Q, and P 00 is where that line meets p.

Any point P on the segment BK of p corresponds to a point R on thesegment LC for which \PQR = 120�. From \P0QR0 = 120� it followsthat \P0QP = \R0QR and PP0 = RR0. In this case a = BP , andc = CR = PK, so that the desired relation is

a+ c = b:

Any point P 0 on p beyondK corresponds to a point R0 on r beyond Lfor which \R0QP 0 = 120�. As before, because \R00QP

00 = 120� it follows

that \P 00QP0 = \R00QR

0 and, therefore, that the pencils QP 0 and QR0

are congruent so that there is a projectivity that takes P 0 to R0. [That is,P 0 � P 0Q � R0Q � R0.] This projectivity takes P 01 (the point at in�nity of p)to L, andK toR01 (the point at in�nity of r). Since cross ratios are preservedit follows that KP 0 � LR0 = constant = KP 00 � LR00 = KQ � LQ = b2: [Ifyou prefer, assign the coordinates 0, 1,1 to the pointsK, P 00, P1 on p, andassign1, 1, 0 to their images | R01, R00, L | on r. Then the projective

368

S

CB

R

R0

L

R0

R0

0

P0

P

K

P0

0

P0

rp

q

q

x

x

q

� �

q

transformation that takes P 0 to R0 takes the corresponding coordinate x to1=x.] Thus (a� b)(c� b) = b2; or ac = ba+ cb; or

1

b=

1

c+

1

aor b =

ca

c+ a:

II. Solutionby David Doster, Choate Rosemary Hall, Wallingford, Con-necticut, USA.

[Editor's comment: Most solvers used either the sine law or the cosinelaw. Here is an example of the former for the case when the middle triangle isthe smallest or, in Doster's words, when pentagon APQRD is not convex.]

In this case, \CQR = 60�+ �, \CRQ = 60�� �. By the law of sineswe have

sin �

sin(120�� �)=QB

PB=

b

a;

sin(60� � �)

sin(60� + �)=QC

RC=b

c:

Since sin(120� � �) = sin(60� + �) and cos(30� � �) = sin(60� + �),we have

b

a+b

c=

sin� + sin(60� � �)

sin(60� + �)=

2 sin 30� cos(30� � �)sin(60� + �)

= 1.

Therefore,1

b=

1

a+

1

c:

369

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; CLAUDIO ARCONCHER, Jundia��, Brazil; �SEFKET ARSLANAGI �C, Ber-lin, Germany; MANSUR BOASE, student, St. Paul's School, London, England;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGELCABEZ �ON OCHOA, Logro ~no, Spain; THEODORE CHRONIS, student, Aristo-tle University of Thessaloniki, Greece; TIM CROSS, King Edward's School,Birmingham, England; C. DIXON, Newcastle upon Tyne, England; HANSENGELHAUPT, Franz{Ludwig{Gymnasium,Bamberg, Germany; J.K. FLOYD,Newnan, Georgia; ROBERT GERETSCHL �AGER, Bundesrealgymnasium, Graz,Austria; DAVID HANKIN, Hunter College Campus Schools, New York, NY,USA; RICHARD I.HESS, Rancho Palos Verdes, California, USA; CYRUSHSIA,student, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursul-inengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State Uni-versity, Big Rapids, Michigan, USA; MITKO CHRISTOV KUNCHEV, Rousse,Bulgaria; KEE-WAI LAU, Hong Kong; P. PENNING, Delft, the Netherlands;NEIL REID, Mississauga, Ontario; JOEL SCHLOSBERG, student, Hunter Col-lege High School, New York NY, USA; D.J. SMEENK, Zaltbommel, the Neth-erlands(2 solutions); PANOS E. TSAOUSSOGLOU, Athens, Greece; and theproposer. One solution was submitted without a name attached, and one so-lution was awed. Several solvers dealt with only one case, but their methodwas su�cient to deal with the omitted case had they considered it.

A few solvers combined the cases in the pretty, but less informativerelation

abc = (a� b)(c� b)(a+ b):

2092. [1995: 343] Proposed by K.R.S. Sastry, Dodballapur, India.I take a three-digit base-ten integer (in which the �rst digit is non-zero)

and consider it as a number in a di�erent base. If I convert this new numberinto base ten, I �nd that it is exactly twice the original number. In what basedoes this happen?

Solution by Theodore Chronis, student, Aristotle University of Thessa-loniki, Greece.Let k be the desired base. Then

ak2 + bk+ c = 2(100a+ 10b+ c)() a(k2 � 200) + b(k� 20) = c

with a 2 f1; 2; : : : ; 9g and b; c 2 f0; 1; : : : ; 9g.If k � 14 then a(k2 � 200) + b(k� 20) � �4a� 6b < 0, which is false.

If k � 16 then a(k2�200)+b(k�20) � 56a�4b � 20, which is also false.

If k = 15 then

25a� 5b = c) c 2 f0; 5g ) 5a� b 2 f0; 1g:

So k = 15, and (150)15 = 300, (145)15 = 290, (295)15 = 590 are all thesolutions.

370

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; TIM CROSS, KingEdward's School, Birmingham, England; CHARLES R. DIMINNIE, AngeloState University, San Angelo, TX, USA; HANS ENGELHAUPT, Franz{Lud-wig{Gymnasium, Bamberg, Germany; ROBERT GERETSCHL �AGER, Bundes-realgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall,North Bay, Ontario; DAVID HANKIN, Hunter College Campus Schools, NewYork, NY, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA;CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Fer-ris State University, Big Rapids, Michigan, USA; KATHLEEN E. LEWIS, SUNYOswego. Oswego, NY, USA; DAVID E. MANES, State University of NewYork, Oneonta, NY, USA; JOHN GRANT McLOUGHLIN, Okanagan Univer-sity College, Kelowna, British Columbia; P. PENNING, Delft, the Nether-lands; HRISTOS SARAGHIOTES, student, Aristotle University of Thessa-loniki, Greece; JOEL SCHLOSBERG, student, Hunter College High School,New York NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany;LAWRENCE SOMER, Catholic University of America, Washington, DC;DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA;KENNETH M. WILKE, Topeka, Kansas, USA; and the proposer. There werefour incorrect solutions.

Most solvers had variations on the above solution. Some did not �ndall the solutions, but were still able to identify the base. STONE asks theopposite question as well: what is the answer if you interchange the roles of10 and the unknown base, and provides the answer, which we will leave tothe reader. Indeed, several of the incorrect solutions treated this probleminstead of the one above. JANOUS and GRANT McLOUGHLIN both con-sider a generalization where the factor above is replaced by larger positiveintegers. JANOUS has written a short computer program to look for smallexamples and found many. GRANT McLOUGHLIN also observed that 121was a solution for the factor 4 with base 21, 441 was a solution for the factor9 with base 31, and 961 was a solution for the factor 16 with base 41; is thisextendable? JANOUS further asks:

1. Let n be an arbitrary natural number in base 10 representation (n)10.Does there exist necessarily a further base b � 11 such that (n)b is aninteger multiple of (n)10?

2. Let n be a natural number satisfying the above question in the positive.Do there exist in�nitely many such bases b? If \yes", what is theirdensity d(n) in the set of all natural numbers?

3. Does there exist for any base b � 11 a natural number n such that (n)bis an integer multiple of (n)10?

371

2093?. [1995: 343] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

Let A;B;C be the angles (in radians) of a triangle. Prove or disprovethat

(sinA+ sinB + sinC)

�1

� � A+

1

� � B+

1

� � C

�� 27

p3

4�:

Solution by Kee-Wai Lau, Hong Kong.The inequality is true.

Let X =� � A

2, Y =

� � B

2, Z =

� � C

2, so that X, Y , Z are also

the angles of a triangle. So, we have

(sinA+ sinB + sinC)

�1

� �A +1

� � B+

1

� �C

�

= 4 cos

�A

2

�cos

�B

2

�cos

�C

2

��1

� �A +1

� � B+

1

� � C

�

= 2 sinX sinY sinZ

�1

X+

1

Y+

1

Z

�

=sinX

X

sinY

Y

sinZ

Z

n(X + Y + Z)

2 � �X2 + Y 2 + Z2�o

� sinX

X

sinY

Y

sinZ

Z

((X + Y + Z)

2 � (X + Y + Z)2

3

)

by the arithmetic mean inequality

=2�2

3

sinX

X

sinY

Y

sinZ

Z

=2�2

3

3

rsinX

X

sinY

Y

sinZ

Z

!3

� 2�2

3

0BB@

sinX

X+

sinY

Y+

sinZ

Z

3

1CCA3

; (1)

by the arithmetic mean{geometric mean inequality.

According to CRUX 1216 [1987: 53, 1988: 120], we have

sinX

X+

sinY

Y+

sinZ

Z� 9p3

2�: (2)

The required inequality now follows immediately from (1) and (2).

372

Also solved by THEODORE CHRONIS, student, Aristotle Universityof Thessaloniki, Greece; V �ACLAV KONE �CN �Y, Ferris State University, BigRapids, Michigan,USA and BOBPRIELIPP,University ofWisconsin{Oshkosh,Wisconsin, USA. One incorrect solution was received.

2094. [1995: 343] Proposed by Paul Yiu, Florida Atlantic University,Boca Raton, Florida, USA.

The problemist Victor Th �ebault has noted (American MathematicalMonthly Vol. 66 (1959), p. 65) an interesting Pythagorean triangle in whichthe two perpendicular sides are integers having the same digits in reverseorder, viz., 88209 and 90288, with hypotenuse 126225.

(a) Can such a Pythagorean triangle be primitive?

(b) Find an example of a primitive Pythagorean triangle in which thehypotenuse and one other side are integers having the same digits in reverseorder.

Solution by Kathleen E. Lewis, SUNY Oswego, Oswego, NY.(a) No, it cannot be primitive. If a and b have the same digits in reverse

order, then they are congruent modulo 3. If they were not divisible by 3, theywould be congruent to 1 or 2, so their squares would have to be congruent to1mod 3. Thus, if c2 = a2+b2, then c2 would have to be congruent to 2mod3. But no perfect square is congruent to 2. Thus a and b are both divisibleby 3 and so are not relatively prime.

(b) One example of such a triple is 33, 56, 65.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; THEODORE CHRONIS, student, Aristotle University of Thessaloniki,Greece; TOBY GEE, student, the John of Gaunt School, Trowbridge, England;RICHARD I. HESS, Rancho PalosVerdes, California, USA; DAVID E.MANES,SUNY at Oneonta, Oneonta, NY, USA; P. PENNING, Delft, the Netherlands;DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA;and the proposer.

Solutions to part (a) only were received from JOEL SCHLOSBERG, stu-dent, Hunter College High School, New York NY, USA; CYRUS HSIA, stu-dent, University of Toronto, Toronto, Ontario; and to (b) only from CHARLESASHBACHER, Cedar Rapids, Iowa, USA.

One reader pointed us to another reference to Th �ebault's type of Py-thagorean triangle in the Reader Re ections section of the MathematicsTeacher [Vol. 82, No. 1, January 1989, p. 8]. The �rst additional such triplegiven there with legs 125928 and 829521 is Pythagorean (but of course notprimitive); however, the second pair of numbers, 725068 and 860527 can-not be the legs of a Pythagorean triangle since the unit digit of (725068)2+(860527)2 is 3, while perfect squares can only have unit digits of 0, 1, 4, 5,6 or 9.

373

One other such triangle \with hypotenuse of 1164481" (no legs given)is mentioned in this article. Is this really the hypotenuse of such a triangleand, if so, what are the legs?

Hess provides another (non-primitive) triple of Th �ebault's type withlegs 2328480 and 0848232.

The type of triple in (b) seems to be quite rare; Penning reports that\going up to numbers above 10000, no others solutions were found". Allprimitive Pythagorean triples have the form (2mn;m2�n2;m2+n2), wherem;n are relatively prime and of opposite parity. Hess reports no other (b)type triples were found for n < m � 999:

2095. [1995: 344] Proposed by Murray S. Klamkin, University ofAlberta.

Prove that

ax(y � z) + ay(z � x) + az(x� y) � 0

where a > 0 and x > y > z.

Solution: a shortened form of that submitted by Juan-Bosco RomeroM�arquez, Universidad de Valladolid, Valladolid, Spain.

The solution is an immediate consequence of the convexity off(x) = ax, x > 0, x 2 R.In Mitrinovi�c's \Elementary Inequalities" (North-Holland, Holland, 1964),pages 22{33, in the section on Jensen's Inequality, we have the followinggeometric interpretation | an equivalent characterization of the convexityof a function:

Consider x1, x2, x3 2 [�; �], where x1 < x2 < x3, and thecorresponding functional values f(x1), f(x2), f(x3). The area ofthe triangle with coordinates (x1; f(x1)), (x2; f(x2)), (x3; f(x3)),is given by �P , where

P =1

2

������x1 f(x1) 1

x2 f(x2) 1

x3 f(x3) 1

������ :Then f is convex if P > 0, and f is concave if P < 0.

We write z = x1, y = x2, x = x3 and f(x) = ax, x > 0, x 2 R, and theresult follows.

Also solved by �SEFKET ARSLANAGI �C, Berlin, Germany;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MANSUR BOASE,student, St. Paul's School, London, England; THEODORE CHRONIS, stu-dent, Aristotle University of Thessaloniki, Greece; TOBY GEE, student, theJohn of Gaunt School, Trowbridge, England; ROBERT GERETSCHL �AGER,

374

Bundesrealgymnasium, Graz, Austria; RICHARD I. HESS, Rancho PalosVerdes, California, USA; JOE HOWARD, New Mexico Highlands University,Las Vegas, NM, USA; CYRUS HSIA, student, University of Toronto, Toron-to, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;KEE-WAI LAU, Hong Kong; DAVID E.MANES, State University of New York,Oneonta, NY, USA; CORY PYE, student, Memorial University, St. John's,Newfoundland; JOEL SCHLOSBERG; student, Hunter College, NY, USA;HEINZ-J�URGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU,Athens, Greece; JOHANNES WALDMANN, Friedrich{Schiller{Universit�at,Jena, Germany; CHRIS WILDHAGEN, Rotterdam, the Netherlands; and theproposer.

Klamkin comments that the problem is given in S. Barnard and J.M.Child's \Higher Algebra" (MacMillan, London, 1949, p. 226) without theconstraint condition x > y > z. He notes (as observed in the highlightedsolution) that the inequality is reversed if x > z > y.

Some of our \senior" solvers knew the more general result, whereasour \junior" solvers tended to prove the result ab initio.

Janous asks how this result can be generalized to four or more vari-ables. [Ed: I will leave it thus and so challenge our readers to generalize itthemselves. Please send me your \nice" generalizations.]

2096. [1995: 344] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.

Triangle A1A2A3 has circumcircle �. The tangents at A1; A2; A3 to �

intersect (the productions of)A2A3; A3A1; A1A2 respectively inB1; B2; B3.The second tangent to � through B1; B2; B3 touches � at C1; C2; C3 respec-tively. Show that A1C1; A2C2; A3C3 are concurrent.

Solution by Christopher J. Bradley, Clifton College, Bristol, UK.By Pascal's theorem, B1; B2; B3 are collinear. The polars ofB1; B2; B3

with respect to � are therefore concurrent. A1C1 is the polar of B1, A2C2

that of B2, and A3C3 that of B3. Hence A1C1; A2C2; A3C3 are concurrentat the pole of B1B2B3

Comment by Chris Fisher.To avoid invoking Pascal's theorem, you can assume, without loss of

generality, that the given triangle is equilateral (because the projective group�xing a conic is transitive on the triangles), in which case, the problem re-duces to an easy observation.

Solved using projective geometry by FRANCISCO BELLOT ROSADO,I.B. Emilio Ferrari, Valladolid, Spain; P. PENNING, Delft, the Netherlands;GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; and JOELSCHLOSBERG, student, Hunter College High School, New York NY, USA.Solved without using projective geometry by FRANCISCO BELLOT ROSADO,

375

I.B. Emilio Ferrari, Valladolid, Spain; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; HOE TECK WEE, student, Hwa Chong JuniorCollege, Singapore: and the proposer.

2097. [1995: 344] Proposed by Federico Ardila, student, Massachu-setts Institute of Technology, Cambridge, Massachusetts, USA.

Let n be a positive integer and p a prime number. Prove that

pn(pn � 1)(pn�1 � 1) : : : (p2 � 1)(p� 1)

is divisible by n! .

Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.The desired result actually holds for all positive integers p.

Let Ap(n) denote the given product. From a well-known formula, we

have n! =Y

q prime

qeq , where eq =

1Xj=1

�n

qj

�.

[Ed: Sei�ert called this \de Polignac's formula".]

Let q be a prime. Since eq �

6664 1Xj=1

n

qj

7775 =

�n

q� 1

�=: aq, it su�ces to

show that qaq divides Ap(n). This is clearly true if q divides p. Otherwise,

q divides pk(q�1) � 1 for all k 2 N by Fermat's Little Theorem. However,aqYk=1

�pk(q�1) � 1

�is a factor ofAp(n) since aq(q�1) � n. Our claim follows.

Also solved by MANSUR BOASE, student, St. Paul's School, London,England; THEODORE CHRONIS, student, Aristotle University of Thessa-loniki, Greece; RICHARD I. HESS, Rancho Palos Verdes, California, USA;CYRUS HSIA, student, University of Toronto, Toronto, Ontario; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, HongKong; JOEL SCHLOSBERG, student, Hunter College High School, New YorkNY, USA; DAVID R. STONE, Georgia Southern University, Statesboro, Geor-gia, USA; CHRIS WILDHAGEN, Rotterdam, the Netherlands; KENNETH S.WILLIAMS, Carleton University, Ottawa, Ontario; and the proposer.

Most submitted solutionswere similar to the one given above. BesidesSei�ert, onlyChronis and Hess noted that the statement holds for all positiveintegers. Janous remarked that the result can be sharpened to: n!jAp(n�1),or equivalently, (n+ 1)!jAp(n).

376

2098. [1995: 344] Proposed by John Magill, Brighton, England.At the conclusion of our �rst inter-species soccer tournament, in which

each team played each of the others once, the scoresheets, prepared by theZecropians and the Valudians, were, respectively,

Won Drawn Lost Goals against Goals for Points

Zecropia c b b cc ffh d

Earth Station b b c fbe ff b

Valudia f b f ah db c

Won Drawn Lost Goals against Goals for Points

Valudia p x p pxq ppr r

Zecropia r x x mm mxr q

Earth Station x x r rqp pq x

Each scoresheet is equivalent to the other in that both give the correctvalues. Each, however, is in the �xed-base positional number systems ofthose who prepared the scoresheets, each base being less than 10 and greaterthan 1. Both Valudians and Zecropians use the same operations of addition,subtraction, division and multiplication, and rules of manipulation, as areused by Earth. I have substituted letters for the symbols originally used.Each letter represents a digit, the same digit wherever it appears. Two pointswere awarded for a win and one for a draw.

As the answer to this puzzle, state the total number of goals scored byeach team, and the total number of goals scored against each team | in thebase 10 number system.

Solution by Kathleen E. Lewis, SUNY Oswego. Oswego, NY, USA.Looking at the scoresheet prepared by the Zecropians, we see that Earth

won b games and drew b games and so should have 3� b points. Since Earthhas b points, b must be zero. Each team played two games, so c = 2 andf = 1. Zecropia won two games and got d points so d = 4. On the Valudians'scoresheet, we get the corresponding values: x = 0, p = 1, r = 2, and q = 4.

Let s denote the base used by the Zecropians and t the base used bythe Valudians. Comparing the goals columns on the two scoresheets, we �ndthat the number of goals scored for Earth Station is equal to (11)s and (14)t,so s + 1 = t + 4. Comparing the goals scored for Valudia, we see that(40)s = (112)t, so t

2 + t+ 2 = 4s = 4(t+ 3). Therefore t = 5 and s = 8.

Zecropia had (22)8 = 18 goals scored against it, so (mm)5 = 18, andthusm is 3. The number of goals scored for Zecropia equals (302)5 = 77, so(11h)8 = 77, and h must be 5. The number of goals against Valudia comesout to (104)5 = 29, so (a5)8 = 29, and a = 3.

The scoresheet prepared by the Earthlings will therefore look like:

Won Drawn Lost Goals against Goals for Points

Earth Station 0 0 2 71 9 0

Valudia 1 0 1 29 32 2

Zecropia 2 0 0 18 77 4

377

Also solved by MANSUR BOASE, student, St. Paul's School, London,England; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg,Germany; JEFFREY K. FLOYD, Newnan, Georgia, USA; ROBERTGERETSCHL �AGER, Bundesrealgymnasium, Graz, Austria; SHAWN GODIN,St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS, RanchoPalos Verdes, California, USA; DAVID E. MANES, State University of NewYork, Oneonta, NY, USA; HRISTOS SARAGHIOTES, student, Aristotle Uni-versity of Thessaloniki, Greece; and the proposer.

Geretschl�ager suggests that Earth Station's dismal performance couldbe \an early example of the Prime Directive at work, as in not interfering(too much) between other worlds' goalposts"! Another explanation mightbe found in the bases 5 and 8, which suggest that Valudians probably havea total of 5 toes and Zecropians a total of 8 toes. Assuming a constant num-ber of toes per foot, we could conclude that Valudians most likely have �ve(one-toed) legs and that Zecropians may very well have four or even eightlegs, which would make both teams a real handful (so to speak) for the un-derlegged Earthlings.

2099. [1995: 345] Proposed by Proof, Warszawa, Poland.The tetrahedron T is contained inside the tetrahedron W . Must the

sum of the lengths of the edges of T be less than the sum of the lengths ofthe edges of W ?

Solution by Nikolai Dolbilin, Fields Institute, Toronto, Ontario and theSteklov Mathematical Institute, Moscow, Russia.

Let W have a \small" triangular base and a summit that is \far" fromits base. Let T have two vertices insideW that are \close" to the base, andtwo insideW that are \close" to the summit.

Then T has four \long" edges while W has only three \long" edges.The parameters can therefore be adjusted so that T has a longer edge-sum.

Also solved by ROBERT GERETSCHL �AGER, Bundesrealgymnasium,Graz, Austria; 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; VICTOROXMAN, University of Haifa, Haifa, Israel; GOTTFRIED PERZ, Pestalozzi-gymnasium, Graz, Austria; and the proposer.

2100. [1995: 345] Proposed by Iliya Bluskov, student, Simon FraserUniversity, Burnaby, BC.

Find 364 �ve{element subsets A1; A2; : : : ; A364 of a 17-element setsuch that jAi \ Aj j � 3 for all 1 � i < j � 364.

378

Solution by the proposer.Without loss of generality, letB = f0, 1, 2, : : : , 16g be the 17{element

set. Let X(5) be the family of all 5{element subsets of B. Then we have���X(5)��� = �

17

5

�. For each k = 0, 1, 2, : : : , 16, de�ne

X(5)

k =

(fa1; a2; a3; a4; a5g 2 X(5)

�����5X

i=1

ai � k (mod 17)

):

Clearly, the classesX(5)

k , k = 0, 1, 2, : : : , 16, are disjoint, and hence, by thepigeon-hole principle, at least one of the classes must contain no less than�17

5

�=17 = 364 \elements".

It remains to show that jAi \ Aj j � 3 for all distinct 5{element subsetsAi and Aj which belong to the same class. Clearly, it su�ces to show thatjAi \ Aj j 6= 4. This is obvious since, if Ai = fa1, a2, a3, a4, a5g andAj = fb1, b2, b3, b4, b5g are such that ai = bi for i = 1, 2, 3, 4, then5X

i=1

ai �5X

i=1

bi (mod 17) would imply that a5 � b5 (mod 17), which is

impossible unless a5 = b5. This completes the proof.

Also solved by Kee-Wai Lau, Hong Kong, who exhibited a family of 3655{element subsets satisfying the given condition. He obtained this examplewith the help of a PC and a program using GW BASIC.

Proposer's remarks. This is not a new problem, but (hopefully) it isnot too well-known. In general, if we let M denote the maximum possiblecardinality of a familyF ofm{element subsets of an n{element set, with thecondition that jAi \ Aj j � m � 2 for all Ai, Aj 2 F such that Ai 6= Aj ,

then using the same argument, one can easily see that M �&�

n

m

�n

', where

dxe denotes the least integer greater than or equal to x. This result is due toGraham and Sloane [1]. For the present case (n = 17; m = 5), the estimatehas been improved to M � 424 [2]. On the other hand, it is known thatM � 476. Finding a structure withM = 476 would resolve a long-standingopen question in design theory: the existence of a Steiner system S(4;5; 17).

References

[1] R.L. Graham and N.J.A. Sloane, Lower Bounds for Constant WeightCodes, IEEE Transactions on Information Theory, IT{26, No. 1 (1980),37{43.

[2] A.E. Brouwer, J.B. Shearer, N.J.A. Sloane andW.D. Smith, A New Tableof Constant Weight Codes, IEEE Transactions on Information Theory,Vol. 6, No. 6 (1990), 1334{1380.

379

Problemist of the Year!

In the index to volume 22, you will �nd a listing of all those who haveparticipated in the Solutions section of CRUX in 1996. Based on these statis-tics, we are pleased to declare

� �

� �

� �

CHRISTOPHER J. BRADLEY

as CRUX Problemist of the Year for 1996.

A close runner up is

� �

� �

WALTHER JANOUS

The following (in alphabetical order) deserve an Honourable Mention sinceeach has participated in at least one third of the solutions in 1996:

CARL BOSLEY RICHARD I. HESS

V �ACLAV KONE �CN �Y KEE-WAI LAU

P. PENNING HEINZ-J�URGEN SEIFFERT.

We o�er our thanks and congratulations to all our participants.

Readers will note that we have now published solutions to all problemsup to 2100.

380

YEAR END FINALE

How a year can y by! It only seems like yesterday that I was persuaded to

take on the mammoth task of being Editor-in-Chief of CRUX. Yet it has been a job of

great satisfaction, getting to know better many of the world's keenest mathematical

problem solvers.

CRUX has gone on-line! This is an exciting development and has led to renewed

interest and new subscribers. Thanks are due to LOKI JORGENSON, NATHALIE SIN-

CLAIR, and the rest of the team at SFU who are responsible for this. As mentioned

earlier in this issue, CRUX and Mathematical Mayhem have joined forces. I am

delighted to welcome NAOKI SATO and CYRUS HSIA to the reconstituted Editorial

Board. I have known both these young men for many years and have had the privilege

or working with them when they were on Canada's IMO team.

There are many people that I wish to thank most sincerely for particular con-

tributions. First and foremost is BILL SANDS. Bill is of such value to me and to

the continuance of CRUX that his code name in my computer �les is CRUXWISE.

As well, I thank most sincerely, CATHY BAKER, ROLAND EDDY, CHRIS FISHER,

BILL SANDS, JIM TOTTEN, and EDWARD WANG, for their regular yeoman ser-

vice in assessing which solutions should be highlighted; DENIS HANSON, HARVEY

ABBOTT, DOUG FARENICK, CHRIS FISHER, ALLENHERMAN,MURRAY KLAMKIN,

JOANNE MCDONALD, JUDI MCDONALD, RICHARDMCINTOSH, DIETER RUOFF,

JIM TOMKINS, and MICHAEL TSATSOMEROS, for ensuring that we have quality ar-

ticles; ANDY LUI, RICHARD GUY, KATHERINE HEINRICH, CLAUDE LAFLAMME,

MURRAY KLAMKIN, MARCIN KUCZMA, and JACK MACKI, for ensuring that we

have quality book reviews, ROBERT WOODROW (and JOANNE LONGWORTH), who

carries the heavy load of two corners, one somewhat new and the other of long stand-

ing, and RICHARD GUY for sage advice whenever necessary. The quality of these

people are vital parts of what makes CRUX what it is. Thank you one and all.

As well, I would like to give special thanks to our retiring Associate Editor,

COLIN BARTHOLOMEW, for keeping me from printing too many typographical

errors; and my colleagues, ERIC JESPERS, P.P. NARAYANASWAMI, MAURICE

OLESON, MIKE PARMENTER and DONALD RIDEOUT for their occasional sage ad-

vice. I have also been helped by someMemorial University students, RON HAYNES,

KIM PENDERGAST and CORY St. CROIX, as well as a WISE Summer student, CINDY

HISCOCK. The staff of the Department of Mathematics and Statistics at Memo-

rial University deserve special mention for their excellent work and support: ROS

ENGLISH,MENIEFRENCH,WANDAHEATH,LEONCEMORRISSEY and KAY SCOTT;

as well as the computer and networking expertise of RANDY BOUZANE. Not to men-

tion GRAHAM WRIGHT, Managing Editor, would be a travesty. 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 my Head of De-

partment, BRUCE WATSON, and the Dean of Science of Memorial University, ALAN

LAW, without whose support and understanding, I would not be able to try to do the

job of Editor-in-Chief.

Last but not least, I send my thanks to you, the readers of CRUX. Without you,

CRUX would not be what it is. Keep those contributions and letters coming in. I do

enjoy knowing you all.

381

INDEX TO VOLUME 22, 1996

Articles

Dissecting Squares into Similar Rectangles

B-Y. Chun, A. Liu & D. van Vliet : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 241

Algebraic Integers and Tensor Products of Matrices

Shaun M. Fallat : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 341

A Note on the Mean Value Theorem

Finbarr Holland : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 290

On the sum of n Dice

J.B. Klerlein : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 145

Dissecting Triangle into Isosceles Triangles

D. Robbins, S. Sivapalan, M. Wong : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 97

Ceva meets Pythagoras!

K.R.S. Sastry : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 193

Unitary Divisor Problems:

K.R.S. Sastry : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 4

On the Generalized Ptolemy Theorem

Shailesh Shirali : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 49

The Skoliad Corner R.E. Woodrow

February No. 11 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 10

March No. 12 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 54

April No. 13 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 101

May No. 14 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 149

September No. 15 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 198

October No. 16 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 249

November No. 17 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 294

December No. 18 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 344

The Olympiad Corner R.E. Woodrow

February No. 171 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 21

March No. 172 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 57

April No. 173 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 107

May No. 174 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 157

September No. 175 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 202

October No. 176 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 251

November No. 177 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 299

December No. 178 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 347

The Academy Corner Bruce Shawyer

February No. 1 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 28

April No. 2 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 117

May No. 3 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 165

September No. 4 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 213

October No. 5 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 268

November No. 6 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 311

December No 7 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 355

Miscellaneous

Citation | Professor Ron Dunkley, OC : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 106

Congratulations to Andy Liu : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 201

Congratulations to some solvers : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 240

In memoriam { P �al Erd }os : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 339

IMO95 Puzzles : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 70

Letter from the Editor / Lettre des r �edacteurs : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1

Letter from the Editor : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 289

Letter from the Editor / Lettre des r �edacteurs : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 381

Year end roundup : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 380

382

Book Reviews Andy Liu

The Monkey and the Calculator;

Aladdin's Sword (Le sabre d'Aladin).

Reviewed by Claude La amme, University of Calgary.: : : : : : : : : : : : : : : : : : : : : : : : : 30

Tournament of the Towns, 1980{1984, Questions and Solutions;

Tournament of the Towns, 1989{1993, Questions and Solutions,

both edited by Peter J. Taylor.

Reviewed by Murray S. Klamkin, University of Alberta. : : : : : : : : : : : : : : : : : : : : : : : 31

Assessing Calculus Reform Efforts : A Report to the Community,

edited by James Leitzel and Alan C. Tucker. Preparing for a New Calculus,

edited by Anita Solow.

Reviewed by Jack Macki, University of Alberta. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 71

e The Story of a Number,

by Eli Maor.

Reviewed by Richard Guy, University of Calgary. : : : : : : : : : : : : : : : : : : : : : : : : : : : : 119

All the Math That's Fit to Print,

by Keith Devlin.

Reviewed by A. Sharma, University of Alberta. : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 167

She Does Math! real-life problems from women on the job,

edited by Marla Parker.

Reviewed by Katherine Heinrich, Simon Fraser University. : : : : : : : : : : : : : : : : : : : 215

Experience in Problem Solving - a W. J. Blundon Commemorative,

edited by R. H. Eddy and M. M. Parmenter.

Reviewed by Murray S. Klamkin, University of Alberta. : : : : : : : : : : : : : : : : : : : : : : 271

Five Hundred Mathematical Challenges,

by Edward J. Barbeau, Murray S. Klamkin and William O.J. Moser.

Reviewed by Marcin E. Kuczma, University of Warsaw. : : : : : : : : : : : : : : : : : : : : : : : 313

The Universe in a Handkerchief,

by Martin Gardner.

Reviewed by Richard Guy, University of Calgary. : : : : : : : : : : : : : : : : : : : : : : : : : : : : 359

Problems

February 2101{2113 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 33

March 2114{2124 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 75

April 2125{2137 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 122

May 2138{2150 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 169

September 2151{2163, 2139 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 217

October 2164{2176 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 273

November 2137, 2177{2188 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 317

December 2189{2200, 220A : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 361

Solutions

February 1827, 2006{2010, 2012{2015 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 36

March 1827, 2000, 2011, 2016{24, 2027{28 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 78

April 2015, 2025{2026, 2029{2034, 2036{2039 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 125

May 2035, 2040{2047, 2049{2051, 2053{2055, 2057 : : : : : : : : : : : : : : : : : : : : : : : : : : : 172

September 1823, 2044, 2048, 2052, 2056, 2058{2066 : : : : : : : : : : : : : : : : : : : : : : : : : : 220

October 1987, 2067, 2069{2071, 2073{2075, 2077 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 276

November 1940, 2068, 2072, 2078{2086, 2088 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 321

December 2076, 2087, 2089{2100 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 364

383

Proposers, solvers and commentators in the PROBLEMS and SOLUTIONSsections for 1996 are:

Hayo Ahlburg: 2009, 2016, 2026, 2056, 2060;Miguel Amengual Covas: 2021, 2035, 2037, 2047, 2053, 2055,

2061, 2063, 2067, 2077, 2079, 2081, 2089, 2091;Claudio Arconcher: 2012, 2017, 2019, 2088, 2091;Federico Ardila: 2007, 2008, 2009, 2025, 2097;�Sefket Arslanagi �c: 2014, 2015, 2016, 2020, 2021, 2031, 2032,

2035, 2038, 2050, 2059, 2060, 2064, 2067, 2068, 2069, 2070, 2076, 2077,

2078, 2082, 2083, 2084, 2088, 2089, 2090, 2091, 2095;Charles Ashbacher: 2016, 2094;Seung-Jin Bang: 2060, 2068;Ed Barbeau: 2030;Gerd Baron: 1823;Jerzy Bednarczuk: 2013;Niels Bejlegaard: 2016, 2018, 2019, 2020, 2032, 2043, 2044,

2045;Martha Bell: 2050;Francisco Bellot Rosado: 2008, 2015, 2016, 2035, 2038, 2047,

2050, 2060, 2061, 2063, 2067, 2068, 2071, 2077, 2096;Manuel Benito Mu~noz: 2073, 2078, 2081, 2082, 2083, 2084,

2085, 2086, 2087, 2088, 2089;Iliya Bluskov: 2085, 2100;Mansur Boase: 2089, 2091, 2095, 2097, 2098;Carl Bosley: 2009, 2012, 2016, 2018, 2019, 2020, 2021, 2022, 2023,

2024, 2026, 2031, 2032, 2034, 2036, 2037, 2038, 2040, 2042, 2045, 2048,

2050, 2052, 2053, 2056, 2058, 2060, 2064, 2065, 2067, 2068, 2069, 2070,

2076, 2083, 2084, 2088;Paul Bracken: 2068;Christopher J. Bradley: 2008, 2009, 2010, 2011, 2012, 2014,

2016, 2017, 2019, 2020, 2021, 2022, 2024, 2027, 2030, 2031, 2032, 2033,

2035, 2037, 2038, 2040, 2041, 2042, 2043, 2045, 2047, 2050, 2052, 2053,

2054, 2055, 2056, 2058, 2059, 2060, 2061, 2063, 2064, 2065, 2066, 2067,

2068, 2070, 2071, 2072, 2073, 2074, 2075, 2077, 2078, 2081, 2082, 2083,

2084, 2089, 2091, 2092, 2094, 2095, 2096;Jim Braselton: 2050;Miguel Angel Cabez �on Ochoa: 2014, 2016, 2022, 2031, 2032,

2035, 2037, 2038, 2042, 2045, 2055, 2056, 2058, 2059, 2061, 2062, 2063,

2067, 2091, 2092;Robyn M. Carley: 2063;Sabin Cautis: 2023, 2031, 2032, 2034, 2062, 2064,

2066, 2067, 2068, 2070;Adrian Chan: 2014, 2016, 2020, 2031, 2032, 2037, 2038, 2050,

2054, 2058, 2060, 2062, 2064, 2066, 2068, 2072, 2076, 2081, 2083, 2088;Shi-Chang Chi: 2015;Ji Chen: 2015, 2039;Theodore Chronis: 2031, 2032, 2054, 2056, 2058, 2076, 2078,

2083, 2085, 2088, 2090, 2091, 2092, 2093, 2094, 2095, 2097;Jan Ciach: 2030, 2049, 2057, 2073;Paul Colucci: 2009;Curtis Cooper: 2083;Tim Cross: 2014, 2016, 2032, 2062, 2068, 2069, 2081, 2083, 2088,

2089, 2091, 2092;Jayabrata Das: 2014;Emeric Deutsch: 2070;C. Dixon: 2091;Luis Victor Dieulefait: 2081, 2087, 2088, 2090;Charles Diminnie: 2092;Nikolai Dolbilin: 2099;David Doster: 2012, 2031, 2032, 2037, 2038, 2060, 2083, 2088,

2089, 2091;Jordi Dou: 1823, 2035, 2051, 2053, 2055, 2059, 2061, 2063, 2069,

2071, 2079, 2091;Peter Dukes: 2007, 2009;John Duncan: 2006;Roland H. Eddy: 2010;Keith Ekblaw: 2016, 2030, 2040, 2062, 2066, 2074;

Hans Engelhaupt: 2007, 2009, 2012, 2061, 2062, 2063, 2064,

2065, 2066, 2067, 2068, 2069, 2070, 2071, 2072, 2074, 2076, 2080, 2081,

2083, 2088, 2089, 2091, 2092, 2098;Emilio Fern �andez Moral: 2073, 2078, 2081, 2082, 2083, 2084,

2085, 2086, 2087, 2088, 2089;Kurt Fink: 2030;Chris Fisher: 2030, 2096;F.J. Flanigan: 2020, 2030, 2044, 2050, 2058, 2065, 2068, 2077,

2080;J.K. Floyd: 2009, 2016, 2022, 2030, 2042, 2052, 2068, 2072, 2074,

2081, 2088, 2091, 2098;Camilla Fox: 2063;Rudolf Fritsch: 2043;Hidetosi Fukagawa: 2067, 2069;Toby Gee: 2009, 2012, 2016, 2018, 2022, 2032, 2034, 2036, 2038,

2040, 2042, 2045, 2050, 2052, 2060, 2064, 2065, 2067, 2068, 2070, 2075,

2076, 2081, 2087, 2088, 2094, 2095;Robert Geretschl�ager: 2062, 2064, 2066, 2068, 2072, 2076,

2079, 2080, 2081, 2088, 2091, 2092, 2095, 2098, 2099;Wolfgang Gmeiner: 2073, 2076, 2077, 2078;Shawn Godin: 2009, 2020, 2022, 2034, 2042, 2062, 2063, 2064,

2065, 2066, 2068, 2070, 2072, 2074, 2080, 2081, 2083, 2092, 2098;Solomon W. Golomb: 2070;Joaqu��n G �omez Rey: 2070;Douglass L. Grant: 2015, 2036, 2039;Herbert G�ulicher: 2055;Richard K. Guy: 2009, 2037;David Hankin: 2009, 2042, 2043, 2045, 2050, 2053, 2053, 2060,

2091, 2092;Ronald Haynes: 2056;G.P. Henderson: 1823, 2057;Richard I. Hess: 2007, 2009, 2012, 2014, 2016, 2022, 2026, 2030,

2034, 2036, 2037, 2038, 2040, 2042, 2043, 2045, 2046, 2052, 2053, 2054,

2056, 2060, 2062, 2064, 2065, 2066, 2067, 2068, 2070, 2071, 2072, 2074,

2076, 2077, 2078, 2080, 2081, 2082, 2083, 2084, 2087, 2088, 2089, 2092,

2091, 2094, 2095, 2097, 2098, 2099;John G. Heuver: 2038, 2042;Joe Howard: 2064, 2068, 2070, 2078, 2084, 2095;Cyrus Hsia: 2009, 2014, 2031, 2032, 2033, 2034, 2036, 2037, 2038,

2040, 2042, 2050, 2072, 2074, 2075, 2076, 2091, 2092, 2094, 2095, 2097;Jun-hua Huang: 2008, 2029;Peter Hurthig: 2007, 2012, 2019, 2020, 2021, 2032, 2036, 2038,

2056, 2058, 2078, 2090;R. Daniel Hurwitz: 2026;Robert B. Israel: 2007;Douglas E. Jackson: 2036, 2048, 2070;Walther Janous: 2008, 2014, 2015, 2016, 2017, 2022, 2024, 2026,

2027, 2028, 2030, 2031, 2032, 2034, 2037, 2038, 2040, 2042, 2043, 2044,

2049, 2050, 2054, 2056, 2058, 2059, 2060, 2061, 2062, 2063, 2064, 2067,

2068, 2069, 2070, 2071, 2072, 2073, 2074, 2076, 2077, 2078, 2079, 2080,

2081, 2083, 2084, 2085, 2086, 2088, 2089, 2090, 2091, 2092, 2093, 2095,

2096, 2097;Jaosia Jaszunska: 2061;Dag Jonsson: 2064, 2078;Neven Juri �c: 2038, 2060;Jamshid Kholdi: 2012, 2062;Friend H. Kierstead Jr.: 2045, 2062, 2066, 2068;N. Kildonan: 2016;Murray S. Klamkin: 2014, 2024, 2030, 2034, 2043, 2044, 2045,

2049, 2050, 2054, 2064, 2073, 2078, 2084, 2086, 2087, 2090, 2095, 2099;V�aclav Kone�cn �y: 2012, 2014, 2016, 2020, 2032, 2034, 2035, 2038,

2042, 2043, 2044, 2045, 2050, 2061, 2062, 2063, 2064, 2066, 2067, 2068,

2070, 2071, 2072, 2073, 2074, 2075, 2077, 2078, 2079, 2082, 2083, 2088,

2090, 2091, 2092, 2093, 2095, 2099;Joe Konhauser: 2006;

384

Polly Kontopoulou: 2076;Jisho Kotani: 2042, 2053;Hiroshi Kotera: 2026;Marcin E. Kuczma: 2010, 2018, 2028, 2048, 2057, 2080;Mitko ChristovKunchev: 2055, 2056, 2067, 2068, 2069, 2091;Sai C. Kwok: 2032, 2064, 2068;Jari Lappalainen: 2041;Kee-Wai Lau: 2010, 2011, 2016, 2020, 2021, 2022, 2027, 2029,

2031, 2032, 2034, 2035, 2037, 2050, 2054, 2056, 2060, 2062, 2064, 2065,

2068, 2070, 2072, 2073, 2074, 2075, 2076, 2077, 2078, 2079, 2080, 2082,

2087, 2088, 2089, 2090, 2091, 2093, 2095, 2097, 2100;Thomas Leong: 2064, 2068, 2070;Katheleen E. Lewis: 2034, 2042, 2045, 2081, 2092, 2094, 2098;David Lindsey: 2081;Andy Liu: 2070;Gary MacGillivray: 2085;Maria Ascensi �on L �opez Chamorro: 2008, 2016, 2035,

2038, 2045, 2050, 2051, 2053, 2056, 2059, 2063, 2071;John Magill: 2066, 2076, 2098;David E.Manes: 2007, 2016, 2020, 2066, 2068, 2081, 2083, 2084,

2088, 2090, 2092, 2094, 2095, 2098;Beatriz Margolis: 2020, 2050, 2062, 2068;J.A. McCallum: 2009, 2014, 2016, 2022, 2050, 2066, 2068, 2074,

2076, 2081;John Grant Mcloughlin: 2066, 2092;Bill Meisel: 2022;Stewart Metchette: 2022, 2074;P. Molenbroek: 2017;Pieter Moree: 2007;Vedula N. Murty: 2023, 2056, 2060, 2064, 2067, 2068;J.E. Oliver: 2016;Victor Oxman: 2031, 2032, 2036, 2038, 2067, 2099;Michael Parmenter: 2022, 2042, 2052, 2058;Dan Pedoe: 2017;P. Penning: 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2027, 2030,

2035, 2038, 2040, 2042, 2043, 2045, 2047, 2055, 2059, 2060, 2071, 2072,

2074, 2075, 2076, 2079, 2081, 2082, 2083, 2086, 2088, 2089, 2091, 2094,

2096;Gottfried Perz: 2007, 2016, 2017, 2032, 2036, 2038, 2041, 2042,

2045, 2055, 2060, 2061, 2063, 2064, 2066, 2067, 2068, 2069, 2075, 2076,

2081, 2096, 2099;Waldemar Pompe: 2008, 2013, 2023, 2047, 2050, 2061, 2063,

2064, 2067, 2069, 2070, 2089;Bob Prielipp: 2015, 2064, 2068, 2083, 2093;Proof: 2099;Cory Pye: 2060, 2062, 2066, 2068, 2081, 2088, 2095;Stanley Rabinowitz: 2046, 2056, 2065, 2074, 2083;Neol Reid: 2091;Juan-Bosco RomeroM�arquez: 2068, 2078, 2095;Jawad Sadek: 2030;Crist �obal S �anchez{Rubio: 2035, 2038, 2043, 2062, 2063, 2064,

2067, 2068, 2069, 2079, 2081, 2088;Bill Sands: 1823, 2009;Hristos Saraghiotes: 2076, 2088, 2092, 2098;K.R.S. Sastry: 2012, 2022, 2033, 2042, 2052, 2062, 2068, 2072,

2081, 2092;Joel Schlossberg: 2091, 2092, 2094, 2095, 2096, 2097;Robert P. Sealy: 2009, 2074, 2080, 2088;

Harry Sedinger: 2070;Heinz-J�urgen Sei�ert: 2009, 2012, 2016, 2020, 2028, 2030,

2031, 2032, 2034, 2034, 2037, 2038, 2045, 2050, 2052, 2060, 2062, 2064,

2065, 2067, 2068, 2070, 2072, 2073, 2074, 2076, 2077, 2078, 2081, 2083,

2084, 2085, 2088, 2090, 2092, 2095, 2097;Toshio Seimiya: 2008, 2011, 2017, 2021, 2027, 2031, 2033, 2038,

2041, 2043, 2044, 2047, 2051, 2055, 2059, 2061, 2063, 2067, 2069, 2069,

2071, 2073, 2075, 2082, 2083, 2089, 2091;Catherine Shevlin: 2025;Koji Shoda: 2026;Ashish Kr. Singh: 2008, 2021, 2027, 2034, 2043, 2045, 2047,

2048, 2062, 2064, 2067, 2068, 2069;D.J. Smeenk: 2008, 2010, 2017, 2019, 2021, 2027, 2031, 2033,

2035, 2041, 2043, 2047, 2055, 2069, 2071, 2075, 2079, 2082, 2089, 2091,

2096;Digby Smith: 2032, 2050, 2052, 2056, 2068, 2072, 2076, 2078,

2083;Lawrence Somer: 2012, 2092;Frederick Stern: 2040;David R. Stone: 2016, 2022, 2040, 2042, 2045, 2050, 2072, 2074,

2076, 2081, 2092, 2094, 2097;Moshe Stupel: 2067;M.V. Subbarao: 2034;David Tascione: 2076;Panos E. Tsaoussoglou: 2014, 2016, 2022, 2031, 2032, 2036,

2038, 2042, 2044, 2045, 2050, 2056, 2058, 2060, 2062, 2064, 2067, 2068,

2071, 2072, 2074, 2076, 2081, 2082, 2083, 2084, 2091, 2095;So�ya Vasina: 2028;David C. Vella: 2081;Dan Velleman: 2006;John Vlachakis: 2016, 2020;Stan Wagon: 2006, 2050;Johannes Waldmann: 2064, 2073, 2095;Edward T.H. Wang: 2016, 2032, 2064, 2068, 2078;Hoe Teck Wee: 2008, 2010, 2021, 2022, 2023, 2027, 2028, 2031,

2032, 2038, 2041, 2042, 2048, 2054, 2059, 2071, 2080, 2096;Chris Wildhagen: 2007, 2008, 2012, 2014, 2016, 2022, 2032,

2038, 2045, 2050, 2060, 2062, 2064, 2066, 2068, 2070, 2072, 2074, 2076,

2077, 2080, 2081, 2087, 2088, 2090, 2095, 2097;Susan Schwartz Wildstrom: 2042, 2050, 2066, 2068;Kenneth M. Wilke: 2062, 2066, 2068, 2070, 2072, 2074, 2076,

2081, 2092;John B. Wilker: 2030;Kenneth S. Williams: 2097;Ana Witt: 2063;Aram A. Yagubyants: 2043, 2063, 2086;Paul Yiu: 2037, 2040, 2042, 2060, 2065, 2067, 2068, 2070, 2072,

2074, 2094;Joseph Zaks: 2077;Dong Zhou: 2039;

Austin Academy Problem Solvers: 2021;Science Academy Problem Solvers: 2022, 2027, 2032,

2035, 2038, 2040;Skidmore College Problem Group: 2070;University of Arizona Problem Solving Lab: 2026;

anonymous: 2064, 2065, 2067, 2068, 2069, 2071, 2077.

We would like to identify anonymous! Thank you.