editorial board - cms-smc · 2020. 2. 25. · editorial board editor-in-chief kseniya garaschuk...

Crux Mathematicorum is a problem-solving journal at the secondary and university undergraduate levels,

published online by the Canadian Mathematical Society. Its aim is primarily educational; it is not a research

journal. Online submission:

https://publications.cms.math.ca/cruxbox/

Crux Mathematicorum est une publication de resolution de problemes de niveau secondaire et de premier

cycle universitaire publiee par la Societe mathematique du Canada. Principalement de nature educative,

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Editorial Board

Editor-in-Chief Kseniya Garaschuk University of the Fraser Valley

MathemAttic Editors John McLoughlin University of New Brunswick

Shawn Godin Cairine Wilson Secondary School

Kelly Paton Quest University Canada

Olympiad Corner Editors Alessandro Ventullo University of Milan

Anamaria Savu University of Alberta

Articles Editor Robert Dawson Saint Mary’s University

Associate Editors Edward Barbeau University of Toronto

Chris Fisher University of Regina

Edward Wang Wilfrid Laurier University

Dennis D. A. Epple Berlin, Germany

Magdalena Georgescu BGU, Be’er Sheva, Israel

Shaun Fallat University of Regina

Chip Curtis Missouri Southern State University

Allen O’Hara University of Western Ontario

Guest Editors Vasile Radu Birchmount Park Collegiate Institute

Aaron Slobodin University of Victoria

Ethan White University of British Columbia

Editor-at-Large Bill Sands University of Calgary

IN THIS ISSUE / DANS CE NUMERO

495 MathemAttic: No. 9

495 Problems: MA41–MA45

497 Solutions: MA16–MA20

500 Teaching Problems: No. 6 John McLoughlin

504 Olympiad Corner: No. 377

504 Problems: OC451–OC455

506 Solutions: OC426–OC430

511 Focus On . . . : No. 38 Michel Bataille

518 Problems: 4481–4490

524 Solutions: 4431–4440

536 Snapshot

Crux MathematicorumFounding Editors / Redacteurs-fondateurs: Leopold Sauve & Frederick G.B. Maskell

Former Editors / Anciens Redacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer,

Shawn Godin

Crux Mathematicorum

with Mathematical MayhemFormer Editors / Anciens Redacteurs: Bruce L.R. Shawyer, James E. Totten, Vaclav Linek,

Shawn Godin

MathemAttic /495

MATHEMATTICNo. 9

The problems in this section are intended for students at the secondary school level.

Click here to submit solutions, comments and generalizations to anyproblem in this section.

To facilitate their consideration, solutions should be received by December 30, 2019.

MA41. The diagram shows the densest packing of seven circles in an equi-lateral triangle.

Determine the exact fraction of the area of the triangle that is covered by thecircles.

MA42. Find all functions of the form f(x) =a+ bx

b+ xwhere a and b are

constants such that f(2) = 2f(5) and f(0) + 3f(−2) = 0.

MA43. If n is not divisible by 4, prove that 1n + 2n + 3n + 4n is divisible by5 for any positive integer n.

MA44. Find the largest positive integer which divides all expressions of theform n5 − n3 where n is a positive integer. Justify your answer.

MA45. A sequence s1, s2, . . . , sn is harmonic if the reciprocals of the terms arein arithmetic sequence. Suppose s1, s2, . . . , s10 are in harmonic sequence. Givens1 = 1.2 and s10 = 3.68, find s1 + s2 + · · ·+ s10.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Copyright © Canadian Mathematical Society, 2019


496/ MathemAttic

Les problemes proposes dans cette section sont appropries aux etudiants de l’ecole sec-ondaire.

Cliquez ici afin de soumettre vos solutions, commentaires ougeneralisations aux problemes proposes dans cette section.

Pour faciliter l’examen des solutions, nous demandons aux lecteurs de les faire parvenirau plus tard le 30 decembre 2019.

La redaction souhaite remercier Rolland Gaudet, professeur titulaire a la retraite al’Universite de Saint-Boniface, d’avoir traduit les problemes.

MA41. Le schema suivant montre l’empilement le plus compact de septcercles dans un triangle equilateral.

Determiner exactement la fraction de la surface du triangle qui est couverte parles cercles.

MA42. Determiner toutes les fonctions de la forme f(x) =a+ bx

b+ x, ou a et b

sont des constantes, puis telles que f(2) = 2f(5) et f(0) + 3f(−2) = 0.

MA43. Si n n’est pas divisible par 4, demontrer que 1n + 2n + 3n + 4n estdivisible par 5 pour tout entier positif n.

MA44. Determiner le plus grand entier positif qui divise toutes les expressionsdu genre n5 − n3, ou n est un entier positif.

MA45. Une sequence s1, s2, . . . , sn est dite harmonique si les reciproquesdes termes sont en progression arithmetique. Supposer que s1, s2, . . . , s10 est har-monique et que s1 = 1.2 puis s10 = 3.68. Determiner s1 + s2 + · · ·+ s10.

Crux Mathematicorum, Vol. 45(9), November 2019


MathemAttic /497

MATHEMATTICSOLUTIONS

Statements of the problems in this section originally appear in 2019: 45(4), p. 168–169.

MA16. Prove that every year must have a month in which the 13th is aFriday. What is the maximum number of Friday the 13ths in one year?

Problem proposed by John McLoughlin.

We received 3 submissions, all correct. We present the solution of Richard Hess,modified by the editor.

The 13th of the month taken (mod 7) for a normal year is

6, 2, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4, (1)

for each month respectively. The 13th of the month, mod(7) for a leap year is

6, 2, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5, (2)

for each month respectively. In each of the above cases, 0 through 6 occur at leastonce. It follows that every year must have at least one Friday the 13th.

In a normal year, (1) implies that the maximum number of Friday the 13ths is 3:in February, March, and November. In a leap year, by (2) the maximum numberof Friday the 13ths is also 3, occurring in January, April, and July.

MA17. Let f(n) = 25n − 72n− 1. Determine, with proof, the largest integerM such that f(n) is divisible by M for every positive integer n.

Originally Problem 5 from the 2005 Iowa Collegiate Mathematics Competition.

We received 5 correct solutions. We present a composite of the similar solutionsby Henry Ricardo and Amit Kumar Basistha, done independently.

We prove that M = 48 by noting that 48 is the largest divisor of f(1) and by usinginduction to show that 48|f(n) for every positive integer n.

When n = 1, f(1) = −48, which indeed has 48 as its largest divisor, so M ≤ 48.Now suppose that 48|f(k) for some n = k ≥ 1. Then

f(k + 1) = 25k+1 − 72(k + 1)− 1

= 25k + 24(25k)− 72k − 72− 1

= (25k − 72k − 1) + 24(25k)− 72

= (25k − 72k − 1) + 24(25k − 3).

In the last expression, the first summand is assumed divisible by 48; and the secondterm, as the product of 24 and an even number, is also a multiple of 48. Thus,48|f(k + 1), so M = 48.


498/ MathemAttic

MA18. Twenty calculus students are comparing grades on their first twoquizzes of the year. The class discovers that every pair of students received thesame grade on at least one of the two quizzes. Prove that the entire class receivedthe same grade on at least one of the two quizzes.

No solutions were received. The problem remains open.

MA19. Choose N elements of 1, 2, 3, . . . , 2N and arrange them in increasingorder. Arrange the remaining N elements in decreasing order. Let Di be theabsolute value of the difference of the ith elements in each arrangement. Provethat

D1 +D2 + · · ·+DN = N2.


We received 3 submissions, all correct. We present the solution provided by HenryRicardo.

Suppose we have

a1, a2, . . . , aN ∪ b1, b2, . . . , bN = 1, 2, . . . , 2N

such thata1 < a2 < · · · < aN and b1 > b2 > · · · > bN .

For each i ∈ 1, 2, . . . , N, one of the two numbers ai, bi is less than or equal to Nand the other is greater than N . If we assume the contrary, ai ≤ N and bi ≤ N ,then

a1 < · · · < ai ≤ N and bN < · · · < bi ≤ N.Thus we have found

i+ (N − i+ 1) = N + 1

distinct positive integers which are less than or equal toN , which is a contradiction.

Similarly, we cannott have ai > N and bi > N .

It follows that

D =N∑i=1

Di

= |a1 − b1|+ |a2 − b2|+ · · ·+ |aN − bN |= (N + 1) + (N + 2) + · · ·+ (N +N)− 1− 2− · · · −N

= N2.


MathemAttic /499

MA20.

a) A line segment of length 11 is randomly cut into three pieces, each of integerlength. What is the probability that the three pieces can be formed into a(non-degenerate) triangle?

b) Can you find the probability above for a line segment of any odd integerlength n = 2k + 1?


We received 2 submissions, both correct. We present the solution by Richard Hesswith extensive details by the editor.

b) We assume that all 3 pieces have positive integer length. Let

S = (x, y) ∈ Z2 : 1 ≤ x < y ≤ 2k.

A cutting of the line segment corresponds to a pair (x, y) ∈ S, where xand y are the locations of the cuts along the line segment, and so the threepieces will have lengths x, y− x, and 2k+ 1− y. We need to determine howmany elements of S result in pieces forming a triangle. Suppose (a, b) ∈ Srepresents a cut with pieces forming a triangle. Note that the pieces form atriangle if and only if all pieces have length at most k. Thus

1 ≤ a, b− a, 2k + 1− b ≤ k,

which is equivalent to 1 ≤ a ≤ k and k + 1 ≤ b ≤ k + a. It follows that thenumber of pairs representing a triangle is

k∑a=1

|b ∈ Z : k + 1 ≤ b ≤ k + a| =k∑a=1

a =k(k + 1)

2.

Since

|S| = 2k(2k − 1)

2,

the probability of choosing a pair representing a triangle is

k(k + 1)

2k(2k − 1)=

k + 1

4k − 2.

a) Using the above formula, the probability is 1/3.


500/ Teaching Problems

TEACHING PROBLEMSNo.6

John McLoughlin

Numerical Curiousities as Starting Points

Many numbers have curious properties that can become starting points for math-ematical discussion including problem solving. This feature of Teaching Problemsoffers a range of examples for potential use in the development of numerical ap-preciation as well as problem solving.

What is meant by appreciating numbers more? Begin with the number 370 as anexample. Perhaps you are unaware of the curious property of this number. Thesum of the cubes of its digits is the number itself. That is,

370 = 33 + 73 + 03 = 27 + 343 + 0.

Clearly, it follows that 371 would share this property, as only the 0 is swapped for1 with each of these digits being their own cubes. You are challenged to find twoother three-digit numbers that also have this property.

Numerical appreciation can be advanced through skill building also, without nec-essarily engaging in problem solving. For instance, listing the factors of 220 and284 (excluding the numbers themselves) we get 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110and 1, 2, 4, 71, 142, respectively. Summing the numbers in each of these lists weget the other number. The sum of the proper divisors of 220 is 284, and the sumof the proper divisors of 284 is 220. Hence, 220 and 284 are called an amicablepair.

Why does this work?

Likewise, another approach to enhancing wonder is to engage students in an ac-tivity or skill-related sequence of calculations that results in something surprising.The motivation is presented to figure out the “why” of the surprise. Consider thefollowing sequence of steps:

• Take any three different digits.

• Arrange them to form the largest and smallest possible three-digit numbers.

• Subtract these two numbers to create the largest positive difference.

• Now take this result and add it to its reversal (the three digits ordered inthe opposite way).

What is your result? It will always be 1089. The challenge is to figure out whydoes this work. You may find it helpful to consider using the three digits a, b, andc with a > b > c.


John McLoughlin /501

That number 1089 is an interesting number in itself, as 1089× 9 = 9801. There isonly one other four-digit number that when multiplied by a single digit results ina product that is its own reversal, aside from trivial cases such as 3333×1 = 3333.

Product Reversal

This particular example is one that I have used in many classes as a problem foraddressing issues with number sense. The answer is far from obvious, thus ensuringadequate time for processing on the part of students. Also, it lends itself nicely toclass discussion. Generally my approach is to let everyone play with the problemfor just a few minutes, and then move into discussion where any student can offeronly one insight that moves us forward collectively in solving the problem.

Determine values for A, B, C, and D that make the following productcorrect:

A B C D× 4

D C B A

(Note: Implicit to such a problem are two assumptions: the leading digit in anumber is not 0, and each letter represents a distinct digit.)

Typically, the first observation concerns the value of A or the quantity “ABCD”as one notices the restriction imposed by the four digit product. (The use ofquotation marks is to indicate that the number contains four digits and it is notthe product of A, B, C, and D that is being referred to here.) Hence, it followsthat A can not be greater than 2. In fact, A must be 2 as A is also the ones’ digitin a multiple of 4, thus making it necessary for A to be an even number. So wehave A = 2. This implies that D = 8 or 9, but in fact if we looked from the otherpoint of view we see that D × 4 ends in 2 making D either 3 or 8. So it followsthat D = 8. Let us reconsider the product from this stage.

2 B C 8× 4

8 C B 2

There are different ways of seeing this to home. One valid way is to make theobservation that B is less than 2 since 2300× 4 > 9000. Further, B must be oddsince any multiple of 4 ending in 2 must have a tens’ digit that is odd. Alternatively,note that 8 × 4 = 32 will produce a carry of 3 to be added to 4 × C, thus resultingin an odd value for B. So B = 1.

When the value of B has been determined, we can solve to get C = 7. Returningto that carrying idea we have 4 × C + 3 ending in 1. This makes it necessaryfor 4 × C to end in 8, thus, giving C = 2 (inadmissible since A = 2) or C = 7.Checking we find that indeed 2178× 4 = 8712.


502/ Teaching Problems

A Factorial Problem

Find all three-digit numbers that equal the sum of the factorials of their digits.That is, find values of x, y and z (not necessarily different) for which the three-digitnumber “xyz” is equal to x! + y! + z!.

Before commencing to attack this problem, it is worth mentioning that the unusualpresentation almost makes it a certainty that the problem is not going to be solvedquickly. Again this is advantageous in teaching situations as it necessitates timefor processing and understanding the problem. Further, it is unclear at first ifthere will be a unique solution, if any. The best way to begin is to make sureone understands the problem. Make up a three-digit number and try it out. Forexample, 281 does not work. However, something is learned quickly as the valueof 8! is much too large. In fact, this points to the need to identify the values ofthe various factorials.

These are summarized here:

0! = 1, 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720

and 7! > 1000. Our first significant insight is to note that the values of the factorialsof 7, 8, and 9 preclude them from appearing as digits in our number. Further, wesee that a 6 can not appear in the number either as the inclusion of the digit 6would force one of the digits to be at least 7 due to the fact that 6! > 700. Hence,our values of the digits are restricted to 0, 1, 2, 3, 4, and 5.

What do we notice next? If you observe that 4! = 24 and 3×24 < 100, it becomesevident that the number must have at least one 5 in it to enable the sum of thefactorials of the digits to be a three-digit number. We can quickly see that 555is unacceptable as 3 × 120 is not equal to 555. So really there are two cases toconsider, namely the number has one 5 or two 5’s in it. Further, the maximumattainable value would be 5! + 5! + 4!. It follows that the three-digit number is notgreater than 500, and hence, any 5’s do not appear in the hundreds’ place.

Case 1: There are two 5’s and the number is of the form x55.

This possibility can be dismissed by noting that since 5! + 5! = 240, the first digitmust be 2 and checking shows this does not work.

Case 2: There is only one 5.

Since 5! = 120 and it is impossible to achieve a sum greater than 200, the numbermust begin with 1. Therefore, the number could be 15y or 1z5. In either case, wehave 121 as the sum of 1!+5!. It is impossible to reach 150 by adding 4! and hencethe 15y case is not workable. The latter case of 1z5 requires adding a value thatends in 4, thus leaving 24 as that possibility. Checking we find that 145 actuallyworks as 1 + 24 + 120 gives the required sum.

In conclusion, 145 is the unique three-digit number equal to the sum of the facto-rials of its digits.


John McLoughlin /503

Some questions to try.

Keeping with the theme of this feature, here are a couple of challenges for yourconsideration. In each case, find any numbers for which the property holds.

1. Three different digits a, b, and c form a three-digit number. Two of these digitscan be selected and ordered to make six different two-digit numbers. Take eachof these two-digit numbers and multiple them separately by the digit that is notpart of that number. Sum these six values together. Lo and behold, the result isthe original three-digit number.

2. There is a four-digit number with a special property. Suppose its digits arearranged in decreasing order to form the four-digit number M . Instead, if its digitsare arranged in increasing order, we get the four-digit number N . The curious factis that M −N results in the original number.


504/ OLYMPIAD CORNER

OLYMPIAD CORNERNo. 377

The problems featured in this section have appeared in a regional or national mathematicalOlympiad.

Click here to submit solutions, comments and generalizations to anyproblem in this section


OC451. Determine the least natural number a such that

a ≥n∑k=1

ak cos(a1 + · · ·+ ak)

for any nonzero natural number n and for any positive real numbers a1, a2, . . . , anwhose sum is at most π.

OC452. Let ABCD be a square. Consider the points E ∈ AB, N ∈ CDand F,M ∈ BC such that triangles AMN and DEF are equilateral. Prove thatPQ = FM , where P = AN ∩DE and Q = AM ∩ EF .

OC453. Let n ≥ 2 be an integer and let A,B ∈ Mn(C). If (AB)3 = On, isit true that (BA)3 = On? Justify your answer.

OC454. Find all the functions f : N → N having the following property foreach natural number m: if d1, d2, . . . , dn are all the divisors of the number m, then

f(d1)f(d2) · . . . · f(dn) = m.

OC455. Let D be a point on the base AB of an isosceles triangle ABC.Select a point E so that the ADEC is a parallelogram. On the line ED, take apoint F such that E ∈ DF and EB = EF . Prove that the length of the chordthat the line BE cuts on the circumcircle of triangle ABF is twice the length ofthe segment AC.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .



OLYMPIAD CORNER /505

Les problemes presentes dans cette section ont deja ete presentes dans le cadre d’uneolympiade mathematique regionale ou nationale.

Cliquez ici afin de soumettre vos solutions, commentaires ougeneralisations aux problemes proposes dans cette section.



OC451. Determiner le plus petit nombre naturel a tel que

a ≥n∑k=1

ak cos(a1 + · · ·+ ak)

pour tout nombre naturel non nul n et pour tous nombres reels positifs a1, a2, . . . , andont la somme est au plus π.

OC452. Soit ABCD un carre. Considerer des points E ∈ AB, N ∈ CD, etF,M ∈ BC tels que les triangles AMN et DEF sont equilateraux. Demontrerque PQ = FM , ou P = AN ∩DE et Q = AM ∩ EF.

OC453. Soit n ≥ 2 un entier et soient A,B ∈Mn(C). Si (AB)3 = On, est-iltoujours le cas que (BA)3 = On ? Justifier votre reponse.

OC454. Determiner toutes les fonctionsf : N → N ayant la propriete quepour tout nombre naturel m, si d1, d2, . . . , dn sont tous les diviseurs de m, alors

f(d1)f(d2) · . . . · f(dn) = m.

OC455. Soit D un point sur la base du triangle isocele ABC. Choisir unpoint E tel que ADEC est un parallelogramme. Sur la ligne ED, soit un point Ftel que E ∈ DF et EB = EF . Puis, soit la corde que la ligne BE taille dans lecercle circonscrit du triangle ABF ; demontrer que la longueur de cette corde estle double de la longueur du segment AC.




OLYMPIAD CORNERSOLUTIONS


OC426. Find all pairs of natural numbers a and k such that for every positiveinteger n relatively prime to a, the number ak

n+1 − 1 is divisible by n.

Originally Moscow MO, 4th Problem, Grade 9, Final Round 2017.

We received 1 correct submission. We present the solution by Sushanth SathishKumar.

We claim that all solutions are of the form a = 1, and k is an arbitrary naturalnumber. Clearly all these work.

Next we assume that a > 1. We show that this is not possible.

Assume k is even. Take n = a+ 1. We obtain 1 ≡ aka+1+1 ≡ (−1)k

a+1+1 moduloa + 1. This reduces to 1 ≡ −1 (mod a+ 1), or 2 ≡ 0 (mod a+ 1) which impliesa = 1; contradiction since we assumed a > 1.

Assume that k is odd. Pick a prime p > 2 satisfying gcd(p, a) = 1. Let g bethe order of a modulo p. Since ap−1 ≡ 1 (mod p) by Fermat’s little theorem andak

p+1 ≡ 1 (mod p) as given, it follows that g|p− 1 and g|kp + 1.

The latter condition implies

k2p ≡ 1 (mod g) ,

and that the possible orders of k modulo g are 1, p, 2, and 2p.

If the order is 1, then k ≡ 1 (mod g), and kp + 1 ≡ 2 (mod g). It follows that g|2,and a2 = (ag)2/g ≡ 1 (mod p). Since k is odd, this shows ak+1 ≡ 1 (mod p).

If the order is p, then g|kp − 1, which in combination with g|kp + 1 forces g|2.Using the same logic as in the above paragraph, we obtain ak+1 ≡ 1 (mod p).

If the order is 2, then kp−1 ≡ (k2)(p−1)/2 ≡ 1 (mod g). So, −1 ≡ k(kp−1) ≡ k(mod g). Thus, g|k + 1, which implies ak+1 ≡ 1 (mod p).

If the order is 2p then 2p|φ(g), where φ is Euler’s Totient function. Therefore2p ≤ φ(g) < g, which is impossible since g|p− 1.

Hence the order of k modulo g can be 1, 2, or p. As shown before ak+1 ≡ 1(mod p), or equivalently p|ak+1 − 1 for every prime p > 2 satisfying gcd(p, a) = 1.However, this is not possible because we can select such a prime p with p > ak+1−1.In conclusion we cannot have a > 1.



OC427. In a scalene triangle ABC, ∠C = 60 and Ω is its circumcircle. Onthe angle bisectors of ∠A and ∠B take two points A′ and B′, respectively suchthat AB′ ‖ BC and BA′ ‖ AC. The line A′B′ intersects Ω at points D and E.Prove that triangle CDE is isosceles.

Originally Russia MO, 7th Problem, Grade 9, Final Round 2017.

We received 1 submission. We present the solution by Andrea Fanchini.

We use barycentric coordinates with reference to 4ABC.

Because ∠C = 60, we have that

c2 = a2 + b2 − ab, c = R√

3, s =√

3(R+ r).

The angle bisectors of ∠A and ∠B are

AI : cy − bz = 0, BI : cx− az = 0.

Then the points A′ and B′ are

A′ = AI ∩BCA∞ = (−c : b : c), B′ = BI ∩ABC∞ = (a : −c : c).

Therefore, the line A′B′ : c(b + c)x + c(a + c)y + (c2 − ab)z = 0 intersects thecircumcircle of 4ABC at points D and E given by

D (a(a+ c)(b− a) : b(b+ c)(a− b) : c(b+ c)(a+ c)) , E(a(b− a) : b(a− b) : c2

).

Finally, we have

CD = DE =»R(R− 2r).



OC428.

(a) Prove that there exist functions f : R → R and g : R → R such thatf g = g f , f f = g g and ∀x ∈ R f(x) 6= g(x).

(b) Prove that if f : R → R and g : R → R are continuous functions such thatf g = g f and ∀x ∈ R f(x) 6= g(x), then ∀x ∈ R (f f)(x) 6= (g g)(x).

Originally Romania MO, 2nd Problem, Grade 11, District Round 2017.

We received 2 submissions, both of which are correct. For part (a), we presenteach solver’s example. For part (b), we present a summary of the similar solutionssubmitted by both solvers.

(a) Solution 1, by Oliver Geupel.

Let f be the identity function and let

g(x) =

1 if x = 0

0 if x = 1

1/x if x /∈ 0, 1.

Solution 2, by Nilendu Das.

Let

f(x) =

®x if x is rational,

1− x if x is irrational,

and

g(x) =

®1− x if x is rational,

x if x is irrational.

In both examples, we can check that

(f g)(x) = (g f)(x),

(f f)(x) = x = (g g)(x),

and f(x) 6= g(x) for all x ∈ R.

(b) Suppose f and g are functions that satisfy the given conditions. Since f andg are continuous functions, so is their difference h(x) = f(x)− g(x).

Since ∀x ∈ R f(x) 6= g(x), the function h has no root, and it never changes itssign by Bolzano’s theorem. Hence,

∀x ∈ R f(x) > g(x) or ∀x ∈ R g(x) > f(x).

By symmetry, there is no loss of generality in assuming that f is always greaterthan g. Hence,

f(f(x)) > g(f(x)) = f(g(x)) > g(g(x)).

Thus, (f f)(x) > (g g)(x), from which (f f)(x) 6= (g g)(x) is immediate.



OC429. Let A ∈ Mn(C) (n ≥ 2) with detA = 0 and let A∗ be its adjoint.Prove that (A∗)2 = tr(A∗)A∗, where tr(A∗) is the trace of the matrix A∗.

Originally Romania MO, 4th Problem, Grade 11, District Round 2017.

We received 2 correct submissions. We present the solution by Brian Bradie.

Let A ∈ Mn(C) (n ≥ 2) with detA = 0. By A∗, the adjoint of A, we mean thetranspose of the cofactor matrix of A. Because detA = 0, rank(A) ≤ n − 1. Weconsider two cases.

Case 1. If rank(A) = n− 1, then rank(A∗) = 1, and there exist non-zero vectorsu, v ∈ Cn such that A∗ = uvT . It follows that

(A∗)2 = (uvT )(uvT ) = u(vTu)vT = (vTu)uvT = tr(A∗)A∗.

Case 2. If rank(A) ≤ n − 2, then rank(A∗) = 0. It follows that A∗ = 0 and(A∗)2 = tr(A∗)A∗ = 0 is trivially satisfied.

Editor’s comments. Oliver Geupel made us aware of the following: the adjoint ofa matrix may refer to its conjugate transpose. If A∗ is the conjugate transpose ofA, then the desired result does not generally hold. A counterexample is the matrix

A =

Ñ1 1 01 0 00 1 0

é.

Then tr(A∗) = 1, detA = 0, and

(A∗)2 =

Ñ2 1 11 1 00 0 0

é6= tr(A∗)A∗.

OC430.

(a) Prove that for any choice of n rational numbers ai/bi from the interval (0,1)with distinct pairs (ai, bi) of positive integers, the sum of the denominators

is at least2√

2

3· n 3

2 .

(b) Prove that if we add the restriction that the rational numbers are distinct,

then the sum of the denominators is at least 2 ·Å

2

3n

ã 32

.

Originally Hungary MO, 3rd Problem, 2nd Category, Final Round 2017.

We received 1 correct submission for part (a) and none for part (b). We presentthe solution for part (a) by Oliver Geupel.



(a) Let m be the unique positive integer with the property that

1 + 2 + · · ·+m ≤ n < 1 + 2 + · · ·+m+ (m+ 1).

Then n has the form n =1

2m(m + 1) + ` for some natural number ` ≤ m. In

every choice of n rational numbers from (0, 1) with distinct pairs of numerator anddenominator, there are at most k occurrences of the denominator k + 1 for anyk ≥ 1. Hence, the sum S of denominators of the choice is at least

m∑k=1

k(k + 1) + `(m+ 2) =m∑k=1

(k2 + k) + `(m+ 2)

=1

6m(m+ 1)(2m+ 1) +

1

2m(m+ 1) + `(m+ 2)

=1

3(m+ 2)(m(m+ 1) + 3`).

Thus,

(3S)2

(2n)3≥Åm(m+ 1) + 3`

m(m+ 1) + 2`

ã2· (m+ 2)2

m(m+ 1) + 2m> 1.

Consequently, 3S > (2n)3/2, and the result is immediate.


Michel Bataille /511

FOCUS ON...No. 38

Michel Bataille

Solutions to Exercises from Focus On... No. 32 – 36

From Focus On... No. 32

1. Through a point P exterior to a given circle pass a secant and a tangent to thecircle. The secant intersects the circle at A and B, and the tangent touches thecircle at C on the same side of the diameter through P as A and B. The projectionof C onto the diameter is Q. Prove that QC bisects ∠AQB.

We observe that the line CQ is the polar of P with respect to the circle. It followsthat if AB intersects CQ at U , then P,U divide A,B harmonically. As a result,the pencil of lines QP,QU,QA,QB is harmonic. Thus, QP and QU , which areperpendicular, are the bisectors of ∠AQB. In particular, QC bisects ∠AQB.

2. The standard construction for bisecting a line segment involves the use of twoarcs and one straight line. Show that it can, in fact, be done with straight linesand just one arc.

Let AB be the given segment.

(1) Use the allowed arc to obtain the point C symmetrical of A about B;

(2) Join a point I (not on the line AB) to points A,B,C and obtain J such thatIA, IC, IB, IJ is a harmonic pencil (see Figure 1). Note that IJ is parallel to AC(since IB is the median from I in ∆AIC).

(3) Draw the harmonic conjugate of IJ with respect to IA, IB. This line meets


512/ Focus On... Solutions to Exercises from Focus On... No. 32 – 36

AB at the midpoint O of the line segment AB (since IJ is parallel to AB). SeeFigure 2.

Figure 1

Figure 2


1. Evaluate a) sinπ

14sin

3π

14sin

5π

14b) tan

6π

7+ 4 sin

5π

7.

a) Let p denote the given product. Then we have

p cosπ

14=

1

2sin

π

7sin

3π

14cos

π

7=

1

4sin

2π

7cos

2π

7=

1

8sin

4π

7=

1

8cos

π

14

and p = 18 .

(Note that 5π14 = π

2 − π7 ,

3π14 = π

2 − 2π7 ,

4π7 = π

2 + π14 .)

b) We show that tan 6π7 + 4 sin 5π

7 =√

7. Let u = exp(2πi/7). Then, we have

tan6π

7=

2i

1 + u6− i = i · 1 + (u6)7

1 + u6− i = i(−u6 + u12 − u18 + u24 − u30 + u36)

= i(−u6 + u5 − u4 + u3 − u2 + u)



and

4 sin5π

7= 4 sin

2π

7= −2i(u− u6).

It follows that

tan6π

7+ 4 sin

5π

7= i(−u− u2 + u3 − u4 + u5 + u6) = i(−i

√7) =

√7

(by Gauss’s Theorem since for k ∈ 1, 2, 3, 4, 5, 6,(k7

)= 1 if and only if k ∈

1, 2, 4).

2. Prove that a) cos7π

15= 4 sin

2π

15cos

2π

5cos

13π

30

b) 1 + 6 cos2π

7= 2√

7 cos

Å1

3cos−1

1

2√

7

ãa) On the one hand,

cos

Å7π

15

ã= cos

Å2π

3− π

5

ã= −1

2cos(π

5

)+

√3

2sin(π

5

)(1)

and on the other hand,

4 sin

Å2π

15

ãcos

Å2π

5

ãcos

Å13π

30

ã= 2 sin

Å2π

15

ãÇcos( π

30

)−√

3

2

å= −

√3 sin

Å2π

15

ã+ 2 sin

Å2π

15

ãcos( π

30

)= −

√3 sin

Å2π

15

ã+

1

2+ sin

( π10

)=

1

2+ sin

( π10

)− 3

2cos(π

5

)+

√3

2sin(π

5

).

(note that 2π15 = π

3 − π5 ).

Comparing with (1), we see that all boils down to proving that

cos(π

5

)=

1

2+ sin

( π10

)or 1− 2 sin2

( π10

)=

1

2+ sin

( π10

).

But

sin( π

10

)= cos

(π2− π

10

)= cos

Å2π

5

ãand the latter follows from 4 cos2

(2π5

)+ 2 cos

(2π5

)= 1 (see Focus On... No 34 p.

28).

b) Let θ = cos−1 12√7. Then, θ ∈ (0, π2 ) and

4 cos3θ

3− 3 cos

θ

3= cos θ =

1

2√

7. (2)



The numbers cos 2π7 = − cos 5π

7 , cos 4π7 = − cos 3π

7 , cos 6π7 = − cos π7 are the roots

of the polynomial P (x) = 8x3 + 4x2 − 4x − 1 (ibid. p. 29) and it follows that1 + 6 cos 2π

7

2√

7,

1 + 6 cos 4π7

2√

7,

1 + 6 cos 6π7

2√

7are the roots of

Q(x) = P

Ç2x√

7− 1

6

å=

14√

7

27

Å4x3 − 3x− 1

2√

7

ã.

Recalling (2), we see that Q(cos θ3

)= 0.

In addition, since 0 < θ3 < θ < π

2 , we have cos θ3 > cos θ, hence 2√

7 cos θ3 − 1 > 0.

Thus2√

7 cos θ3 − 1

6is a positive root of P (x) and since the only positive root of

P (x) is cos 2π7 , we must have

2√

7 cos θ3 − 1

6= cos 2π

7 . The result follows.


1. Let Xn =

∫ π4

0

(tanx)n dx. Compute Xn +Xn+2 and deduce that Xn ∼ 12n .

We calculate

Xn +Xn+2 =

∫ π4

0

(tanx)n(1 + tan2 x) dx =

ï(tanx)n+1

n+ 1

òπ/40

=1

n+ 1.

For 0 ≤ x ≤ π4 we have 0 ≤ tanx ≤ 1, hence (tanx)n+1 ≤ (tanx)n and therefore

Xn+1 ≤ Xn for n ∈ N. For n > 2, we deduce that

1

n+ 1= Xn+2 +Xn ≤ 2Xn ≤ Xn +Xn−2 =

1

n− 1

so that Xn satisfies nn+1 ≤ 2nXn ≤ n

n−1 . The squeeze principle now giveslimn→∞

2nXn = 1, as desired.

2. Let Yn =

∫ 1

0

xn lnx

xn − 1dx. Show that Yn ∼ α

n2 for some positive α.

We observe that

Yn =1

n

∫ 1

0

xn · ln(xn)

xn − 1dx =

1

n2

Çn ·∫ 1

0

xn · f(xn) dx

åwhere f is the continuous function defined on [0, 1] by f(x) = ln x

x−1 if x 6= 1,f(1) = 1. Theorem 1(i) in Focus On... No 36 p. 138 then gives

limn→∞

n ·∫ 1

0

xn · f(xn) dx =

∫ 1

0

f(x) dx =

∫ 1

0

lnx

x− 1dx =

∫ 1

0

− ln(1− u)

udu =

π2

6



and finally, Yn ∼ αn2 with α = π2

6 .

3. Let f1(t) =1

1 + tand for k ≥ 2, let fk(t) =

d

dt(tfk−1(t)). Prove that for any

integer m ≥ 1,∫ 1

0

tn

1 + tdt =

f1(1)

n− f2(1)

n2+ · · ·+ (−1)m−1

fm(1)

nm+ o(1/nm)

as n→∞.

Integrating by parts, we obtain the recursion∫ 1

0

tnfk−1(t) dt =fk−1(1)

n− 1

n

∫ 1

0

tnfk(t) dt

from which we readily deduce that for m ≥ 1,∫ 1

0

tn

1 + tdt =

f1(1)

n− f2(1)

n2+ · · ·+ (−1)m−1

fm(1)

nm+

(−1)m

nm

∫ 1

0

tnfm+1(t) dt.

Since limn→∞

n ·∫ 1

0

tnfm+1(t) dt = fm+1(1) (Theorem 1(ii), ibid.), we have

(−1)m

nm

∫ 1

0

tnfm+1(t) dt = o

Å1

nm

ãand the result follows.

4. Let Zn =

∫ 1

0

(ax2 + bx+ c)n dx where a, b are negative numbers and a + b + c

is positive. Use this number of Focus On... to obtain Zn ∼ cn+1

−nb already foundin No 10.

Let φ be the function defined on [0, 1] by φ(x) = ax2 + bx+ c. Since a, b < 0, φ isstrictly decreasing on [0, 1] with φ(0) = c, φ(1) = a+ b+ c. Thus, φ is a bijectionfrom [0, 1] onto [a+ b+ c, c]. It is easy to obtain that for y ∈ [a+ b+ c, c], we have

φ−1(y) =−b−

√b2 − 4ac+ 4ay

2aand (φ−1)′(y) =

−1√b2 − 4ac+ 4ay

.

As a result, the change of variables y = ax2 + bx+ c = φ(x) leads to

Zn =

∫ c

a+b+c

yn · 1√b2 − 4ac+ 4ay

dy.

Now, we remark that 0 < a+ b+ c < c and that the function y 7→ 1√b2−4ac+4ay

is



continuous on [0, c]. Thus, we can transform Zn as follows:

Zn =

∫ c

0

yn · 1√b2 − 4ac+ 4ay

dy −∫ a+b+c

0

yn · 1√b2 − 4ac+ 4ay

dy

= cn+1

∫ 1

0

un · 1√b2 − 4ac+ 4acu

du

− (a+ b+ c)n+1

∫ 1

0

un · 1√b2 − 4ac+ 4a(a+ b+ c)u

du.

Using Theorem 1(ii) again, we see that∫ 1

0

un · 1√b2 − 4ac+ 4acu

du ∼ 1

−nband ∫ 1

0

un · 1√b2 − 4ac+ 4a(a+ b+ c)u

du ∼ 1

−n(2a+ b)

and because(a+ b+ c)n+1

n(2a+ b)= o

Åcn+1

nb

ãas n→∞, we can conclude that

Zn ∼cn+1

−nb .


1. Given an acute triangle ABC, let O be its circumcenter, let M be the inter-section of lines AO and BC, and let D be the other intersection of AO with thecircumcircle of ABC. Let E be that point on AD such that M is the midpoint ofED. Let F be the point at which the perpendicular to AD at M meets AC. Provethat EF is perpendicular to AB.



(We use the conventions and notations set in the first part of Focus On... No. 36).

Clearly, d = −a and since d+ e = 2m, we have e = 2m+a. Since F is the point ofintersection of the lines AC and the perpendicular to AO at M , f satisfies their

respective equations z + acz = a+ c andz −ma

+z −ma

= 0. We readily obtain

f =a2(cm− 1) + c(m− a)

c− a .

Now, lines EF and AB are perpendicular if and only if f − e = ab(f − e), that is,

a2cm+m(2a− c)− 2ac = 2ab−mb+ ab(a− 2c)m.

Because M is on OA, m satisfies am − am = 0, that is, m = a2m. Using thisexpression of m, the above condition easily becomes m+ bcm = b+ c. Since M ison BC, the latter holds and therefore EF and AB are perpendicular.

Note that the hypothesis “ABC acute” is not necessary.

2. Let H, I,O be the orthocenter, incenter, circumcenter of a triangle ABC andlet J be the reflection of I about O. Prove that the line through the midpoints ofJH and BC is parallel to AI.

(We use the conventions and notations set in the second part of Focus On... No36).

Recalling that the affixes of A,B,C, I are a2, b2, c2,−(ab + bc + ca), we see thatthe affixes of J and H are ab+ bc+ ca and a2 + b2 + c2, respectively. Let m andn be the respective affixes of the midpoints of BC and JH. Since the bisector AIintersects the circumcircle at A and A′ with affixes a2 and −bc, we have to showthat n−m = a2bc(n−m). Because 2m = b2+c2 and 2n = a2+b2+c2+ab+bc+ca,the result directly follows from 2n− 2m = a2 + ab+ bc+ ca and

a2bc(2n− 2m) = a2bc

Å1

a2+

1

ab+

1

bc+

1

ca

ã= bc+ ca+ a2 + ab.


518/ Problems

PROBLEMS

Click here to submit problems proposals as well as solutions, commentsand generalizations to any problem in this section.


4481. Proposed by Warut Suksompong.

Find all functions f : R→ R such that

f(x2 + y2) = f(x+ y)f(x− y) + 2f(y)y

for all x, y ∈ R.

4482. Proposed by Miguel Ochoa Sanchez and Leonard Giugiuc.

Let ABC be a triangle with incenter I. The line AI intersects BC at D. A linel passes through I and intersects the sides AB and AC at P and Q, respectively.Show that

AC · [BDIP ]

[API]+AB · [CDIQ]

[AQI]= 2 ·BC +

BC2

AB +AC,

where square brackets denote area.

4483. Proposed by Paul Bracken.

For non-negative integers m and n, evaluate the following sum in closed form

m∑j=0

j2Çj + n

j

å.



Problems /519

4484. Proposed by Leonard Giugiuc and Michael Rozenberg.

Let a, b, c ∈ [0, 2] such that a+ b+ c = 3. Prove that

4(ab+ bc+ ac) ≤ 12− ((a− b)(b− c)(c− a))2

and find when the equality holds.

4485. Proposed by Jonathan Parker and Eugen J. Ionascu.

For every square matrix with real entries A = [ai,j ]i=1..n,j=1,2...n, we define thevalue

GM(A) = maxπ∈Sn

mina1π(1), a2π(2), ..., anπ(n)

where Sn is the set of all permutations of the set [n] := 1, 2, 3, ..., n.Given the 6× 6 matrix

A :=

20 9 7 26 27 1319 18 17 6 12 2522 24 21 11 20 1120 8 9 23 5 1422 17 4 10 36 3321 16 23 35 15 34

find the value GM(A).

4486. Proposed by Marian Cucoanes and Marius Dragan.

Let a, b > 0, c > 1 such that a2 ≥ b2c. Compute

limn→∞

(a− b√c)(a− b 3√c) · · · (a− b n√c).

4487. Proposed by Martin Lukarevski.

Let a, b, c be the sides of a triangle ABC, ma, mb, mc the corresponding mediansand R, r its circumradius and inradius respectively. Prove that

a2

m2b +m2

c

+b2

m2c +m2

a

+c2

m2a +m2

b

≥ 4r

R.

4488. Proposed by George Apostolopoulos.

Let ABC be an acute-angled triangle. Prove that

√cotA+

√cotB +

√cotC ≤

…cot

A

2· cot

B

2· cot

C

2.


520/ Problems

4489. Proposed by Arsalan Wares.

Regular hexagon A has its vertices at points A1, A2, A3, A4, A5 and A6. Sixcircular congruent arcs are drawn inside hexagon A and all six pass through thecenter of A. The terminal points of each of the six arcs divide the sides of A in theratio 3 : 7. The six regions within A that are bounded only by circular arcs havebeen shaded. Find the ratio of the area of A to the area of the shaded region.

4490. Proposed by Borislav Mirchev and Leonard Giugiuc.

A line ` through the orthocenter H of the acute triangle ABC meets the cir-cumcircle at points K on the smaller arc AC and L on the smaller arc BC. IfM,N , and P are the feet of the perpendiculars to ` from the vertices A,B, andC, respectively, prove that PH = |KM − LN |.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


Problems /521

Cliquez ici afin de proposer de nouveaux problemes, de meme que pouroffrir des solutions, commentaires ou generalisations aux problemes

proposs dans cette section.



4481. Propose par Warut Suksompong.

Determiner toute fonction f : R→ R telle que

f(x2 + y2) = f(x+ y)f(x− y) + 2f(y)y

pour tous x, y ∈ R.

4482. Propose par Miguel Ochoa Sanchez et Leonard Giugiuc.

Soit ABC un triangle et I le centre de son cercle inscrit. La ligne AI intersecteBC en D. Une ligne l passe par I et coupe AB et AC en P et Q, respectivement.Demontrer que

AC · [BDIP ]

[API]+AB · [CDIQ]

[AQI]= 2 ·BC +

BC2

AB +AC,

ou le crochet denote la surface.

4483. Propose par Paul Bracken.

Pour m et n des entiers non negatifs, evaluer en forme close la somme

m∑j=0

j2Çj + n

j

å.



522/ Problems

4484. Propose par Leonard Giugiuc et Michael Rozenberg.

Soient a, b, c ∈ [0, 2] tels que a+ b+ c = 3. Demontrer que

4(ab+ bc+ ac) ≤ 12− ((a− b)(b− c)(c− a))2

et determiner toute condition donnant egalite.

4485. Propose par Jonathan Parker et Eugen J. Ionascu.

Pour A = [ai,j ]i=1..n,j=1,2...n une matrice carree avec elements reels, on pose

GM(A) = maxπ∈Sn

mina1π(1), a2π(2), ..., anπ(n)

ou Sn est l’ensemble des permutations de l’ensemble [n] := 1, 2, 3, ..., n. Pour A,matrice 6× 6 donnee par

A :=

20 9 7 26 27 1319 18 17 6 12 2522 24 21 11 20 1120 8 9 23 5 1422 17 4 10 36 3321 16 23 35 15 34

,

determiner GM(A).

4486. Propose par Marian Cucoanes et Marius Dragan.

Soient a, b > 0, c > 1, tels que a2 ≥ b2c. Calculer

limn→∞

(a− b√c)(a− b 3√c) · · · (a− b n√c).

4487. Propose par Martin Lukarevski.

Soient a, b, c les cotes du triangle ABC, ma, mb, mc les medianes, puis R et r lesrayons du cercle circonscrit et du cercle inscrit, respectivement. Demontrer que

a2

m2b +m2

c

+b2

m2c +m2

a

+c2

m2a +m2

b

≥ 4r

R.

4488. Propose par George Apostolopoulos.

Soit ABC un triangle acutangle. Demontrer que

√cotA+

√cotB +

√cotC ≤

…cot

A

2· cot

B

2· cot

C

2.


Problems /523

4489. Propose par Arsalan Wares.

Un hexagone regulier A possede les sommets A1, A2, A3, A4, A5 et A6. Six arcscirculaires congrus sont traces a l’interieur de A, passant tous par le centre de A.Les points terminaux de ces six arcs divisent les cotes de A en ratio 3 : 7. Les 6regions bornees par deux arcs circulaires voisins sont ombragees. Determiner leratio de cette surface par rapport a la surface ombragee.

4490. Propose par Borislav Mirchev et Leonard Giugiuc.

Une ligne ` passant par l’orthocentre du triangle acutangle ABC intersecte lecercle circonscrit, au point K dans le petit arc AC et au point L dans le petit arcAC. Si M,N et P sont les pieds des perpendiculaires vers l a partir des sommetsA,B et C, respectivement, demontrer que PH = |KM − LN |.


524/ Solutions

SOLUTIONSNo problem is ever permanently closed. The editor is always pleased to consider forpublication new solutions or new insights on past problems.


4431. Proposed by Sefket Arslanagic.

Let x, y ≥ 0 and x+ y = 2. Prove that√x2 + 8 +

√y2 + 8 +

√xy + 8 ≥ 9.

We received 18 submissions, of which 17 were correct, and the other one incorrect.We present the solution by Kee-Wai Lau.

By the AM-GM inequality, we have√x2 + 8 +

√y2 + 8 +

√xy + 8 ≥ 3 6

»(x2 + 8)(y2 + 8)(xy + 8),

so it suffices to show that

(x2 + 8)(y2 + 8)(xy + 8) ≥ 729. (1)

Since

(x2 + 8)(y2 + 8) = (xy)2 + 8(x+ y)2 − 16xy + 64 = (xy)2 − 16xy + 96,

we have

(x2 + 8)(y2 + 8)(xy + 8) =((xy)2 − 16xy + 96

)(xy + 8)

= (xy)3 − 8(xy)2 − 32xy + 768

=((xy)2 − 7xy − 39

)(xy − 1) + 729.

Now,

0 ≤ xy =(x+ y)2 − (x− y)2

4= 1− (x− y)2

4≤ 1

gives xy − 1 ≤ 0 and (xy)2 − 7xy − 39 < 0, so that((xy)2 − 7xy − 39

)(xy − 1) + 729 ≥ 729,

so (1) follows.

Editor’s comments. Many submitted solutions used convexity and Jensen’s In-equality involving tedious computations. The solution presented above is the onlyone using nothing more than elementary algebra.


Solutions /525

4432. Proposed by Marian Cucoanes and Marius Dragan.

Let x1, x2, . . . , xk ≥ 0, n ∈ N be such that

(x1 + x2)(x2 + x3) . . . (xk + x1) ≥ (x1 + 1)(x2 + 1) . . . (xk + 1)

Prove that

xn+11

x2 + 1+

xn+12

x3 + 1+ · · ·+ xn+1

k

x1 + 1≥ xn1x2 + 1

+xn2

x3 + 1+ · · ·+ xnk

x1 + 1.

We received 3 correct solutions. One additional solution contained statements thatneeded justification. We present the solution of the proposers.

Setting xk+1 = x1, using the arithmetic-geometric means inequality along withthe hypothesis, we have that

k∑i=1

xn+1i

xi+1 + 1−

k∑i=1

xnixi+1 + 1

=k∑i=1

(xi − 1)(xni − 1)

xi+1 + 1+

k∑i=1

xi − 1

xi+1 + 1

=

[k∑i=1

(xi − 1)(xni − 1)

xi+1 + 1

]+

[k∑i=1

(xi + xi+1)− (xi+1 + 1)

xi+1 + 1

]

=

[k∑i=1

(xi − 1)(xni − 1)

xi+1 + 1

]+

[k∑i=1

xi + xi+1

xi+1 + 1

]− k

≥ 0 + k

[k∏i=1

xi + xi+1

xi+1 + 1

]1/k− k ≥ 0 + k − k = 0,

as desired.

Editor’s comments. Variants of this solution came from Prithwijit De and MarianDinca. Let

f(m) =m∑i=1

xmi (xi+1 + 1)−1.

De used the Cauchy-Schwarz inequality to show that f(m+1)f(m−1) ≥ (f(m))2

for m ≥ 1. From a proof of f(1) ≥ f(0), this allowed an induction argumentto show that f(m) increases with m. Dinca obtained f(k + 1) ≥ f(k) from theweighted Chebyshev inequality(

k∑i=1

wi

)(k∑j=1

wjajbj

)≥(

k∑i=1

wiai

)(k∑j=1

wjbj

)

with ai and bi monotonic in the same direction. For 1 ≤ i ≤ k, we takewi = (xi+1 + 1)−1, ai = xni and bi = xi.


526/ Solutions

4433. Proposed by Leonard Giugiuc.

Let ABCD be a tetrahedron and let M be an interior point of ABCD. IfAB ≥ AC ≥ AD ≥ BC ≥ BD ≥ CD, then prove that

MA+MB +MC +MD < AB +AC +AD.

We received only 1 correct submission, from the proposer, whose solution wepresent.

LetVMBCD

VABCD= x,

VMACD

VABCD= y,

VMABD

VABCD= z, and

VMABC

VABCD= t.

Then x, y, z, t > 0 and x+ y+ z + t = 1. Further, for any point P from space wehave −−→

PM = x−→PA+ y

−−→PB + z

−−→PC + t

−−→PD.

Therefore, −−→AM = y

−−→AB + z

−→AC + t

−−→AD (1)

−−→BM = x

−−→BA+ z

−−→BC + t

−−→BD (2)

−−→CM = x

−→CA+ y

−−→CB + t

−−→CD (3)

−−→DM = x

−−→DA+ y

−−→DB + z

−−→DC (4)

We apply in (1) the triangle inequality. Because no two of y−−→AB, z

−→AC, t

−−→AD are

collinear we obtain the strict inequalities

|−−→AM | = |y−−→AB + z−→AC + t

−−→AD| < y|−−→AB|+ z|−→AC|+ t|−−→AD|,

andMA < yAB + zAC + tAD.

Similarly, from relations (2), (3), and (4) we obtain

MB < xAB + zBC + tBD,

MC < xAC + yBC + tCD,

MD < xAD + yBD + zCD.

Adding the last four inequalities, we have

MA+MB +MC +MD

< x(AB +AC +AD) + y(BA+BC +BD) + z(CA+ CB + CD) + t(DA+DB +DC)

< x(AB +AC +AD) + y(AB +AC +AD) + z(AB +AC +AD) + t(AB +AC +AD)

= (x+ y + z + t)(AB +AC +AD)

= AB +AC +AD,

as required. Above we used the inequalities provided between the sides of thetetrahedron, AB ≥ AC ≥ AD ≥ BC ≥ BD ≥ CD.


Solutions /527

4434. Proposed by Michel Bataille.

Let ABC be a scalene triangle with BC = a,CA = b, AB = c. The internalbisectors of ∠A,∠B,∠C meet the opposite sides at D,E, F and the perpendicularbisectors of AD,BE,CF intersect the lines BC,CA,AB at U, V,W , respectively.Prove that the sum

(b2 − c2)PU2 + (c2 − a2)PV 2 + (a2 − b2)PW 2

is independent of the point P chosen in the plane of ABC and find its value interms of a, b, c.

We received 5 submissions, all of which were correct, and present the solution byLeonard Giugiuc.

Let the external bisector of the angle at A of ∆ABC intersect BC at A′, and letM be the midpoint of AD. Because the angle bisectors at a vertex divide theopposite side externally and internally in the ratio c : b, we have

A′ =bB − cCb− c and D =

bB + cC

b+ c.

Since both AA′ and MU are perpendicular to AD, M the midpoint of AD impliesthat U must be the midpoint of A′D; it follows that

U =A′ +D

2=b2B − c2Cb2 − c2 ,

and for any point P in the plane, therefore,

−−→PU =

b2−−→PB − c2−−→PCb2 − c2 .

Applying the cosine law to ∆PBC, we find that

2−−→PB · −−→PC = 2|−−→PB||−−→PC| cos∠BPC = PB2 + PC2 − a2,

and we deduce that

(b2 − c2)PU2 = (b2 − c2)

Çb2−−→PB − c2−−→PCb2 − c2

å2

=b4PB2 + c4PC2 − b2c2(PB2 + PC2 − a2)

b2 − c2

= b2PB2 − c2PC2 +a2b2c2

b2 − c2 .

Analogously,

(c2 − a2)PV 2 = c2PC2 − a2PA2 +a2b2c2

c2 − a2 ,

(a2 − b2)PW 2 = a2PA2 − b2PB2 +a2b2c2

a2 − b2 .


528/ Solutions

Adding the three equations, we find that the sum we seek equals

a2b2c2Å

1

a2 − b2 +1

b2 − c2 +1

c2 − a2ã,

which is independent of the choice of P , as claimed.

Editor’s comment. The proposer added, in passing, that U, V,W are collinear byMenelaus’s theorem.

4435. Proposed by Dao Thanh Oai.

Let ABC be a triangle with circumcenter O. Let CBC , CCA, CAB be three arbi-trary circles through B and C, C and A, A and B with centers OA, OB and OC ,respectively. Suppose OA,OB and OC meet OBOC , OCOA and OAOB at A′, B′

and C ′, respectively. Show that OAA′, OBB

′ and OCC′ are concurrent.

We received 8 solutions. We present the solution by Prithwijit De.

Observe that OOA, OOB and OOC are the perpendicular bisectors of BC, CAand AB respectively. Since O is the circumcenter of ABC, it follows that

∠A′OOC = ∠AOOC =1

2∠AOB = ∠ACB = ∠C,

∠A′OOB = ∠AOOB =1

2∠AOC = ∠ABC = ∠B.

Therefore,

OBA′

A′OC=

[OBOA′]

[OCOA′]=OBO · sin∠A′OOBOCO · sin∠A′OOC

=

ÅOBO

OCO

ã·Å

sin∠Bsin∠C

ã.


Solutions /529

Similarly, we show that

OCB′

B′OA=

ÅOCO

OAO

ã·Å

sin∠Csin∠A

ã,

OAC′

C ′OB=

ÅOAO

OBO

ã·Å

sin∠Asin∠B

ã.

Thus we obtain

OBA′

A′OC· OCB

′

B′OA· OAC

′

C ′OB= 1

and hence, by the converse of Ceva’s theorem, OAA′, OBB

′ and OCC′ are either

concurrent or parallel.

Editor’s comment. As pointed out by Michel Bataille, it is possible to choose OA,OB and OC such that the lines OAA

′, OBB′ and OCC

′ are parallel instead ofconcurrent. An example is given below:

4436. Proposed by Leonard Giugiuc.

Find all positive solutions to the following equation:

(x+ y + z)2(10x+ 4y + z) =243xyz

2.

We received 8 solutions, all correct. We present the solution by Sefket Arslanagic.

Note first that (x, y, z) = (2, 3, 4) is a solution since

(x+ y + z)2(10x+ 4y + z) = (81)(36) = (243)(12) =243xyz

2.

We show that all solutions are given by k(2, 3, 4) for some k > 0.


530/ Solutions

Let x = 2u, y = 3v, and z = 4w. Then u > 0, v > 0, w > 0. By AM-GMinequality, we have

x+ y + z = 2u+ 3v + 4w = u+ u+ v + v + v + w + w + w + w ≥ 99√u2v3w4,

so(x+ y + z)2 ≥ 81

9√u4v6w8. (1)

Similarly,10x+ 4y + z = 4(5u+ 3v + w) ≥ (4)(9)

9√u5v3w. (2)

Multiplying (1) and (2), we then obtain

(x+ y + z)2(10x+ 4y + z) ≥ (81)(36)uvw

= (36)(22)uvw

= (36)(22)(x

2

)(y3

)(z4

)=

243

2xyz.

Since

(x+ y + z)2(10x+ 4y + z) =243

2xyz,

we conclude that u = v = w = k for some k > 0, so (x, y, z) = k(2, 3, 4), completingthe proof.

4437. Proposed by Leonard Giugiuc and Shafiqur Rahman.

Let a be a real number with a > 1. Find limn→∞(n n√an− n− lnn).

We received 10 correct solutions. An additional solution involved a step that seemedas difficult as the problem posed. We present two solutions.

Solution 1, by Michel Bataille.

The result holds for all positive values of a. Since

ex = 1 + x+x2

2+ o(x2) as x→ 0,

it follows that

n n√an = nelnn/neln a/n

= n

Ç1 +

lnn

n+

ln2 n

2n2+ o

Çln2 n

n2

ååÇ1 +

ln a

n+

ln2 a

2n2+ o

Å1

n2

ãå= n+ lnn+ ln a+ o(1)

as n→∞. Hencelimn→∞

(n n√an− n− lnn) = ln a.


Solutions /531

Solution 2, by Henry Ricardo.

Note that limx→∞ x1/x = 1, limx→0(ax − 1)/x = ln a. Then

n n√an− n− lnn = n1/n · n( n

√a− 1) + n( n

√n− 1)− lnn

= n1/nña1/n − 1

1/n

ô+ n(elnn/n − 1)− lnn

= n1/nña1/n − 1

1/n

ô+

Çlnn+

ln2 n

2n+ o

Çln2 n

n

åå− lnn.

Letting n→∞ yields the required limit, ln a.

Editor’s comments. Florentin Visescu gave a variant of the second solution inwhich he rendered the limit expression as

n1/nña1/n − 1

1/n

ô+

Çln2 n

n

å[elnn/n − 1− lnn

n(lnnn

)2],

and used the fact that limx→0(ex − 1− x)x−2 = 12 .

4438. Proposed by Michel Bataille.

Let a, b, c be nonzero distinct real numbers and let A be the matrixÑ2(a2 − bc) b2 − a2 (c+ b)(c− a)

(a+ c)(a− b) 2(b2 − ca) c2 − b2a2 − c2 (b+ a)(b− c) 2(c2 − ab)

é.

Prove that A is invertible.

The statement as written is not true.

We received 8 submissions. Of those, 2 were incorrect proving the result as stated,3 provided a counterexample to the claim, 2 found an additional condition thatensures the conclusion holds. We present the solution by Marie-Nicole Gras.

We will prove that A is invertible if a, b and c are of the same sign and notnecessarily invertible otherwise.

Let a, b, c be nonzero distinct real numbers; the matrix A is invertible if and onlyif its determinant ∆ is nonzero. We have

∆ =

∣∣∣∣∣∣2(a2 − bc) b2 − a2 (c+ b)(c− a)

(a+ c)(a− b) 2(b2 − ca) c2 − b2a2 − c2 (b+ a)(b− c) 2(c2 − ab)

∣∣∣∣∣∣ .We first add the last two columns to the first one, and we obtain

2(a2 − bc) + (b2 − a2) + (c+ b)(c− a) = a2 + b2 + c2 − ab− bc− ca(a+ c)(a− b) + 2(b2 − ca) + c2 − b2 = a2 + b2 + c2 − ab− bc− caa2 − c2 + (b+ a)(b− c) + 2(c2 − ab) = a2 + b2 + c2 − ab− bc− ca .


532/ Solutions

Then ∆ = (a2 + b2 + c2 − ab− bc− ca)D, with

D =

∣∣∣∣∣∣1 b2 − a2 (c+ b)(c− a)1 2(b2 − ca) c2 − b21 (b+ a)(b− c) 2(c2 − ab)

∣∣∣∣∣∣ =

∣∣∣∣∣∣0 −b2 − a2 + 2ac b2 + bc− ac− ab1 2b2 − 2ac c2 − b20 −b2 + ab+ ac− bc c2 + b2 − 2ab

∣∣∣∣∣∣and we deduce

D = (b2 + a2 − 2ac)(b2 + c2 − 2ab)− (b2 + bc− ab− ac)2

= (b2 + a2)(b2 + c2)− (b2 + b2c2 + a2b2 + a2c2)− 2ac(b2 + c2)

− 2ab(b2 + a2) + 4a2bc− 2b2(bc− ab− ac)− 2(−ab2c− abc2 + a2bc)

= −2(a3b+ b3c+ c3a) + 2abc(a+ b+ c).

Then we obtain

∆ = −(

(a− b)2 + (b− c)2 + (c− a)2)(

(a3b+ b3c+ c3a)− abc(a+ b+ c)).

We put δ = a3b+ b3c+ c3a− abc(a+ b+ c) and we compute

(a+ b+ c) δ = (a+ b+ c)(a3b+ b3c+ c3a)− abc(a+ b+ c)2

= a4b+ ab3c+ a2c3 + a3b2 + b4c+ abc3 + a3bc+ b3c2 + ac4

− a3bc− ab3c− abc3 − 2abc(a+ b+ c)

= a4b+ b4c+ c4a+ a3b2 + b2c3 + c3a2 − 2a2bc− 2ab2c− 2abc2

= b(a2 − bc)2 + c(b2 − ac)2 + a(c2 − ab)2.

If a > 0, b > 0, c > 0 or a < 0, b < 0, c < 0, then a+ b+ c 6= 0; whence, from theabove identity,

δ = 0 ⇐⇒ a2 = bc, b2 = ac, c2 = ab ⇐⇒ a = b = c, since a, b, c ∈ R .

However, by assumption, a, b, c are distinct; then ∆ 6= 0, and the matrix A isinvertible.

If a, b and c are not of the same sign, it is possible that ∆ = 0, with distincta, b, c; for example, if a = 1, b = 2 and c is the real root of x3 − 2x2 + 2x + 2,c ≈ −0.5747430, then ∆ = 0.

4439?. Proposed by Cristi Savescu.

Let M be a set of at least six points in the plane such that any of its six pointsubsets represents, in a certain order, the vertices of two triangles with the samecentroid. Prove that M has exactly six points.

We received one solution by Roy Barbara, which is presented below, modified bythe editor.

It suffices to show that M cannot have seven points. Suppose to the contrary thatM is a set of seven points A0, A1, . . . , A6 in the plane. Since the property in the


Solutions /533

question is invariant under translations and rotations of the plane, we may assumewithout loss of generality that A0 is the origin O and that all the x-coordinates ofthe points are distinct.

The centroid of a triangle ABC can be calculated as 13 (A+B+C). Therefore the

hypothesis can be formulated as seven equations

εk0A0 + εk1A1 + · · ·+ εk6A6 = O

with k ∈ 0, 1, . . . , 6, εkk = 0, and the property that in each equation three of thecoefficients are +1 and three are −1. Setting ai to be the first coordinate of Ai(note that a0 = 0) we obtain that xi = ai, i = 1, . . . , 6 is a solution to the system(S) of seven linear equations in six variables:

εk1x1 + · · ·+ εk6x6 = 0

for k ∈ 0, 1, . . . , 6. We will show that (S) does not have a non-zero solution,contradicting the hypothesis.

Suppose (S) has a non-zero solution. Since all the coefficients in (S) are integers,we can express the solution set (after using Gauss-Jordan reduction) in terms ofrational constants and some parameters (at least one). Assigning a rational valueto each parameter then gives a rational solution to (S). We can multiply thissolution by the least common multiple of its denominators to obtain an integersolution to (S). Finally, we divide this solution by the greatest common divisor ofits values to obtain an integer solution xi = ni to (S) such that gcd(n1, . . . , n6) = 1.We observe that the first equation of (S)

ε01x1 + ε02x2 + · · ·+ ε06x6 = 0

implies that6∑i=1

ni is even, since all the coefficients are ±1. The second equation

implies that6∑i=2

ni is even and thus n1 must also be even. Similarly, nk is even for

k = 2, . . . , 6, contradicting that the greatest common divisor of the ni is 1.

4440. Proposed by Mihaela Berindeanu.

Let H be the orthocenter of triangle ABC and let AA1, BB1, CC1 be the altitudes;define the points X to be the intersection of AA1 with B1C1 and Y to be wherethe perpendicular from X to AC intersects AB. Prove that the line Y A1 passesthrough the midpoint of BH.

We received 11 submissions, of which one was flawed. We feature two solutions.

Solution 1, by Michel Bataille.

Let H1 be the reflection of H in BC; because ∠H1BC = ∠H1AC = 90 − ∠C,H1 lies on the circumcircle Γ of ∆ABC (see figure on the next page).


534/ Solutions

We first prove thatAX

AH=

AA1

AH1. To this aim, we consider the inversion I with

centre A such that I(A1) = H. This inversion transforms the line BC into thecircle with diameter AH. Observing that B1 and C1 lie on this circle (since∠AB1H = ∠AC1H = 90), we see that I(B) = C1 and I(C) = B1 and, therefore,the line B1C1 inverts into the circle Γ. It follows that I(X) = H1, and so AX ·AH1 = AH ·AA1, as desired.

Now, the line Y X being perpendicular to AC is parallel to BB1, henceAY

AB=AX

AH.

As a result, we haveAY

AB=

AA1

AH1, and so the lines Y A1 and BH1 are parallel.

Since A1 is the midpoint of HH1, the line A1Y intersects HB in its midpoint.

Solution 2, a generalization by Chenxu Li.

The result is, in fact, a theorem of affine geometry:

If P is an arbitrary point in the plane of ∆ABC but not on any of thelines joining a pair of vertices, define A1, B1, C1 to be the feet of thecevians AP,BP,CP (on the lines BC,CA,AB). As in the problem,define X to be the intersection of AA1 with B1C1, and Y to be wherethe line parallel to BB1 through X intersects AB. Then the line Y A1

passes through the midpoint of BP .

Proof. Let M be the intersection of Y A1 and BB1; we are to prove that M is themidpoint of BP . Because A and P are diagonal points of the complete quadrangleB1C1BC, A1 and X are harmonic conjugates with respect to A and P . (See, forexample, [Crux 39:4, “Harmonic Sets Part 2: Quadrangles and Quadrilaterals”(Apr. 2013), pages 174-177] or [44:7 (Sep. 2018) “Harmonic Ranges and Pencils”,pages 291-296].)


Solutions /535

The points A,A1, P,X project from Y to the points B,M,P,∞ of the line BB1.Thus M and ∞ are harmonic conjugates with respect to B and P , which impliesthat M is the midpoint of BP , as desired. In the original problem, of course, Pis taken to be the orthocenter H. (When P = H, both lines XY and BB1 areperpendicular to AC, so they are parallel, as required by the generalization.)

Editor’s comments. Zvonaru reports that our problem 4440 appeared as a contestproblem in Hungary. He saw it on Andrei Eckstein’s site:

https://pregatirematematicaolimpiadejuniori.wordpress.com/problema-saptamanii

(problem of the week nr. 100: July 9, 2018). The site provides five solutions (inRomanian), two of which come with English translations; all five are different fromany of the solutions submitted to Crux. One is Zvonaru’s own solution to whichhe added the same generalization as our solution 2 (featured above).


SNAPSHOT

The home of Michel Bataille is Rouen. Situated on river Seine and engulfed ingreenery, Rouen is the capital of the northern French region of Normandy.

editorial board - cms-smc · 2020. 2. 25. · editorial board editor-in-chief kseniya garaschuk...

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