Crux Mathematicorum - cms.math.ca

Crux MathematicorumVOLUME 41, NO. 8 October / Octobre 2015

Editorial Board

Editor-in-Chief Kseniya Garaschuk University of British Columbia

Contest Corner Editor John McLoughlin University of New Brunswick

Olympiad Corner Editor Carmen Bruni University of Waterloo

Book Reviews Editor Robert Bilinski College Montmorency

Articles Editor Robert Dawson Saint Mary’s University

Problems Editors Edward Barbeau University of Toronto

Chris Fisher University of Regina

Anna Kuczynska University of the Fraser Valley

Edward Wang Wilfrid Laurier University

Dennis D. A. Epple Berlin, Germany

Magdalena Georgescu University of Toronto

Assistant Editors Chip Curtis Missouri Southern State University

Lino Demasi Ottawa, ON

Allen O’Hara University of Western Ontario

Guest Editors Joseph Horan University of Victoria

Cameron Morland Simon Fraser University

Kelly Paton University of British Columbia

Alessandro Ventullo University of Milan

Mallory Flynn University of British Columbia

Editor-at-Large Bill Sands University of Calgary

Managing Editor Denise Charron Canadian Mathematical Society

Copyright c© Canadian Mathematical Society, 2016

324

IN THIS ISSUE / DANS CE NUMERO

325 Editorial Kseniya Garaschuk

326 The Contest Corner: No. 38 John McLoughlin

326 Problems: CC186–CC190

329 Solutions: CC136–CC140

332 The Olympiad Corner: No. 336 Carmen Bruni

332 Problems: OC246–OC250

334 Solutions: OC186–OC190

339 Book Reviews Robert Bilinski

341 Solutions to Inequality Problems from the Chinese

Mathematical Olympiad Huawei Zhu

346 Angle Bisectors in a Triangle I. F. Sharygin

352 Problems: 4071–4080

356 Solutions: 3971–3980

367 Solvers and proposers index

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

Crux Mathematicorum, Vol. 41(8), October 2015

325

EDITORIALThe future is here yet again for Crux : our problem proposals and solution sub-missions will now be processed using an electronic online form called Cruxbox!Our new automated system will allow for more efficient and reliable process. Forall submissions, visit

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

The system accepts various file formats, will inform you of the successful uploadand will send a confirmation email once the problem gets filed into our system.For problem proposals, your confirmation email will also contain the proposal’stracking number whose status you can now check at

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

As we would like to fully move to this system as quickly as possible, we nowdiscourage submissions by email (in fact, we will simply redirect you to the newsystem if you try to submit by email). So I urge you to familiarize yourself withCruxbox and email me at [email protected] if you have any commentsabout it.

If Cruxbox is our future, then inequalities are our past, present and future. WhenI first took over this job, the previous editor Shawn Godin warned me about thebacklog of inequalities he had acquired. I did not quite understand the issue(namely, why not just publish the accepted ones as they get accepted?) andthen I saw the overwhelming number of inequality submissions we receive. Yes,inequalities are popular and I understand why: they are like little puzzles thatallow us to figure out which tools from our problem-solving toolbox to use when,how, in which order, to what extent. But we receive a truly staggering numberof inequalities; in fact, nowadays a vast majority of our proposals are inequalities.So let me remind you that variety is the spice of life and encourage you to spiceup your submissions by stepping outside the inequality comfort zone. After all, Ido not believe that anyone wants our Problems section to become the Inequalitiessection.

Kseniya Garaschuk


326/ THE CONTEST CORNER

THE CONTEST CORNERNo. 38

John McLoughlin

Les problemes presentes dans cette section ont deja ete presentes dans le cadre d’unconcours mathematique de niveau secondaire ou de premier cycle universitaire, ou enont ete inspires. Nous invitons les lecteurs a presenter leurs solutions, commentaires etgeneralisations pour n’importe quel probleme. S’il vous plaıt vous referer aux regles desoumission a l’endos de la couverture ou en ligne.

Pour faciliter l’examen des solutions, nous demandons aux lecteurs de les faire parvenirau redacteur au plus tard le 1 octobre 2016 ; toutefois, les solutions recues apres cettedate seront aussi examinees jusqu’au moment de la publication.

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

CC186. Soit n un entier positif. Faire le decompte des nombres k ∈ {0, 1, . . . , n}pour lesquels

(nk

)est impair. Demontrer que ce nombre est une puissance de deux,

c’est-a-dire est de la forme 2p pour un certain entier non negatif p.

CC187. Dans le diagramme, la surface du triangle ABC est 1, AD = 13AB,

EC = 13AC et DF = FE. Determiner la surface du triangle en gris BFC.

CC188. Un plan divise l’espace en deux regions. Deux plans intersectant enune ligne divisent l’espace en quatre regions. Supposer maintenant que douze planssont donnes dans l’espace tel que

a) deux d’entre eux intersectent toujours en une ligne,

b) trois d’entre eux intersectent toujours en un point, et

c) quatre d’entre eux n’ont jamais un point en commun.

Dans combien de regions l’espace est-il divise ? Justifier votre reponse.


THE CONTEST CORNER /327

CC189. Des pieces de monnaie sont placees sur certains des 100 carres d’unegrille 10×10. Tout carre est voisin a un autre carre couvert d’une piece de monnaie.Determiner le nombre minimum de pieces de monnaie. (Deux carres distincts sontvoisins s’ils partagent un cote.)

CC190. Les lettres du mot TRIANGLE forment un arrangement tel qu’in-dique.

Determiner le nombre de facons d’epeler le mot TRIANGLE, utilisant des lettresadjacentes dans cet arrangement et allant vers le haut, la gauche ou la droite.

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

CC186. Let n be a positive integer. Count the number of k ∈ {0, 1, . . . , n} forwhich

(nk

)is odd. Prove that this number is a power of two, i.e. it is of the form

2p for some non-negative integer p.

CC187. In the diagram the area of the triangle ABC is 1, AD = 13AB,

EC = 13AC and DF = FE. Find the area of the shaded triangle BFC.

CC188. A plane divides space into two regions. Two planes that intersect ina line divide space into four regions. Now suppose that twelve planes are given inspace so that three conditions are met :

a) every two of them intersect in a line,

b) every three of them intersect in a point, and



c) no four of them have a common point.

Into how many regions is space divided ? Justify your answer.

CC189. Coins are placed on some of the 100 squares in a 10× 10 grid. Everysquare is next to another square with a coin. Find the minimum possible number ofcoins. (We say that two squares are next to each other when they share a commonedge but are not equal.)

CC190. An arrangement of the letters from the word TRIANGLE is shown.

Find the number of ways that the word TRIANGLE can be spelled out, usingadjacent letters, going up or left or right, in this arrangement.



CONTEST CORNERSOLUTIONS

Les enonces des problemes dans cette section paraissent initialement dans 2014 : 40(8),p. 319–320.

CC136. A spiderweb is a square 100×100 grid with knots at each intersection.The spider sits at one corner of his spiderweb ; there are 100 flies caught in theweb with at most one fly per knot. Can the spider get all the flies in no more than2000 moves, if in one move it crawls to an adjacent knot ?

Originally from Tournament of Towns Fall Round 2014, A-level, Juniors.

We received no solutions to this problem.

CC137. An Emperor invited 2015 wizards to a festival. Each wizard, but notthe Emperor, knows which wizards are good and which ones are evil. A goodwizard always tells the truth, while an evil wizard can either tell the truth orlie. At the festival, the Emperor gives every wizard a card with one “yes-or-no”question (questions might be different for different wizards), learns all the answersand then expels one wizard through a magic door which shows if this wizard isgood or evil. Then the Emperor makes new cards and repeats the procedure withthe remaining wizards until he wants to stop (with or without expelling a wizard).Prove that the Emperor can devise his questions so that all the evil wizards areexpelled while expelling at most one good wizard.

Originally from Tournament of Towns, 2015, Spring, A-level, Senior.

We received and present one solution by Josh Zukewich and others (unnamed).

This solution is in two parts. In the first part (Part A) we find a living good wizard,expelling at most one good wizard in the process. In the second (Part B), we usethis fact to expel all remaining bad wizards.

Part A : Pick a wizard at random (called Jim). Ask every wizard : “Is Jim good ?”Jim’s answer does not matter for our analysis. For the remaining wizards’ answers,consider the two exhaustive possibilities :

(1) Every non-Jim wizard says ‘no.’

(2) Some non-Jim wizard says ‘yes.’

Our actions are as follows for each possibility :

(1) Expel Jim. Jim goes through the door and he is either :

(1.1) Good. Everyone else lied and is therefore bad. Expel everyone else inturn. Then we are done.

(1.2) Bad. You expelled a bad wizard. Return to A.



(2) Expel a wizard (called Jane) that said ‘yes.’ When Jane goes through thedoor, she is either :

(2.1) Good. Then you have identified that Jim is good. You are done step A.

(2.2) Bad. You have expelled a bad wizard. Return to A.

In every iteration of step A we either expel a bad wizard, and repeat, or expel agood wizard and identify a good wizard still in the group.

Part B : Arrange all the wizards in a line facing forward, with Jim at the back.Ask every wizard : “Is the wizard directly ahead of you good ?” If Jim says

(1) ‘No,’ expel the wizard in front of Jim. You have expelled a bad wizard.Return to B.

(2) ‘Yes,’ the wizard in front of Jim is also good. Keep moving your attentionahead until a wizard says ‘no.’ The first wizard in front of Jim that says‘no’ identifies the wizard in front of them as bad. Expel that bad wizard andreturn to B.

Goodness is transitive : Each wizard that says ‘yes’ moving ahead from Jimconfirms the goodness of another wizard. Once the second-to-front wizard in theline says ‘yes,’ you have confirmed that all the remaining wizards are good, andyou are done. You choose not to expel a wizard once this condition is met.

CC138. Prove that the integer

2n−1∑i=1

(2i− 1)2i−1 = 11 + 33 + 55 + · · ·+ (2n − 1)2n−1

is a multiple of 2n but not a multiple of 2n+1.

Originally from Tournament of Towns, Fall Round 2011, A-level, Senior.

No solutions were received. We present a hint to encourage readers.

Consider the problem modulo 2n+2. Specifically, prove that k2n ≡ 1 (mod 2n+2)

for any odd positive integer k.

CC139. It is well-known that if in a quadrilateral the circumcircle and theincircle have the same centre, then the quadrilateral is a square. Is the similarstatement true in 3 dimensions ? Namely, if a cuboid is inscribed into a sphere andcircumscribed around a sphere and the centres of these spheres coincide, does itimply that the cuboid is a cube ? (A cuboid is a polyhedron with 6 quadrilateralfaces such that each vertex belongs to 3 edges.)

Originally from Tournament of Towns, Spring Round 2015, A-level, Seniors.

We received one correct solution. We present the solution of the Missouri StateUniversity Problem Solving Group.



The answer is no.

Consider the polyhedron whose vertices are

A = (a2; 1; a), B = (a2;−1; a), C = (−a2;−1; a), D = (−a2; 1; a),

E = (1; a2;−a), F = (1;−a2;−a), G = (−1;−a2;−a), H = (−1; a2;−a).

These clearly lie on the sphere of radius√a4 + a2 + 1 centered at the origin.

The rectangular faces ABCD and EFGH are tangent to the sphere of radius acentered at the origin. The other four faces, ABFE, BCGF , CDHG, and DAEHare congruent trapezoids lying in the planes

2ax− (a2 − 1)z = a3 + a,

−2ay + (a2 − 1)z = a3 + a,

2ax+ (a2 − 1)z = −a3 − a,2ay + (a2 − 1)z = a3 + a,

respectively.

We claim that the sphere of radius a centered at the origin is tangent to theseplanes at the points( 2a2

a2 + 1, 0,

a− a3

a2 + 1

),(

0,− 2a2

a2 + 1,a3 − aa2 + 1

),(

− 2a2

a2 + 1, 0,

a− a3

a2 + 1

),(

0,2a2

a2 + 1,a3 − aa2 + 1

)respectively. To see this, note that each point lies on its corresponding plane andits position vector has length a and is parallel to the normal vector of that plane.This cuboid satisfies the conditions of the problem, but is not a cube unless a = 1.

CC140. Let P (x) be a polynomial with real coefficients so that the equationP (m) + P (n) = 0 has infinitely many pairs of integer solutions (m,n). Prove thatthe graph of y = P (x) has a centre of symmetry.

Originally from Tournament of Towns, Fall Round 2008, A-level, Seniors.

We received no solutions to this problem.


332/ THE OLYMPIAD CORNER

THE OLYMPIAD CORNERNo. 336

Carmen Bruni

Les problemes presentes dans cette section ont deja ete presentes dans le cadre d’uneolympiade mathematique regionale ou nationale. Nous invitons les lecteurs a presenterleurs solutions, commentaires et generalisations pour n’importe quel probleme. S’il vousplaıt vous referer aux regles de soumission a l’endos de la couverture ou en ligne.


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

OC246. Determiner tous les entiers positifs x, y and z tels que x3 = 3y7z + 8.

OC247. Soit a1 ≤ a2 ≤ · · · une suite non decroissante d’entiers positifs. Un

entier positif n est dit bon s’il existe i tel que n =i

ai. Demontrer que si 2013 est

bon, 20 l’est aussi.

OC248. Soit B et C deux points specifiques sur un cercle centre a O, maispas sur un meme diametre. Soit A qui bouge sur le cercle tout en etant distinctde B et C et n’appartenant pas a la bissectrice perpendiculaire de BC. Soit Hl’orthocentre de 4ABC et soit M et N les mi points des segments BC et AHrespectivement. La ligne AM intersecte le cercle de nouveau a D ; enfin, NM etOC intersectent au point P . Determiner le locus des points P au fur et a mesureque A bouge sur le cercle.

OC249. Determiner toutes les fonctions f : R→ R verifiant

f(xf(x) + f(x)f(y) + y − 1) = f(xf(x) + xy) + y − 1.

OC250. Alberte et Bertrand s’amusent a un jeu de nombres. Commencantavec un entier positif n, ils prennent leur tour a modifier ce nombre, Alberteallant premiere. Chaque joueur change le nombre courant k a k − 1 ou dk/2e. Lapersonne qui change le nombre a 0 ou 1 gagne. Determiner n ou Alberte a unestrategie gagnante.

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


THE OLYMPIAD CORNER /333

OC246. Find all positive integers x, y and z such that x3 = 3y7z + 8.

OC247. Let a1 ≤ a2 ≤ · · · be a non-decreasing sequence of positive integers.

A positive integer n is called good if there is an index i such that n =i

ai. Prove

that if 2013 is good, then so is 20.

OC248. Let B and C be two fixed points on a circle centered at O that arenot diametrically opposite. Let A be a variable point on the circle distinct fromB and C and not belonging to the perpendicular bisector of BC. Let H be theorthocenter of 4ABC, and M and N be the midpoints of the segments BC andAH, respectively. The line AM intersects the circle again at D, and finally, NMand OD intersect at P . Determine the locus of points P as A moves around thecircle.

OC249. Determine all the functions f : R→ R that satisfies the following

f(xf(x) + f(x)f(y) + y − 1) = f(xf(x) + xy) + y − 1.

OC250. Alice and Bob play a number game. Starting with a positive integern they take turns changing the number with Alice going first. Each player maychange the current number k to either k − 1 or dk/2e. The person who changes 1to 0 wins. Determine all n such that Alice has a winning strategy.



OLYMPIAD SOLUTIONSLes enonces des problemes dans cette section paraissent initialement dans 2014 : 40(6),p. 240–241.

OC186. Show that if for non-negative integers m,n,N, k, the equation

(n2 + 1)2k

· (44n3 + 11n2 + 10n+ 2) = Nm

holds, then necessarily m = 1.

Originally problem 1 of the 2013 Austria Federal Competition for Advanced Stu-dents.

We received two correct submissions. We present the solution by Oliver Geupel.

We consider the cases for even and odd numbers n in succession.

If n = 2q is an even number, then (n2 + 1)2k

is odd and

44n3 + 11n2 + 10n+ 2 = 2(22n3 + 22q2 + 10q + 1) ≡ 2 (mod 4) .

Hence Nm ≡ 2 (mod 4), that is, the multiplicity of the prime factor 2 in Nm is 1,so that m = 1.

It remains to consider n to be an odd number. Then n2 + 1 is even. Since n2 6≡ −1(mod 4), we have n2 + 1 ≡ 2 (mod 4). Thus, the multiplicity of the prime factor

2 in (n2 + 1)2k

is 2k. Moreover, the number

44n3 + 11n2 + 10n+ 2 = 2(22n3 + 5n+ 1) + 11n2

is odd. Therefore the multiplicity of the prime factor 2 in Nm is 2k. As a conse-quence, m = 2` with ` ≤ k, so that the number 44n3 + 11n2 + 10n + 2 has the

form r2`

with a rational number r. It is enough to prove that ` = 0.

The proof is by contradiction. Suppose ` ≥ 1. We deduce that the odd number44n3 + 11n2 + 10n+ 2 is a perfect square, which implies that it is congruent to 1(mod 4). On the other hand, we have

44n3 + 11n2 + 10n+ 2 ≡ −1 (mod 4) ,

a contradiction. The proof is complete.

OC187. Let PA and PB be tangents to a circle centered at O and C a pointon the minor arc AB. The perpendicular from C to PC intersects internal anglebisectors of ∠AOC and ∠BOC at D and E. Show that CD = CE.

Originally problem 6 of the 2013 China Western Mathematical Olympiad.

We received four correct submissions. We present the solution by Oliver Geupel.



Consider the problem in the plane of complex numbers where the circle is the unitcircle and a, b, c, . . . are the coordinates of points A,B,C, . . . . We have a = 1/a,b = 1/b, c = 1/c. The intersection P of the tangents from A and B has coordinatep = 2ab/(a + b). Hence p = 2/(a + b). Since d and e are on the angle bisectors,there are real numbers λ, µ such that d = λ(a+ c) and e = µ(b+ c).

Since CD is perpendicular to CP , we obtain

0 = (c− d) (c− p) +(c− d

)(c− p) = 2− cp− cp− cd− cd+ dp+ dp

= 2− 2c

a+ b− 2ab

(a+ b)c− λÅa+ c

a+a+ c

c− 2(a+ c)

a+ b− 2(a+ c)b

(a+ b)c

ã=

2(c− b)(a− c)(a+ b)c

− λ (a+ c)(a− b)(a− c)(a+ b)ac

,

whence

λ =2a(c− b)

(a− b)(a+ c).

Similarly,

µ =2b(c− a)

(b− a)(b+ c).

We compute

d+ e = λ(a+ c) + µ(b+ c) =2a(c− b)a− b

+2b(c− a)

b− a= 2c,

that is c = (d + e)/2. Consequently, C is the midpoint of the line segment DE.The proof is complete.



OC188. Find the maximum value of M for which for all positive real numbersa, b, c, we have

a3 + b3 + c3 − 3abc ≥M(ab2 + bc2 + ca2 − 3abc).

Originally problem 2 from day 2 of the 2013 Turkey Mathematical Olympiad SecondRound.

We received one correct submission and four incorrect submissions. We presentthe solution by Michel Bataille.

Editor’s note. Some solutions correctly determined the maximum value M provi-ded the inequality holds but failed to prove that when M = 3 · 2−2/3, then theinequality does indeed hold for all positive real numbers a, b and c. The solutionbelow justifies this.

We show that the required maximal value is α =3 3√

2

2.

First, suppose that M satisfies

a3 + b3 + c3 − 3abc ≥M(ab2 + bc2 + ca2 − 3abc)

for all a, b, c > 0. Taking the limit as c→ 0 on both sides, we must have a3 + b3 ≥Mab2 for all a, b > 0. Taking x = b

a , it follows that M ≤ φ(x) for all x > 0 whereφ(x) = x+ x−2. However, for any positive x,

φ(x) =x

2+x

2+

1

x2≥ 3

3

…x

2· x

2· 1

x2=

33√

4= α,

with equality ifx

2=

1

x2, that is, x = 3

√2. In particular, M ≤ φ( 3

√2) = α.

To complete the proof, we consider a, b, c > 0 and show that

a3 + b3 + c3 − 3abc ≥ α(ab2 + bc2 + ca2 − 3abc). (1)

Without loss of generality, we suppose that c = min(a, b, c). It is easily checkedthat

2(a3 + b3 + c3 − 3abc) = (a+ b+ c)((a− b)2 + (b− c)2 + (c− a)2),

2(ab2 + bc2 + ca2 − 3abc) = a(b− c)2 + b(c− a)2 + c(a− b)2 + (a− b)(b− c)(c− a).

As a result, the inequality (1) is equivalent to

(a+ b− (α− 1)c)(a− b)2 + (b+ c− (α− 1)a)(b− c)2 + (c+ a− (α− 1)b)(c− a)2

≥ α(a− b)(b− c)(c− a). (2)

Now, let L(a, b, c) be the left-hand side of (2) and R(a, b, c) be its right-hand side.If a1 = a− c, b1 = b− c, and c1 = 0, then

a1 − b1 = a− b, b1 − c1 = b− c, c1 − a1 = c− a,



hence R(a1, b1, c1) = R(a, b, c). On the other hand, we observe that

a+ b− (α− 1)c− (a1 + b1 − (α− 1)c1) = b+ c− (α− 1)a− (b1 + c1 − (α− 1)a1)

= c+ a− (α− 1)b− (c1 + a1 − (α− 1)b1)

= (3− α)c > 0

and so L(a, b, c) ≥ L(a1, b1, c1). To conclude, it is sufficient that the inequality

L(a1, b1, c1) ≥ R(a1, b1, c1)

holds. But, since c1 = 0, the latter is equivalent to

a31 + b31 ≥ αa1b21,

which holds if a1 = 0 or b1 = 0 and also if a1, b1 > 0 since α ≤ φ( b1a1 ), so we aredone.

OC189. Let n, k be positive integers with n ≥ k. There are n people, eachperson belongs to exactly one of group 1, group 2, · · · , group k and more than orequal to one person belong to any groups. Show that n2 candies can be deliveredto n persons in such way that all of the following conditions are satisfied.

1. At least one candy is delivered to each person.

2. The integers ai are candies that are delivered to each person belonging togroup i (1 ≤ i ≤ k) such that if 1 ≤ i < j ≤ k, then ai > aj .

Originally problem 1 of the 2013 Japan Mathematical Olympiad Finals.

We present the solution by Oliver Geupel. There were no other submissions.

Suppose that there are bj persons in group j (1 ≤ j ≤ k), so that

k∑j=1

bj = n.

We shall show that the numbers

a1 = n+k∑j=2

bj , ai+1 = ai − bi − bi+1 (1)

for i ≥ 2 satisfy the conditions of the problem. We have for any j ∈ N,

aj = n+k∑

i=j+1

bi −j−1∑i=1

bi

anda1 > a2 > · · · > ak = bk ≥ 1.



Moreover,

k∑j=1

ajbj =k∑j=1

(n+

k∑i=j+1

bi −j−1∑i=1

bi

)bj

= nk∑j=1

bj +k∑j=1

k∑i=j+1

bjbi −k∑j=1

j−1∑i=1

bibj

= n2 +∑

1≤j<i≤k

bjbi −∑

1≤i<j≤k

bibj

= n2.

Consequently, (1) is a solution of the problem.

OC190. Let a,m and n be positive integers with m ≤ n. Prove that if one ofthe numbers am and an is divisible by m, then the other number must also bedivisible by n.

Originally problem 4 of the 2013 Philippines Mathematical Olympiad.

We present the solution by Oliver Geupel. There were no other submissions.

Editor’s note. The above statement contains a typo. It should read

Let a,m and n be positive integers with m ≤ n. Prove that if one ofthe numbers am and an is divisible by m, then the other number mustalso be divisible by m.

Thanks to Oliver Geupel for the catch !

By assumption we have eitherm | am orm | an. Ifm | am, thenm ≤ n immediatelyimplies m | am | an and we are done. Now assume m | an. The number am has thesame prime divisors as an, each with multiplicity at least m. Since the multiplicityof every prime divisor of m is not greater than m, we conclude m | am. Thiscompletes the proof.


BOOK REVIEWS /339

BOOK REVIEWSRobert Bilinski

The Ellipse : A Historical and Mathematical Journey by Arthur MazerISBN 978-0-470-58718-8, hardcover, 303 pagesPublished by Wiley, 2010

Reviewed by Robert Bilinski, College Montmorency.

Arthur Mazer started his career at Utah State University as an assistant profes-sor and then transferred to the energy industry. He currently works in industryas a manager of the Quantitative Analytics Department at Southern CaliforniaEdison. He has written three books : a manuscript on power planning, and twomathematical history books that came out roughly at the same time. The first isthe subject of this book review and the second is titled Quest to understand themotion of the universe. In the preface, we see Mazer’s motivation :

There is of course the opposite argument [...] that is, mathematicsand science are the drivers of historical evolution and in Darwinian fa-shion, philosophies and political entities that promote scientific excel-lence flourish, while those that do not fade away. This latter argumentprovides the perspective for this book.

This book seemed to be a bit of a paradox to me. In fact, that is what attracted meto it. The title implies that the book is specialized ; yet, it is published by Wiley as ageneral interest book. After reading the book, it becomes clear that the book is notabout the ellipse. Rather, it uses the ellipse as its main thread to explore the worldof mathematics. The chapters are organized roughly in chronological order as theauthor explores the evolution of thought in society and the interactions betweenthe society and mathematical ideas. Mazer gives witty titles to his chapters, suchas “Chapter 6 : The Slayer Calculus” or “Chapter 7 : Eight minutes that changedhistory”.

The first two chapters focus mainly on history with a bit of a primer on numbersand modular arithmetic that is explored through cycles on graphs. The followingfive chapters start off with a short introduction, which is then followed by about50 pages of math. Each of these chapters is an exploration of small math morselsthat range in complexity and subject. The link to the ellipse is not always clear atthe beginning, but all the threads are neatly tied together into a tapestry aboutthe ellipse by the end of the book.

It may make for some intense reading. I will not go into detail for each chap-ter, but to give an idea, I will provide a synopsis of “Chapter 4 : The languageAlgebra”. We start off with Cartesian coordinates, then continue to translations,polynomials, conics and the quadratic formula, the circle and the four dimensionalsphere, Cavalieri’s theorem, Pyramids, higher dimensions, areas and volumes ofspheres, finite series, induction, linear algebra, transformations, determinants, andthe ellipse. No spoilers !


340/ BOOK REVIEWS

The book is well written. It is abundantly illustratedand, to the author’s credit, not superfluously. Withthe hustle and bustle of teaching and the mass ofinformation presented, I am pretty sure I missed athing or two and a second re-read is a must. A mi-nus for me was that the knowledge about the ellipseexposed in the book ends around the 18th centurywithout any mention of more recent results. It makesit a side project for me to look into in the comingmonths, unless a Crux reader sends me useful re-ferences to more recent results and more technicalmanuscripts about the ellipse. As an avid Crux rea-der, I also hoped the book would make me a betterproblem solver, but there are no posed problems tobe found.

The range of mathematics presented in the book is far-reaching, but at an introduc-tory level. It passes through some number theory, classical geometry, stereometry,calculus, combinatorics, analysis, differential geometry, projective geometry, vec-tor calculus, etc. It could serve as a great run through or “transversal” review of astudent’s mathematical studies. If teacher colleges across the country have takenthe same path as Quebec’s, this book should be a mandatory read for all newsecondary level teachers as it would widen their mathematical horizons greatly,yet at a manageable pace. It could be used in a high school honours math class orfor a math camp. If you have read a lot of general interest math books, it dependsmore on you and what you are looking for in a book. Personally, I am happy tohave it on my bookshelf as a reference book and I will definitely try to find thetime to re-read it when I am less pressed for time. Happy reading !


Huawei Zhu /341

Solutions to Inequality Problems from

the Chinese Mathematical OlympiadHuawei Zhu

Guangzhou Institute of Educational Research

In this article, we present solutions to the problems in Inequality Problems fromthe Chinese Mathematical Olympiad appearing in Crux 41(7), p. 299–302.

Problem 2005.

Let a1 = 2116 and 2an − 3an−1 =

3

2n+1for n ≥ 2. Prove that for any integers

m ≥ n ≥ 2, Åan +

3

2n+3

ã 1m

(m−

Å2

3

ãn(m−1)m

)<

m2 − 1

m− n+ 1.

Solution to Problem 2005.

The given recurrence relation may be rewritten as

2nan +3

8= 3

Å2n−1an−1 +

3

8

ã.

Let bn = 2nan + 38 . Then b1 = 2a + 3

8 = 3 and for n ≥ 2, bn = 3bn−1. Iterationyields bn = 3n, so that an = ( 3

2 )n − 32n+3 . The desired inequality is equivalent toÅ

1− n

m+ 1

ãÅ3

2

ã nm

(m−

Å2

3

ãn(m−1)m

)< m− 1.

Since m ≥ 2, (1 + 1m )m ≥ 1 +

(m1

)1m +

(m2

)1m2 ≥ ( 3

2 )2. From the Arithmetic-Geometric Means Inequality,Å

1− n

m+ 1

ãm=

Å1− n

m+ 1

ãm1m(n−1)

<

Çm(1− n

m+1 ) +m(n− 1)

mn

åmn=

Åm

m+ 1

ãmn=

Ç1

(1 + 1m )m

ån<

Å2

3

ã2n.


342/ SOLUTIONS TO INEQUALITY PROBLEMS FROM THE CHINESEMATHEMATICAL OLYMPIAD

The desired inequality will follow from

m− 1 >

Å2

3

ã 2nmÅ

3

2

ã nm

(m−

Å2

3

ãn(m−1)m

)

=

Å2

3

ã nm

(m−

Å2

3

ãn(m−1)m

)= t(m− tm−1)

where t = ( 23 )

nm < 1. This certainly holds since

mt− tm −m+ 1 = (t− 1)(m− (tm−1 + tm−2 + · · ·+ t+ 1)) < 0.

Problem 2006.

The real numbers a1, a2, . . . , an have sum 0. Prove that

max1≤k≤n

a2k ≤n

3

n−1∑i=1

(ai − ai+1)2.


For 1 ≤ k ≤ n− 1, let dk = ak − ak+1. For a specific k, we have

a1 = ak + dk−1 + · · ·+ d2 + d1,

a2 = ak + dk−1 + · · ·+ d2,

· · · = · · ·ak−1 = ak + dk−1,

ak = ak,

ak+1 = ak − dk,· · · = · · ·an = ak − dk − dk−1 − · · · − dn−1.

Summation yields

0 = nak − (n− 1)dk − (n− k− 1)dk+1 − · · · − dn−1 + (k− 1)dk−1 + · · ·+ 2d2 + d1.

By Cauchy’s Inequality,

(nak)2

=

(n−1∑i=k

(n− i)di −k−1∑i=1

idi

)2

≤

(n−1∑i=1

i2

)(n−1∑i=1

d2i

)

=n(n− 1)(2n− 1)

6

(n−1∑i=1

d2i

)

≤ n3

3

(n−1∑i=1

d2i

).


Huawei Zhu /343

The desired result follows immediately.

Problem 2007.

Let a, b and c be given complex numbers. Let m = |a+ b| and n = |a− b|. Provethat if mn 6= 0, then

max{|ac+ b|, |bc+ a|} ≥ mn√m2 + n2

.


We have

max{|ac+ b|, |bc+ a|} ≥ |b||ac+ b|+ |a||bc+ a||a|+ |b|

≥ |b(ac+ b)− a(bc+ a)||a|+ |b|

≥ |b+ a||b− a|√2(|a|2 + |b|2

=mn√m2 + n2

.

Problem 2008.

For a positive integer n, let the real numbers x1 ≤ · · · ≤ xn and y1 ≥ · · · ≥ yn be

such thatn∑i=1

ixi ≥n∑i=1

iyi. Prove that for any real number λ,

n∑i=1

biλcxi ≥n∑i=1

biλcyi.


First note that we must have xn ≥ yn. Otherwise, y1 ≥ y2 ≥ · · · ≥ yn > xn ≥

xn−1 ≥ · · · ≥ x1, which contradictsn∑i=1

ixi ≥n∑i=1

iyi. We use mathematical induc-

tion on n. For n = 1, we have x1 ≥ y1. Hence bλcx1 ≥ bλcy1 for any positivereal number λ. Suppose the result holds for some n ≥ 1, so that there existsa real number λ with the desired properties. Consider the case n + 1. We havexn+1 ≥ yn+1. For 1 ≤ i ≤ n, define ai = xi + xn+1 and bi = yi + yn+1. Clearly, wehave a1 ≤ a2 ≤ · · · ≤ an and b1 ≥ b2 ≥ · · · ≥ bn. Moreover,

n∑i−1

iai =n∑i=1

ixi +n(n+ 1)

2xn+1 ≥

n∑i=1

iyi +n(n+ 1)

2yn+1 =

n∑i=1

ibi.


344/ SOLUTIONS TO INEQUALITY PROBLEMS FROM THE CHINESEMATHEMATICAL OLYMPIAD

By the induction hypothesis,n∑i=1

biλcai ≥n∑i=1

biλcbi. Now

n+1∑i=1

biλcxi −n∑i=1

biλcai = xn+1

(b(n+ 1)λc −

n∑i=1

biλc

)

= yn+1

(b(n+ 1)λc −

n∑i=1

biλc

)

=n+1∑i=1

biλcyi −n∑i=1

biλcbi.

It follows thatn+1∑i=1

biλcxi −n+1∑i=1

biλcyi ≥n∑i=1

biλcai −n∑i=1

biλcbi ≥ 0.

Problem 2009.

The real numbers a1, a2, . . . , an, n ≥ 3, are such that

min1≤i<j≤n

|ai − aj | = 1.

Determine in terms of n the minimum value ofn∑k=1

|ak|3.


We may assume that a1 < a2 < · · · < an. For 1 ≤ k ≤ n,

|ak|+ |an−k+1| ≥ an−k+1 − ak= (an−k+1 − an−k) + (an−k − an−k−1) + · · ·+ (ak+1 − ak)

≥ |(n− k + 1)− k|.

It follows thatn∑k=1

|ak|3 =1

2

n∑k=1

(|ak|+ |an−k+1|)3

=1

2

n∑k=1

(|ak|+ |an−k+1|)Å

1

4(|ak|+ |an−k+1|)2 +

3

4(|ak| − |an−k+1|)2

ã≥ 1

8

n∑k=1

(|ak|+ |an−k+1|)3

≥ 1

8

n∑k=1

|n− 2k + 1|3

=1

8((n− 1)3 + (n− 3)3 + · · ·+ (−(n+ 1))3).


Huawei Zhu /345

When n is odd,

n∑k=1

|ak|3 ≥1

4

Ñ23bn−1

2 c∑i=1

i3

é=

(n2 − 1)2

32.

When n is even,

n∑k=1

|ak|3 =1

4

bn2 c∑i=1

(2i− 1)3 =1

4

Ñn∑j=1

j3 − 23bn2 c∑i=1

i3

é=n2(n− 2)2

32.

The minimum value in each case can be attained by taking ai = i − n+12 for

1 ≤ i ≤ k.

Problem 2011.

Let n ≥ 4 be an integer. Let ak and bk, 1 ≤ k ≤ n be non-negative real numberssuch that

∑nk=1 ak =

∑nk=1 bk > 0. Determine the maximum value of∑n

k=1 ak(ak + bk)∑nk=1 bk(ak + bk)

.


Since the given expression is homogeneous, we may take∑nk=1 ak =

∑nk=1 bk = 1.

We may also assume that b1 ≤ bk for 2 ≤ k ≤ n. Then

(n− 1)n∑k=1

b2k + (n− 2)n∑k=1

akbk ≥ (n− 1)b21 + (n− 1)n∑k=2

b2k + (n− 2)b1

n∑k=1

ak

≥ (n− 1)b21 +

(n∑k=2

bk

)2

+ (n− 2)b1

= (n− 1)b21 + (1− b1)2 + (n− 2)b1

= nb21 + (n− 4)b1 + 1

≥ 1

=n∑k=1

ak ≥n∑k=1

a2k.

It follows thatn∑k=1

ak(ak + bk) ≤ (n− 1)n∑k=1

bk(ak + bk),

so that the maximum value of the given expression is n− 1. This is attained when

a1 = 1, a2 = a3 = · · · = an = 0 = b1 and b2 = b3 = · · · = bn =1

n− 1.


346/ ANGLE BISECTORS IN A TRIANGLE

Angle Bisectors in a TriangleI. F. Sharygin

In this article, we have collected some geometric facts which are directly or tan-gentially related to the angle bisectors in a triangle. These results vary from easylemmas to serious theorems, but we will not classify them ; rather, we will justnumber them. Every statement that occurs without a proof is considered as anexercise for the reader. In fact, even the presented proofs are rather concise, whichallows the reader to fill in the details.

Everyone should know this

Let us first recall some standard notation : let ABC be the given triangle, SABCis its area, |BC| = a, |CA| = b, |AB| = c, 2p = a + b + c, O and R are thecircumcentre and the circumradius, I and r are the incentre and the inradius.Furthermore, the triangle has three excircles, each of which touches one side of atriangle and the extensions of the other two sides. Their centres and their radii willbe denoted by Ia, Ib, Ic, ra, rb, rc (Ia denotes the centre of the excircle touchingthe side BC and extensions of the sides AB and AC with ra being its radius).Further notation will be presented as needed.

1. Suppose the internal bisector of ∠A intersects the side BC at a point A1. Then

|BA1||A1C|

=|BA||AC|

=c

b.

2. Suppose the external angle bisector of ∠A intersects the line BC at the pointA2. Then

|BA2||A2C|

=|BA||AC|

=c

b.

3. SABC = pr.

4. SABC = (p− a)ra.

5. Let M be the point of tangency of the incircle with the side AB. Then |AM | =p− a.

6. Let M be the point of tangency of the excircle with the centre Ia and the lineAB. Then |AM | = p.

7. Points B and C lie on the circle with diameter IIa and the centre of that circlelies on a circumcircle (see Figure 1.)

Therefore, the centre I of the incircle has the following property : the lines AI,BI and CI (that is, the angle bisectors of the triangle) go through the centres ofthe circumcircles of triangles BIC, CIA and AIB, respectively. The converse istrue as well, namely :


I. F. SHARYGIN /347

Figure 1: See Problem 7. Figure 2: See Problem 8.

8. If the lines AM , BM and CM go through the centres of the circumcircles oftriangles BMC, CMA and AMB, then M is the centre of the incircle of ABC.

Indeed, let Ma, Mb and Mc be the points of intersection (different from M) of thelines AM , BM and CM with the corresponding circles (see Figure 2). Then MMa,MMb and MMc are diameters of those circles ; therefore, MaA, MbB and McCare altitudes of the triangle MaMbMc. This implies that ∠BAM = ∠BMcM =90◦ − ∠BMaC = ∠CMbM = ∠CAM , which means that M lies on the anglebisector of angle A and, analogously, on the angle bisectors of angles B and C.

Distances between centres of special circles

9. |OI|2 = R2 − 2Rr (Euler’s formula).

10. |OIa|2 = R2 + 2Rra.

11. |IIa|2 = 4R(ra − r).

Figure 3: See Problems 9 and 10.

For the proof of 9 and 10, recall that if M and N are points of intersection of aline passing through an arbitrary point P with the circle of radius R and centreO, then |PM | · |PN | = |R2 − |OP |2| ; this follows from the similarity of triangles



PMM ′ and PNN ′, where M ′ and N ′ are points of intersection of the line OPwith the circle (see Figure 3). It implies that R2 − |OI|2 = |IA| · |IL|, where L isthe point of intersection of the angle bisector of angle A and the circumcircle (seeFigure 4). But |IA| = r/ sin (∠A/2) and, by 7, |IL| = |LB| = 2R× sin (∠A/2), soR2 − |OI|2 = 2Rr. Analogously,

|OIa|2 −R2 = |IaL| · |IaA| = 2R× sin (∠A/2)× rasin (∠A/2)

= 2Rra.

Finally,

|IIa|2 = 2|IL| · (|IaA| − |IA|) = 4R sin (∠A/2) · ra − rsin (∠A/2)

= 4R(ra − r).

Figure 4: See Problems 9 and 10.

12. Consider the points symmetric to the centres of the excircles with respect tothe centre of the circumcircle. These points lie on the circle of radius 2R with thecentre I.

Two extremal properties of the centre of the incircle

Consider an arbitrary point M inside the triangle ABC. There are many inequa-lities concerning the distances between M and the vertices of the triangle. We willconsider two such inequalities.

13. Let A1 be the point of intersection of the line AM and the circumcircle. Then

|BM | · |CM ||A1M |

≥ 2r

and equality holds if M coincides with I.

Suppose that the smallest value of f(M) = |BM |·|CM ||A1M | is achieved when M is some

point inside ABC. We will show that M = I. Then, since f(I) = 2r (this follows,for example, from similar triangles BID and IaIC in Figure 4), it would imply


I. F. SHARYGIN /349

that if f(M) achieves its minimum inside ABC, then f(M) ≥ 2r. The italicizedstatement is far from trivial and should be carefully proven.

Construct the circumcircle of triangle AMC (see Figure 5). Consider trianglesCMA1 formed by moving the point M along the arc AC — they are all similar(why ?) and hence the ratio |CM |/|A1M | is constant for all of them. Therefore,if the minimum of f(M) is achieved at M , then the line BM must go throughthe circumcentre of the triangle AMC (otherwise, we could reduce |BM | whilekeeping |CM |/|A1M | constant). Now, let B1 and C1 be the points of intersectionof lines BM and CM with the circumcircle of ABC. Then, as we saw in the proofof 9, we have |MA| · |MA1| = |MB| · |MB1| = |MC| · |MC1| and hence

|BM | · |CM ||A1M |

=|CM | · |AM ||B1M |

=|AM | · |BM ||C1M |

.

Therefore, lines AM and CM must also pass through the circumcentres of trianglesBMC and AMB respectively. Then M is the circumcentre of ABC (by 8).

Figure 5: See Problem 13. Figure 6: See Problem 14.

One must always be careful when using indirect proofs like the one above (wherewe did not directly prove that f(M) ≥ f(I) for all points M inside ABC) since afunction does not always achieve its minimum and maximum.

14. Show that

|AM | sin∠BMC + |BM | sin∠CMA+ |CM | sin∠AMB ≤ p,

and equality holds if M coincides with I.

The proof of this statement will also be indirect : we will show that the point Mwhere the left-hand side achieves its maximum (if it exists !) coincides with I.

Construct the circumcircle of triangle BMC and extend the line AM until thesecond point of intersection A2 (see Figure 6). Apply Ptolemy’s theorem to thequadrangle BMCA2 to get :

|BM | · |A2C|+ |CM | · |A2B| = |BC| · |A2M |.



Since the lengths of chords of a circle are proportional to the sines of anglessubtended by those chords, we have

|BM | sin∠A2MC + |CM | sin∠A2MB = |A2M | sin∠BMC

or|BM | sin∠AMC + |CM | sin∠AMB = |A2M | sin∠BMC.

Comparing the last equation to 14, we see that the left side of the inequality equals|AA2| sin∠BMC. Therefore, the line AM must go through the circumcentre ofBMC since otherwise we can increase the value of the left-hand side of 14 bymoving M along the arc BC. The rest of the proof is similar to that of 13.

We leave it to the reader to prove that if M = I, then |AA2| sin∠BMC = p. Tosee that, you can use 6 and 7 and the fact that ∠BIC = 90◦ + ∠A/2.

When intuition fails

When two similar elements of a triangle are equal (such as two angles or two me-dians), it seems natural to expect the triangle to be isosceles. Among the problemsof this type, one of the hardest to prove is the Steiner-Lehmus theorem.

15. If a triangle has two angle bisectors of equal lengths, then it is isosceles.

This problem is well-known, whereas the following amusing variation is not usuallyfamiliar even to geometry buffs.

16. Suppose a triangle ABC has angle bisectors AA1, BB1 and CC1. If the triangleA1B1C1 is isosceles, is ABC isosceles as well ?

Experiments with the graphic software suggest that the answer is no ; 4A1B1C1

can be isosceles while4ABC is not. We do not know of any brief, elegant construc-tion of a counterexample.

And for now, some more problems.

17. Prove that the angle bisector in a triangle bisects the angle between the cir-cumradius and the altitude from the same vertex.

18. Let AA1 be the angle bisector of ∠A in a triangle ABC. Show that

|AA1| =»bc− |BA1| · |CA1| =

2bc cos (∠A/2)

b+ c.

19. Suppose a triangle ABC has angle bisectors AA1, BB1 and CC1. Show thatthe altitudes of ABC are angle bisectors of A1B1C1.

20. Let M and N be the projections of the point of intersection of the altitudesof ABC onto the internal and external angle bisector of angle A. Show that theline MN divides the side BC in half.

21. Let S be the sum of the areas of the three triangles whose vertices are thepoints where an excircle touches the sides (or their extensions) of the given triangle


I. F. SHARYGIN /351

ABC. Let T be the area of the triangle formed by the points where the incircletouches the sides of ABC. Prove that S = SABC + T .

22. Suppose a triangle ABC has angle bisectors AA1, BB1 and CC1 ; let L andK be the points of intersection of the lines AA1 with B1C1 and CC1 with A1B1,respectively. Show that BB1 bisects angle LBK.

23. Let M and N be the midpoints of the diagonals AC and BC of a cyclicquadrilateral ABCD. Prove that if BC bisects angle ANC, then AC bisects angleBMD.

24. In a triangle ABC, let M be the point of intersection of the angle bisectorof angle B with the line passing through the midpoint of AC and the midpointof the altitude from the vertex B. Let N be the midpoint of the angle bisector ofangle B. Show that the angle bisector of angle C also bisects angle MCN .

25. Suppose a triangle ABC has angle bisectors AA1, BB1 and CC1, and constructthe circle O through the points A1, B1 and C1. Consider the three chords of Oformed by the segments of the sides of ABC lying inside O. Prove that the lengthof one of these chords is equal to the sum of the other two.

26. In a triangle ABC, letK and L be points on the sides AB andBC, respectively,such that |AK| = |KL| = |LC|. Draw the line parallel to the angle bisector of angleB through the point of intersection of the lines AL and CK. Let M be the pointof intersection of this line with the line AB. Show that |AM | = |BC|.

27. Let ABCD be a cyclic quadrilateral. Let K be the point of intersection ofthe extensions of the sides AB and CD ; let L be the point of intersection of theextensions of the sides BC and AD. Show that the bisectors of the angles BKCand BLA are perpendicular and intersect on the line connecting the midpoints ofAC and BD.

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

This article appeared in Russian in Kvant, 1983(8), p. 32–36. It has been translatedand adapted with permission.


352/ PROBLEMS

PROBLEMSNous invitons les lecteurs a presenter leurs solutions, commentaires et generalisationspour n’importe quel probleme presente dans cette section. De plus, nous les encourageonsa soumettre des propositions de problemes. S’il vous plaıt vous referer aux regles desoumission a l’endos de la couverture ou en ligne.


La redaction souhaite remercier Andre Ladouceur, Ottawa, ON, d’avoir traduit lesproblemes.

4071. Proposed by Leonard Giugiuc and Daniel Sitaru.

Prove that if a, b, c ∈ (0, 1), then aa+1bb+1cc+1 < e2(a+b+c)−6.

4072. Proposed by Michel Bataille.

Let a, b be distinct positive real numbers and A = a+b2 , G =

√ab, L = a−b

ln a−ln b .Prove that

L

G>

4A+ 5G

A+ 8G.

4073. Proposed by Daniel Sitaru.

Solve the following system :®sin 2x+ cos 3y = −1,√

sin2 x+ sin2 y +√

cos2 x+ cos2 y = 1 + sin(x+ y).

4074. Proposed by Abdilkadir Altinas.

Consider the triangle ABC with the following measures :

Show that a+ b = c ; that is, |AE|+ |AC| = |AB|.


PROBLEMS /353

4075. Proposed by Leonard Giugiuc and Daniel Sitaru.

Prove that in any triangle ABC with BC = a, CA = b, AB = c the followinginequality holds :

3√abc ·

√a2 + b2 + c2 ≥ 4[ABC],

where [ABC] is the area of triangle ABC.

4076. Proposed by Mehtaab Sawhney.

Prove that(x2 + y2 + z2)3 ≥ (x3 + y3 + z3 + 3(

√3− 1)xyz)2

for all nonnegative reals x, y, and z.

4077. Proposed by George Apostolopoulos.

Let ABC be a triangle. Prove that

sinA

2· sinB · sinC + sinA · sin B

2· sinC + sinA · sinB · sin C

2≤ 9

8.


Given θ such that π3 ≤ θ ≤ 5π

3 , let M0 be a point of a circle with centre O andradius R and Mk its image under the counterclockwise rotation with centre O andangle kθ. If M is the point diametrically opposite to M0 and n is a positive integer,show that

n∑k=0

MMk ≥ (2n+ 1) · R2.

4079. Proposed by Mihaela Berindeanu.

Let x, y, z > 0 and x+ y + z = 2016. Prove that :

x

…yz

y + 2015z+ y

…xz

z + 2015x+ z

…xy

x+ 2015z≤ 2016√

3.

4080. Proposed by Alina Sıntamarian and Ovidiu Furdui.

Let a, b ∈ R, with ab > 0. Calculate∫ ∞0

x2e−(ax− bx )

2

dx.

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


354/ PROBLEMS

4071. Propose par Leonard Giugiuc et Daniel Sitaru.

Soit a, b, c ∈ (0, 1). Demontrer que

aa+1bb+1cc+1 < e2(a+b+c)−6.

4072. Propose par Michel Bataille.

Soit a et b des reels strictement positifs distincts et soit A = a+b2 , G =

√ab et

L = a−bln a−ln b . Demontrer que

L

G>

4A+ 5G

A+ 8G.

4073. Propose par Daniel Sitaru.

Resoudre ce systeme d’equations :®sin 2x+ cos 3y = −1,√

sin2 x+ sin2 y +√

cos2 x+ cos2 y = 1 + sin(x+ y).

4074. Propose par Abdilkadir Altinas.

On considere le triangle ABC dans la figure suivante :

Demontrer que a+ b = c.

4075. Propose par Leonard Giugiuc et Daniel Sitaru.

Soit un triangle ABC avec BC = a, CA = b et AB = c. Demontrer que

3√abc ·

√a2 + b2 + c2 ≥ 4[ABC],

[ABC] etant l’aire du triangle ABC.

4076. Propose par Mehtaab Sawhney.

Demontrer que pour tous reels non negatifs x, y et z,

(x2 + y2 + z2)3 ≥ (x3 + y3 + z3 + 3(√

3− 1)xyz)2.


PROBLEMS /355

4077. Propose par George Apostolopoulos.

Etant donne un triangle ABC, demontrer que

sinA

2· sinB · sinC + sinA · sin B

2· sinC + sinA · sinB · sin C

2≤ 9

8.

4078. Propose par Michel Bataille.

Soit θ tel que π3 ≤ θ ≤

5π3 , soit M0 un point sur un cercle de centre O et de rayon

R et soit Mk son image par une rotation de centre O et d’angle kθ dans le senscontraire des aiguilles d’une montre . Sachant que M est le point diametralementoppose a M0 et que n est un entier strictement positif, demontrer que

n∑k=0

MMk ≥ (2n+ 1) · R2.

4079. Propose par Mihaela Berindeanu.

Soit x, y et z des reels strictement positifs tels que x + y + z = 2016. Demontrerque

x

…yz

y + 2015z+ y

…xz

z + 2015x+ z

…xy

x+ 2015z≤ 2016√

3.

4080. Propose par Alina Sıntamarian et Ovidiu Furdui.

Soit a et b des reels tels que ab > 0. Determiner la valeur de∫ ∞0

x2e−(ax− bx )

2

dx.


356/ SOLUTIONS

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

Statements of the problems in this section originally appear in 2014 : 40(8), p. 345–348.

3971. Proposed by Marcel Chirita.

Let x, y, z be positive real numbers such that 4(x+y+z)2 = 9(x2 +y2 +z2). Showthat

4−√

15 ≤ x

z≤ 4 +

√15.

We received 21 submissions of which 20 were correct and complete. We presentthe solution by Roy Barbara.

By the hypothesis we have

8(xy + yz + zx) = 5(x2 + y2 + z2). (1)

Set k = xz . Replacing x by kz in (1) and rearranging yields :

5y2 − 8z(k + 1)y + z2(5k2 − 8k + 5) = 0. (2)

Looking at (2) as a quadratic equation in y (with root y a real number), we seethat its discriminant must be non-negative, that is,

z2[16(k + 1)2 − 5(5k2 − 8k + 5)] ≥ 0.

Hence, since z2 > 0, 16(k + 1)2 − 5(5k2 − 8k + 5) ≥ 0, which simplifies to

k2 − 8k + 1 ≤ 0. (3)

Inequality (3) shows that k must lie between the two zeros of k2−8k+1 = 0, thatis,

4−√

15 ≤ x

z≤ 4 +

√15.


Find all functions f : (0,∞)→ (0,∞) such that

f

Åxf

Å1

y

ãã= xf

Å1

x+ y

ãfor all positive real numbers x, y.

We received five solutions, of which three were correct and complete. We presenttwo solutions.


SOLUTIONS /357

Solution 1, by Roy Barbara.

The function f(x) = xx+1 is a unique solution. Indeed, let f be any solution and

let y be any positive real number. Set

x =

y2

4+ f

Å1

y

ã−1− y

2. (1)

Then, x > 0 and x+ yx− fÄ1y

ä−1= 0, whence

xf

Å1

y

ã=

1

x+ y.

Hence,

f

Åxf

Å1

y

ãã= f

Å1

x+ y

ã.

Since fÄxfÄ1y

ää= xf

Ä1

x+y

äby hypothesis, we conclude that x = 1 and therefore

by (1) y2

4+ f

Å1

y

ã−1=y

2+ 1.

Squaring yields fÄ1y

ä−1= y + 1, that is, f

Ä1y

ä= 1

y+1 . Since y > 0 is arbitrary,

replacing y by 1y yields f(y) = y

y+1 .

Solution 2, by Mohammed Aassila.

Let

(x, y) =

Ñ−1 +

»1 + 4t2

f(t)

2t,

1

t

é,

where t > 0. Clearly, x > 0 and y > 0. Then we obtain

1 =−1 +

»1 + 4t2

f(t)

2t.

Thus, f(t) = tt+1 and it is easy to check that it satisfies the given functional

equation.

Editor’s Comments. We don’t know anything about the injectivity of f , so itcannot be used to solve the problem ! If f were injective, the problem would betrivial because if we set x = 1, then we get

f

Åf

Å1

y

ãã= f

Å1

1 + y

ã=⇒ f

Å1

y

ã=

1

1 + y,

which yields f(y) = yy+1 .


358/ SOLUTIONS

3973. Proposed by Dragoljub Milosevic.

Let ma,mb,mc be the medians, ra, rb, rc the exradii, R the circumradius and rthe inradius of a triangle. Prove that

ma

ra+mb

rb+mc

rc≤ 2R

r− 1.

We received four correct solutions. We present the one by M. Bello, M. Benito,O. Ciaurri, E. Fernandez, and L. Roncal, expanded slightly by the editor.

Let pa, pb, and pc be the distances from the circumcentre O of the triangle tothe sides BC, CA, and AB, respectively. Then it is known (see page 73 of O.Bottema et al., Geometric inequalities, Wolters-Noordhoff Publishing, Groningen,The Netherlands, 1969) that

ma ≤ R+ pa, mb ≤ R+ pb, mc ≤ R+ pc. (1)

Now let K be the foot of the perpendicular from O to BC. Then

pa = R cos(∠BOK) = R cos

Å1

2∠BOC

ã= R cosA,

and similarly, pb = R cosB and pc = R cosC. (2)

From (1) and (2) we have

ma ≤ R(a+ cosA), mb ≤ R(1 + cosB), mc ≤ R(1 + cosC). (3)

It is well known that

rs = (s− a)ra = (s− b)rb = (s− c)rc, (4)

where s = 12 (a+ b+ c). From (3) and (4) we then obtain∑

cyclic

ma

ra≤ R

rs

∑cyc

(1 + cosA)(s− a)

=R

r

∑cyc

(1 + cosA)− R

rs

(2s+

∑cyc

a cosA

)

=R

r

(3 +

∑cyc

cosA

)− 2R

r− R

rs

∑cyc

a cosA

=R

r+R

r

∑cyc

cosA− R

rs

∑cyc

a cosA. (5)

Now, ∑cyc

cosA = 1 + 4 sinA

2sin

B

2sin

C

2= 1 +

4(s− a)(s− b)(s− c)abc

= 1 + 4

ÅF 2/s

4RF

ã= 1 +

F

sR= 1 +

rs

sR= 1 +

r

R, (6)


SOLUTIONS /359

where F denotes the area of the triangle. Using (2) we have∑cyc

a cosA =1

R

∑cyc

apa =2F

R. (7)

Substituting (6) and (7) into (5) we finally have∑cyc

ma

ra≤ R

r+R

r

(1 +

r

R

)− R

rs

Å2F

R

ã=

2R

r+ 1− 2F

rs=

2R

r− 1

since F = rs, and our proof is complete.


Let a, b and c be positive real numbers such that a+ b+ c = 3. Prove that…a

b+

1

a+

…b

c+

1

b+

…c

a+

1

c≥ 3√

2.

We received 22 solutions and 5 incorrect submissions (mostly due to arithmeticerrors). We present the solution by Mohammed Aassila.

By the power mean inequality we have

a+ b+ c

3≥Ça−1/4 + b−1/4 + c−1/4

3

å−4.

Using the hypothesis a+ b+ c = 3 and rearranging, we get

a−1/4 + b−1/4 + c−1/4 ≥ 3.

Finally, applying the AM-GM inequality to each term, we have»ab + 1

a +»

bc + 1

b +»

ca + 1

c ≥√

2»

1b +

√2»

1c +

√2»

1a

=√

2(a−1/4 + b−1/4 + c−1/4

)≥ 3√

2.

Note that equality holds if and only if a = b = c = 1.

3975. Proposed by Ovidiu Furdui.

Let k ≥ 1 be an integer. Calculate∫ ∞0

ex − 1

ex + 1lnkÅex + 1

ex − 1

ãdx.

We received six correct solutions. We present the solution by Anastasios Kotronis.


360/ SOLUTIONS

With the change of the order of integration and summation being justified by the

constant sign of the summand, the change of variables y = lnÄex+1ex−1

ägives∫ ∞

0

ex − 1

ex + 1lnkÅex + 1

ex − 1

ãdx = 2

∫ ∞0

e−2yyk

1− e−2ydy

= 2

∫ ∞0

∞∑n=0

e−2(n+1)yyk dy

= 2∞∑n=0

∫ ∞0

e−2(n+1)yyk dy.

With the substitution z = 2(n+ 1)y the above is equal to

2−k∞∑n=1

1

nk+1

∫ ∞0

e−zzk dz = 2−kΓ(k + 1)ζ(k + 1).

3976. Proposed by Cristinel Mortici and Leonard Giugiuc.

Find positive integer solutions for the following equation :

1

x− 1

y+

1

z=

x

y − z.

We received two correct and one incorrect solution. Presented is the solution byMadhav R. Modak, slightly modified by the editor.

First, let positive integers x, y, z with y 6= z, satisfy the given equation. Then

(yz − xz + xy)(y − z) = x2yz. (1)

Let d = (y, z) so that y = ad, z = bd, (a, b) = 1, where a, b are positive integers.Substituting into (1) and dividing by d2 yields

(dab− xb+ xa)(a− b) = x2ab. (2)

Since (a, b) = 1, a is co-prime to b and a − b. As a divides the left hand side,it must also divide x. Similarly, b|x. Thus ab|x. Hence let x = kab where k is apositive integer. Substituting into (2) and dividing by ab yields

(d− kb+ ka)(a− b) = k2a2b2. (3)

Since a − b is co-prime to both a and b, (3) shows that (a − b)|k2. Hence (3)determines the integer d as follows :

d = −k(a− b) + a2b2 · k2

a− b. (4)


SOLUTIONS /361

Note that |a − b| < max{|a|, |b|}. and therefore |a − b|2 < a2b2. It follows thatthe magnitude of the left summand in (4) is less than the magnitude of the rightsummand. As d is positive, it must be a > b.

Thus a solution (x, y, z) of (1) necessarily satisfies the following conditions. Thereexist positive integers a, b, k with

x = kab, y = ad, z = bd, where

a > b, (a, b) = 1, (a− b)|k2, and

d = −k(a− b) + a2b2k2

a− b.

(5)

Conversely it is easy to check that substituting x, y, z from (5) into the origi-nal equation yields a feasible solution. Hence (5) gives all solutions to the givenequation.

As an example, for a = 2, b = 1, k = 1, we obtain the solution (x, y, z) = (2, 6, 3).

Editor’s comments. The solution can be easily extended to all integers x, y, z byallowing a, b, k to be negative. In that case the condition a > b can also be omittedin (5) without changing the solution set.

3977. Proposed by Dragoljub Milosevic.

Given a triangle ABC, prove that

1

3− 2 cosA+

1

3− 2 cosB+

1

3− 2 cosC≥ 3

2.

We received 15 correct solutions and three incorrect submissions. We present thejoint solution by M. Bello, M. Benito, O. Ciaurri, E. Fernandez, and L. Roncal,which is one of only a few using elementary methods.

Let S denote the semiperimeter of ∆ABC. Then using the fact that

sin2 A

2=

(s− b)(s− c)bc

,

we have

3− 2 cosA = 1 + 2(1− cosA) = 1 + 4 sin2 A

2

= 1 +4(s− b)(s− c)

bc=bc+ 4(s− b)(s− c)

bc.

Hence,∑cyc

1

3− cosA=∑cyc

bc

bc+ 4(s− b)(s− c)= 3−

∑cyc

4(s− b)(s− c)bc+ 4(s− b)(s− c)


362/ SOLUTIONS

so it suffices to show that ∑cyc

4(s− b)(s− c)bc+ 4(s− b)(s− c)

≤ 3

2.

By the AM-GM inequality, we have bc + 4(s − b)(s − c) ≥ 2√

4bc(s− b)(s− c),and thus,

∑cyc

4(s− b)(s− c)bc+ 4(s− b)(s− c)

≤∑cyc

(s− b)(s− c)

bc=∑cyc

sinA

2.

Since it is well known (see item 2.9 on p.20 of [1] O. Bottema et al., Geometricinequalities, Wolters-Noordhoff Publishing, Groningen, The Netherlands, 1969)that

∑cyc sin A

2 ≤32 , our proof is complete. The equality holds if and only if

∆ABC is equilateral.

Editor’s comments. The proofs of most other solvers made use of various ‘hea-vy’ machineries. These include : convexity of a function ; Lagrange’s Multipliers ;Muirhead’s Inequality ; Sturm method ; Gerretsen’s Inequality ; and Half-convexFunction Theorem.

3978. Proposed by Billy Jin and Edward T.H. Wang.

Let n > 2 be a positive integer. A permutation σ = (a1, a2, . . . , an) of S(n) ={1, 2, . . . , n} is called a zigzag permutation if, when reading from left to right,the ai’s alternately change their parity and increase/decrease in magnitudes. Forexample, (5, 6, 3, 4, 1, 2) is a zigzag permutation of S(6). Determine the number ofzigzag permutations of S(n).

We received six correct submissions. We present the solution by M. Bello, M.Benito, O. Ciaurri, E. Fernandez, and L. Roncal.

Let us prove that there exist 2 zigzag permutations of S(n) if n is even and noneif n is odd.

For 1 < j < n, we say aj is a local minimum for the permutation (a1, a2, . . . , an)if aj < min{aj−1, aj+1}. A local maximum is defined analogously, and these defi-nitions can be adapted for a1 and an in an obvious manner.

Because in a zigzag permutation the parity alternates, we deduce that numbers ofthe same parity must be all local minima or all local maxima. But 1 is a minimum,whence the odd numbers must all be local minima while the even numbers must allbe local maxima. Note, as an immediate consequence, that n (which is the largestelement of S(n)) must automatically be a local maximum, which means that n isnecessarily even. Because there is only one element of S(n) less than 2, 2 musttake an extreme position and 1 must be next to it ; that is, we either have a1 = 2or an = 2. Let us first examine the case in which 2 comes first, so that a2 = 1. Thesame observation now holds for the remaining elements of S(n) : 3 is the minimumof those remaining numbers and, therefore, the only remaining number less than


SOLUTIONS /363

4. Because 4 must be surrounded by the two smallest odd numbers 1 and 3, wededuce that a3 = 4 and a4 = 3. When n is even, this construction will continueup to the largest even number n (and therefore smallest remaining even number)as an−1 and with n− 1 appearing last as an. This explicit construction proves theexistence and uniqueness of a zigzag permutation starting with a1 = 2. It remainsto observe that the reverse of a zigzag permutation is also a zigzag permutation(having an = 2), from which we conclude that when n is even there exist exactlytwo zigzag permutations of S(n).

Editor’s Comments. Trey Smith of Angelo State University observed that ourproblem bears a striking resemblance to “Problem 4755 : A Permutation Problem”,which appeared in The American Mathematical Monthly, proposed by ChandlerDavis (64 (1957), p. 596) with a solution by W.J. Blundon (65 (1958), p. 533-534). That problem calls for the number of permutations of S(n) in which relativeminima alternate with relative maxima (without requiring alternating parity). Itis quite remarkable how dropping the parity requirement turns our elementaryproblem into the rather difficult Monthly problem, whose solution involves theBernoulli and Euler numbers ! For that problem, the number of permutations forn = 1, 2, 3, 4, 5, 6, 7 . . . turns out to be 1, 2, 4, 10, 32, 122, 544, . . . , which is sequenceA001250 in the On-Line Encyclopedia of Integer Sequences.


Let AD,BE and CF be the internal bisectors of the triangle ABC. Prove that

[DEF ]

[ABC]≤ R1

4r,

where R1 denotes the circumradius of DEF , r denotes the inradius of ABC and[·] represents the area of the corresponding triangle.

We received eight correct solutions. We present the solution by M. Bello, M. Be-nito, O. Ciaurri, E. Fernandez, and L. Roncal.

The bisector theorem gives, for example, BF = aca+b and BD = ac

b+c . Then, deno-ting [BDF ] = Sb and [ABC] = S, we have

Sb =1

2BF ·BD · sinB =

1

2

a2c2

(a+ b)(b+ c)· sinB.

Since S = 12ac sinB,

Sb =ac

(a+ b)(b+ c)S.

Analogously,

Sa := [AFE] =bc

(a+ b)(a+ c)S and Sc := [CDE] =

ab

(a+ c)(b+ c)S.


364/ SOLUTIONS

From this,

[DEF ]

[ABC]=S − Sa − Sb − Sc

S

= 1−∑cyc

ac

(a+ b)(b+ c)=

2abc

(a+ b)(a+ c)(b+ c). (1)

On the other hand, denoting FD = e, the law of cosines on 4BDF gives

e2 =( ac

b+ c

)2+( ac

a+ b

)2− 2

a2c2

(a+ b)(b+ c)· cosB

=a2c2

(a+ b)2(b+ c)2((b+ c)2 + (a+ b)2 − 2(a+ b)(b+ c) cosB

)=

a2c2

(a+ b)2(b+ c)2(a2 + c2 − 2ac cosB + 2b(a+ b+ c)(1− cosB)

).

Now the law of cosines on 4ABC gives b2 = a2 + c2 − 2ac cosB, so

e2 =a2c2

(a+ b)2(b+ c)2(b2 + 2b(a+ b+ c)(1− cosB)

)=

a2c2

(a+ b)2(b+ c)2· bÅb+ 8p sin2 B

2

ã=

abc

(a+ b)2(b+ c)2(abc+ 8p(p− a)(p− c)

),

where p is the semiperimeter of the triangle ABC and we have used the well known

identity sin2 B2 = (p−a)(p−c)

ac . Analogously, denoting DE = f and EF = d, we have

f2 =abc

(a+ c)2(b+ c)2(abc+ 8p(p− a)(p− b)

)and

d2 =abc

(a+ b)2(a+ c)2(abc+ 8p(p− b)(p− c)

).

From this,

d2e2f2 =a3b3c3

(a+ b)4(a+ c)4(b+ c)4

∏cyc

(abc+ 8p(p− a)(p− b)

). (2)

By using the identities [DEF ] = def4R1

and [ABC] = pr, we obtain

R1

4r=

defp

16[DEF ] [ABC]

and the proposed inequality becomes

16[DEF ]2 ≤ defp.


SOLUTIONS /365

Now, applying (1) and Heron’s formula [ABC]2 = p(p − a)(p − b)(p − c) in theprevious inequality, we have

64a2b2c2(p− a)(p− b)(p− c)(a+ b)2(a+ c)2(b+ c)2

≤ def,

which becomes

212abc(p− a)2(p− b)2(p− c)2 ≤∏cyc


), (3)

after using (2).

In order to prove this inequality (3), we will use the following estimation obtainedfrom the AM-GM inequality :

abc+ 8p(p− a)(p− b) = abc+ 4(a+ b+ c)(p− a)(p− b)= abc+ 4a(p− a)(p− b) + 4b(p− a)(p− b) + 4c(p− a)(p− b)

≥ 4 4

»43a2b2c2(p− a)3(p− b)3.

In a similar way, we can prove that

abc+ 8p(p− a)(p− c) ≥ 4 4

»43a2b2c2(p− a)3(p− c)3

and

abc+ 8p(p− b)(p− c) ≥ 4 4

»43a2b2c2(p− b)3(p− c)3.

Multiplying the three inequalities, we obtain the estimation∏cyc


)≥ 221/2

(abc(p− a)(p− b)(p− c)

)3/2,

and to deduce (3) it is sufficient prove the inequality

221/2(abc(p− a)(p− b)(p− c)

)3/2 ≥ 212abc(p− a)2(p− b)2(p− c)2,

or, equivalently,abc ≥ 8(p− a)(p− b)(p− c). (4)

But abc = 4R[ABC], where R denotes the circumradius of ABC, and

(p− a)(p− b)(p− c) = r[ABC],

and (4) becomesR ≥ 2r,

which is well known and true.


366/ SOLUTIONS

3980. Proposed by S. Viswanathan.

Let a, b, c be distinct real numbers. Prove that(∑cyc

a+ b

a− b

)(∏cyc

a+ b

a− b

)<

1

3.

We received seven correct solutions. We present the solution by M. Bello, M. Be-nito, O. Ciaurri, E. Fernandez, and L. Roncal.

Let us prove that (∑cyc

a+ b

a− b

)(∏cyc

a+ b

a− b

)<

1

4.

If a = 0, the previous inequality becomes

−Åb+ c

b− c

ã2<

1

4,

which is obviously true, and we can assume that abc 6= 0. With the change ofvariables

α =a+ b

a− b, β =

b+ c

b− c, γ =

c+ a

c− a,

the inequality to be proved is equivalent to

αβγ(α+ β + γ) <1

4,

with the restriction αβ+βγ+γα+ 1 = 0. The above inequality is of course true ifαβγ = 0, so we can assume without loss of generality that αβγ 6= 0, and αβ > 0.From the restriction, we obtain

αβ

Ç−1− αβ +

Å1 + αβ

α+ β

ã2å<

1

4.

Since (α+ β)2 ≥ 4αβ, setting t = αβ, it is enough to prove

αβ(1 + αβ)

Å−1 +

1 + αβ

4αβ

ã<

1

4⇐⇒ (1 + t)(1− 3t) < 1, t > 0,

which is obviously true.


AUTHORS’ INDEX /367

AUTHORS’ INDEXSolvers and proposers appearing in this issue

(Bold font indicates featured solution.)

Proposers

Abdilkadir Altinas, Turkey : 4074George Apostolopoulos, Messolonghi, Greece : 4077Michel Bataille, Rouen, France : 4072, 4078Mihaela Berindeanu, Bucharest, Romania : 4079Leonard Giugiuc and Daniel Sitaru, Romania : 4071, 4075Mehtaab Sawhney, Commack High School, Commack, NY, USA : 4076Alina Sıntamarian and Ovidiu Furdui, Technical University of Cluj-Napoca,

Cluj-Napoca, Romania : 4080Daniel Sitaru, Severin, Romania : 4073

Solvers - individuals

Mohammed Aassila, Strasbourg, France : 3972, 3974Arkady Alt, San Jose, CA, USA : 3971, 3973, 3977, 3979, 3980George Apostolopoulos, Messolonghi, Greece : 3974, 3979Sefket Arslanagic, University of Sarajevo, Sarajevo, Bosnia and Herzegovina : 3973, 3974,

3977, 3980Fernando Ballesta Yague, I.E.S. Infante don Juan Manuel, Murcia, Spain : 3971Roy Barbara, Lebanese University, Fanar, Lebanon : 3971, 3972, 3977Michel Bataille, Rouen, France : OC187, OC188, 3971, 3972, 3974, 3975, 3977, 3978,

3979, 3980Marcel Chirita, Bucharest, Romania : 3971Matei Coiculescu, East Lyme High School, East Lyme, CT, USA : 3974Prithwijit De, Homi Bhabha Centre for Science Education, Mumbai, India : 3971, 3974Joseph DiMuro, Biola University, La Mirada, CA, USA : 3971, 3978Marian Dinca, Bucharest, Romania : 3977, 3979Andrea Fanchini, Cantu, Italy : OC187, 3977Ovidiu Furdui, Technical University of Cluj-Napoca, Romania : 3975Oliver Geupel, Bruhl, NRW, Germany : OC186, OC187, OC189, OC190, 3971, 3974,

3978, 3979Leonard Giugiuc, Drobeta Turnu, Severin, Romania : 3980Vaclav Konecny, Big Rapids, Michigan : 3971, 3975Anastasios Kotronis, Athens, Greece : 3975Kee-Wai Lau, Hong Kong, China : 3971, 3974, 3977Salem Malikic, student, Simon Fraser University, Burnaby, BC : 3974Phil McCartney, Northern Kentucky University, Highland Heights, KY, USA : 3974Dragoljub Milosevic, Gornji Milanovac, Serbia : 3971, 3973, 3977Madhav R. Modak, formerly of Sir Parashurambhau College, Pune, India : 3971, 3974,

3976Theoklitos K. Parayiou, High School Lyceum Linopetras Limassol Cyprus,

Agioi Theodoroi, Greece : 3971, 3974, 3979Ricard Peiro i Estruch, IES “Abastos Valencia, Spain : 3974


368/ AUTHORS’ INDEX

Paolo Perfetti, Dipartimento di Matematica, Universita degli studi di Tor Vergata Roma,Rome, Italy : 3974, 3975, 3977

Angel Plaza, University of Las Palmas de Gran Canaria, Spain : 3971, 3974, 3977Wolfgang Remmel, Vienna, Austria : 3974Henry Ricardo, Tappan, NY, USA : 3974Joel Schlosberg, Bayside, NY, USA : 3971Digby Smith, Mount Royal University, Calgary, AB : 3971, 3974, 3978Trey Smith, Angelo State University, San Angelo, TX, USA : 3971Daniel Vacaru, Pitesti, Romania : 3971S. Viswanathan, Central Indian Research Institute, Maharashtra, India : 3980Joseph Walegir, Nixa High School, Nixa, MO, USA : 3977Henry Xia, Vancouver, BC : OC187Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA : OC186Titu Zvonaru, Comanesti, Romania : 3974, 3977, 3979, 3980

Solvers - collaborationsDionne Bailey, Elsie Campbell, and Charles R. Diminnie, Angelo State University,

San Angelo, USA : 3971, 3974, 3977D. M. Batinetu-Giurgiu, Bucharest, and Neculai Stanciu, Buzau, Romania : 3974Leonard Giugiuc and Cristinel Mortici, Romania : 3976M. Bello, M. Benito, O. Ciaurri, E. Fernandez, and L. Roncal, Logrono, Spain : 3971,

3973, 3974, 3975, 3977, 3978, 3979, 3980Billy Jin, University of Waterloo, and Edward T.H. Wang, Wilfred Laurier University,

Waterloo, ON : 3978Missouri State University Problem Solving Group : CC139Haohao Wang and Jerzy Wojdylo, Southeast MO State University, MO, USA : 3977Josh Zukewich and others (unnamed), La Posada : CC137


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