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Crux Mathematicorum

VOLUME 41, NO. 2 February / Fevrier 2015

Editorial Board

Editor-in-ChiefKseniya GaraschukUniversity of British Columbia Contest Corner EditorRobert BilinskiCollege Montmorency Olympiad Corner EditorCarmen BruniUniversity of Waterloo Book Reviews EditorRobert BilinskiCollege Montmorency Articles EditorRobert DawsonSaint Mary's University Problems EditorsEdward BarbeauUniversity of Toronto

Chris FisherUniversity of Regina

Anna KuczynskaUniversity of the Fraser Valley

Edward WangWilfrid Laurier University

Dennis D. A. EppleBerlin, Germany

Magdalena GeorgescuUniversity of Toronto

Assistant EditorsChip CurtisMissouri Southern State University

Lino DemasiOttawa, ON

Allen O'HaraUniversity of Western Ontario

Guest EditorsAlejandro EricksonDurham University

Joseph HoranUniversity of Victoria

Amanda MallochUniversity of Victoria

Mallory FlynnUniversity of British Columbia

Editor-at-LargeBill SandsUniversity of Calgary

Managing EditorDenise CharronCanadian Mathematical Society

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46

IN THIS ISSUE / DANS CE NUM

ERO

47 EditorialKseniya Garaschuk

48 The Contest Corner: No. 32Robert Dawson

48 Problems: CC156{CC160

51 Solutions: CC106{CC110

55 The Olympiad Corner: No. 330Carmen Bruni

55 Problems: OC216{OC220

57 Solutions: OC156{OC160

60 Extending a TetrahedronI. Sharygin

65 Graphs and Edge ColouringDavid A. Pike

71 Problems: 3960, 4011{4020

75 Solutions: 2832b, 3911{3920

91 Solvers and proposers indexCrux Mathematicorum

Founding 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 Mayhem Former Editors / Anciens Redacteurs: Bruce L.R. Shawyer, James E. Totten, Vaclav Linek, Shawn GodinCrux Mathematicorum, Vol. 41(2), February 2015 47

EDITORIAL

As I was reviewing my last editorial, I thought of how mathematics is taught across the globe. Surely, math is math everywhere, but students' and teachers' approaches to it vary culture to culture. I have experienced this rst-hand: I moved to Canada when I was 18 having gone through the school system and even one year of university in Belarus (yes, I was born in the USSR). For the sake of strengthening my English, I decided to start university from scratch in Canada. And now that I work in mathematical education, I often compare my two freshman years. Belorussian rst-year calculus was Canadian third-year analysis: rigorous, technical, precise. We could nd limits only usingand, we dealt with functions purely analytically and we memorized a lot of proofs. In Canada, my experience was the complete opposite: all we did was direct computations, graph functions and generalize patterns. The former was theoretical, the latter | practical. Which one is better? Clearly, you need both. My theoretical experience prepared me for doing math thoroughly, with great attention to detail and with forethought; my practical ex- perience built up my intuition for math and taught me to always look for and make connections between various representations of the same mathematical object. But habits are persistent and we tend to stick to what we are more comfortable with. So my Canadian students resist the theoretical side of math calling it \dry" and \irrelevant", while my Belorussian types often object to graphing and describing math in words since it is \watered-down" and \not mathy". You can't please everyone. AtCrux, it is also clear that people (both amongst our subscribers and within the Editorial Board) have dierent preferred methods and tend to be faithful to their favourite approaches. Whatever your taste, you will nd something to your liking on the pages of our journal. And if not, then clearly we are missing your submissions! So send them along tocrux-psol@cms.math.cafor problem proposals and tocrux-articles@cms.math.cafor articles.

Kseniya Garaschuk

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48/ THE CONTEST CORNER

THE CONTEST CORNER

No. 32

Robert Dawson

Les problemes presentes dans cette section ont deja ete presentes dans le cadre d'un concours mathematique de niveau secondaire ou de premier cycle universitaire, ou en ont ete inspires. Nous invitons les lecteurs a presenter leurs solutions, commentaires et generalisations pour n'importe quel probleme. S'il vous pla^t vous referer aux regles de soumission a l'endos de la couverture ou en ligne. Pour faciliter l'examen des solutions, nous demandons aux lecteurs de les faire parvenir au redacteur au plus tard le1 avril 2016; toutefois, les solutions recues apres cette date seront aussi examinees jusqu'au moment de la publication. La redaction souhaite remercier Andre Ladouceur, Ottawa, ON, d'avoir traduit les problemes.CC156. Decrire et realiser un croquis precis de la region qui represente l'en- semble f(x;y;z) :jxj+jyj61;jyj+jzj61;jzj+jxj61g: CC157.Etant donne une matrice 55 dont chacun des nombres est un 0 ou un 1, demontrer qu'il doit exister une sous-matrice 22 (c'est-a-dire l'intersection de la reunion de deux rangees avec la reunion de deux colonnes) dont tous les nombres sont soit 0, soit 1. CC158. On considere un point mobileAsur la partie positive de l'axe des abscisses, un point mobileBsur la partie positive de l'axe des ordonnees et l'origine Ode maniere que le triangleABOait toujours une aire de 4. Determiner l'equation d'une courbe, denie dans le premier quadrant, qui est tangente a chacun des segmentsAB. CC159. La disposition de huit tuiles carrees, ci-dessous a gauche, peut ^etre divisee en deux groupes congruents de quatre tuiles, comme sur la droite. (On remarque qu'un groupe est le re et de l'autre dans un miroir, ce qui est permis.) Determiner une facon de diviser la disposition suivante de 100 tuiles en deux

Crux Mathematicorum, Vol. 41(2), February 2015

THE CONTEST CORNER /49

groupes congruents de 50 tuiles, ou demontrer qu'il est impossible de le realiser. (19 tuiles dans la 10erangee) CC160. Determiner tous les triplets (f;g;h) de fonctions continues a valeurs reelles denies surRtelles pour tout nombre reelx, f(g(x)) =g(h(x)) =h(f(x)) =x:

CC156. Describe and accurately sketch the region

f(x;y;z) :jxj+jyj61;jyj+jzj61;jzj+jxj61g: CC157. Show that if a 55 matrix is lled with zeros and ones, there must always be a 22 submatrix (that is, the intersection of the union of two rows with the union of two columns) consisting entirely of zeros or entirely of ones. CC158. Suppose movable pointsA,Blie on the positivex-axis andy-axis, respectively, in such a way that4ABO, whereOis the origin, always has area 4. Find an equation for a curve in the rst quadrant which is tangent to each of the line segmentsAB. CC159. The following pattern of eight square tiles can be divided into two congruent sets of four tiles as shown. (Note that one set is the mirror image of the other | this is legal.)

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50/ THE CONTEST CORNER

Find a way to divide the following pattern of 100 tiles into two congruent sets of fty tiles, or show it cannot be done. (19 tiles in the 10th row) CC160. Find all triples of continuous functionsf;g;h:R7!Rsuch that, for allx2R, f(g(x)) =g(h(x)) =h(f(x)) =x :Three really magic squares Having received his yearly salary in silver coins, the royal Mathematician arranged the coins into vertical stacks and placed them on a33square so that the numbers representing the amount of coins in each stack formed a magic square, that is a square such that the sum of the numbers along every row, column and diagonal of the square is the same. Some stacks came out being quite tall, but none were higher than 300 coins tall. The King liked the arrangement but lamented over the fact that all the numbers came out being composite. \If your majesty gives me 9 more coins, I will add one to each stack; the magic square property will be preserved but all the new numbers will be prime", replied the Mathematician. The King nearly agreed, but was interrupted by the Joker, who took away one coin from each stack and the new numbers all became prime (and the square, of course, remained a magic square). What was the original magic square composed by the Mathematician? From Kvant, 1981 (9), p.31.Crux Mathematicorum, Vol. 41(2), February 2015

THE CONTEST CORNER /51

CONTEST CORNER

SOLUTIONS

Les enonces des problemes dans cette section apparaissent initialement dans 2014: 40(2), p. 51{52.CC106. At each summit of a regular tetrahedron of side length 3, we cut o a pyramid such that the cut-o surface makes an equilateral triangle. The four equilateral triangles thus obtained have all dierent dimensions. What is the total length of the edges of the solid thus truncated? Provide a proof. Originally problem 29 from Demi-nale du Concours Maxi de Mathematiques de

Belgique 2008.

We received two incomplete submissions to this problem, neither of which ade- quately proved that the four tetrahedral corners that are removed from the original tetrahedron must be regular. We present an editor's solution. LetA;B;C, andDbe the vertices of the given tetrahedron, and letP;Q, andR be the points onDA;DB, andDC, respectively, such thatPQRis the equilateral triangle formed by cutting o the pyramid containing the vertexD. Suppose that the polyhedron has been labeled so thatDPDQDR. We will rst prove that these segments must, in fact, be equal. Compare trianglesDPRandDQR. We haveRP=RQand the angles atDare both 60. Because we assume that RDis no larger than eitherDPorDQ, the angles atPandQmust be acute. From the sine law (applied to both triangles) we have sin\DQR=DRsin60RQ =DRsin60RP = sin\DPR; from which we conclude that the two triangles are congruent, whenceDQ=DP.

Focusing now on the 60

anglePDQ, we note that the length of the segmentPQ increases monotonically as the lengthsDP=DQincrease, so there will be exactly one position ofPandQfor whichPQ=QR(=RP), namely where the lengths DP;DQ;DRare all equal. Because the angles atDare all 60, the three faces at Dare equilateral triangles, and the tetrahedronDPQRis therefore regular. Returning to the problem, because all edges of a regular tetrahedron have the same length,DP+DQ+DR=PQ+QR+RPand we conclude that the truncation at the vertexDdoes not change the sum of the edge lengths. Of course, the same can be said about the truncation at the other vertices, so the total length of the edges of the truncated tetrahedron must equal 18. CC107. In a right triangleABCwith right angle atBandBC= 1, we place

Don sideACsuch thatAD=AB=12

. What is the length ofDC?

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52/ THE CONTEST CORNER

Originally problem 6 from Demi-nale du Concours Maxi de Mathematiques de

Belgique 2002.

There were eight solution submitted for this question, all essentially the same.

By the Pythagorean Theorem we have

AC=pBC

2+AB2=É1 +

14 =p5 2 and

DC=ACAD=p5

2 12 =p512 CC108. In an orthonormal system, the line with equationy= 5xcrosses the parabola with equationy=x2in pointA. The perpendicular toOAatO intersects the parabola atB. What is the area of triangleAOB? Originally problem 20 from Demi-nale du Concours Maxi de Mathematiques de

Belgique 2009.

We received six correct solutions, and one incorrect solution. We present the solution of Titu Zvonaru. It is easy to deduce thatA(5;25). The slope ofOBis1=5. Solving the system y=15 x,y=x2we obtainB(15 ;125

It follows thatOA=p5

2+ 252= 5p26,OB=È1

5 2+125 2=p26 25
. Hence the area of the triangle isAOB=OAOB2 =2610 =135 CC109. LetEbe the set of realsxfor which the two sides of the following equality are dened: cot8xcot27x=sinkxsin8xsin27x: If this equality holds for all the elements ofE, what is the value ofk? Originally problem 21 from Demi-nale du Concours Maxi de Mathematiques de

Belgique 2009.

We received seven submitted solutions to this problem, one of which was incorrect and ve were incomplete. We present the only correct solution by Paolo Perfetti modied by the editor.

Note rst thatE=fx2Rjx6=m8

andx6=m27 for anym2Z. Forx2E, the given equality is equivalent to sin8xsin27x(cot8xcot27x) = sinkx:(1) We shall prove that the only value ofkfor which (1) holds for allx2Eisk= 19.

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THE CONTEST CORNER /53

Since sin8xsin27x(cot8xcot27x) = sin27xcos8xcos27xsin8x = sin(27x8x) = sin19x; k= 19 satises (1). Next, suppose (1) holds for allx2Eand somek2Zwithk6= 19. Ifk=19, then from (1) we have 2sin19x= 0 for allx2E, which is false (for example, ifx=38 , thenx2E, but sin19x= sin2 = 16= 0). Hencek6=19:

From (1), we also have

2sin x‹ x‹ = 0:(2)

Since sin

19k2 x= 0 if and only if19k2 x=morx=2m19kand cos19+k2 x= 0 if and only if 19+k2 x= (m+12 )orx=(2m+1)19+kfor somem2Z, there must be somex2Ethat does not satisfy (2). (To be more precise, the set of allxsuch thatx=2m19korx=(2m+1)19+kfor somem2Zis countable whileEis clearly uncountable.) This is a contradiction and our proof is complete. CC110. What is the number of real solutions to the equation: j1 +x jx j1xjjj=j x jx1jj: Originally problem 26 from Demi-nale du Concours Maxi de Mathematiques de

Belgique 2009.

We have received four correct solutions and one incorrect submission. We present the solution by Henry Ricardo. We compute the left-hand side (LHS) and the right-hand side (RHS) on three intervals that cover the real number line.

Case 1.Suppose that 0x1. Then

RHS =j x(1x)j=j x1 +xj= 1:

Whenx2[;12

j1 +x jx(1x)jj=j1 +x(12x)j= 3x and whenx2(12 ;1], j1 +x j2x1jj=j1 +x(2x1)j=j2xj= 2x so that

LHS =§3xif 0x12

2xif12

< x1:

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54/ THE CONTEST CORNER

Thus LHS = RHS when either 3x= 1 or 2x= 1, which impliesx=13 and x= 1 forx2[0;1].

Case 2.Ifx >1, we have

LHS =j1 +x jx(x1)jj=j1 +x1j=x

and

RHS =j x(x1)j=j 2x+ 1j= 2x1:

Butx >1 implies that (2x1)x=x1>0, so RHS>LHS and there are no solutions to the equation in the interval (1;1). Case 3.Finally, forx2(1;0), RHS = 1 and LHS =j3xj=3x, so LHS =

RHS if and only if3x= 1, which impliesx=13

Thus the only solutions of the given equation arex=13 ;1;13 .Math Quotes The solution of problems is one of the lowest forms of mathematical research. Yet, its educational value cannot be overestimated. It is the ladder by which the mind ascends into higher elds of original research and investigation. Many dormant minds have been aroused into activity through the mastery of a single problem. Benjamin Franklin Finkel.Crux Mathematicorum, Vol. 41(2), February 2015

THE OLYMPIAD CORNER /55

THE OLYMPIAD CORNER

No. 330

Carmen Bruni

Les problemes presentes dans cette section ont deja ete presentes dans le cadre d'une olympiade mathematique regionale ou nationale. Nous invitons les lecteurs a presenter leurs solutions, commentaires et generalisations pour n'importe quel probleme. S'il vous pla^t vous referer aux regles de soumission a l'endos de la couverture ou en ligne. Pour faciliter l'examen des solutions, nous demandons aux lecteurs de les faire parvenir au redacteur au plus tard le1 avril 2016; toutefois, les solutions recues apres cette date 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.OC216. Soitp=n2+ 1 un nombre premier. Determiner toutes les solutions

entieres a l'equation suivante : x

2(n2+ 1)y2=n2:

OC217. SoitGle centrode du triangle rectangleABCou\BCA= 90. Soit Ple point sur le rayonAGtel que\CPA=\CAB, et soitQle point sur le rayon BGtel que\CQB=\ABC. Demontrer que les cercles circonscrits deAQGet

BPGse rencontrent a un point sur le c^oteAB.

OC218. Determiner toute fonctionf:N!Nsatisfaisant

f(mn) = lcm(m;n)gcd(f(m);f(n)) pour tous les entiers positifsmetn. OC219. Pourmetndes entiers positifs donnes, demontrer qu'il existe un entierctel que les nombrescmetcnont le m^eme nombre d'occurences de chaque chire non nul, lorsqu'ils sont exprimes en base 10. OC220. SoitA1A2:::A8un octagone convexe ou tous les c^otes sont de m^eme longueur et ou les c^otes opposes sont paralleles. Pour chaquei= 1;:::;8, posons B ile point d'intersection des segmentsAiAi+4etAi1Ai+1, ouAj+8=Ajet B j+8=Bjpour toutj. Fournir un nombrei, parmi 1, 2, 3, et 4, satisfaisant A iAi+4B iBi+4632

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56/ THE OLYMPIAD CORNER

OC216. Letp=n2+ 1 be a given prime number. Find the set of integer solutions to the following equation : x

2(n2+ 1)y2=n2:

OC217. LetGbe the centroid of a right-angled triangleABCwith\BCA= 90
. LetPbe the point on rayAGsuch that\CPA=\CAB, and letQbe the point on rayBGsuch that\CQB=\ABC. Prove that the circumcircles of trianglesAQGandBPGmeet at a point on sideAB.

OC218. Find all functionsf:N!Nsatisfying

f(mn) = lcm(m;n)gcd(f(m);f(n)) for all positive integersm;n. OC219. Given positive integersmandn, prove that there is a positive integer csuch that the numberscmandcnhave the same number of occurrences of each non-zero digit when written in base ten. OC220. LetA1A2:::A8be a convex octagon such that all of its sides are equal and its opposite sides are parallel. For eachi= 1;:::;8, deneBias the intersection between segmentsAiAi+4andAi1Ai+1, whereAj+8=AjandBj+8=Bjfor all j. Show that some numberi, amongst 1, 2, 3, and 4 satises A iAi+4B iBi+4632 :Crux Mathematicorum, Vol. 41(2), February 2015

THE OLYMPIAD CORNER /57

OLYMPIAD SOLUTIONS

Les enonces des problemes dans cette section apparaissent initialement dans 2014 : 40(2), p. 56{57.OC156. LetABCDbe a tetrahedron. Prove that vertexD, center of ins- phere and centroid ofABCDare collinear if and only if the areas of triangles

ABD;BCD;CADare equal.

Originally question 2 from day 1 of the Poland Math Olympiad. We received four correct solutions to this problem. We present the solution by

Michel Bataille.

LetIandrbe the center and the radius of the insphere. LetV() andA() denote volume and area, respectively.

SinceV(IBCD) =13

r A(BCD) (and similarly for trianglesCDA,DABand

ABC), we can use

(A(BCD) :A(CDA) :A(DAB) :A(ABC)) instead of

V(IBCD) :V(ICDA) :V(IDAB) :V(IABC))

for the barycentric coordinates ofIrelative to (A;B;C;D). It follows that

I= (A(BCD))A+ (A(CDA))B+ (A(DAB))C+ (A(ABC))D

whereis the sum of the areas of the faces ofABCDand the bold face letters represent the coordinates of the points. In particular, we have !DI= (A(BCD))!DA+ (A(CDA))!DB+ (A(DAB))!DC:(1)

LetGbe the centroid ofABCD. Then,

4G=A+B+C+D;

from which we deduce

4!DG=!DA+!DB+!DC:(2)

Now,D;I;Gare collinear if and only if

!DI=(4!DG) (3) for some real number. Since!DA;!DB;!DCare not coplanar, (1) and (2) show that (3) occurs if and only if

A(BCD) =A(CDA) =A(DAB):

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OC157. Find all functionsf:R!Rsuch that

f(f(x)2+f(y)) =xf(x) +y;8x;y2R: Originally question 4 from Kyrgyzstan National Olympiad. We received three correct solutions and one incorrect submission. We present the solution by Henry Ricardo.

Lettingx= 0 in the given equation, we have

f(f(0)2+f(y)) =y;8y2R(1) sofisonto. Thus there existsa2Rsuch thatf(a) = 0. Takingx=ain the original equation gives us f(f(y)) =y;8y2R;(2) which shows thatfis one to one sincef(x) =f(y) impliesf(f(x)) =f(f(y)), or by (2),x=y. Now replacexbyf(x) in the original equation to obtain f(x2+f(y)) =f(x)f(f(x)) +y=xf(x) +y=f(f(x)2+f(y))

Usnig the fact thatfis one to one, we have

x

2+f(y) =f(x)2+f(y):

Thereforex2=f(x)2andf(x) =x.

To eliminate the possibility thatf(x) =xfor somexandf(y) =yfor some y6=x, suppose thatxy6= 0 andf(x) =x;f(y) =y. The original equation gives usf(x2y) =x2+y, but we know thatf(x) =xand so(x2y) =x2+y implies eitherx= 0 ory= 0. Now it is clear thatf(x) =xfor allx2R andf(x) =xfor allx2Rsatisfy the original equation and are the only such functions. OC158. Prove that a nite simple planar graph has an orientation so that every vertex has out-degree at most 3. Originally question 4 from day 1 of the Romania TST.

We received no solutions to this problem.

OC159. Letpbe an odd prime number. Prove that there exists a natural numberxsuch thatxand 4xare both primitive roots modulop. Originally question 3 from the 2012 Iran National Math Olympiad Third Round. We received one correct solution and one incorrect submission. We present the solution by Oliver Geupel.

Crux Mathematicorum, Vol. 41(2), February 2015

THE OLYMPIAD CORNER /59

The existence of a primitive root modulopis a well-known fact. Suppose thata is a primitive root modulop. Then there is a positive integerrsuch that 2ar (modp) and therefore 4a2r(modp). Letp1;p2;:::;p`denote the distinct prime divisors ofp1. For 16k6`, letskbe an integer such thatsk60 (modpk) and s k6 2r(modpk). Find via the Chinese Remainder Theorem, a natural number msuch that msk(modpk);16k6`: Then, neithermnorm+2ris divisible by anypk. Hence, each of the numbersm andm+ 2ris coprime withp1.

We obtain

p1-m;2m;3m; :::;(p2)m; p1-m+ 2r;2(m+ 2r);3(m+ 2r); :::;(p2)(m+ 2r): Thus, a m; a2m; a3m; :::; a(p2)m61 (modp); a m+2r; a2(m+2r); a3(m+2r); :::; a(p2)(m+2r)61 (modp); sinceais a primitive root modulo the primep. We have obtained thatamand a m+2r4am(modp) are primitive roots modulop. Therefore,x=amhas the required property. OC160. The incircle of triangleABC, is tangent to sidesBC;CAandAB atD;ErespectivelyF. LetTandSbe the re ection ofFwith respect toB respectively the re ection ofEwith respect toC. Prove that the incenter of triangleASTis inside or on the incircle of triangleABC. Originally question 3 from day 2 of the Iran National Math Olympiad Second

Round.

No solutions were received.Copyright

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60/ FROM THE ARCHIVES : EXTENDING A TETRAHEDRON

Extending a tetrahedron

I. Sharygin

One of the most beautiful tools that can be used when solving geometrical problems consists of replacing the geometric gure in question with another one, a more convenient one in some sense. For example, if the problem involves a triangle with a median, often it is helpful to use this triangle to construct a parallelogram therefore extending the median to turn it into the parallelogram's diagonal. In this article, we will consider several problems involving a triangular pyramid, the so-called tetrahedron, that can be solved by extending the tetrahedron to another solid, often a parallelepiped. The rst way to extend the tetrahedron is presented in Figure 1. Here,AA1BDquotesdbs_dbs24.pdfusesText_30

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