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Crux Mathematicorum

VOLUME 44, NO. 2 Fevrier / February 2018

Editorial Board

Editor-in-ChiefKseniya GaraschukUniversity of the Fraser Valley Contest Corner EditorJohn McLoughlinUniversity of New Brunswick Articles EditorRobert DawsonSaint Mary's University Problems EditorsEdward BarbeauUniversity of Toronto

Chris FisherUniversity of Regina

Edward WangWilfrid Laurier University

Dennis D. A. EppleBerlin, Germany

Magdalena GeorgescuBGU, Be'er Sheva, Israel

Shaun FallatUniversity of Regina

Assistant EditorsChip CurtisMissouri Southern State University

Allen O'HaraUniversity of Western Ontario

Guest EditorsKelly PatonUniversity of British Columbia

Alessandro VentulloUniversity of Milan

Editor-at-LargeBill SandsUniversity of Calgary

Managing EditorDenise CharronCanadian Mathematical Society

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45

IN THIS ISSUE / DANS CE NUM

ERO

46 EditorialKseniya Garaschuk

47 The Contest Corner: No. 62John McLoughlin

47 Problems: CC306{CC310

49 Solutions: CC256{CC260

52 The Olympiad Corner: No. 360

52 Problems: OC366{OC370

54 Solutions: OC306{OC310

60 Problem Solving 101: No. 3Shawn Godin

63 On the Centres of Root-Mean-Square Triangles

Michel Bataille

69 Problems: 4311{4320

73 Solutions: 4211{4220

84 From the Archives

85 Solvers and proposers index for issues 1 and 2Crux 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 GodinCopyright

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46

EDITORIAL

In problem solving, I am often reminded of a Russian proverb: \Do not shoot a sparrow with a cannon." The meaning is subtle. It is not just about overkill and exerting more power than needed. The fact is, you might not actually be able to accomplish the task at hand since the chosen weapon, while powerful, is simply ill-suited: a heavy awkward cannon to aim with versus a quick little bird that is fast to get away. I thought of this saying after my recent calculus exam. Here is a part of one problem from it: Rainbow trout in Deer Lake can no longer reproduce due to habitat destruction, so city ocials consider stocking the lake with sh and allowing locals to sh them out. As such, the sh population satises the dierential equation dFdt =srF;whereF(t) is the number of sh at timet(in months),sis the stocking rate (in number of sh per month) andris the shing rate (proportion of sh population that gets shed out every month). a) Fishing is p rohibitedb etweenOctob er1st and Marc h1st, but the stocking continues at the rate of 100 sh per month (occurring always in the 2nd of a month). If there are an estimated 1500 trout in the lake on October 1st, how many sh will there be on

March 1st?

b)::: My class was stumped! They all realized that they can plug inr= 0 but that left them with a form of a dierential equation we haven't yet studied (in this course, dierential equations come before antiderivatives). They pulled out just about every weapon from their dierential equations ammunition: I saw phase diagrams, analysis of steady states, slope elds, ... All of that for a poor little linear growth, which in the end successfully escaped many of their attacks. Lesson for my students and the rest of us: read and think before reaching for a bazooka. In math and otherwise.

Kseniya Garaschuk

Crux Mathematicorum, Vol. 44(2), February 2018

THE CONTEST CORNER /47

THE CONTEST CORNER

No. 62

John McLoughlin

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 plus tard le1er juillet 2018. La redaction souhaite remercier Andre Ladouceur, Ottawa, ON, d'avoir traduit les problemes.CC306. On considere un cube 555 dont la surface exterieure est peinte en bleu. Biz coupe le cube en 5

3cubes unites, puis il en prend un au hasard. Sachant

que le cube a au moins une face bleue, quelle est la probabilite que ce cube ait exactement deux faces bleues? CC307. Determiner toutes les solutions entieres (x;y) de l'equation x

2xy+ 2017y= 0:

CC308. On denit lamatrice de Pascalnncomme suit :a1j=ai1= 1; a ij=ai1;j+ai;jilorsquei;j >1. Par exemple, la matrice de Pascal 33 est 2

41 1 1

1 2 3

1 3 63

5 Demontrer que toute matrice de Pascal est inversible. CC309. SoitP(x) etQ(x) des polyn^omes avec coecients reels. Determiner des conditions necessaires et susantes surNde maniere que si le polyn^ome P(Q(x)) est de degreN, il existe une valeur reelle dextelle queP(x) =Q(x).

CC310. On donne

tanx+ cotx+ secx+ cscx= 6:

Determiner la valeur de

sinx+ cosx:

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CC306. Consider a 555 cube with the outside surface painted blue. Buzz cuts the cube into 5

3unit cubes, then picks a cube at random. Given that the

cube Buzz picked has at least one painted blue face, what is the probability that the cube has exactly two blue faces? CC307. Find (with proof) all integer solutions (x;y) to x

2xy+ 2017y= 0:

CC308. Dene thennPascal matrixas follows :a1j=ai1= 1, while a ij=ai1;j+ai;jifori;j >1. So, for instance, the 33 Pascal matrix is 2

41 1 1

1 2 3

1 3 63

5

Show that every Pascal matrix is invertible.

CC309. SupposeP(x) andQ(x) are polynomials with real coecients. Find necessary and sucient conditions onNto guarantee that if the polynomial P(Q(x)) has degreeN, there exists realxwithP(x) =Q(x).

CC310. Suppose

tanx+ cotx+ secx+ cscx= 6:

Find the value of

sinx+ cosx:Crux Mathematicorum, Vol. 44(2), February 2018

THE CONTEST CORNER /49

CONTEST CORNER

SOLUTIONS

Les enonces des problemes dans cette section paraissent initialement dans 2017 : 43(2), p. 44{45.CC256. All vertices of a polygonPlie at points with integer coordinates in the plane (that is to say, both their co-ordinates are integers), and all sides ofP have integer lengths. Prove that the perimeter ofPmust be even. Originally question 5 from The University of Melbourne Department of Mathema- tics and Statistics School Mathematics Competition, 2012 (Senior Division). We received ve correct solutions. We present the one by Steven Chow. Letnbe the number of vertices ofP, and denote by (xj;yj) the vertices ofPin clockwise order (where 1jn), with the additional convention thatxn+1=x1 andyn+1=y1. Ifais any integer thenaa2(mod 2). Using this observation, as well as the fact that the side lengths ofPare integers, the perimeter ofPis n X j=1È(xjxj+1)2+ (yjyj+1)2nX j=1€ (xjxj+1)2+ (yjyj+1)2Š nX j=1((xjxj+1) + (yjyj+1)) (mod 2): In the last line, all coordinates appear once with a positive sign and once with a negative sign, so the sum is equal to zero. Therefore, the perimeter ofPis even. CC257. It is asserted that one can nd a subsetSof the nonnegative integers such that every nonnegative integer can be written uniquely in the formx+2yfor x;y2S. Prove or disprove the assertion. Originally question 6 from The University of Melbourne Department of Mathema- tics and Statistics School Mathematics Competition, 2012 (Senior Division). We received three correct solutions. We present the solution of the Missouri State

University Problem Solving Group.

We prove that there is such a set, and our construction shows there is only one such setS. The only way to write 0 asx+ 2yfor nonnegative integersx,yis x=y= 0:Therefore 02S. The only way to write 1 in this form isx= 1,y= 0; and therefore 12S. We may write 2 = 2+2(0) or 2 = 0+2(1);but if 22S, then we do not have uniqueness. Therefore, 2=2S. Now 3 = 3 + 2(0) and 3 = 1 + 2(1); so to ensure uniqueness we must have 3=2S. Continuing in this way we see thatS

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50/ THE CONTEST CORNER

must contain 0;1;4;5;16;17;20;21;64;65;and we notice that, other than 0, each of these numbers is a sum of powers of 4. For example, 17 = 4

2+40;20 = 42+41;

and 21 = 4

2+ 41+ 40:LetSbe the set consisting of 0 together with all possible

sums of powers of 4 :

S=f0;1;4;5;16;17;20;21;64;65;g

To see that every nonnegative integerNcan be written in the formN=x+ 2y forx,y2S, consider the binary representation ofN: N=tX i=0b i2iwith eachbi2 f0;1g: Now,

N=bt=2cX

i=0b i22i+b(t1)=2cX i=0b i22i+1=bt=2cX i=0b i22i+ 2b(t1)=2cX i=0b i22i:

Taking

x=bt=2cX i=0b i22iandy=b(t1)=2cX i=0b

2i+122i;

we see thatN=x+2y, andx;y2S. The uniqueness follows from the uniqueness of the base 2 representation. CC258. The three pointsA,BandCin the diagram are vertices of an equilateral triangle. Given any pointPon the circle containingA,BandC, consider the three distancesAP,BPandCP. Prove that the sum of the two

shorter distances gives the longer distance.Originally question 7 from The University of Melbourne Department of Mathema-

tics and Statistics School Mathematics Competition, 2015 (Intermediate Division). We received 25 correct solutions, representing 12 solvers. We present the solution by Andrea Fanchini. By Ptolemy's theorem,ABCP=BCAP=CABP;butAB=BC=CA, so

CP=AP+BP:

Crux Mathematicorum, Vol. 44(2), February 2018

THE CONTEST CORNER /51

CC259. If you are told that a rectangle has areaAand perimeterP, is that sucient information to determine its side lengths? Originally question 2 from The University of Melbourne Department of Mathema- tics and Statistics School Mathematics Competition, 2013 (Senior Division). We received ten correct solutions. We present an amalgamation of many. Letxandybe the side lengths of the rectangle. ThenA=xyandP= 2(x+y) so thaty=P2 x. Plugging into the area x‹ ()x2P2 x+A= 0:

Using the quadratic formula, we obtain

x=P4 pP 216A4
andy=P4 pP 216A4
Since

P216A= 4(x+y)216xy= 4(xy)20;

there is always a real solution to our quadratic equation, and furthermore, there is only one solution for eachAandP, once we account for rotation. Hence, the side lengths are uniquely determined as a pair from the specied area and perimeter of a rectangle. CC260. Assume you have a 9-faced die, appropriately constructed so that when the die is thrown, each of the faces (which are numbered 1 to 9) occurs with equal probability. Determine the probability that afternthrows of the die, the product of all the numbers thrown will be divisible by 14. Originally question 6 from The University of Melbourne Department of Mathema- tics and Statistics School Mathematics Competition, 2013 (Senior Division). We received six correct solutions and one incorrect solution. We present the solu- tion of the Missouri State University Problem Solving Group. LetAbe the set ofntosses that do not contain an even number andBthe set ofntosses that do not contain 7. The product of numbers shown inntosses is a multiple of 14 if and only if both an even number and 7 appear among then tosses. Thus we need to to countjAc\Bcj. Now,jAj= 5n(since each toss can be any of the ve outcomes, 1;3;5;7;9),jBj= 8n;jA\Bj= 4n. Thus, jA[Bj=jAj+jBj jA\Bj= 5n+ 8n4n; and jAc\Bcj=j(A[B)cj= 9n jA[Bj= 9n5n8n+ 4n:

Thus the required probability is

9 n5n8n+ 4n9 n:Copyright c

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52/ THE OLYMPIAD CORNER

THE OLYMPIAD CORNER

No. 360

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 plus tard le1er juillet 2018. La redaction souhaite remercier Andre Ladouceur, Ottawa, ON, d'avoir traduit les problemes.OC366. Demontrer qu'il existe un nombre inni de triplets (a;b;c) d'entiers strictement positifs tels quea;betcsoient premiers entre eux deux a deux et ab+c;bc+aetca+bsoient premiers entre eux deux a deux. OC367. Un concours de mathematiques est compose de 3 problemes, chacun pouvant recevoir une note entiere de 0 a 7. On sait qu'etant donne n'importe quels deux concurrents, il existe au plus un probleme pour lequel les concurrents ont recu la m^eme note (par exemple, il n'y a pas deux concurrents qui ont recu, dans l'ordre, les notes 7;1;2 et 7;1;5, mais il peut y avoir deux concurrents qui ont recu, dans l'ordre, les notes 7;1;2 et 7;2;1). Determiner le nombre maximal de concurrents. OC368. Soitnun entier strictement positif. Determiner, en fonction den, le nombre de solutions de l'equation x

2+ 2016y2= 2017n:

OC369. SoitIle centre du cercle inscrit dans le triangleABC. SoitDle point d'intersection deAIavec le c^oteBCetSle point d'intersection deAIavec le cercle circonscrit au triangleABC(S6=A). SoitKetLles centres des cercles inscrits dans les triangles respectifsDSBetDCS. SoitPl'image deIpar une re exion par rapport a l'axeKL. Demontrer queBP?CP. OC370. Soit deux entiersnetktels quenk2. Vous jouez au jeu suivant contre un genie maleque. Le genie tient 2ncartes, numerotees d'un c^ote de 1 a n, deux cartes pour chaque valeur dei,i= 1;:::;n. Au depart, le genie aligne les cartes a l'envers dans un ordre quelconque. Vous montrez du doigt n'importe quelleskcartes. Le genie remet alors ces cartes a l'endroit. Si deux des cartes ont le m^eme numero, le jeu est termine et vous avez gagne. Autrement, vous devez

Crux Mathematicorum, Vol. 44(2), February 2018

THE OLYMPIAD CORNER /53

fermer les yeux pendant que le genie permute leskcartes choisies et les remet a l'envers. C'est ensuite votre tour a nouveau. On dit que ce jeu est gagnable s'il existe un entier strictement positifmet une strategie qui garantit une victoire enmtours ou moins, peu importe comment le genie joue. Pour quelles valeurs denet dekle jeu est-il gagnable? OC366. Prove that there exist innitely many positive integer triples (a;b;c) such thata;b;care pairwise relatively prime, andab+c;bc+a;ca+bare pairwise relatively prime. OC367. A mathematical contest had 3 problems, each of which was given a score between 0 and 7, inclusive. It is known that, for any two contestants, there exists at most one problem in which they have obtained the same score (for example, there are no two contestants whose ordered scores are 7;1;2 and 7;1;5, but there might be two contestants whose ordered scores are 7;1;2 and 7;2;1).

Find the maximum number of contestants.

OC368. Letnbe a positive integer. Find the number of solutions of x

2+ 2016y2= 2017n

as a function ofn. OC369. LetIbe the incenter of4ABC. LetDbe the point of intersection ofAIwithBCand letSbe the point of intersection ofAIwith the circumcircle ofABC(S6=A). LetKandLbe incenters of4DSBand4DCS. LetPbe a requotesdbs_dbs24.pdfusesText_30

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