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Crux Mathematicorum

VOLUME 38, NO. 6

JUNE / JUIN 2012

IN THIS ISSUE / DANS CE NUM

ERO

215 EditorialShawn Godin

216 Mathematical MayhemShawn Godin

216 Mayhem Solutions: M495{M500

222 The Contest Corner: No. 6Shawn Godin

224 The Olympiad Corner: No. 304Nicolae Strungaru

224 The Olympiad Corner Problems: OC86{OC90

226 The Olympiad Corner Solutions: OC26{OC30

232 Book ReviewsAmar Sodhi

232A Wealth of Numbers: An Anthology of 500

Years of Popular Mathematics Writing

Edited by Benjamin Wardhaugh

233The Irrationals: a Story of the Numbers

You Can't Count On

by Julian Havil

235 Problem Solvers Toolkit: No. 1Shawn Godin

238 Recurring Crux Congurations 7:J. Chris Fisher

241 Problems: 3751{3760

245 Solutions: 3651{3660

Published by Publie par

Canadian Mathematical Society Societe mathematique du Canada

209 - 1725 St. Laurent Blvd. 209 - 1725 boul. St. Laurent

Ottawa, Ontario, Canada K1G 3V4 Ottawa (Ontario) Canada K1G 3V4

FAX: 613{733{8994 Telec : 613{733{8994

email: subscriptions@cms.math.ca email : abonnements@smc.math.ca 215

EDITORIAL

Shawn Godin

We have been working on a few new features this volume. TheContest Cornerstarted earlier this year featuring problems that have appeared in math- ematics contests at the high school and undergraduate levels. We are starting to receive solutions to these problems and we look forward to more solutions from our readers. The solutions will start to appear next volume. Last issue saw the rstDepartment Highlight. These will appear every second issue and give our readers some insight into the programs and activities at various mathematics departments across Canada. Last issue we also re-launched theProblem of the Month. A column of the same name appeared in theMathematical Mayhemsection for years and presented the discussion of a problem from a high school mathematics contest. With the re-launch, we are keeping the same format, yet we will be including any problem that would be of interest to readers ofCrux Mathematicorum. This column will be dedicated to the former Editor-in-Chief ofCrux Mathematicorumand my personal friend and mentor, the late Jim Totten. Jim had a great love of mathematics and shared that love with his students and others through his problem of the week, his work on mathematics contests and outreach programs. We will feature problems in this section that we hope Jim would have enjoyed. This issue we introduce a new column, theProblem Solver's Toolkit. This will be similar to Michel Bataille'sFocus On ..., which also started to appear this volume, although it will be a bit more elementary in nature.Problem Solver's Toolkitwill feature techniques that will be of use to mathematical problem solvers at all levels. We hope that you enjoy our new problem column and other features. We always welcome your feedback.

Shawn GodinCopyright

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216/ MAYHEM SOLUTIONS

MAYHEM SOLUTIONS

Mathematical Mayhemis being reformatted as a stand-alone mathematics journal for high school students. Solutions to problems that appeared in the last volume ofCruxwill appear in this volume, after which timeMathematical Mayhemwill be discontinued inCrux. NewMayhemproblems will appear when the journal is relaunched in 2013.M495.Proposed by the Mayhem Sta. All possible lines are drawn through the point (0;0) and the points (x;y), wherexandyare whole numbers with 1x;y10. How many distinct lines are drawn?

Solution by Florencio Cano Vargas, Inca, Spain.

Since the lines pass by (0;0), each line is characterized by a single parameter: the slopemand counting the lines amounts to counting all possible values of the slope. Sincem=yx , the dierent values of the slope are the irreducible fractions yx with 1x;y10. The boundaries are110 m101 Let us callNthe number of lines,N[m <1] the number of lines withm <1 andN[m >1] the number of lines withm >1. By symmetry around the line x=y(i.e.m= 1), we haveN[m <1] =N[m >1] and the requested number can be written as:

N= 2N[m <1] + 1

where the last term accounts for the casem= 1 which is considered separately. To evaluateN[m <1] we still have to count the number of irreducible fractions yx with 1y < x10. We study the dierent cases forx: x= 1:This case gives no values ofm <1. x= 2:We look for irreducible fractionsy2 , i.e., values ofyrelatively prime to 2, which is justy= 1. x= 3:We look for irreducible fractionsy3 , i.e., values ofyrelatively prime to 3, which are two valuesy= 1;2. From these cases it can be inferred that we look for the number of values of yrelatively prime toxand which are smaller thanx. This is just Euler's totient function'(x) (Euler's totient function'(n) is dened as the number of positive integers less than or equal tonthat are relatively prime ton. We are looking for the number of positive integers strictly less thannwhich are relatively prime to n, but this subtlety makes no dierence since a number is not relatively prime to itself.) Then we can write:

N[m <1] =10?

x=2'(x) = 1 + 2 + 2 + 4 + 2 + 6 + 4 + 6 + 4 = 31 and the nal answer isN= 231 + 1 = 63 dierent lines.

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MAYHEM SOLUTIONS / 217

Also solved by KONSTANTINOS DAGIADAS, Agrinio, Greece; GESINE GEUPEL, student, Max Ernst Gymnasium, Bruhl, NRW, Germany; RICARD PEIRO, IES \Abastos", Valencia, Spain; CASSIO DOS SANTOS SOUSA, Instituto Tecnologico de Aeronautica, S~ao Paulo, Brazil; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA. M496.Proposed by Sally Li, student, Marc Garneau Collegiate Institute,

Toronto, ON.

Show that if we write the numbers from 1 tonaround a circle, in any order, then, for allx= 1;2;:::;n, we are guaranteed to nd a block ofxconsecutive numbers that add up to at least?x(n+ 1)2 . Heredyeis the ceiling function, that is, the least integer greater than or equal toy. Sod6:2e= 7,de= 4,d8:3e=8 andd10e= 10.

Solution by Florencio Cano Vargas, Inca, Spain.

Let us rst note that the sum of all the numbers of the circle is given by n(n+1)2 Letx < nbe the number of consecutive numbers we take at a time. For a givenxwe havendierent combinations of consecutive numbers and each number of the circle enters inxcombinations. If we deneSkas the sum of the numbers in thekthcombination (k= 1;:::;n), then the sum of the numbers contained in all the combinationsfS1;S2;:::;Sngisxtimes the sum of the numbers of the circle: S

1+S2++Sn=xn(n+ 1)2

(1) hence S

1+S2++Snn

=x(n+ 1)2 Then the problem is equivalent to prove that in an arithmetic mean of pos- itive integers, there is at least one which is larger than or equal to the mean. Let us suppose that it is not so, and that allSksatisfy: S k1+S2++Sn< xn(n+ 1)2 which contradicts (1). Therefore, there must be at least someSkx(n+1)2

SinceSkis integer for anyk, butx(n+1)2

isn't necessarily an integer, then we can strengthen the claim, that is, there will be at least onekfor which S k?x(n+ 1)2 It remains to prove the casex=n. In that case, the only block is the whole set of numbers, whose sum is n(n+1)2 =?x(n+1)2 ?, which completes the proof.

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218/ MAYHEM SOLUTIONS

Also solved by the proposer.

M497.Proposed by Pedro Henrique O. Pantoja, student, UFRN, Brazil. Find all integersa,b,cwherecis a prime number such thatab+candabc are both perfect squares. Solution by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA. Assumecis an odd prime andab+c=m2andabc=n2for some integers mandn. Subtracting the two equations yields 2c=m2n2= (m+n)(mn). Unique factorization then impliesmn= 2 andm+n=c, which are contradictory equations sincemnalways have the same parity. Hence, ifcis an odd prime, there are no integersa;bsuch thatab+candabcare both perfect squares.

However, ifc= 2, thenab+c=m2andabc=n2implym2n2=

(m+n)(mn) = 4. Therefore, eitherm+n= 4 andmn= 1 orm+n=mn= 2 by unique factorization. As noted above, the equationsm+n= 4 andmn= 1 are contradictory. Ifm+n=mn= 2, thenm= 2 andn= 0. Thereforeab= 2, whencea= 2 andb= 1. Thus, (a;b;c) = (2;1;2) is the unique solution.

Also solved by the proposer.

M498.Proposed by Bruce Shawyer, Memorial University of Newfoundland,

St. John's, NL.

Right triangleABChas its right angle atC. The two sidesCBandCAare of integer length. Determine the condition for the radius of the incircle of triangle

ABCto be a rational number.

Solution by Cassio dos Santos Sousa, Instituto Tecnologico de Aeronautica, S~ao

Paulo, Brazil.

TakeBC=a,AC=b, andAB=c. To calculate the radiusrof the incircle of triangleABC, we may use the gure below.A BCabc rr r As the incenter is the intersection of the angle bisectors, there will be three pairs of congruent triangles formed (one shaded from each pair), each can be proved usingAAScomparison (one of the angles comes from the bisection, the other angle is 90 , and the common side has lengthr). Then, from the triangles

Crux Mathematicorum, Vol. 38(6), June 2012

MAYHEM SOLUTIONS / 219

formed with vertexA: br=c(ar) r=a+bc2 We were given thataandbare integers, hence, if we want a rational value forr, thencmust be rational. The Pythagorean theorem givesc2=a2+b2, an integer, socis either integer or irrational. If we wantrto be rational, thencmust be an integer.

To ndaandbsuch thatpa

2+b2will be an integer value, we may use

Euclid's Formula:

a=k(m2n2) b=k(2mn) c=k(m2+n2) Ifk;mandnare integers, andm > n, then (a;b;c) will be a Pythagorean triple, and thenrwill be a rational value. Also solved by KONSTANTINOS DAGIADAS, Agrinio, Greece; GESINE GEUPEL, student, Max Ernst Gymnasium, Bruhl, NRW, Germany; RICARD PEIRO, IES \Abastos", Valencia, Spain; CASSIO DOS SANTOS SOUSA, Instituto Tecnologico de Aeronautica, S~ao Paulo, Brazil; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposer. M499.Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buzau, Romania.Two circles of radius 1 are drawn so that each circle passes through the centre of the other circle.

Find the area of the goblet like region contained

between the common radius, the circumferences and one of the common tangents as shown in the diagram to the right. Solution by Gesine Geupel, student, Max Ernst Gymnasium, Bruhl, NRW,

Germany.FABC

DLetA,B,C, andDbe the points as shown

in the gure. LetFbe the darkly shaded area, that is, half of the upper part of the goblet. Tri- anglesABCandABDare equilateral, since each side is a radius of one of the circles. By sym- metry, the second half of the upper part of the goblet, the part besideF, is equal to the shaded area aboveF.

Hence, the area of the upper part of the

goblet, 2F, is equal to the area of a 120sector, minus the area of triangleACD.

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Note that trianglesAEC,BEC,AEDandBEDare congruent by hypotenuse- side, hence the [ACD] = [AEC] + [AED] = [AEC] + [BEC] = [ABC] (where [XY Z] represents the area of triangleXY Z). Thus 2F=3 p3 4 :GEB HIDLetE,HandIbe the points as shown in the diagram. LetGbe the darkest shaded area, that is, half of the lower part of the goblet. Looking at the rectangleBEHIwith sides 1 and 12 , it is broken into three parts: half of the base of the goblet (G), triangleBEDand sectorBDI. The triangle is half of an equilateral triangle, and the sector has angle 30 , hence G=12 p3 8 12

Thus the area of the goblet is

2F+ 2G=3

p3 4 + 2?12 p3 8 12 = 1 +6 p3 2 := 0:6576: Also solved by FLORENCIO CANO VARGAS, Inca, Spain; ANDHIKA GILANG, student, SMPN 8, Yogyakarta, Indonesia; THARIQ SURYA GUMELAR, student, SMPN 8, Yogyakarta, Indonesia; RICARD PEIRO, IES \Abastos", Valencia, Spain; ALMER FANDRIYANTO, student, SMAN 25, Bandung, Indonesia; KONSTANTINE ZELATOR, Uni- versity of Pittsburgh, Pittsburgh, PA, USA; and the proposer. One correct solution with no name on it and one incorrect solution were also received. M500.Proposed by Edward T.H. Wang and Dexter S.Y. Wei, Wilfrid Laurier

University, Waterloo, ON.

LetNdenote the set of natural numbers.

(a) Sho wt hatif n2N, there do not exista;b2Nsuch that[a;b]a+b=n, where [a;b] denotes the least common multiple ofaandb. (b) Sho wthat for an yn2N, there exist innitely many triples (a;b;c) of natural numbers such that [a;b;c]a+b+c=n, where [a;b;c] denotes the least common multiple ofa,bandc.

Solution by Florencio Cano Vargas, Inca, Spain.

(a) Let us supp osethat for a giv enn2N, there exista;b2Nthat satisfy the condition given in the problem. We can writea=da0;b=db0where d=gcd(a;b). Then gcd(a0;b0) = 1 and [a;b] =da0b0so the condition of the problem can be rewritten as: a 0b0a

0+b0=n:

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MAYHEM SOLUTIONS / 221

First of all note that sincea0andb0are relatively prime and we cannot have a

0+b0= 1, thena0+b06= 1 must dividea0b0.

Letp >1 be any prime common factor ofa0b0anda0+b0. Sincea0andb0 are relatively prime they don't share any prime factor, and thereforepis a factor either ofa0orb0. Let us assume without loss of generality that it is a factor ofa0, i.e.a0=psfor some integers. Then for some integerq a

0+b0=pq)ps+b0=pq)b0=p(qs)

and thenpis also a factor ofb0which contradicts the fact that gcd(a0;b0) = 1. This means thata0b0and (a0+b0) are relatively prime and then the fraction a 0b0a

0+b0is irreducible andn =2N, which contradicts the initial assumption.

(b) W ecan lo okfor triples ( a;b;c) such thata=a0d;b=b0d;c=c0dwith gcd(a;b;c) =dand witha0;b0;c0pairwise relatively prime. Then we can write lcm(a;b;c) =da0b0c0and the condition of the problem can be rewritten as: a

0b0c0a

0+b0+c0=n:

To enforce this property let us chooseb0=a0+ 1, which is always relatively prime witha0andc0= 1. We end up with a condition fora0: a

0(a0+ 1)a

0+ (a0+ 1) + 1=n,a0(a0+ 1)2(a0+ 1)=n,a0= 2n

which givesb0= 2n+ 1 andc0= 1. Hence a solution for a givenn2Nis the inntie set of triples: (a;b;c) = (2nd;2nd+d;d); d2N: Also solved by DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; CASSIO DOS SANTOS SOUSA, Instituto Tecnologico de Aeronautica, S~ao Paulo, Brazil; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposers.Copyright c

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222/ THE CONTEST CORNER

THE CONTEST CORNER

No. 6

Shawn Godin

The Contest Cornerest une nouvelle rubrique oerte pasCrux Mathematicorum, comblant ainsi le vide suite a la mutation en 2013 deMathematical MayhemetSkoliad vers une nouvelle revue en ligne. Il s'agira d'un amalgame deSkoliad,The Olympiad Corneret l'ancienAcademy Cornerd'il y a plusieurs annees. Les problemes en vedette seront tires de concours destines aux ecoles secondaires et au premier cycle universi- taire; les lecteurs seront invites a soumettre leurs solutions; ces solutions commenceront a para^tre au prochain numero. Les solutions peuvent ^etre envoyees a :Shawn Godin, Cairine Wilson S.S., 975 Orleans Blvd., Orleans, ON, CANADA, K1C 2Z5ou par couriel a crux-contest@cms.math.ca. Toutes solutions aux problemes dans ce numero doivent nous parvenir au plus tard le1 decembre 2013. Chaque probleme sera publie dans les deux langues ocielles du Canada (anglais et francais). Dans les numeros 1, 3, 5, 7 et 9, l'anglais precedera le francais, et dans les numeros 2, 4, 6, 8 et 10, le francais precedera l'anglais. Dans la section des solutions, le probleme sera publie dans la langue de la principale solution presentee. La redaction souhaite remercier Rolland Gaudet, de Universite de Saint-Boniface,

Winnipeg, MB, d'avoir traduit les problemes.CC26. Une fonctionfverief(1) = 2, et aussi, pour tout entier positifn >1,

f(1) +f(2) +f(3) ++f(n) =n2f(n):

Determiner la valeur def(2013).

CC27. Sur chacune des faces d'un cubennn, on dessine un grillage de n

2petits carres. On trace alors un chemin de (0;0;0) a (n;n;n) en utilisant, sans

reculs, des c^otes des petits carres. Determiner le nombre de tels chemins. CC28. Le polyn^ome quartiqueP(x) verieP(1) = 0. Aussi, il atteint sa valeur maximale de 3 au deux valeursx= 2 etx= 3. CalculerP(5). CC29. Considerer trois parallelogrammesP1,P2etP3. Le parallelogrammeP3 se situe a l'interieur du parallelogrammeP2, et ses sommets se trouvent sur les c^otes deP2. De facon similaire, le parallelogrammeP2se situe a l'interieur du parallelogrammeP1, et ses sommets se trouvent sur les c^otes deP1. Enn, les c^otes deP3sont paralleles aux c^otes deP1. Demontrer qu'un des c^otes deP3a une longueur au moins la moitie de la longueur du c^ote parallele deP1.

Crux Mathematicorum, Vol. 38(6), June 2012

THE CONTEST CORNER / 223

CC30. Deux enfants egostes s'amusent au jeu suivant. Ils commencent avec un bol contenantNbonbons, le nombreNetant connu des deux enfants. Tour a tour, chaque enfant prend (si possible) un bonbon ou plus, mais sujet a la restriction de ne jamais prendre plus que la moitie des bonbons restants. Le gagnant n'est pas l'enfant ayant le plus grand nombre de bonbons a la n, mais le dernier a pouvoir en prendre.A titre d'exemple, s'il y a 3 bonbons au depart, le premier joueur doit en prendre un seul, car deux depasse la moitie des bonbons disponibles a ce moment; ensuite, le deuxieme joueur est force d'en prendre un seul, laissant un bonbon dans le bol; le premier joueur ne peut plus en prendre et perd le jeu. (a)Demontrer que si le jeu demarre avec 2000 bonbons, alors le premier joueur gagne. (b)Demontrer que si le jeu demarre avec 999999 (2000 neufs) bonbons, alors le premier joueur gagne. CC26. A functionfis dened in such a way thatf(1) = 2, and for each positive integern >1, f(1) +f(2) +f(3) ++f(n) =n2f(n):

Determine the value off(2013).

CC27. Annncube has its faces ruled inton2unit squares. A path is to be traced on the surface of the cube starting at (0;0;0) and ending at (n;n;n) moving only in a positive sense along the ruled lines. Determine the number of distinct paths. CC28. The quartic polynomialP(x) satisesP(1) = 0 and attains its maxi- mum value of 3 at bothx= 2 andx= 3. ComputeP(5). CC29. Consider three parallelogramsP1,P2,P3. ParallelogramP3is inside parallelogramP2, and the vertices ofP3are on the edges ofP2. ParallelogramP2 is inside parallelogramP1, and the vertices ofP2are on the edges ofP1. The sides ofP3are parallel to the sides ofP1. Prove that one side ofP3has length at least half the length of the parallel side ofP1. CC30. Two polite but vindictive children play a game as follows. They start with a bowl containingNcandies, the number known to both contestants. In turn, each child takes (if possible) one or more candies, subject to the rule that no child may take, on any one turn, more than half of what is left. The winner is not the child who gets most candy, but the last child who gets to take some. Thus, if there are 3 candies, the rst player may only take one, as two would be more than half. The second player may take one of the remaining candies; and the rst player cannot move and loses. (a)Show that if the game begins with 2000 candies the rst player wins. (b)Show that if the game begins with 999999 (2000 9's) candies, the rst player wins.

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224/ THE OLYMPIAD CORNER

THE OLYMPIAD CORNER

No. 304

Nicolae Strungaru

Toutes solutions aux problemes dans ce numero doivent nous parvenir au plus tard le1 decembre 2013. Chaque probleme sera publie dans les deux langues ocielles du Canada (anglais et francais). Dans les numeros 1, 3, 5, 7 et 9, l'anglais precedera le francais, et dans les numeros 2, 4, 6, 8 et 10, le francais precedera l'anglais. Dans la section des solutions, le probleme sera publie dans la langue de la principale solution presentee. La redaction souhaite remercier Jean-Marc Terrier, de l'Universite de Montreal,

d'avoir traduit les problemes.OC86. Lors d'une reunion contenant un nombre ni de participants, certains

se trouvaient ^etre des amis. Parmi tout groupe de 4 personnes, ou bien il y en avait

3 tous amis entre eux ou alors 3 tous inconnus l'un de l'autre. Montrer qu'on peut

alors separer tous les participants en deux groupes de sorte que le premier groupe ne contienne que des amis entre eux et que le second groupe ne contienne que des inconnus l'un de l'autre. (L'amitie est une relation mutuelle). OC87. Appelons un nombre naturelndele, s'il existe des nombres naturels a < b < ctels queajb;bjcetn=a+b+c. (i) Mon trerque tous les nom bresnaturels son td elessauf un nom breni d'en tre eux.quotesdbs_dbs24.pdfusesText_30

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