[PDF] Crux Mathematicorum OC151. Soit ABC un triangle

View - Download Crux Mathematicorum

Previous PDF

Next PDF

Crux Mathematicorum

VOLUME 39, NO. 9 November / Novembre 2013

Editorial Board

Editor-in-ChiefKseniya GaraschukUniversity of British Columbia Olympiad EditorNicolae StrungaruGrant MacEwan University Book Reviews EditorJohn McLoughlinUniversity of New Brunswick 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

Assistant EditorsChip CurtisMissouri Southern State University

Lino DemasiOttawa, ON

Allen O'HaraUniversity of Western Ontario

Peter O'HaraLondon, ON

Mohamed OmarHarvey Mudd College

Guest EditorsJoseph HoranUniversity of Victoria

Amanda MallochUniversity of Victoria

Editor-at-LargeBill SandsUniversity of Calgary

Managing EditorJohan RudnickCanadian Mathematical Society

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

Copyright

c

Canadian Mathematical Society, 2014

IN THIS ISSUE / DANS CE NUM

ERO

391 EditorialKseniya Garaschuk

392 The Contest Corner: No. 19Kseniya Garaschuk

392 Problems: CC87, CC91{CC95

394 Solutions: CC41{CC45

397 The Olympiad Corner: No. 317Nicolae Strungaru

397 Problems: OC151{OC155

399 Solutions: OC91{OC95

402 Book ReviewsJohn McLoughlin

404 Focus On ...: No. 9Michel Bataille

409 Problem Solver's Toolkit: No. 8Gerhard J. Woeginger

413 Problems: 3881{3890

418 Solutions: 3781{3790

431 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 Godin

391

EDITORIAL

DearCruxreaders,

Welcome to issue 9 of Volume 39.

First of all, I have some very exciting news!

The CMS oce and Michael Doob have been hard at work scanning oldCrux issues and now all the volumes are available online - all the way from the very rst issue of Eureka (journal's rst name) in March 1975. Now, you can witness rst-hand the evolution and the growthCruxhas gone through over the past 39 years; to guide you in your explorations ofCrux's early years, you can consult CruxChronology by J. Chris Fisher (Volume 37, issue 2). I strongly recommend ipping through some early issues as you can get easily inspired by the materials and Leo Sauve's personality present on each page of the journal. Just for fun, here

is the very rst problem (that is 3889 problems ago):While you ponder that, I will take care of some typos and omissions of the past 2

issues: There was a typo in the problem 3872 and the new corrected version is presented in this issue. Anastasios Kotronis should be included in the list of solvers of problem 3780. There is a small typo on page 338 of Volume 39: in the solution of problem

3770, in line 2 \ab+ 2a+b+c" should be \ab+a+b+c".

Please do not hesitate to send your corrections to the typos you see to me at crux-editors@cms.math.ca. Please note that the new address for submission of problem proposals and numbered problems iscrux-psol@cms.math.ca. Since our journal is very audience driven, please email me your comments and feedback on our materials. I always look forward to your emails.

Kseniya Garaschuk

Copyright

c

Canadian Mathematical Society, 2014

392/ THE CONTEST CORNER

THE CONTEST CORNER

No. 19

Kseniya Garaschuk

The problems featured in this section have appeared in, or have been inspired by, a math- ematics contest question at either the high school or the undergraduate level. Readers are invited to submit solutions, comments and generalizations to any problem. Please email your submissions tocrux-contest@cms.math.caor mail them to the address inside the back cover. Electronic submissions are preferable. Submissions of solutions.Each solution should be contained in a separate le named using the conventionLastNameFirstNameCCProblemNumber(example DoeJaneOC1234.tex). It is preferred that readers submit a L

ATEX le anda pdf le

for each solution, although other formats are also accepted. Submissions by regular mail are also accepted. Each solution should start on a separate page and name(s) of solver(s) with aliation, city and country should appear at the start of eachsolution. To facilitate their consideration, solutions should be received by the editor by1 March

2015, although late solutions will also be considered until a solution is published.

Each problem is given in English and French, the ocial languages of Canada. In issues

1, 3, 5, 7, and 9, English will precede French, and in issues 2, 4, 6, 8, and 10, French

will precede English. In the solutions' section, the problem will be stated in the language of the primary featured solution.

The editor thanks Andre Ladouceur, Ottawa, ON, for translations of the problems.CC87.Correction. In issue 8, we accidentally re-printed CC33 as CC87. This

is the corrected version of CC87. LetABCDEbe a regular pentagon with each side of length 1. The length ofBE isand the angleFEAis, whereFis the intersection ofACandBE. Find and cos. CC91. A line segment of constant length 1 moves with one end on thex-axis and the other end on they-axis. The region swept out (that is, the union of all possible placements) isR. Find the equation of the boundary ofR. CC92. Each of the positive integers 2013 and 3210 has the following three properties: 1. it is an in tegerb etween1000 and 10000, 2. its four d igitsare consecutiv ein tegers,and 3. it is divisib leb y3. In total, how many positive integers have these three properties?

Crux Mathematicorum, Vol. 39(9), November 2013

THE CONTEST CORNER / 393

CC93. Ifx;y;z >0 andxyz= 1, nd the range of all possible values of x

3+y3+z3x3y3z3x+y+zx1y1z1:

CC94. If log2x;(1 + log4x) and log84xare consecutive terms of a geometric sequence, determine the possible values ofx. CC95. Positive integersx;y;zsatisfyxy+z= 160. Determine the smallest possible value ofx+yz. CC87.Correction. Dans le numero 8 de la revue, on a presente le probleme CC33 a la place du probleme CC87. Voici le vrai probleme CC87. SoitABCDEun pentagone regulier ayant des c^otes de longueur 1. Soitla longueur deBE,Fle point d'intersection deACetBEetla mesure de l'angle

FEA. Determineret cos.

CC91. Un segment de droite de longueur 1 se deplace de maniere qu'une de ses extremites soit toujours sur l'axe des abscisses et l'autre sur l'axe des ordonnees. SoitRla region balayee par le segment (c'est-a-dire la reunion de tous les points sur les positions du segment a mesure qu'il se deplace). Determiner l'equation de la frontiere deR. CC92. Chacun des entiers 2013 et 3210 satisfait aux trois proprietes suivantes : 1. il est un en tieren tre1000 et 10000, 2. ses quatre c hiresson tdes en tierscons ecutifset 3. il est divisible par 3. Combien y a-t-il d'entiers positifs qui satisfont a ces trois proprietes? CC93. Sachant quex;y;z >0 etxyz= 1, determiner l'etendue de toutes les valeurs possibles de l'expression x

3+y3+z3x3y3z3x+y+zx1y1z1:

CC94. Sachant que log2x;(1 + log4x) et log84xsont des termes consecutifs d'une suite geometrique, determiner toutes les valeurs possibles dex. CC95. Les entiers strictement positifsx;y;zverient l'equationxy+z= 160. Determiner la plus petite valeur possible dex+yz.Copyright c

Canadian Mathematical Society, 2014

394/ THE CONTEST CORNER

CONTEST CORNER

SOLUTIONS

CC41. Ace runs with constant speed and Flash runsxtimes as fast,x >1. Flash gives Ace a head start ofymetres, and, at a given signal, they start o in the same direction. Find the distance Flash must run to catch Ace. Originally problem 7 of 2005 W.J. Blundon Mathematics Contest. Solved by R. I. Hess; and D. Vacaru. We present the solution by Daniel Vacaru. Letvbe Ace's speed, then Flash has speedvx. Lettbe the amount of time it takes for Flash to catch Ace. When Flash catches up to Ace, Ace isvt+ymetres from the start and Flash isxvtmetres from the start. Thus,vt+y=xvt. Solving fort, we gett=yv(x1). At that time, Flash has run xvyv(x1)=xyx1metres. CC42.4ABChas its vertices on a circle of radiusr. If the lengths of two of the medians of4ABCare equal tor, determine the side lengths of4ABC. Originally 2012 Canadian Senior Mathematics Contest, problem B3c.

Solved by M. Amengual Covas;

S. Arslanagic; M. Bataille; M. Coiculescu; R. Hess; and D. Vacaru. We present the solution by Miguel Amengual Covas. LetGbe the centroid of4ABCand suppose that the two equal medians are the medianADto sideBCand the median to sideCA. Clearly, then,4ABC is isosceles withBC=CA. Thus the medianCMto sideABlies along the perpendicular bisector of chordABand it passes through the circumcentreOof

4ABC. Therefore, we have

AO

2OM2=AG2GM2:(1)

LetGM=x. SinceGtrisects each median of4ABC, we haveOM=OA

MC=r3xandAG=23

AD=23 r. When these are substituted into (1), we getr2(r3x)2= (23 r)2x2. Solving forx, we obtainx=23 r(which is not admissible) andx=r12 . Hence,

AB= 2AM= 2Êr

2 =rp7 2 and

BC=CA=pAM

2+MC2=̂

rp7 4 2 +r4 2=rp2 2

Crux Mathematicorum, Vol. 39(9), November 2013

THE CONTEST CORNER / 395

CC43. A circle has diameterAB.Pis a xed point ofABlying betweenA andB. A pointX, distinct fromAandB, lies on the circumference of the circle. Prove that tan(\AXP)tan(\XAP) is constant for all values ofX. Originally Question 6 of 2005 APICS Math Competition.

Solved by M. Amengual Covas;

S. Arslanagic; M. Bataille; R. I. Hess; J. G. Heu- ver; and T. Zvonaru. We present the solution of Michel Bataille modied by the editor. For simplicity, let=\XAPand=\AXP. SinceABis a diameter,\AXB= 90
and hence\BXP= 90. Since triangleAXBis right-angled with right angle atX,\PBX= 90. We now apply Law of Sines on4AXPand4PXB.

On4AXPwe havePAsin=PXsin, so

sinsin=PAPX :(1)

On4PXB,PBsin(90

)=PXsin(90

Since sin(90

) = cosand sin(90) = cos, this implies coscos=PXPB :(2)

Equations (1) and (2) imply

PXPB =PAPB which is constant for all values ofX.

CC44. Leta0= 1 and forn0 letan+1=an12

a2n. Find limn!1nan, if it exists. Originally Question 6 on 2009 University of Waterloo Big E Contest. Solved by M. Bataille; and D. Vacaru. We present Michel Bataille's solution.

We show that lim

n!1nan= 2.

Sincean+1an=12

a2n<0 for alln0, the sequencefangis decreasing. It follows thatana0= 1 for alln0:Froman+1=an2 (2an);an easy induction shows thatan>0 for alln0. Being decreasing and bounded below, the sequence fangis convergent. Let`= limn!1an:Since`is also the limit offan+1g;we must have`=`12 `2and so`= 0:Becausean+1a n= 112 an;we have limn!1a n+1a n= 1:Now, we calculate

Copyright

c

Canadian Mathematical Society, 2014

396/ THE CONTEST CORNER

1a n+11a n=anan+1a nan+1=12 a2na nan+1=12 ana n+1 and so the sequence 1a n+11a nis convergent towards12 :The same is true of its

Cesaro meanfCngdened by

C n=1n n X 1a n1a n1‹ But C n=1n 1a n1‹ =1na n1n and so lim n!1nan= limn!11C n+1n = 2:

CC45. Thebaseball sumof two rational numbersab

andcd is dened to bea+cb+d.

Starting with the rational numbers

01 and11 as Stage 0, the baseball sum of each consecutive pair of rational numbers in a stage is inserted between the pair to arrive at the next stage. The rst few stages of this process are shown below :

STAGE 0 :

01 11

STAGE 1 :01

12 11

STAGE 2 :01

13 12 23
11

STAGE 3 :01

14 13 25
12 35
23
34
11

Prove that :

i) no rational n umberwill b einserted more than once, ii) no inserted fraction is reducible, and iii) ev eryrational n umberb etween0 and 1 will b einserted in the pattern at some stage. Originally 2006 Canadian Open Mathematics Challenge, problem B4 b). One incorrect solution was received.Crux Mathematicorum, Vol. 39(9), November 2013

THE OLYMPIAD CORNER / 397

THE OLYMPIAD CORNER

No. 317

Nicolae Strungaru

The problems featured in this section have appeared in a regional or national mathematical Olympiad. Readers are invited to submit solutions, comments and generalizations to any problem. Please email your submissions tocrux-olympiad@cms.math.caor mail them to the address inside the back cover. Electronic submissions are preferable. Submissions of solutions.Each solution should be contained in a separate le named using the conventionLastNameFirstNameOCProblemNumber(example DoeJaneOC1234.tex). It is preferred that readers submit a L

ATEX le anda pdf le

for each solution, although other formats are also accepted. Submissions by regular mail are also accepted. Each solution should start on a separate page and name(s) of solver(s) with aliation, city and country should appear at the start of eachsolution. To facilitate their consideration, solutions should be received by the editor by1 March

2015, although late solutions will also be considered until a solution is published.

Each problem is given in English and French, the ocial languages of Canada. In issues

1, 3, 5, 7, and 9, English will precede French, and in issues 2, 4, 6, 8, and 10, French

will precede English. In the solutions' section, the problem will be stated in the language of the primary featured solution. The editor thanks Rolland Gaudet, de l'Universite Saint-Boniface a Winnipeg, for trans- lations of the problems.OC151. LetABCbe a triangle. The tangent atAto the circumcircle inter- sects the lineBCatP. LetQandRbe the symmetrical ofPwith respect to the linesABandAC, respectively. Prove thatBC?QR. OC152. Find all non-constant polynomialsP(x) =xn+an1xn1++a1x+ a

0with integer coecients whose roots are exactly the numbersa0;a1;:::;an1

each with multiplicity 1. OC153. Find all non-decreasing functions from the set of real numbers to itself such that for all real numbersx;ywe have f(f(x2) +y+f(y)) =x2+ 2f(y):

OC154. Forn2Z+we denote

x n:=‚2n nŒ

Copyright

c

Canadian Mathematical Society, 2014

398/ THE OLYMPIAD CORNER

Prove there exist innitely many nite setsA;Bof positive integers, such that

A\B=;, andQ

i2AxiQ j2Bxj= 2012: OC155. There are 42 students taking part in the Team Selection Test. It is known that every student knows exactly 20 other students. Show that we can divide the students into 2 groups or 21 groups such that the number of students in each group is equal and every two students in the same group know each other. OC151. SoitABCun triangle et soitPle point d'intersection de a ligneBC et de la tangente du cercle circonscrit au pointA. SoitQetRsymetriques aP par rapport aux lignes AB et AC respectivement. Demontrer queBC?QR. OC152. Determiner tous les polyn^omes non constantsP(x) =xn+an1xn1+ +a1x+a0avec coecients entiers dont les racines sont exactement les nombres a

0;a1;:::;an1avec les m^emes multiplicites.

OC153. Determiner toutes les fonctions non decroissantes des nombres reels aux nombres reels telles que pour toutx;yreels on a f(f(x2) +y+f(y)) =x2+ 2f(y):

OC154. Pourn2Z+, denotons

x n:=‚2n nŒ Demontrer qu'il existe un nombre inni d'ensembles nis d'entiers positifsAetB, tels queA\B=;etQ i2AxiQ j2Bxj= 2012: OC155. Soit 42 etudiants, ou on sait que tout etudiant connait exactement 20 autres etudiants. Demontrer qu'il est possible de repartir l'ensemble des etudiants en 2 sous-ensembles ou en 21 sous-ensembles de facon a ce que chaque sous- ensemble ait le m^eme nombre d'etudiants et que tous les etudiants dans un sous- ensemble se connaissent.Crux Mathematicorum, Vol. 39(9), November 2013

THE OLYMPIAD CORNER / 399

OLYMPIAD SOLUTIONS

OC91. Prove that no integer consisting of one 2, one 1 and the rest of digits

0 can be written neither as the sum of two perfect squares nor the sum of two

perfect cubes. Originally question 8 from the 2011 Estonian National Olympiad. Solved by O. Geupel; D. E. Manes; and T. Zvonaru. We give the solution by Titu

Zvonaru.

Since the sum of the digits ofnis 3, it follows thatnis divisible by 3 but not divisible by 9. Suppose by contradiction that there existsa;bsuch thatn=a2+b2. As the quadratic residues modulo 3 are 0 and 1 anda2+b20 (mod 3) it follows that ab0 (mod 3):quotesdbs_dbs24.pdfusesText_30

[PDF] Cercle circonscrit au triangle rectangle 62 Carrés - Radio

[PDF] Cercle Commercial Suisse - franzoesisch-lernen

[PDF] Cercle Concorde - Vigo, cabinet d`avocats

[PDF] cercle condorcet-voltaire - France

[PDF] Cercle de Bourem Mali - Fusions Et Acquisitions

[PDF] Cercle de discussion - Vendredi 15 novembre de 12h à 14h au Best

[PDF] CERCLE DE DROIT FRANÇAIS CERCULUI DE DREPT FRANCEZ

[PDF] Cercle de garde Nyon et environs: Planning 2016

[PDF] Cercle de Judo et d`Arts Martiaux d`Ermont

[PDF] CERCLE DE LA VOILE DE BORDEAUX

[PDF] Cercle de la Voile de Bordeaux 12 COUPE REGION AQUITAINE

[PDF] Cercle de Lecture Bibliothèque Pour Tous Bondues - France

[PDF] Cercle de Lettres sur Bonne Volonté l`astrologie spirituelle /3 - Antiparasitaire

[PDF] Cercle de Lettres sur Bonne Volonté l`astrologie spirituelle /3 Bélier - Antiparasitaire

[PDF] Cercle de Lettres sur Bonne Volonté l`astrologie spirituelle /4 Lion - Antiparasitaire