65SKOLIAD No. 115Lily Yen and Mogens HansenPlease send solutions to problems in this Skoliad by September 1, 2009.Solutions should be sent to Lily Yen andMogensHansen at the address insidethe bak over. A opy of MATHEMATICAL MAYHEM will be presentedto the pre-university reader who sends in the best solution(s) before thedeadline. The deision of the editors is �nal.The Skoliad is in transition and, unfortunately, some submitted solu-tions have been lost. Our apologies for this inonveniene. Please resubmitany solutions to ontests appearing in Skoliad in or after the Marh 2008issue of CRUX.

Our featured ontest for this month is the British Columbia SeondaryShool Mathematis Contest 2007, Final Round, Part B. Our thanks go toClint Lee, Okanagan College, Vernon, BC, for permission to use it.La r �edation souhaite remerier Rolland Gaudet, de Coll �ege universi-taire de Saint-Bonifae, Winnipeg, MB, d'avoir traduit e onours.Conours math �ematique des �eoles seondairesde la Colombie-Britannique 2007Ronde �nale, partie B1. Jeanne a une olletion de pi �ees de monnaie de 5|, de 10| et de 25|, dontla valeur totale est de 2, 00$. Si les pi �ees de 5| �etaient de 10| et les pi �ees de10| �etaient de 5|, alors la valeur de la olletion serait de 1, 70$. D �eterminertoutes les possibilit �es de nombres de pi �ees de 5|, de 10| et de 25| dans laolletion de Jeanne.2. Un ube 3 × 3 × 3 est form�e en empilant des ubes 1× 1× 1. D �eterminerle nombre total de ubes, dont les ôt �es sont de taille enti �ere, ontenus dansle ube 3 × 3 × 3.3. Les longueurs des ôt �es d'un triangle sont13, 13 et 10. Le erle ironsrit de e tri-angle est le erle passant par les trois som-mets du triangle et ayant ii son entre �al'int �erieur du triangle (voir le diagramme �adroite). D �eterminer le rayon du erle irons-rit.

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4. Le jeu de Latin se joue sur un tableau form�e d'une grille quatre par quatre,ave une rang �ee additionnelle en haut omme en bas, puis une olonne addi-tionnelle �a gauhe omme �a droite. Les lettres A, B et C sont pla �ees dans les

66

ases de la grille quatre par quatre, de fa�on �a e que haque lettre se retrouveexatement une fois dans haque rang �ee et exatement une fois dans haqueolonne. En ons �equene haque rang �ee, de même que haque olonne, auraexatement une ase vide. Des lettres sont pla �ees, omme indies de solu-tion, �a ertains endroits dans les rang �ees et olonnes additionnelles ; ellesindiquent la lettre la plus pr �es se retrouvant dans la rang �ee ou olonne de lagrille ontenant l'indie. Le diagramme suivant donne une position de d �epartdu jeu de Latin et la solution qui en r �esulte.

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BAAC

Debut

rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

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..B CB

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BCA

BACACB

CABSolution

Le diagramme d'un autre jeu de Latin estfourni �a droite.Compl �eter e tableau, en fournissant unesolution ompl �ete. Donner une justi�-ation des �etapes permettant d'obtenirvotre solution...............................................................................................................................................................................

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AC C

B5. D �eterminer toutes les solutions x et y, enti �eres et positives, �a l' �equation

1

x− 1

y=

1

12.

British Columbia Seondary Shool MathematisContest 2007Final Round, Part B1. Joan has a olletion of nikels, dimes, and quarters worth $2.00. If thenikels were dimes and the dimes were nikels, the value of the oins wouldbe $1.70. Determine all of the possibilities for the number of nikels, dimes,and quarters that Joan ould have.

672. A 3 × 3 × 3 ube is formed by staking 1 × 1 × 1 ubes. Determine thetotal number of ubes with sides of integral length that are ontained in the3 × 3 × 3 ube.3. The lengths of the sides of a triangle are13, 13, and 10. The irumsribed irle of atriangle is the irle that goes through eah ofthe three verties of the triangle and here hasits entre inside the triangle (see the diagramat right). Find the radius of the irumsribedirle.

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4. The game of End View onsists of a tableau with a four by four grid, oneadditional row at the top and at the bottom, and one additional olumn onthe right and on the left. The letters A, B, and C are plaed in the four byfour grid in suh a way that every letter appears exatly one in eah row andeah olumn. This means that there will be exatly one empty square in eahrow and eah olumn. Letters are plaed in the additional rows and olumnsas hints, at the end of some rows and olumns of the four by four grid, toindiate the nearest letter that an be found by reading that row or olumnof the grid. The diagram below shows the starting tableau and the resultingsolution tableau for a game of End View.

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.B CB

BAAC

Startrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

rrrrrrrrrrrr

rrrrrrrrrrrrrrr

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.B CB

BAAC

BCA

BACACB

CABSolution

The diagram for another game of EndView is shown at right.Fill in this tableau with the omplete so-lution. Give a justi�ation of the stepsthat you used to �nd the solution..............................................................................................................................................................................

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AC C

B

685. Determine all of the positive integer solutions, x and y, to the equation

1

x− 1

y=

1

12.

Now we will give solutions to the British Columbia Seondary ShoolMathematis Contest 2006, Junior Final Round, Part B [2008 : 129-130℄. Thistime all the solutions are based on in-lass work by students of the editors.Unfortunately, the solutions below were already typeset when we reeived abath of (reovered) solutions from our readers. Thus, we ould not featureany reader's solution this time.1. Equilateral triangles I, II, III, and IV are suh that the altitude of triangleI is the side of triangle II, the altitude of triangle II is the side of triangleIII, and the altitude of triangle III is the side of triangle IV. If the area oftriangle I is 2, �nd the area of triangle IV.Solution.If the side length of triangle I is x1, then bythe Pythagorean Theorem the height, h1, of trian-gle I is √32

x1. But then the side length, x2, of tri-angle II is √32

x1 and the height, h2, of triangle IIis √32

h1. Therefore the area, A2, of triangle II isx2h2

2=

(√3

2

)2x1h1

2=

3

4A1, where A1 is the area

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

h x

x/2of triangle I. Similarly, A3 = 34A2 and A4 = 3

4A3, where A3 and A4 are theareas of triangle III and triangle IV, respetively. Sine A1 = 2, we �nallyobtain A4 = (3

4

)3

A1 =27

32.Also solved by NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia;KARTHIK NATARAJAN; and LUYUN ZHONG-QIAO, Columbia International College, Hamil-ton, ON.2. A square has an area of 3 square units, and a ube has a volume of 5 ubiunits. Whih is larger, the edge length of the square or the edge length ofthe ube? Justify your answer using the exat values of the two quantities.Solution.Let s be the side length of the square, and let c be the side length ofthe ube. Then s2 = 3 and c3 = 5, so s6 = 33 = 27 and c6 = 52 = 25,whene s > c.Also solved by NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia;KARTHIK NATARAJAN; and LUYUN ZHONG-QIAO, Columbia International College, Hamil-ton, ON.

693. A ertain positive integer has \6" as its last (rightmost) digit. This num-ber is transformed into a new number by moving the \6" to the beginningof the number (leftmost position). For example, the number 1236 wouldbe transformed to 6123, while 51476 beomes 65147. What is the smallestsuh positive integer for whih this transformation inreases the value of thenumber by a fator of 4?Solution.Say the integer is . . . A6, that is the rightmost two digits are A and 6.Then 4(. . . A6) = 6 . . . A. But sine 4 · 6 = 24, the unit digit of 4(. . . A6)must be 4. Thus A = 4 and the integer must be . . . B46. Sine 4 · 46 = 184and 4(. . . B46) = 6 . . . B4, you have that B = 8 and the integer must be. . . C846. Now, 4 · 846 = 3384 and 4(. . . C846) = 6 . . . C84, so C = 3and the integer is . . . D3846. Again, 4 · 3846 = 15384 and so we have4(. . . D3846) = 6 . . . D384, so D = 5 and the integer is . . . E53846. Mus-tering all our patiene, 4·53846 = 215384 and 4(. . . E53846) = 6 . . . E5384,so E = 1 and the integer is . . . F153846. Sine 4 · 153846 = 615384|that'sit, the rightmost six digits of the required integer must be 153846. Hene,the smallest suh integer is 153846.Also solved by KARTHIK NATARAJAN.4. The members of a ommittee sit at a irular table so that eah ommitteemember has two neighbours. Eah member of the ommittee has a ertainnumber of dollars in his or her wallet. The hairperson of the ommittee hasone more dollar than the vie hairperson, who sits on his right and has onemore dollar than the member on her right, who has one more dollar than theperson on his right, and so on, until the member on the hair's left is reahed.The hairperson now gives one dollar to the vie hair, who gives two dollarsto the member on her right, who gives three dollars to the member on hisright, and so on, until the member on the hair's left is reahed. There arethen two neighbours, one of whom has four times as muh as the other.(a) What is the smallest possible number of members of the ommittee? Inthis ase, how muh did the poorest member of the ommittee have at�rst?(b) If there are at least 12 members of the ommittee, what is the smallestpossible number of members of the ommittee? In this ase, how muhdid the poorest member of the ommittee have at �rst?Solution.Say the ommittee has n members and the poorest member has x dol-lars. Then the hair has x+n−1 dollars, the vie hair has x+n−2 dollars,and so on. Eah person, exept the poorest, reeives one fewer dollar thanhe or she gives away, so eah person, exept the poorest, loses one dollar.Consequently, the poorest person gains n − 1 dollars. The hair therefore

70now has x+n−2 dollars, the vie hair has x+n−3 dollars, and so on untilthe person who used to be the seond poorest now has x dollars, while theperson who used to be poorest �nds him- or herself with x + n − 1 dollars.Thus the dollar amounts have simply shifted by one person around the table.The di�erene in dollar amounts between neighbours is therefore always onedollar, exept between the poorest and the wealthiest person. A di�erene ofone dollar annot quadruple an integer amount of dollars, so x+n−1 = 4x.But then n − 1 = 3x, so n − 1 is a multiple of 3.For part (a), you an use n = 4 and x = 1, so the smallest possibleommittee has four members and the poorest member has one dollar.For part (b), you an use n = 13 and x = 4, so the smallest ommitteewith 12 or more members has 13 members.Also solved by KARTHIK NATARAJAN.5. An equilateral triangle, 20 entimetres on a side, isinsribed in a square, as shown in the diagram. Find thelength of the side of the square.

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20

20

20

Solution.The two smaller triangles in the �g-ure both have a right angle, a hypotenuse oflength 20, and a side shared with the square.Therefore the third sides of those triangles areequal, and you may label the piees as shown.Using the Pythagorean Theorem twie,2y2 = 202 and (x + y)2 + x2 = 202,so y = ±10√2 and 2x2 + 2xy + y2 = 400. ........................................................................................................................................................................................................

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20

20

20

x + y

x

y

y x

x + y

Sine y is a length, y = 10√2 and x2 +xy −100 = 0. The quadrati formulanow yieldsx =

−y ±√

y2 + 400

2=

−10√

2 ± 10√

6

2= −5

√2 ± 5

√6 .Sine x is also a length, x = 5√6 − 5√2, and therefore the side length ofthe square is x + y = 5√6 + 5√2.Also solved by NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia;KARTHIK NATARAJAN; and LUYUN ZHONG-QIAO, Columbia International College, Hamil-ton, ON.

We have also reovered a orret solution to problem #4 of the Mathe-matis Assoiation of Quebe Contest 2006 at [2008 : 66, 68℄ by Luyun Zhong-Qiao, Columbia International College, Hamilton, ON.

71MATHEMATICAL MAYHEMMathematial Mayhem began in 1988 as a Mathematial Journal for and byHigh Shool and University Students. It ontinues, with the same emphasis,as an integral part of Crux Mathematiorum with Mathematial Mayhem.The Mayhem Editor is Ian VanderBurgh (University of Waterloo). Theother sta� members are Monika Khbeis (Asension of Our Lord SeondaryShool, Mississauga) and Eri Robert (Leo Hayes High Shool, Frederiton).

Mayhem ProblemsVeuillez nous transmettre vos solutions aux probl �emes du pr �esent num�eroavant le 15 juin 2009. Les solutions re�ues apr �es ette date ne seront prises en ompteque s'il nous reste du temps avant la publiation des solutions.Chaque probl �eme sera publi �e dans les deux langues oÆielles du Canada(anglais et fran�ais). Dans les num�eros 1, 3, 5 et 7, l'anglais pr �e �edera le fran�ais,et dans les num�eros 2, 4, 6 et 8, le fran�ais pr �e �edera l'anglais.La r �edation souhaite remerier Jean-Mar Terrier, de l'Universit �e deMontr �eal, d'avoir traduit les probl �emes.M382. Propos �e par l' �Equipe de Mayhem.D �eterminer toutes les paires (x, y) d'entiers tels que 4x2 − y2 = 480.M383. Propos �e par l' �Equipe de Mayhem.Dans un retangle ABCD, P est sur BC et Q est sur DC de sorte queBP = 1, AP = PQ = 2 et l'angle APQ = 90◦. D �eterminer la longueurde QD.M384. Propos �e par Kunal Singh, �etudiant, Kendriya Vidyalaya Shool,Shillong, Inde.Dans la �gure i-ontre, le pointE est sur AB et le point D est sur ACde sorte que AE = EB = DC = 1et AD = 2. D �eterminer le rapport del'aire du quadrilat �ere BCDE �a elledu triangle ABC. ................................................................................

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

CA D

B

E

1

1

2 1M385. Propos �e par Mih�aly Benze, Brasov, Roumanie.En base 10, l'entier N = 1 · · · 114 · · · 44 ommene ave 2009 hi�res 1ons �eutifs suivis de 4018 hi�res 4 ons �eutifs. Montrer que N n'est pas unarr �e parfait.

72M386. Propos �e par Neulai Staniu, �Eole Tehnique Sup �erieure de SaintMueni Sava, Bera, Roumanie.D �eterminer tous les nombres r �eels x pour lesquels

√

2 + 4x − 2x2 +√

6 + 6x − 3x2 = x2 − 2x + 6 .M387. Propos �e par John Grant MLoughlin, Universit �e du Nouveau-Brunswik, Frederiton, NB.On peut mesurer la temp �erature en degr �es Fahrenheit (F ) ou en degr �esCelsius (C). Les deux �ehelles sont reli �ees par la formule F = 1.8C + 32.Lorsqu'on onvertit en Fahrenheit une temp �erature exprim �ee en Celsius parun nombre de deux hi�res, on onstate parfois que, une fois les Fahrenheitarrondis �a l'entier le plus prohe, les hi�res des dizaines et des unit �es ont�et �e permut �es. Trouver toutes les valeurs enti �eres de deux hi�res en C pourlesquelles ei arrive.

.................................................................M382. Proposed by the Mayhem Sta�.Determine all pairs (x, y) of integers for whih 4x2 − y2 = 480.M383. Proposed by the Mayhem Sta�.In retangle ABCD, P is on side BC and Q is on side DC so thatBP = 1, AP = PQ = 2 and ∠APQ = 90◦. Determine the length of QD.M384. Proposed by Kunal Singh, student, Kendriya Vidyalaya Shool,Shillong, India.In the diagram at right, the pointE is on AB and the point D is on ACsuh that AE = EB = DC = 1and AD = 2. Determine the ratioof the area of quadrilateral BCDE tothe area of triangle ABC. ................................................................................

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CA D

B

E

1

1

2 1M385. Proposed by Mih�aly Benze, Brasov, Romania.The base 10 integer N = 1 · · · 114 · · · 44 starts o� with 2009 onse-utive digits 1 followed by 4018 onseutive digits 4. Prove that N is not aperfet square.

73M386. Proposed by Neulai Staniu, Saint Mueni Sava TehnologialHigh Shool, Bera, Romania.Determine all real numbers x for whih

√

2 + 4x − 2x2 +√

6 + 6x − 3x2 = x2 − 2x + 6 .M387. Proposed by John Grant MLoughlin, University of New Bruns-wik, Frederiton, NB.Temperature an be measured in degrees Fahrenheit (F ) or in degreesCelsius (C). The two sales are related by the formula F = 1.8C +32. Whena two-digit integer degree temperature in Celsius is onverted to Fahrenheitand rounded to the nearest integer degree, it turns out the ones and tensdigits of the original Celsius temperature C sometimes swith plaes to givethe rounded Fahrenheit equivalent. Find all two-digit integer values of Cfor whih this o

urs.

Mayhem SolutionsM344. Proposed by the Mayhem Sta�.Consider the square array

1 2 34 5 67 8 9formed by listing the numbers 1 to 9 in order in onseutive rows. The sumof the integers on eah diagonal is 15. If a similar array is onstruted usingthe integers 1 to 10 000, what is the sum of the numbers on eah diagonal?Solution by Jos �e Hern �andez Santiago, student, Universidad Tenol �ogia dela Mixtea, Oaxaa, Mexio, modi�ed by the editor.Let S1 represent the sum of the integers on the diagonal that runs fromtop left to bottom right. Let S2 represent the sum of the integers on thediagonal that runs from top right to bottom left.Our array is 100 by 100. When we move one olumn to the right in thesame row, the number inreases by 1; when we move one olumn to the leftin the same row, the number dereases by 1. When we move one row downin the same olumn, the number inreases by 100.On the top left to bottom right diagonal, eah number is one olumn tothe right and one row down, so is 1 + 100 = 101 greater than the previous

74number on the diagonal. Thus,

S1 = 1 + 102 + 203 + · · · + 9899 + 10000= (100 · 0 + 1) + (100 · 1 + 2) + · · · + (100 · 98 + 99)

+ (100 · 99 + 100)= 100(0 + 1 + · · · + 98 + 99) + (1 + 2 + · · · + 99 + 100)= 100

(

1

2(99)(100)

)

+1

2(100)(101)

= 100(99)(50) + 50(101) = 495000 + 5050 = 500050 .On the top right to bottom left diagonal, eah number is one olumn tothe left and one row down, so is −1 + 100 = 99 greater than the previousnumber on the diagonal. Thus,S2 = 100 + 199 + 298 + · · · + 9802 + 9901

= (100(0) + 100) + (100(1) + 99) + · · · + (100(98) + 2)+ (100(99) + 1)

= 100(0 + 1 + · · · + 98 + 99) + (100 + 99 + · · · + 2 + 1) .Therefore, S2 = S1 = 50050.Also solved by EDIN AJANOVIC, student, First Bosniak High Shool, Sarajevo, Bosniaand Herzegovina; CAO MINH QUANG, Nguyen Binh Khiem High Shool, Vinh Long, Vietnam;JACLYN CHANG, student, Western Canada High Shool, Calgary, AB; PETER CHIEN, student,Central Elgin Collegiate, St. Thomas, ON; NATALIA DESY, student, SMA Xaverius 1,Palembang, Indonesia; RICARD PEIR �O, IES \Abastos", Valenia, Spain; KUNAL SINGH,student, Kendriya Vidyalaya Shool, Shillong, India; and MRIDUL SINGH, student, KendriyaVidyalaya Shool, Shillong, India.M345. Proposed by John GrantMLoughlin, University of New Brunswik,Frederiton, NB.The area of isoseles △ABC is q√15. Given that AB = 2BC, expressthe perimeter of △ABC in terms of q.Solution by Kunal Singh, student, Kendriya Vidyalaya Shool, Shillong, In-dia. Let BC = x and AB = 2x. Sine △ABC is isoseles, then AC = BCor AC = AB. Sine △ABC is presumably not a degenerate triangle, wetake AC = AB = 2x (otherwise AC = BC = x and AB = 2x, whihwould mean that the triangle was a straight line segment of length 2x).Let D be the midpoint of BC. Sine △ABC is isoseles, then AD isperpendiular to BC. The length of this altitude is thusAD =

√

AB2 −(

1

2BC

)2

=

√

(2x)2 −(

1

2x)2

=

√

4x2 − 14x2 =

√

15

4x2 =

√15

2x .

75The area of △ABC is q√15 and also equal to 1

2(BC)(AD) =

1

2x

(√15

2x

),so that q√15 = √154

x2, or x2 = 4q and hene x = 2√q.Thus, AB + BC + CA = 2x + x + 2x = 5x = 10√q is the perimeterof the triangle.Also solved by EDIN AJANOVIC, student, First Bosniak High Shool, Sarajevo, Bosniaand Herzegovina; CAO MINH QUANG, Nguyen Binh Khiem High Shool, Vinh Long, Vietnam;JACLYN CHANG, student, Western Canada High Shool, Calgary, AB; NATALIA DESY, student,SMA Xaverius 1, Palembang, Indonesia; JOS �E HERN�ANDEZ SANTIAGO, student, UniversidadTenol �ogia de laMixtea, Oaxaa, Mexio; RICARD PEIR �O, IES \Abastos", Valenia, Spain; andMRIDUL SINGH, student, Kendriya Vidyalaya Shool, Shillong, India. There were 2 inorretsolutions submitted.M346. Proposed by the Mayhem Sta�.Without using a alulator, �nd the number of digits in the integer 280.Solution by Edin Ajanovi, student, First Bosniak High Shool, Sarajevo,Bosnia and Herzegovina.We determine the positive integer n for whih 10n ≤ 280 < 10n+1.This will imply that the integer 280 has n + 1 digits.Sine 1000 = 103 < 210 = 1024, raising both sides to the exponent 8we obtain 1024 < 280.Also, 65536 = 216 < 105 = 100000, hene 280 < 1025 upon raisingboth sides to the exponent 5.Thus, 1024 < 280 < 1025, whene 280 has 25 digits.Also solved by RICARD PEIR �O, IES \Abastos", Valenia, Spain; and MRIDUL SINGH,student, Kendriya Vidyalaya Shool, Shillong, India. There were 5 inomplete solutions sub-mitted.Singh submitted a solution that inluded more of a thought proess that led to the sameoutome as that of Ajanovi.M347. Proposed by the Mayhem Sta�.Four positive integers a, b, c, and d are suh that(a + b + c)d = 420 ,(a + c + d)b = 403 ,(a + b + d)c = 363 ,(b + c + d)a = 228 .Find the four integers.Solution by Riard Peir �o, IES \Abastos", Valenia, Spain.First, we write out the prime fatorizations of the right sides of theequations:

420 = 22 · 3 · 5 · 7 , 403 = 13 · 31 ,363 = 3 · 112 , 228 = 22 · 3 · 19 .

76Next, we subtrat the seond equation from the �rst equation to obtain(a + c)d + bd − (a + c)b − bd = 420 − 403, or (a + c)(d − b) = 17.Sine a, b, c, and d are positive integers, then a + c > 1. Sine (a + c) is adivisor of 17, whih is a prime number, then a + c = 17 and so d − b = 1.From the third equation, c is odd sine it is a divisor of 363. Sinea + c = 17, then a = 17 − c so a is even. Also, 1 ≤ c ≤ 15 and 2 ≤ a ≤ 16.Sine a is an even divisor of 228 = 22 ·3·19 between 2 and 16 inlusive,then a ould be 2, 4, 6, or 12. Sine c is an odd divisor of 363 = 3 · 112between 1 and 15 inlusive, then c ould be 1, 3, or 11. Sine a + c = 17,then a must equal 6 and c must equal 11.So we know that a = 6, c = 11, and d = b+1. Substituting these intothe seond equation, we obtain

(6 + 11 + b + 1)b = 403 ;(b + 18)b = 403 ;

b2 + 18b − 403 = 0 ;(b − 13)(b + 31) = 0 .Sine b is a positive integer, we have b = 13, whene d = 14.Therefore, (a, b, c, d) = (6, 13, 11, 14), whih we an hek satis�eseah of the four equations.Also solved by CAO MINH QUANG, Nguyen Binh Khiem High Shool, Vinh Long,Vietnam; and KUNAL SINGH, student, Kendriya Vidyalaya Shool, Shillong, India. There were2 inomplete solutions and 1 inorret solution submitted.M348. Proposed by the Mayhem Sta�.The perimeter of a setor of a irle is 12 (the perimeter inludes thetwo radii and the ar). Determine the radius of the irle that maximizes thearea of the setor.Solution by Natalia Desy, student, SMA Xaverius 1, Palembang, Indonesia,modi�ed by the editor.Let r be the radius of the irle, let θ be the entral angle of the setormeasured in radians, and let A be the area of the setor.The length of the ar of the setor is θ

2π(2πr) = θr. S