449

MATHEMATICAL MAYHEM

Mathematial Mayhem began in 1988 as a Mathematial Journal for and by

High 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). The

other sta members are Monika Khbeis (Asension of Our Lord Seondary

Shool, Mississauga), Eri Robert (LeoHayesHigh Shool, Frederiton), Larry

Rie (University of Waterloo), and Ron Lanaster (University of Toronto).

Mayhem Problems

Veuillez nous transmettre vos solutions aux probl emes du pr esent numero

avant le 15 Mars 2009. Les solutions reues apr es ette date ne seront prises en

ompte que s'il nous reste du temps avant la publiation des solutions.

Chaque probl eme sera publi e dans les deux langues oielles du Canada

(anglais et franais). Dans les numeros 1, 3, 5 et 7, l'anglais pr e edera le franais,

et dans les numeros 2, 4, 6 et 8, le franais pr e edera l'anglais.

La r edation souhaite remerier Jean-Mar Terrier, de l'Universit e de

Montr eal, d'avoir traduit les probl emes.

M369. Propos e par l'

Equipe de Mayhem.

Soit A(0, 0), B(6, 0), C(6, 4) et D(0, 4) les sommets d'un retangle.Par le point P (4, 3), on trae d'une part une droite horizontale oupant BCen M et AD en N et d'autre part une droite vertiale oupant AB en Q etCD en R. Montrer que AP , DM et BR passent toutes par le me^me point.

M370. Propos e par l'

Equipe de Mayhem.

(a) Montrer que cos(A+B) + cos(AB) = 2 cosA cosB pour tous lesangles A et B.

(b) Montrer que cosC + cosD = 2 cos(C +D

2

)cos

(C D

2

)pour tous

les angles C et D.

() Trouver la valeur exate de cos 20 + cos 60 + cos 100 + cos 140,sans l'aide d'une alulatrie.

M371. Propos e par Panagiote Ligouras,

Eole Seondaire L eonard de

Vini, Noi, Italie.

Un segmentAB de longueur 3 ontient un pointC tel queAC = 2. On

onstruit d'un me^me o^t e deAB deux triangles equilat erauxACF et CBE.D eterminer l'aire du triangle AKE si K est le point milieu de FC.

450

M372. Propos e par l'

Equipe de Mayhem.

Soit x un nombre r eel satisfaisant x3 = x+1. Trouver des entiers a, bet c de sorte que x7 = ax2 + bx+ c.

M373. Propos e par Kunal Singh, etudiant, Kendriya Vidyalaya Shool,

Shillong, Inde.

Les o^t es d'un triangle sont mesur es par trois nombres entiers ons eu-

tifs et le plus grand angle est le double du plus petit. D eterminer la longueur

des o^t es du triangle.

M374. Propos e par Mihaly Benze, Brasov, Roumanie.

Soit p un nombre premier x e, ave p 3. Trouver le nombre de solu-tions de x3 + y3 = x2y + xy2 + p2009, o u x et y sont des entiers.

M375. Propos e par Neulai Staniu,

Eole Tehnique Sup erieure de Saint

Mueni Sava, Bera, Roumanie.

D eterminer toutes les solutions r eelles du syst eme d' equations

1

x2+

4

y2+

9

z2= 4 ; x2 + y2 + z2 = 9 ; xyz =

9

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M369. Proposed by the Mayhem Sta.

A retangle has verties A(0, 0), B(6, 0), C(6, 4), and D(0, 4). A hor-izontal line is drawn through P (4, 3), meeting BC at M and AD at N . Avertial line is drawn through P , meeting AB atQ and CD atR. Prove thatAP , DM , and BR all pass through the same point.

M370. Proposed by the Mayhem Sta.

(a) Prove that cos(A + B) + cos(A B) = 2 cosA cosB for all anglesA and B.

(b) Prove that cosC + cosD = 2 cos(C +D

2

)cos

(C D

2

)for all angles

C and D.

() Determine the exat value of cos 20 + cos 60 + cos 100 + cos 140,without using a alulator.

M371. Proposed by Panagiote Ligouras, Leonardo da Vini High Shool,

Noi, Italy.

Suppose that the line segment AB has length 3 and C is on AB withAC = 2. Equilateral triangles ACF and CBE are onstruted on the sameside of AB. If K is the midpoint of FC, determine the area ofAKE.

451

M372. Proposed by the Mayhem Sta.

A real number x satises x3 = x + 1. Determine integers a, b, and cso that x7 = ax2 + bx+ c.

M373. Proposed by Kunal Singh, student, Kendriya Vidyalaya Shool,

Shillong, India.

The side lengths of a triangle are three onseutive positive integers

and the largest angle in the triangle is twie the smallest one. Determine the

side lengths of the triangle.

M374. Proposed by Mihaly Benze, Brasov, Romania.

Suppose that p is a xed prime number with p 3. Determine thenumber of solutions to x3 + y3 = x2y + xy2 + p2009, where x and y areintegers.

M375. Proposed by Neulai Staniu, Saint Mueni Sava Tehnologial

High Shool, Bera, Romania.

Determine all real solutions to the system of equations

1

x2+

4

y2+

9

z2= 4 ; x2 + y2 + z2 = 9 ; xyz =

9

2.

Mayhem Solutions

M332. Proposed by Dionne Bailey, Elsie Campbell, and Charles R.

Diminnie, Angelo State University, San Angelo, TX, USA.

A losed right irular ylinder has an integer radius and an integer

height. The numerial value of the volume is four times the numerial value

of its total surfae area (inluding its top and bottom). Determine the small-

est possible volume for the ylinder.

Solution by Cao Minh Quang, Nguyen Binh Khiem High Shool, Vinh Long,

Vietnam.

Let r and h be the radius and the height of the losed right irular

ylinder. The volume of suh a ylinder is V = r2h and the surfae area isA = 2r2 + 2rh.

From the hypotheses, r2h = 4(2r2 + 2rh), or rh = 8r + 8h, orrh 8r 8h+64 = 64, or (r 8)(h 8) = 64. Note that r 8 > 8 andh 8 > 8. This gives us the following possibilities:

452

r 8 1 2 4 8 16 32 64h 8 64 32 16 84 4 2 1r 9 10 12 16 24 40 72h 72 40 24 16 12 10 9V 5832 4000 3456 4096 6912 16000 46656

Thus, the smallest possible volume for the ylinder is 3456.

Also solved by DENISE CORNWELL, student, Angelo State University, San Angelo,TX, USA; RICHARD I. HESS, Ranho Palos Verdes, CA, USA; RICARD PEIR

O, IES \Abastos",

Valenia, Spain; BILLY SUANDITO, Palembang, Indonesia; and TITU ZVONARU, Comanesti,

Romania. There was 1 inomplete solution submitted.

M333. Proposed by the Mayhem Sta.

Anne and Brenda play a game whih begins with a pile of n toothpiks.They alternate turns with Anne going rst. On eah player's turn, she must

remove 1, 3, or 6 toothpiks from the pile. The player who removes the lasttoothpik wins the game. For whih of the values of n from 36 to 40 inlusivedoes Brenda have a winning strategy?

Solution by Rihard I. Hess, Ranho Palos Verdes, CA, USA, modied by the

editor.

We an build a table of winning and losing positions for Anne. Her

winning positions start with 1, 3, or 6, sine she an immediately win byremoving all of the toothpiks.

Starting with 2 toothpiks, Anne must remove 1 toothpik, leavingBrenda with 1, and so Brenda wins. Starting with 4 toothpiks, Anne mustremove 1 or 3 toothpiks, leaving Brenda with 3 or 1 (respetively), and soBrenda wins by removing all of the toothpiks.

Starting with 5 toothpiks, Anne an remove 3 toothpiks, thus leavingBrenda with 2 toothpiks. Sine 2 is a losing position for whoever goes rst,then Brenda loses, so Anne wins.

So far, 1, 3, 5, and 6 are winning positions for Anne, while 2 and 4 arelosing positions for Anne.

Starting with a pile of size n, Anne must redue the pile to one of sizen 1, n 3, or n 6 and pass to Brenda. If the person who goes rst hasa winning strategy starting with a pile of eah of these sizes, then Anne will

lose. In other words, if Anne has a winning strategy starting with piles of

size n 1, n 3, and n 6, then Anne will lose starting with a pile of sizen, as Brenda an implement Anne's strategy for the smaller pile and win, nomatter what Anne does. If one or more of these pile sizes are suh that the

rst person does not have a winning strategy, then Anne should redue to

this size, thus preventing Brenda from being able to win, and so Anne herself

will win.

We an examine the ases from n = 7 to n = 40, obtaining the follow-ing lists:

453

Winning positions for Anne: 1, 3, 5, 6, 7, 8, 10, 12, 14, 15, 16, 17, 19, 21,23, 24, 25, 26, 28, 30, 32, 33, 34, 35, 37, 39.

Losing positions for Anne: 2, 4, 9, 11, 13, 18, 20, 22, 27, 29, 31, 36, 38, 40.

Therefore, Brenda wins for n = 36, 38, 40.

Also solved by JACLYN CHANG, student, Western Canada High Shool, Calgary, AB.

See the Problem of the Month olumn in [2007 : 15{17 for a similar problem with a

more detailed explanation.

M334. Proposed by the Mayhem Sta.

(a) Determine all integers x for whihx 33x 2 is an integer.

(b) Determine all integers y for whih3y3 + 3

3y2 + y 2 is an integer.

I. Solution by Edin Ajanovi, student, First Bosniak High Shool, Sarajevo,

Bosnia and Herzegovina.

Let A be an integer suh that A =x 33x 2 . Then 3A is an integer and

3A =3x 93x 2 =

3x 2 73x 2 = 1

7

3x 2 .

Thus,

7

3x 2 is an integer; that is, 3x 2 is a divisor of 7, so 3x 2 isone of 1, 7. Sine x is an integer, x = 1 or x = 3. This answers part (a).

Now let B be an integer suh that

B =3y3 + 3

3y2 + y 2 = y y2 2y 33y2 + y 2

= y (y 3)(y + 1)(y + 1)(3y 2) = y

y 33y 2 .

Sine y is an integer,y 33y 2 is an integer. From the solution to part (a),

y = 1 or y = 3, whih answers part (b).

II. Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON.

We show that the only integer solutions to part (a) are x = 1 and x = 3.

Let f(x) =x 33x 2 . Then f(0) =

3

2, f(1) = 2, f(2) = 1

4, and

f(3) = 0. Of these, only f(1) and f(3) are integers.If x > 3, then f(x) is not an integer, sine 3x 2 > x 3 > 0 for

x > 3 and so 0 s + 3 > 0 for s 1, f(s) is not an integer by a similarargument so, f(x) is not an integer.

454

Therefore, f(x) is an integer for integer values of x if and only if x = 1or x = 3.

Also solved by RICARD PEIR O, IES \Abastos", Valenia, Spain; CAO MINH QUANG,Nguyen Binh Khiem High Shool, Vinh Long, Vietnam; and TITU ZVONARU, Comanesti,

Romania. There was one inorret and one inomplete solution submitted.

M335. Proposed by the Mayhem sta.

In a sequene of four numbers, the seond number is twie the rst

number. Also, the sum of the rst and fourth numbers is 9, the sum of theseond and third is 7, and the sum of the squares of the four numbers is 78.Determine all suh sequenes.

Solution by Denise Cornwell, student, Angelo State University, San Angelo,

TX, USA.

Let a, b, c, and d represent the rst, seond, third and fourth number,respetively. We an now write the given information as b = 2a, a+ d = 9,b+ c = 7 and a2 + b2 + c2 + d2 = 78.

The rst three equations allow us to rewrite b, c, and d in terms of a,obtaining b = 2a, c = 7 b = 7 2a, and d = 9 a.

Therefore,

a2 + (2a)2 + (7 2a)2 + (9 a)2 = 78 ,a2 + 4a2 + 49 28a+ 4a2 + 81 18a+ a2 78 = 0 ,

5a2 23a+ 26 = 0 ,(5a 13)(a 2) = 0 ,

hene a =13

5or a = 2.

Therefore, the sequenes are a =13

5, b =

26

5, c =

9

5, d =

32

5and

a = 2, b = 4, c = 3, d = 7. Both sequenes satisfy the given requirements.

Also solved by EDIN AJANOVIC, student, First Bosniak High Shool, Sarajevo,Bosnia and Herzegovina; JACLYN CHANG, student, Western Canada High Shool, Calgary,

AB; RICHARD I. HESS, Ranho Palos Verdes, CA, USA; CAO MINH QUANG, Nguyen Binh

Khiem High Shool, Vinh Long, Vietnam; KUNAL SINGH, student, Kendriya Vidyalaya Shool,

Shillong, India; BILLY SUANDITO, Palembang, Indonesia; and TITU ZVONARU, Comanesti,

Romania. There was one inorret and one inomplete solution submitted.

M336. Proposed by the Mayhem Sta.

A lattie point is a point (x, y) in the oordinate plane with eah of xand y an integer. Suppose that n is a positive integer. Determine the numberof lattie points inside the region |x|+ |y| n.

Solution by Edin Ajanovi, student, First Bosniak High Shool, Sarajevo,

Bosnia and Herzegovina, modied by the editor.

We an rewrite the given inequality as the equations |x|+ |y| = 0 and|x|+ |y| = k for 1 k n, where x, y Z.

455

The equation |x| + |y| = 0 has one integer solution only, namely(x, y) = (0, 0).

Consider next |x| + |y| = k, for an integer k with 1 k n. We anremove the absolute values by splitting into four ases:

Case 1. The integers x and y satisfy x+ y = k, where x 0 and y 0.This has solutions (k, 0), (k1, 1), . . . , (1, k1), (0, k), for a total of k+1solutions.

Case 2. The integers x and y satisfy x y = k, where x 0 and y < 0.This has solutions (k 1,1), (k 2,2), . . . , (1,(k 1)), (0,k), fora total of k solutions.

Case 3. The integers x and y satisfy x+ y = k, where x < 0 and y 0.This ase is the same as Case 3, but with the roles of x and y swithed, sothere are a total of k solutions here as well.

Case 4. The integers x and y satisfy x y = k, where x < 0 and y < 0.This has solutions

( 1,(k 1)), ( 2,(k 2)), . . . , ( (k 2),2),( (k 1),1), for a total of k 1 solutions.Thus, for eah k with 1 k n, the equation |x| + |y| = k has

(k + 1) + k + k + (k 1) = 4k solutions.Therefore, the number of lattie points inside the region is

1 +

nk=1

4k = 1 + 4

nk=1

k = 1 + 4 (1 + 2 + + n)

= 1 + 4 n(n+ 1)2

= 2n2 + 2n + 1 .

Also solved by RICARD PEIR

O, IES \Abastos", Valenia, Spain. There were one inorret

and two inomplete solutions submitted.

M337. Proposed by the Mayhem Sta.

On sides AB and CD of retangle ABCD with AD < AB, points Fand E are hosen so that AFCE is a rhombus.

(a) If AB = 16 and BC = 12, determine EF .

(b) If AB = x and BC = y, determine EF in terms of x and y.

Solution by Kunal Singh, student, Kendriya Vidyalaya Shool, Shillong, India.

We present the solution to (b), whih is a general version of the spei

ase in (a).

Suppose that AF = FC = CE = EA = m. Let O be the point ofintersetion of diagonals AC and EF of rhombus AFCE. Note that ACand EF are perpendiular and biset eah other at O.

456

By the Pythagorean Theorem,

CF 2 FB2 = CB2 ,m2 (xm)2 = y2 ,

m2 x2 + 2mxm2 = y2 ,2mx = x2 + y2 ,

m =x2 + y2

2x.

Now, AF = m =x2 + y2

2xand AC =

AB2 +BC2 =

x2 + y2. Also,

OA = 12AC. Thus, by the Pythagorean Theorem again,

OF 2 = AF 2 OA2

=

(x2 + y2

2x

)2(

x2 + y2

2

)2

=

(x4 + y4 + 2x2y2

4x2

)

(x2 + y2

4

)

=x4 + y4 + 2x2y2 x4 x2y2

4x2

=y4 + x2y2

4x2.

Therefore,

OF =

y4 + x2y2

4x2=

y2(y2 + x2)

2x=

yx2 + y2

2x,

and

EF = 2OF =2yx2 + y2

2x=

yx2 + y2

x.

In part (a), this yields

EF =12162 + 122

16=

12(20)

16= 15 .

Also solved by EDIN AJANOVIC, student, First Bosniak High Shool, Sarajevo, Bosniaand Herzegovina; JACLYN CHANG, student, Western Canada High Shool, Calgary, AB (part

(a) only); RICHARD I. HESS, Ranho Palos Verdes, CA, USA; RICARD PEIR

O, IES \Abastos",

Valenia, Spain; CAO MINH QUANG, Nguyen Binh Khiem High Shool, Vinh Long, Vietnam;

BILLY SUANDITO, Palembang, Indonesia; LUYAN ZHONG-QIAO, Columbia International

College, Hamilton, ON; and TITU ZVONARU, Comanesti, Romania.

457

Problem of the Month

Ian VanderBurgh

Here is a problem that might seem to be not very interesting initially,

but turns out to have a whole lot of unexpeted solutions.

Problem (2005 Canadian Open Mathematis

Challenge) In the grid shown, eah row has

a value assigned to it and eah olumn has a

value assigned to it. The number in eah ell

is the sum of its row and olumn values. For

example, the \8" is the sum of the value as-

3 0 5 6 22 5 0 1 y5 2 x 8 00 3 2 3 5

4 7 2 1 9signed to the 3rd row and the value assigned to the 4th olumn. Determinethe values of x and y.

It is tempting rst of all to give labels to the values that are assigned

to the rows and olumns in order to be able to dive into some algebra. Let's

label the values assigned to the ve olumns A, B, C, D, E and the valuesassigned to the ve rows a, b, c, d, e.

Eah entry in the table gives us an equation involving two of these

variables. For example, the 3 in row 4, olumn 2 gives us d + B = 3,and the 9 in row 5, olumn 5, gives us e+E = 9. We ould proeed andwrite down 25 equations, one for eah entry in the table. These equationswould inlude 12 variables { the 10 that label the rows and olumns togetherwith x and y. We ould then spend pages and pages wading through algebratrying to ome up with the answers. At this point, we would hope that there

has to be a better way. Maybe we should have looked before we leapt!

Here are three neat ways to approah this. (As a point of interest, I

was reently talking about this problem with a friend while driving and so

neither of us really wanted to do any algebra, and so were fored to ome up

with better ways to do it.)

Solution 1. If we hoose ve entries from the table whih inlude one from

eah row and one from eah olumn, then the sum of these entries is onstant

no matter how we hoose the entries, as it is always equal to

A+B + C +D + E + a+ b+ c+ d+ e .

Can you see why? Here are three ways in whih this an be done (looking at

the underlined numbers in the two grids below and the grid on the following

page):

3 0 5 6 22 5 0 1 y5 2 x 8 00 3 2 3 5

4 7 2 1 9

3 0 5 6 22 5 0 1 y5 2 x 8 00 3 2 3 5

4 7 2 1 9

458

Therefore,

3 + (5) + 2 + 8 + (9)= (4) + (3) + x+ 1 + (2)= 3 + y + 2 + (2) + 3 ,

or 1 = x 8 = y + 6. Thus, x = 7 andy = 7.

3 0 5 6 22 5 0 1 y5 2 x 8 00 3 2 3 5

4 7 2 1 9

Solution 2. Consider the rst two entries in row 1. From the labels above, wehave 3 = A+ a and 0 = B + a. Subtrating these, we obtain the equation3 = 3 0 = (A+ a) (B + a) = AB.

Notie that whenever we take entries in olumns 1 and 2 from the samerow, their dierene will always equal AB, whih is equal to 3. Similarly,sine the dierene between the 0 and the 5 in the rst row is 5, then everyentry in olumn 3 will be 5 greater than the entry in olumn 2 from the samerow. In row 3, we see that x = 2 + 5 = 7.

Also, sine the dierene between the 6 and the 2 in the rst row is8, then every entry in olumn 5 is 8 less than the entry in olumn 4 from thesame row. In row 2, we see that y = 1 8 = 7. Thus, x = 7 and y = 7.

Solution 3. Consider the sub-grid

0 1x 8

.

Sine the 0 is in row 2 and olumn 3, then 0 = b + C. Similarly,1 = b+D, 8 = c+D, and x = c+ C.

But then 0+ 8 = (b+C) + (c+D) = (c+C) + (b+D) = x+1, orx = 7.

In a similar way, by looking at the sub-grid

1 y8 0

we an show that

1 + 0 = y + 8, or y = 7. Thus, x = 7 and y = 7.So there are three dierent but neat solutions to the problem. One

footnote to the nal solution is that in fat, in any sub-grid of the form

p qr s

, we must have p+ s = q + r. Can you see why?

Another interesting point about this problem is that it might be easier

for those who know less! If we replaed the x and the y with \?" and gaveit to someone who didn't know a lot of algebra, they might nd the answers

faster than those of us who go immediately to algebra. Sometimes, the extra

mahinery that we have an get in the way.

As 2008 draws to a lose, the Mayhem Editor has three enormous sets of

thanks to oer. First, to the Mayhem Sta, espeially to Monika Khbeis and

Eri Robert, for all of their help over the past year. Seond, to the Editor-

in-Chief of CRUX with MAYHEM, V alav Linek, for all of his help and en-

ouragement over the past year (as well as for his sharp eyes!). Third, to

the Mayhem readership for their support. Please keep those problems and

solutions oming!

459

THE OLYMPIAD CORNER

No. 274

R.E. Woodrow

With the Winter break oming up, I have deided to fous this issue

mainly on providing problems for your puzzling pleasure, and to give some

time for the mails to deliver the solutions to problems from 2008 numbers

of the Corner to restore the readers' solutions le, whih is partiularly thin

for the February 2008 number, as you will see later in the olumn.

To start you o we have the problems proposed but not used at the

47

th

IMO in Slovenia 2006. My thanks go to Robert Morewood, Canadian

Team Leader at the IMO for olleting them for our use.

47

th

INTERNATIONAL MATHEMATICAL OLYMPIAD

SLOVENIA 2006

Problems Proposed But Not Used

Contributing Countries. Argentina, Australia, Brazil, Bulgaria, Canada,

Colombia, Czeh Republi, Estonia, Finland, Frane, Georgia, Greee, Hong

Kong, India, Indonesia, Iran, Ireland, Italy, Japan, Republi of Korea,

Luxembourg, Netherlands, Poland, Peru, Romania, Russia, Serbia and

Montenegro, Singapore, Slovakia, South Afria, Sweden, Taiwan, Ukraine,

United Kingdom, United States of Ameria, Venezuela.

Problem Seletion Committee. Andrej Bauer, Robert Geretshlager, G eza

K os, Marin Kuzma, Sventoslav Savhev.

Algebra

A1. Given an arbitrary real number a0, dene a sequene of real numbersa0, a1, a2, . . . by the reursion

ai+1 = ai {ai} , i 0 ,

where ai is the greatest integer not exeeding ai, and {ai} = ai ai.Prove that ai = ai+2 for suiently large i.

A2. Let a0 = 1 and dene the sequene of real numbers a0, a1, a2, . . .by the reursion

nk=0

ankk + 1

= 0

for n 1. Show that an > 0 for n 1.

460

A3. Let c0 = 1, c1 = 0 and dene the sequene c0, c1, c2, . . . by thereursion cn+2 = cn+1 + cn for n 0. Let S be the set of ordered pairs(x, y) suh that

x =jJ

cj and y =jJ

cj1

for some nite set J of positive integers. Prove that there exist real numbers, , m, and M with the property that an ordered pair of non-negativeintegers (x, y) satises the inequality

m < x+ y < M

if and only if (x, y) S. (By onvention an empty sum is 0.)

A4. Let a1, a2, . . . , an be positive real numbers. Prove thati

461

C3. A ake is shaped as an nn square with n2 unit squares. Strawberrieslie on some of the unit squares so that eah row or olumn ontains exatly

one strawberry; all this arrangement A.Let B be another suh arrangement. Suppose that every grid retangle

with one vertex at the top left orner of the ake ontains no fewer strawber-

ries of arrangement B than of arrangement A. Prove that the arrangementB an be obtained from A by performing a sequene of swaps, where a swap

onsists of seleting a grid retangle with only two strawberries, situated at

its top right orner and bottom left orner, and then moving these two straw-

berries to the other two orners of that retangle.

C4. An (n, k)-tournament is a ompetition with n players held in k roundssuh that

(a) Eah player plays in eah round, and every two players meet at most

one.

(b) If playerAmeets playerB in round i, playerC meets playerD in roundi, and player A meets player C in round j, then player B meets playerD in round j.

Determine all pairs (n, k) for whih there exists an (n, k)-tournament.

C5. A holey triangle is an upward equilateral triangle of side length n withn upward unit triangular holes ut out. A diamond is a unit rhombus withangles of 60 and 120. Prove that a holey triangle T an be tiled withdiamonds if and only if for eah k = 1, 2, . . . , n every upward equilateraltriangle of side length k in T ontains at most k holes.

C6. Let P be a onvex polyhedron with no parallel edges and no edge par-allel to a fae other than the two faes it borders. A pair of points on P areantipodal if there exist two parallel planes eah ontaining one of the points

and suh that P lies between them. Let A be the number of antipodal pairsof verties and letB be the number of antipodal pairs of mid-points of edges.Express AB in terms of the numbers of verties, edges, and faes of P.

Geometry

G1. Let ABCD be a trapezoid withAB||CD and AB > CD. PointsK and L lie on the line segmentsAB and CD, respetively, suh thatAK : KB = DL : LC. Suppose thatthere are points P and Q on the linesegment KL satisfying

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p p

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A B

CD

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L

APB = BCD and CQD = ABC .

Prove that the points P , Q, B, and C are onyli.

462

G2. Let ABCDE be a onvex pen-tagon suh that

BAC = CAD = DAE ;

ABC = ACD = ADE .

The diagonalsBD andCE interset atP . Prove that the line AP bisets theside CD.

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G3. A point D is hosen on the sideAC of a triangle ABC with

ACB < BAC < 90

in suh a way that BD = BA. Theinirle of ABC is tangent to AB andAC at points K and L, respetively.Let J be the inentre of triangleBCD.Prove that the line KL intersets theline segment AJ at its mid-point.

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G4. In triangle ABC, let J be the

entre of the exirle tangent to side

BC at A1 and to the extensions ofsides AC and AB at B1 and C1, re-spetively. Suppose that the lines

A1B1 and AB are perpendiular andinterset at D. Let E be the footof the perpendiular from C1 to lineDJ. Determine the angles BEA1and AEB1.

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E

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G5. Cirles1 and 2 with entresO1and O2 are externally tangent at pointD and internaly tangent to a irle at points E and F , respetively. Linet is the ommon tangent of 1 and 2at D. Let AB be the diameter of perpendiular to t, so that A, E, andO1 are on the same side of t. Provethat the lines AO1, BO2, EF , and tare onurrent.

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12t

463

G6. In a triangle ABC, letMa, Mb,Mc be the respetive mid-points ofthe sides BC, CA, AB and let Ta,Tb, Tc be the mid-points of the arsBC, CA, AB of the irumirle ofABC not ontaining A, B, C, respe-tively. For eah i {a,b,c}, let ibe the irle with diameter MiTi. Letpi be the ommon external tangent toj , k suh that {i,j,k} = {a,b,c}and suh that i lies on one side ofpi while j , k lie on the other side.Prove that the lines pa, pb, pc form atriangle similar to ABC and nd theratio of similitude.

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G7. Let ABCD be a onvex quadri-lateral. A irle passing throughA andD and a irle passing through B andC are externally tangent at the pointP in the interior of the quadrilateral.Prove that if PAB + PDC 90and PBA + PCD 90, thenAB + CD BC +AD.

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G8. PointsA1,B1,C1 are on the sidesBC,CA,AB of a triangleABC, re-spetively. The irumirles of triangles AB1C1, BC1A1, CA1B1 intersetthe irumirle of triangle ABC again at points A2, B2, C2, respetively(that is, A2 6= A, B2 6= B, C2 6= C). Points A3, B3, C3 are symmetrito A1, B1, C1 with respet to the mid-points of the sides BC, CA, ABrespetively. Prove that the triangles A2B2C2 and A3B3C3 are similar.

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A1

464

Number Theory

N1. Given x (0, 1) let y (0, 1) be the number whose nth digit after thedeimal point is the (2n)th digit after the deimal point of x. Prove that if xis a rational number, then y is a rational number.

N2. For eah positive integer n let

f(n) =1

n

(n

1

+

n

2

+ +

n

n

),

where x is the greatest integer not exeeding x.(a) Prove that f(n+ 1) > f(n) for innitely many n.

(b) Prove that f(n+ 1) < f(n) for innitely many n.

N3. Find all solutions in integers of the equation

x7 1x 1 = y

5 1 .

N4. Let a and b be relatively prime integers with 1 < b < a . Dene theweight of an integer c, denoted by w(c), to be the minimum possible valueof |x|+ |y| taken over all pairs of integers x and y suh that

ax+ by = c .

An integer c is alled a loal hampion if w(c) max{w(c a), w(c b)}.Find all loal hampions and determine their number.

N5. Prove that for every positive integer n, there exists an integer m suhthat 2m +m is divisible by n.

A seond problem set is the 2005/06 Swedish Mathematial Contest.

My thanks go to Robert Morewood, Canadian Team Leader at the IMO, for

olleting them for our use.

SWEDISH MATHEMATICAL CONTEST 2005/2006

Final Round

November 19, 2005 (Time: 5 hours)

1. Find all solutions in integers x and y of the equation

(x+ y2)(x2 + y) = (x+ y)3 .

465

2. A queue in front of a ounter onsists of 12 persons. The ounter is then

losed beause of a tehnial problem and the 12 people are redireted toanother one. In how many dierent ways an the new queue be formed if

eah person maintains the same position as before, or is one step loser to

the front, or is one step farther from the front?

3. In the triangle ABC the angle bisetor from A intersets the side BCin the point D and the angle bisetor from C intersets the side AB in thepoint E. The angle at B is greater than 60. Prove that AE + CD < AC.

4. The polynomial f(x) is of degree four. The zeroes of f are real and forman arithmeti progression, that is, the zeroes are a, a+d, a+2d, and a+3dwhere a and d are real numbers. Prove that the three zeroes of f (x) alsoform an arithmeti progression.

5. Eah square on a 20052005 hessboard is painted either blak or white.This is done in suh a way that eah 2 2 \sub-hessboard" ontains an oddnumber of blak squares. Show that the number of blak squares among the

four orner squares is even. In how many dierent ways an the hessboard

be painted so that the above ondition is satised?

6. All the edges of a regular tetrahedron are of length 1. The tetrahedronis projeted orthogonally into a plane. Determine the largest possible area

and the least possible area of the image.

Next we look at an alternate solution to problem 2 of the 17

th

Irish

Mathematial Olympiad disussed at [2007 : 151; 2008 : 88-89.

2. Let A and B be distint points on a irle T . Let C be a point distintfrom B suh that |AB| = |AC| and suh that BC is tangent to T at B.Suppose that the bisetor of ABC meets AC at a point D inside T . Showthat ABC > 72.

Alternate Solution by Luyan Zhong-

Qiao, Columbia International College,

Hamilton, ON.

Let D 6= B be the seond point ofintersetion of BD with the irle T .Let ABD = DBC = . SineAB = AC, we have ACB = 2.Also, BC is tangent to the irle andAB is a hord, hene DAB = .Let ABO = .

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466

Sine BC is tangent to T , we have 2 + = 90. Adding the anglesin the isoseles triangle ABC yields 4+CAB = 180. From these twoequations it follows that CAB = 2. Sine D is inside T we have

= DAB > CAB = 2 ,

and therefore

2> . This last inequality together with 2+ = 90 yields

2 +

2> 90, or 5

2> 90. Hene, > 36 and ABC = 2 > 72.

Next we look at a omment on the solution to problem 6 of the 2007

Italian Olympiad [2007: 149{150; 208: 84{85.

6. Let P be a point inside the triangle ABC. Say that the lines AP , BP ,and CP meet the sides of ABC at A, B, and C, respetively. Let

x =AP

PA, y =

BP

PB, z =

CP

PC.

Prove that xyz = x+ y + z + 2.

Comment by J. Chris Fisher, University of Regina, Regina, SK.

The result is a theorem of Euler featured in the Crux Mathematiorum

artile \Euler's Triangle Theorem" by G.C. Shephard in [1999 : 148{153.

Euler's proof is very neat, even nier than any of the solutions found in Crux

Mathematiorum. It an be loated in Edward Sandifer's book, How Euler

Did It, and on his webpage: http://www.maa.org/news/howeulerdidit.html

lik on \19

th

Century Triangle Geometry".

Now we turn to solutions from our readers to problems given in the

February 2008 number of the Corner. First a solution to a problem of the

11

th

Form, Final Round, XXXI Russian Mathematial Olympiad given in the

Corner at [2008: 20{21.

5. (N. Agakhanov) Does there exist a bounded funtion f : R R withf(1) > 0 suh that

f2(x+ y) f2(x) + 2f(xy) + f2(y)for all x, y R?

Solved by Li Zhou, Polk Community College, Winter Haven, FL, USA.

Suppose that f is suh a funtion. Let a0 = 1 and an = an1 +1

an1for n 1. Then

f2(a1) f2(a0) + 2f(1) + f2(

1

a0

) 2f(1) .

467

As an indution step, assume that f2(an) 2nf(1) for some n 1. Then

f2(an+1) = f2

(an +

1

an

)

f2(an) + 2f(1) + f2(

1

an

) f2(an) + 2f(1) 2(n+ 1)f(1) ,

ompleting the indution. Hene f2(an) 2nf(1) for all n 1, ontra-diting the fats that f(1) > 0 and f is bounded.

And to omplete our les for the Corner, we look at a problem of the

Taiwan Mathematial Olympiad, Seleted Problems 2005, given in [2008:

21{22.

1. A ABC is given with side lengths a, b, and c. A point P lies insideABC, and the distanes from P to the three sides are p, q, and r, respe-tively. Prove that

R a2 + b2 + c2

18 3pqr

,

where R is the irumradius ofABC. When does equality hold?Solved by Arkady Alt, San Jose, CA, USA; Mihel Bataille, Rouen, Frane;

and George Tsapakidis, Agrinio, Greee. We give Bataille's write-up.

Let F denote the area of ABC. We have the well-known relation2F =

abc

2R, but also from the denition of p, q, and r we have the equation

2F = ap+ bq + cr. Thus, the proposed inequality is equivalent to

abc

2(ap+ bq + cr) a

2 + b2 + c2

18 3pqr

or

(a2 + b2 + c2)(ap+ bq + cr) 9abc 3pqr . (1)By the AM{GM Inequality,

a2 + b2 + c2 3 3a2b2c2 and ap+ bq + cr 3 3

abcpqr ,

and the inequality (1) now follows from

(a2 + b2 + c2)(ap+ bq + cr) 9 3a2b2c2 3

abc 3pqr .

That ompletes the Corner for this number, and this Volume. As Joanne

Canape, who has been translating my sribbles into beautiful L

A

T

E

X has de-

ided that twenty-plus years is enough, I want to thank her too for all the

help over the time we've worked together.

468

BOOK REVIEW

John Grant MLoughlin

The Symmetries of Things

By John H. Conway, Heidi Burgiel, and Chaim Goodman{Strauss, published

by AK Peters, Wellesley, MA, 2008

ISBN 978-1-56881-220-5, hardover, 423+xxv pages, US$75.00

Reviewed by J. Chris Fisher, University of Regina, Regina, SK

The authors set themselves the ambitious goal of produing a book that

appeals to everybody. As far as I an tell from a single reading, they have

su

eeded admirably. The rst thing anybody would notie about the book

is that it is lled with beautiful and fasinating photographs and omputer

drawings. No speial knowledge is required for admiring beauty; this book

would be as muh at home on the living room oee table as on the oe

shelf. Of ourse, it is primarily a mathematis book.

The ontents have been organized into three parts. The rst of them

desribes and enumerates the symmetries found in repeating patterns on sur-

faes; it is written at a level suitable for Crux withMayhem readers. This part

might well serve as a textbook for a geometry ourse direted at university

students speializing in mathematis, eduation, physial siene, or om-

puter siene. What makes the authors' approah both novel and elementary

is the introdution of what they all the orbifold signature notation. Groups

are not needed here; the onept an be easily desribed and quikly mas-

tered. Here is the idea. A point in a pattern where two mirrors of symmetry

meet at an angle of

mis alled kaleidosopi and is denoted by m; points

having rotational symmetry of orderm (but no kaleidosopi symmetry) are

alled gyrational and are denoted by m (with no asterisk). If a region hasan oppositely oriented image in the pattern that is not explained by mirrors,

then these two regions must be related by a glide reetion, whih here is

alled a mirale (short for \mirrorless rossing", they say), denoted by .To identify the signature of any repeating

plane pattern one writes down the symbols

starting from the middle and working out-

ward. First loate mirror lines and eah

kind of kaleidosopi point, if any (where two

points are of the same kind if they are related

by a symmetry of the pattern); list them after

the asterisk in dereasing order. Next loate

any gyrational points and order them before

any asterisk. Then look for mirales. Typial

signatures are 632 for a pattern whose sym-metry is explained by three kinds of mirrors

that meet in pairs at angles of

6,

3, and

2;

632 (having no asterisk) for a pattern with

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

632

469

6{fold, 3{fold, and 2{fold gyrational points but no reetions or mirales;222 for a pattern with two kinds of kaleidosopi points where a pair ofmirrors interset at right angles, and one point where there is a half-turn

symmetry but no mirror.

632

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