Đề thi Toán hình học quốc tế năm 1993-1997

Đề thi Toán hình học quốc tế năm 1993-1997

/ Al"- AUSTRALlAN MATHEMATlCS TRUST PUBLlCATlON

TOURNAMENT OF TOWNS

1993 - 1997

QUESTlONS AND SOlUTlONS

DEDlCATED TO THE l/IEMORY OF NB VASSlUEV

PROBLEMS COM AN, TOURNAMENTS 1 TO 18

CDSRJ

AUSTRALlAN MATHEMATlCS T RUST

Australian Mathematics Trust University of Canberra ACT 2601

AUSTRAUA

Copyright © 1998 Australian Mathematics Trust

Telephone: +61 2 6201 5137 AMTOS Pty Ltd ACN 058 370 559

National Library of Australia Card Number and lSSN Australian Mathematics Trust Enrichment Series lSSN 1326-0170

Tournament of the Towns 1993-1997

lSBN 1 876420 03 0

THE AUSTRALIAN MATHEMATlCS TR UST

ED J BARBEAU, Toronto CANADA

GEORGE BERZSENYl, Terra Haute USA

RoN DUNKLEY, Waterloo CANADA

WALTER E MlENTKA, Lincoln USA

NlKOLAY KONSTANTINOV, Moscow RUSSlA

ANDY LlU, Edmonton CANADA

JORDAN B TABOV, Sofia BuLGARlA

JOHN WEBB, Cape Town SOUTH AFRlCA

The books in this series are selected for their motivating, interesting and stimulating sets of quality problems, with a lucid expository style

in their solutions Typically, the problems have occurred in either national or international contests at the secondary school level

They are intended to be sufficiently detailed at an elementary level for the mathematically inclined or interested to understand but, at the same time, be interesting and sometimes challenging to the undergraduate and the more advanced mathematician lt is believed that these mathematics competition problems are a positive influence on the learning and enrichment of mathematics

ENRlCHMENT SERlES

BOOKS lN THE SERlES

ALL THE BEST FROM THE AUSTRAUAN MATHEMATICS COMPETITION

4 AUSTRALIAN MATHEMATICS COMPETITION BOOK 2 1985-1991

PJ O'Halloran, G Pollard, PJ Taylor

5 PROBLEM SOLVlNG VlA THE AMC

PREFACE

The International Mathematics Tournament of Towns has continued to prosper during the period covered by this book, the fourth in the series A complete history of the Tournament is to be found in the earlier books, particularly the first published book which covered the years 1984 to

1988

Results of the Tournament can be found on the Australian Mathematics Trust's web site, whose URL is

http://www.amt.canberra.edu.au This site gives all the contact details and information about how the Tournament is organised A perusal shows that the Tournament has now grown to 99 towns, located in all corners of the earth

The Problems Committee Chairman , Kolya Vassiliev, who had been the pre-eminent Russian problems editor (and significant problem creator)

for decades, sudden ly died in May 1998 Kolya was also for many years the problems editor of the Journal Kvant His loss is- major and it is difficult to imagine how his special style can be replaced by the cen­tral committee An obituary appears eleswhere in this book, which is dedicated to his memory

I would like particularly to acknowledge the work of my colleague, Dr Andrei Storozhev, who took over from me the production of the problems and solutions (and hence this book) about three years ago Andrei is a graduate of the Moscow State University and his presence in Canberra, with his native Russian and history of mathematics enrichment classes

in Russia, has taken a considerable load off me

In keeping with previous books in this series, authors of solutions have been identified Unfortunately, some records were not kept for solu­tion s of Tournament 17 and 18 problems and many solution s are not acknowledged, although it is generally believed that the unacknowledged solution s are due to Andy Liu and Andrei Storozhev

It is also necessary to acknowledge Professor Andy Liu, who continues

to be the other great contributor to the English solutions, and hence the solutions which appear in this book Andy has also supplied great support to the Russian organisers of the Tournament, particularly the President of the Central Organising Committee, Nikolay Konstantinov Peter Taylor

Chairman , Australian IMTOT Committee

19 June 1998

JUN I O R QUESTI O N S

Aut u m n 1993 ( 0 Level )

1 We are given a hexagon with a number written on each of its sides and vertices Each number on a vertex is equal to the sum of the two numbers on neighbouring sides Assume all numbers of the sides and one vertex number were erased Is it possible to find out the number that had been erased from a vertex?

(Folklore, 3 points)

2 Vertices A, B and C of a triangle are connected with points A', B' and C' lying in the opposite sides of the triangle (not at vertices) Can the midpoints of the segments AA', BB' an d CC' lie 'in a straight line? (Folklore, 3 points)

3 A natural number A is given One may add to it one of its divisors

d (1 < d < A) One may then repeat this operation with the new number A+ d and so on Prove that starting from A= 4 one can get any composite number by these operations

(M Vyalyi, 4 points)

4 Three players Alexander, Beverley and Catherine participate in a tournament (all of them play the same number of games with each other) Is it possible that Alexander gets more points than the others, Catherine gets less points than the others, but Alexander has a smaller number of wins than the others and Catherine has a greater number of wins than the others? (A win scores 1 point, a draw scores !-) (A Rubin , 5 points)

4 TOURNAMENT 15

Autum n 1993 (A Level)

1 10 integers are written in a row A second row of 10 integers is formed as follows: the integer written under each integer A of the first row is equal to the total number of integers that stand to the right side of A (in the first row) and are strictly greater than A A third row is formed by the same way under the secon d one, and so

on

(a) Prove that after several steps a "zero row" (i.e a row consist­ing entirely of zeros) appears

(2 points) (b) What is the maximal possible number of non-zero rows (i.e rows in which at least one entry is not zero) ?

(S Tokarev, 2 points)

2 The square PQRS is placed inside the square ABCD in such a way that the segments AP, BQ, CR and DS intersect neither each other nor the square PQRS Prove that the sum of areas of quadrilaterals A BQP and C D SR is equal to the sum of the areas

of quadrilaterals B CRQ an d DAPS (Folklore, 3 points)

3 Three angles of a non-convex, non-self-intersecting quadrilateral are equal to 45 degrees (i.e the last equals 225 degrees) Prove that the midpoints of its sides are vertices of a square

(V Proizvolov, 3 points)

4 Diagon als of a 1 by 1 square are arranged in an 8 by 8 table (one

in each 1 by 1 square) Consider the union W of all 64 diagonals drawn The set W con sists of several connected pieces (two points belong to the same piece if an d on ly if W contains a path between them) Can the number of the pieces be greater than

(a) 15,

(b) 20?

(2 points) (NB Vassiliev, 3 points)

5 Let S(n) denote the sum of digits of n (in decimal representation )

D o there exist three different natural numbers n , p and q such that

n + S( n) = p + S( p) = q + S( q) ?

(M Gerver, 6 points)

6 Construct a set of k integer weights that allows you to get any total integer weight from 1 up to 55 grams even if some of the weights

of the set are lost Consider two versions:

(a) k = 10, and any one of the weights may be lost ; (4 points)

(b) k = 12, and any two of the weights may be lost

(D Zvonkin , 4 points) (In both cases prove that the set found has the property required.)

6 TOURNAMENT 15

S pring 1 994 (0 Level)

1 Construct a convex quadrilateral given the lengths of all its sides and the length of the segment between the midpoints of its

2 60 children participate in a summer camp Among any 10 of the children there are three or more who live in the same block Prove that there must be 15 or more children from the same block

(Folklore, 4 points)

3 Let 0 be a point inside a convex polygon A 1 A2 An such that

LO A 1 An � L O A 1 A2, LOA 2A 1 � L OA 2A3 , , L OAn-lAn-2 � OAn-l An , L OAnAn-1 � L OAnA l and all of these angles are acute Prove that 0 is the centre of the circle inscribed in the

� polygon (V Proizvolov, 4 points)

4 Ten coins are placed in a circle, showing "heads" (the tails are down) Two moves are allowed:

(a) turn over four consecutively placed coins;

(b) turn over four coins placed as X X O X X (X is one of the coins

to be turned over, 0 is not touched)

Is it possible to have all ten coin s showing "tails" after a finite sequence of such moves? (A Tolpygo, 5 points)

Spri ng 1 994 (A Level)

1 A schoolgirl forgot to write a multiplication sign between two digit numbers and wrote them as one number This 6-digit re­sult proved to be 3 times greater than the product (obtained by multiplication) Find these numbers (A Kovaldzhi, 3 points)

3-2 Two circles intersect at the points A and B Tangent lines drawn

to both of the circles at the point A intersect the circles at the points M and N The lines B M and B N intersect the circles once more at the points P and Q respectively Prove that the segments

M P and NQ are equal (I Nagel, 3 points)

3 Each of the 450 members of a parliament gives a slap in the face to exactly one of his colleagues Prove that after that they can choose

a committee consisting of 150 members, none of whom has been slapped in the face by any other member of the committee

standing in different rows and different columns, at least two are

5 Does there exist a convex pentagon from which a similar pentagon can be cut off by a straight line? (S Tokarev, 5 points)

6 At each integer point of the numerical line a lamp with a toggle button is placed If the button is pressed, a lit lamp is turned off, an unlit one is turned on Initially all the lamps are unlit A stencil with a finite set of fixed holes at integer distances is cho­sen The stencil may be moved along the line as a rigid body, and for any fixed position of the stencil, one may push simultaneously all the buttons accessible through the holes Prove that for any stencil it is possible to get exactly two lit lamps after several such

8 TOURNAMENT 15

1 In a 10 by 10 square grid (which we call "the bay" ) you are re­quested to place ten "ships": one 1 by 4 ship, two 1 by 3 ships, three 1 by 2 ships and four 1 by 1 ships The ships may not have common points (even corners) but may touch the "shore" of the bay Prove that

(a) by placing the ships one after the other arbitrarily but in the order indicated above, it is always possible to complete the

(b) by placing the ships in reverse order (beginning with the small­

er ones) , it is possible to reach a situation where the next ship cannot be placed (give an example) (KN Ignatjev, 2 points)

SEN I O R QUEST I O N S

Autu m n 1993 ( 0 Level)

1 Consider the set of solutions of the equation

x2 + y3 = z2

in positive integers Is it finite or infinite? (Folklore, 3 points)

2 Points M and N are taken on the hypotenuse of a right triangle ABC so that BC = BM and AC = AN Prove that the angle MCN is equal to 45 degrees (Folklore, 3 points)

3 Each of the numbers 1, 2, 3, 25 is arranged in a 5 by 5 table In each row they appear in increasing order (left to right) Find the maximal and minimal possible sum of the numbers in the third

4 Peter wants to make an unusual die having different positive inte­gers on each of its faces For neighbouring faces the corresponding numbers should differ by at least two Find the minimal sum of the six numbers (Folklore, 5 points)

10 TOURNAMENT 15

Autu m n 1993 (A Level)

1 Two tangents C A and C B are drawn to a circle (A and B being the tangent points) Consider a "triangle" bounded by an arc A,B

(the smaller one) and segments C A and C B Prove that the length

of any segment inside the triangle is not greater than the length of

2 The decimal representation of all integers from 1 to an arbitrary integer n are written one after another as such:

123 9101 1 99100 (n) Does there exist n such that each of the digits 0, 1 , 2, 9 appears the same number of times in the given sequence?

(A Andzans, 3 points)

3 Consider the hexagon which is formed by the vertices of two equi­lateral triangles (not necessarily equal) when the triangles intersect Prove that the area of the hexagon is unchanged when one of the triangles is translated (without rotating) relative to the other in such a way that the hexagon continues to be defined

(V Proizvolov, 3 points)

4 A convex 1 993-gon is divided into convex 7-gons Prove that there are 3 neighbouring sides of the 1993-gon belonging to one such 7-gon (A vertex of a 7-gon may not be positioned on the interior

of a side of the 1993-gon, and two 7-gons either have no common points, exactly one common vertex or a complete common side )

6 If it is known that the equation

x4 + ax3 + 2x2 + bx + 1 = 0 has a (real) root, prove the inequality

a2 + b2;::: 8

(A Egorov, 8 points)

Spring 1994 (0 Level)

1 A triangle ABC is ins cribed in a circle Let A1 be the point dia­metrically opposed to A, Ao be the midpoint of the side B C and A2 be the point symmetric to A 1 with respect to Ao; the points B2 and C2 are defined in a similar way starting from B and C Prove that the three points A 2, B2 and C2 coincide

(A Jagubjanz, 4 points )

2 The sequence of pos itive integers a1, a2 , is s uch that for each

n = 1 , 2, the quadratic equation an+2x2 + an+lX + an = 0 has

a real root Can the sequence consist of

(a) 10 terms ,

(b) an infinite number of terms ?

(2 points ) (A Shapovalov, 3 points )

3 A chocolate bar has five lengthwise dents and eight crosswise ones, which can be used to break up the bar into sections (one can get a total of 9 x 6 = 54 cells ) Two players play the following game with such a bar At each move (the two players move alternatively) one player breaks off a section of width one from the bar along a single dent and eats it, the other player does the same with what's left of the bar, and so on When one of the players breaks up a section

of width two into two strips of width one, he eats one of the strips and the other player eats the other strip Prove that the player who has the firs t move can play so as to eat at least 6 cells more than his opponent (no matter how his opponent plays )

(R Fedorov, 4 points)

4 Ten coins are placed in a circle, s howing "heads" (the tails are down) Two moves are allowed:

(a) turn over four consecutiyely placed coins ;

(b) turn over four coins placed as X X O X X (X is one of the coins

to be turned over, 0 is not touched)

Is it possible to have all ten coins showing "tails" after a finite sequence of such moves ? (A Tolpygo, 4 points )

(a) if the first number of the sequence is rational, then the se­quence will be periodic (i.e the terms repeat with a certain cycle length after a certain term in the sequence) ;

(2 points) (b) if the sequence is periodic, then the first number is rational

(G Shabat, 2 points)

3 At least one of the coefficients of a polynomial P(x) is negative Can all of the coefficients of all of its powers ( P ( x)) n , n > 1 , be positive? (0 Kryzhanovskij, 4 points)

4 A point D is placed on the side BC of the triangle ABC Circles are inscribed in the triangles ABD and AC D ; their common exterior tangent line (other than BC) intersects AD at the point K Prove that the length of AK does not depend on the position of D (An exterior tangent of two circles is one which is tangent to both circles but does not pass between them.) (I Sharygin, 5 points)

5 Find the maximal integer M with nonzero last digit (in its decimal representation) such that after crossing out one of its digits (not the first one) we can get an integer that divides M ·

(A Galochkin, 5 points)

6 Consider a convex quadrilateral ABC D Pairs of its opposite sides are continued until they intersect: BA and CD at the point P,

BC and AD at the point Q Let K be the intersection point of the exterior bisectors of the angles A and C of the quadrilateral,

L be the intersection point of the exterior bisectors of the angles

B and D of the quadrilateral, and M be the intersection point of the exterior bisectors of the angles P and Q (the exterior bisector

of an angle X is the line passing through X and perpendicular to its ordinary bisector) Prove that the points K , L and M lie on a straight line (S Markelov , 5 points)

7 Consider an arbitrary "figure" F (non convex polygon) A chord

of F is defined to be a segment which lies entirely within F and whose ends are on its boundary

(a) Does there always exist a chord of F that divides its area in

(b) Prove that for any F there exists a chord such that the area

of each of the two parts of F is not less than 1/3 of the area

(c) Can the number 1/3 in (b) be changed to a greater one?

(V Proizvolov, 2 points)

14 TOURNAMENT 15

JUN IO R S O LUTIO N S

Autu m n 1993 ( 0 Level)

1 Consider the numbers before any erasure

Colour the vertex numbers alternately red and blue Then the sum

of the reds and the sum of the blues are both equal to the sum of the side numbers

So after erasure, a missing red or blue number can be recovered

(because red sum = blue sum)

(MS Brooks)

2 Let A", B" , C" , D, E and F be the respective midpoints of AA' , BB' , CC' , BC , C A and (l B Since A' lies between B and C , A" lies between E and F Similarly, B" lies between F and D, and C" lies between D and E

(A Liu)

4 Let Alexander, Beverley and Catherine play six games against one another All six games between Alexander and Beverley are draws Both Beverley and Catherine win three games against each other Alexander beats Catherine twice and loses once while drawing

three So Alexander has 6.5 points, Beverley has 6 and Cather­ine 5 5, while Alexander has 2 wins, Beverley has 3 and Catherine

4

(A Liu)

16 TOURNAMENT 15

Autum n 1993 (A Level)

1 (a) After the first row, all integers are non-negative The last

integer on each row from the second row on must be 0, as there are no integers to the right of the last integer in the preceding row The second last integer on each row from the third row on must be 0, as there are only Os to the right of the second last integer in the preceding row Similarly, the third last integer on each row from the fourth row on must be

0, and so on It follows that the eleventh row must consist of only Os

We use [] to denote area Drop perpendiculars PT, Q U , RV and

SW respectively from P, Q, R and S towards the perimeter of ABCD, say with T on AB, U on B C , V on CD and W on DA Let EFGH be the square with sides parallel to those of ABCD, and with P on EF, Q on FG, R on GH and S on HE Let K , L,

M and N be the fourth vertices of the rectangles AT P K , B U Q L ,

C V R M and DWSN

Then [ATP] = [AKP], [BLQ] = [BUQ], [CVR] = [CMR] and [DWS ] = [DNS] T riangles ESP, FPQ, GQR and HRS are all congruent to one another, so that [F PQ]+[H RS] = [ESP]+[GQR] and ES = FP = GQ = HR

3 Let A B C D be the quadrilateral with angle BCD = 225° Extend

BC to cut AD at 0 Then OAB and O C D are both isosceles triangles It follows that a 90° rotation about 0 will map A into

B and C into D , so that AC = BD and they are perpendicular to each other

Let P, Q, R and S be the respective midpoints of A B , B C , CD and DA

(A Liu)

4 Both parts are answered in the affirmative T he diagram below shows a case with 21 pieces T he dotted diagonals are isolated, all others are connected

Since there are 20 isolated diagonals , we have 21 pieces

(A Liu)

6 (a) Consider a sequence of objects whose respective weights are

the terms of the Fibonacci sequence, defined by f (1) = f (2) =

1 and f(n) = f (n - 1) + f (n- 2) for n;::: 3 Suppose no object

is ever missing We claim that for any integer w, 1 :::; w :::; f(n + 2) - 1, there is a subset of the first n objects whose total weight is w We prove this by induction on n For n = 1 , /(3) - 1 = 1 , and the claim i s trivial Suppose it holds for some n ;::: 1 Consider the first (n + 1)st object By the induction hypothesis, there is a subset of the first n objects having total weight w, where 1 :::; w :::; f (n + 2) - 1 Adding the (n + 1)st object, we will have a subset of total weight w for f (n + 1) :5 w :5 f (n + 2) - 1 + f (n + 1) = f (n + 3) - 1

We have f (n + 1) :::; f (n + 2) - 1 and the claim is justified Suppose now that one arbitrary object may be missing

We claim that for any integer w, 1 :::; w :::; f (n + 1) - 1, there

is a subset of the first n objects whose total weight is w

We prove this by induction on n For n = 1, f (n + 1) - 1 = 0, and the claim is trivial

Suppose it holds for some n � 1 Consider the first n + 1 objects If the (n + 1)st object is missing, then none of the first

n is, and we have already proved that there will be a subset having total weight w for 1 :::; w :::; f (n + 2) - 1 Suppose the (n + 1)st object is not missing By the induction hypothesis, there is a subset of the first n objects, minus an arbitrary one, having total weight w, where 1 :::; w :::; f ( n + 1) - 1 Adding the ( n + 1 )st object, we have a subset of total weight w, where f(n + 1) :5 w :5 f(n + 1) - 1 + f (n + 1) � f (n + 2) - 1 This justifies our claim It follows that if we take 10 objects

of respective weights /(1) = 1, /(2) = 1, /(3) = 2, /(4) = 3,

!(5) = 5, !(6) = 8, !(7) = 13, f (8) = 21, /(9) = 34 and

20 TO U R N A M E N T 15

.f ( 10) = 55, we have a subset of total weight w for 1 :=::; w :=::; .f ( l l ) - 1 = 88, even if an arbitrary object may be missing (b) Consider the sequence of objects whose respective weights are the terms of a generalized Fibonacci sequence which is defined

We claim that for any integer w, 1 � w :=::; g(n + 1) - 1, there

is a subset of the first n objects whose total weight is w

We prove this by induction on n For n = 1, g(2) - 1 = 0 and the claim is trivial

Suppose it holds for some n 2: 1 Consider the first n + 1 objects If the (n + 1)st object is missing, then at most one

of the first n is, and we h ave already indicated that there will

be a subset having total weight w for 1 � w :=::; g(n + 2) - 1 Suppose the ( n + 1)st object i s not missing By the induction hypothesis, there is a subset of the first n objects, minus two arbitrary ones, having total weight w for 1 :=::; w :=::; g(n + 1) - 1 Adding the (n + 1)st object, we have a subset of total weight

w , where g(n + 1) :=::; w :=::; g(n + 1) - 1 +g(n + 1 ) 2: g(n+2) - l This justifies our claim It follows that if we take 12 objects

of respective weights g(1) = 1, g(2) = 1, g(3) = 1, g(4) = 2, g(5) = 3, g(6) = 4, g(7) = 6, g(8) = 9, g(9) = 13, g(10) = 19, g(ll) = 28 and g(12) = 41, we have a subset of total weight

w , 1 :=::; w � g(13) - 1 = 59, even if two arbitrary objects may

be missing

(A Liu)

Spring 1994 ( 0 Level)

1 Let A , B, C and D be the vertices of a quadrilateral and let X and

Y be the midpoints of its diagonals AC and BD respectively

So we may assume that we are given the angles between opposite sides of the quadrilateral Hence we can construct the triangle DEA such that ED is parallel and equal to B C

Similarly, we are able t o construct the triangle EBA a s E B C D is

a parallelogram So we know the size of LDAB and can construct the triangle A D B Therefore, the whole quadrilateral ABCD can

be constructed

(A Storozhev)

2 Let us assume that there are no blocks with 15 or more children

We denote by N the number of blocks with at least two children

N cannot be any greater than 4 since otherwise we could easily find 10 children no three of whom live in the same block

If N is less than 4, then the total number of children living in the big blocks cannot be greater than 14 x 3 = 42 It means that the number of blocks with 1 child is at least 60 -42 = 18 and again

we can pick 10 children any three of whom do not live in the same block

The remaining possibility is N = 4, in which case the number

of blocks with 1 child is at least 60 - 14 x 4 4 and we come

22 TOU RNAM ENT 15

to a contradiction with the terms of the problem, for example, by picking 8 children from the big blocks and 2 children from the small ones

Thus our assumption was wrong

(A Storozhev)

4 Let the coins be numbered consecutively from 1 to 10 We define the weight of the ith coin as 0 if the coin shows "heads" or i if it shows "tails"

Then we define the weight of the whole configuration of coins as the sum of their weights It is easily seen that both the allowed moves

do not change the remainder of the weight of the configuration divided by 2

Since the weight of the initial configuration is equal to 0 we can­not achieve the configuration where all coins are showing "tails" because the weight of the latter is equal to 1 + 2 + + 10 = 55, which is an odd number

(A Storozhev)

24 TOU RNAM ENT 15 Spring 1994 (A Level)

1 Let x and y be the given numbers Then we can interpret the data in terms of an algebraic equation as 3xy = 1000x + y Since two of the three terms of this equation are divisible by x, y is to

be divisible by x That is, y = nx for some integer n Clearly,

1 $ n $ 9 as both x and y are 3-digit numbers The equation 3xy = 1000x + y can be reduced to 3xn = 1000 + n T herefore n

is a divisor of 1000 and simultaneously 1000 + n is divisible by 3, which means that n = 2, 5 or 8 If n = 5 or n = 8, then we have 5x = 335 or 8x = 336 and x cannot be a 3-digit number So n = 2,

x = 167 and y = 334

(A Storozhev)

LAQB = LAM B Similarly, LAPB = LAN B Therefore, the triangles AQN and A M P are similar So the problem will be solved if we show that AN = AP

To show that AN = AP, we prove that LAN P = LAP N We have LAP N = LABN as they are subtended by the same arc But LABN is an exterior angle of the triangle AQB whence LABN = LAQB + LQAB Thus LAPN = LEAN+ LQAB = LQAN

On the other hand, LAN P = LAN B + LBN P By the alter­nate segment theorem LAN B = LM A B We also have LBN P = LEAP because they are subtended by the same arc So LAN P = LMAB + LEAP = LMAP

But LQAN = LM AP as the triangles QAN and MAP are similar Hence LAP N = LAN P and thus the assertion is proved

We put red shirts on those who have slapped members wearing green, and green shirts on those who have slapped members wearing red or yellow

Suppose not all members are wearing coloured shirts, and yet the set of those who do cannot be enlarged Move them aside and look for another slapping circle as before among the remaining members Eventually all 450 members will be wearing red, yellow

or green shirts By the Pigeonhole Principle, at least 150 of them will be wearing shirts of the same colour These members cannot have slapped one another

ping circles with 3 members in each T hen f or each of red, yellow and green, exactly 150 members are wearing shirts of that colour

(A Liu)

4 Let us consider 10 entries of the table standing in different rows and different columns and let ij be the number of the column of the entry in the i-th row Assuming that these entries are all different,

we see that all the differences (i - ij) have different residues modulo

10 Therefore, the sum

(1 - it) + (2 - i2) + + ( 10 -ilQ)

0 + 1 + + 9

5 (mod 10) But simultaneously,

( 1 -il ) + (2- i2) + + ( 10 -ilO) ( 1 + 2 + + 10) -(il + i2 + + i10)

0

AB = a, BC = ka and CD = k2a where � < k < 1

(A Storozhev)

Alternative 2

Let MAN be an equilateral triangle of side 1 + t, where t < 1 is a positive real number to be chosen later Draw equilateral triangles

BAF, NBC and M ED inside ABC, with B on AN , C and D on

MN , and E and F on A M , where BF = 1, B C = t and DE= t3

A

We choose t so that C D = B F - M D = t2• In other words, t is the real root of f (t) = t3 +t2 - l Since f (O) = -1 and f ( l ) = 1, there is such a t between 0 and 1 Moreover, EF = B C - M E = t - t3 = t4, and EA = t4 + 1 The convex pentagons ABC DE and B CDEF have corresponding angles equal In order for them to be similar,

We can obtain BCDEF from ABC DE with a straight cut

(A Liu)

6 To make the terms of the problem more algebraical, we regard the lit lamps as ls placed at the corresponding point of the numerical line and the unlit lamps as Os Also, we consider the holes of the stencil as ls and put Os at integer distances from the holes Now we describe an algorithm to obtain a configuration with only two ls on the line

First, we apply the stencil to the line with Os only If we obtain a configuration with only two ls, then we have finished our process

28 TO U R N A M E N T 15

If it is not the case, we move the stencil to the right so that the first

1 of the stencil is at the same place as the second 1 of the configura­tion and then use the stencil If we do not succeed, we repeat this operation At each step, we obtain a configuration starting with 1, then containing Os and ending at a sequence consisting of 1s and

Os of length n - 1, where n is the length of the stencil Since we may do as many steps as necessary, we can find two configurations with the same ending sequences

Going back step by step, we come to the conclusion that at some stage we had a configuration with the same sequence of the last

n -1 digits as that of the stencil

But according to the algorithm described above, such a configura­tion might arise only if the previous one contained only two 1s So

at some stage, we must obtain a required configuration

(A Storozhev)

7 (a) Clearly we can place the 1 x 4 ship On the bay, mark eight

1 x 3 ships that are parallel to each other and separated by two cells Of these eight marked ships, at most two can have common points with the placed ship

Hence we can place a 1 x 3 ship in the bay so that it does not have common points with the 1 x 4 ship Similarly, if we have one 1 x 4 ship and one 1 x 3 ship, at most four of the marked ships can have common points with the placed ships Therefore we can place another 1 x 3 ship in the bay so that

it does not have common points with the placed ships Now assume that we have already placed one 1 x 4 ship, two

1 x 3 ships and less than three 1 x 2 ships Mark twelve 1 x 2 ships that are parallel to each other and separated by two cells Then there are at most ten marked ships that have common points with the placed ships Therefore we can place one more

1 x 2 ship in the bay

Finally, assume that we have placed one 1 x 4 ship, two 1 x 3 ships, three 1 x 2 ships and less than four 1 x 1 ships in the bay Mark sixteen 1 x 1 ships that are separated by two cells Then there are at most 15 marked ships that have common points with the placed ships Hence we can place one more

1 x 1 ship in the bay

(b) One of many possible configurations is shown on the following

picture

Ill 1 bmd 1, -1-++-�mm+ i

30 TOU RNAM ENT 15

2 Since B G = BM and AG

LANG = goo -A/2

B Hence

3 Let the number in the i-th row and j-th column be denoted by

a(i,j)

We are given that a(i,j) < a(i, k) whenever j < k We may assume that a(i, 3) < a(k, 3) if i < k The maximum value of a(5, 3) is 23 since a(5, 3) < a(5, 4) < a(5, 5) The maximum value of a(4, 3) is

20 since a(4, 3) < a(4, 4) < a(4, 5) and a(4, 3) < a(5, 3) Similarly, the maximum values of a(3, 3), a(2, 3) and a(1 , 3) are 17, 14 and 1 1 respectively It follows that the sum o f the five numbers i n column three is at most

It follows that the best Peter can do is to use 1, 2, 4, 5, 7 and 8,

by putting the consecutive numbers on opposite faces Hence the minimum sum of the six numbers is 27

(A Liu)

32 T O U RNAM ENT 15 Autu m n 1993 (A Level)

1 Let d = C A = C B We may assume that the segment inside the curved triangle has both endpoints on its perimeter If one endpoint is A or B , the other can only be C , and its length is d

c

Suppose the segment is C F with F between A and B on the given circle Since this segment is inside the circle with centre C and radius d, C F < d

Suppose the segment is DE with D on BC and E on CA We may assume that it is tangent to the given circle at a point F Otherwise, a parallel displacement of DE towards the given circle, with D staying on BC and E on CA, will increase its length We now have

2DE = D F+FE+DE < DB +EA +CD +CE = 2d

or D E < d

B c

Finally, suppose the segment is EF with E on CA and F on the arc AB Perform a parallel shift of E F towards C to E' F' , with E' on AC and F' on the arc AB, until either E' = C or E' F'

is tangent to the given circle (as shown in the diagram above)

In the first case, we have EF ::;: C F' < d In the second case,

EF ::;: E' F' < E' D' < d, where D' is the point of intersection of E'F' with BC

m Had we started with 0 instead of 1, and fill in the leading Os so that every number has m digits, we would have a(n) = b(n) = c(n)

at this point by symmetry Since this is not the case, a(n) is less than b(n) = c(n) It follows that there does not exist a positive integer n for which a(n) = b(n) = c(n)

(A Liu)

3 We use [ ] to denote area Let 0 be a point inside both ACE and BDF such that when it is translated to O' along with B' D' F' , O'

is still inside ACE

Let x be the angle between AC and BO Then it is also the angle between AC and B'O'

[OCDE] = [01CD1E]

and

[OEFAJ = [01EF1A] Adding these together,

[ABCDEFJ = [AB1CD1EF1]

(Murray Klamkin)

4 Let n be the number of 7-gons Let v and e be respectively the numbers of vertices and edges in the interior of the 1993-gon The 7-gons among them have 7n sides Each side of the 1993-gon con­tributes 1 to this total, while each interior edge contributes 2 Hence 7n = 1993 + 2e

The sum of the angles of the 7-gons is 5mr The sum of the angles

of the 1993-gon is 199br , while that around each interior vertex is 21r Hence 5n = 1991 + 2v

Eliminating n from these two equations, we have 14v + 3972 = 10e

Let b be the number of vertices of the 1993-gon incident with at least 1 interior edge Since the 7-gons are convex, each interior vertex is incident with at least 3 interior edges It follows that 2e � 3v + b, so that 3972 � 5b + v > 5b or b � 794