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Methods of Problem Solving Book 3

JB TABOV, EM KOLEV & PJ TAYLOR

Methods of Problem Solving Book 3

JB TABOV, EM KOLEV & PJ TAYLOR

Published by

AMT PUBLISHING Australian Maths Trust 170 Haydon Drive Bruce ACT 2617 AUSTRALIA Telephone: +61 2 6201 5136 www.amt.edu.au

Copyright ©2012 Australian Mathematics Trust PDF Edition 2019 AMTT Limited ACN 083 950 341 National Library of Australia Card Number and ISSN Australian Mathematics Trust Enrichment Series ISSN 1326-0170 Methods of Problem Solving Book 3 ISBN 978-1-876420-30-7

ENRICHMENT SERIES EDITORIAL COMMITTEE EDITOR Peter J Taylor, Canberra Australia Warren J Atkins, Newcastle Australia Ed J Barbeau, Toronto Canada George Berzsenyi, Denver USA Ron Dunkley, Waterloo Canada Shay Gueron, Haifa Israel Nikolay Konstantinov, Moscow Russia Andy Liu, Edmonton Canada Walter E Mientka, Lincoln USA Jordan B Tabov, Sofia Bulgaria John Webb, Cape Town South Africa 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. It is believed that these mathematics competition problems are a positive influence on the learning and enrichment of mathematics.

BOOKS IN THE ENRICHMENT SERIES 1

Australian Mathematics Competition 1978–1984 Book 1 WJ Atkins, JD Edwards, DJ King, PJ O’Halloran & PJ Taylor

2

Mathematical Toolchest AW Plank & NH Williams

3

Tournament of Towns questions and solutions 1984–1989 Book 2 PJ Taylor

4 Australian Mathematics Competition 1985–1991 Book 2 PJ O’Halloran, G Pollard & PJ Taylor 5

Problem Solving Via the AMC W Atkins

6 Tournament of Towns questions and solutions 1980–1984 Book 1 PJ Taylor 7

Tournament of Towns questions and solutions 1989–1993 Book 3 PJ Taylor

8

Asian Pacific Mathematics Olympiad 1989-2000 H Lausch & C Bosch Giral

9 Methods Of Problem Solving Book 1 JB s & PJ Taylor 10 Challenge! 1991-1995 JB Henry, J Dowsey, AR Edwards, LJ Mottershead, A Nakos & G Vardaro 11 USSR Mathematical Olympiads 1989-1992 AM Slinko 12 Australian Mathematical Olympiads 1979-1995 H Lausch & PJ Taylor 13 Chinese Mathematics Competitions and Olympiads 1981-1993 A Liu 14 Polish & Austrian Mathematical Olympiads 1981– 1995 ME Kuczma & E Windischbacher 15 International Mathematics Tournament of Towns 1993–1997 Book 4 PJ Taylor & AM Storozhev 16 Australian Mathematics Competition 1992–1998 Book 3 WJ Atkins, JE Munro & PJ Taylor 17 Seeking Solutions JC Burns 18 101 Problems in Algebra T Andreescu & Z Feng 19 Methods of Problem Solving Book 2 JB Tabov & PJ Taylor

20 Hungary-Israel Mathematics Competition: The First Twelve Years S Gueron 21 Bulgarian Mathematics Competition 1992-2001 BJ Lazarov, JB Tabov, PJ Taylor & A Storozhev 22 Chinese Mathematics Competitions and Olympiads 1993-2001 Book 2 A Liu 23 International Mathematics Tournament of Towns 1997-2002 Book 5 AM Storozhev 24 Australian Mathematics Competition Book 1999-2005 Book 4 WJ Atkins & PJ Taylor 25 Challenge! 1999-2006 Book 2 JB Henry, J Dowsey, AR Edwards, LJ Mottershead, A Nakos, G Vardaro & PJ Taylor 26 International Mathematics Tournament of Towns 2002-2007 Book 6 PJ Taylor 27 International Mathematical Talent Search Part 1 G Berzsenyi 28 International Mathematical Talent Search Part 2 G Berzsenyi 29 Australian Mathematical Olympiads 1996-2011 Book 2 H Lausch, A Di Pasquale, DC Hunt & PJ Taylor 30 Methods of Problem Solving Book 3 JB Tabov, EM Kolev & PJ Taylor 31 Australian Mathematics Competition 2006-2012 Book 5 WJ Atkins & PJ Taylor 32 Australian Intermediate Mathematics Olympiads 1999-2013 JB Henry & KL McAvaney 33 Problem Solving Tactics A Di Pasquale, N Do & D Mathews 34 Chinese Mathematics Competitions and Olympiads 2001-2009 Book 3 Y Fu, Z Li & A Liu

PREFACE We published a first book Methods of Problem Solving in 1996, planning to make it the first of a series on this theme. The aim was to have chapters on special facets of problem solving, also presenting for the first time some problems which were not well known in the English language. We followed up with a second book, with 5 more chapters, each on a particular problem-solving theme, in 2002. Now Emil Kolev, in recent years leader of the Bulgarian IMO team, has joined our writing team and we present 5 new chapters in a third book. As in the previous books, the problems have been selected to illustrate the most important and useful ideas, details and techniques of the method discussed. The number of the problems in each chapter is about 10 to 15, and this gives readers the opportunity to understand quickly the essence of the method. Almost every chapter begins with a short theoretical explanation. Jordan Tabov, Emil Kolev and Peter Taylor Sofia and Canberra December 2011

CONTENTS Prefacevii Enumeration3 Combinatorial Games

25

Sum of Segments

53

Ceva's Theorem

73

Functional Equations

93

General References

117

ENUMERATION

1. ENUMERATION Enumeration

The field of combinatorics, and in particular this part of combinatorics (counting techniques) is a popular source of competition problems and other challenges. Partly this is because it is a way of testing innate mathematical ability without dependence on particular results students learn at school. Combinatorics, and in particular enumeration, has elementary problems which can be challenging, easy to understand, and which can guide the student to seeking generalisation. In the first part of this chapter we will guide the student through three such methods, via problems which have been set in competitions. DERANGEMENTS Derangements are re-arrangements of sets of numbers such that none of them remain in their original position. In particular a derangement of the set {1, 2, 3, . . ., n} is the set with the numbers written in a different order, such that the ith number is not i. For example {2, 3, 1} is a derangement of {1, 2, 3} but {3, 2, 1} is not since the number 2 is in its original position. Example 1 (Australian Mathematics Competition, 1981) In how many different ways can a careless office boy place four letters in four envelopes so that no one gets the right letter? (A) 4

(B) 9

(C) 12

(D) 6

(E) 24

Solution 1 We give two solutions. It is easy enough to systematically count the outcome for numbers of this size and Alternative 1 does this without needing much prior knowledge. In the second solution we use an established formula which enables the problem to be generalised so that we could solve the problem if the number of letters was any size. Alternative 1 If we consider the 24 arrangements of the numbers 1, 2, 3, 4 and cross off all those with 1 in the first place (i.e. the first letter in its proper envelope), 2 in the second place, 3 in the third place, or 4 in the fourth place, we have left these nine arrangements: 2 3 4

1 1 1

4 4 2

3 2 3

2 3 4

4 4 3

1 1 1

3 2 2

2 3 4

3 4 3

4 2 2

1 1 1

Therefore there are nine different ways, hence (B).

4Enumeration 4 1. Enumeration

Alternative 2 Here we note the problem is equivalent to a derangement problem with n = 4. The following formula gives the number of derangements D(n), of the numbers 1, 2, 3, . . ., n. 1 1 1 n 1 , D(n) = n! 1 − + − + . . . + (−1) 1! 2! 3! n! where n!, known as n factorial, equals n(n − 1)(n − 2) . . . 3 × 2 × 1 (for example 4! = 4×3×2×1 = 24). For a simple explanation of this formula see Niven [48]. So the “careless office boy with four letters” problem is a special case of the above formula with n = 4, where 1 1 1 1 D(4) = 4! 1 − + − + 1! 2! 3! 4! 1 1 1 = 24 1 − 1 + − + 2 6 24 = 9, hence (B). Partial Derangements We can also devise problems in which a fixed number of members of a set must be deranged while the remainder remain fixed. This can be exemplified by: Example 2 (Australian Mathematics Competition, 1983) In the school band five children each own their own trumpet. In how many different ways can exactly three of the five children take home the wrong trumpet, while the other two take home the right trumpet? (A) 5 (B) 6 (C) 10 (D) 20 (E) 30 Note: Again, this problem is simple enough that we can first work through, counting by first principles. Then we can look for the generalisation. Solution 2 Suppose the students taking home the wrong trumpet are called A, B and C. These can take the wrong trumpets in two ways, e.g. A takes trumpet B, B takes trumpet C and C takes trumpet A, or A takes trumpet C, B takes trumpet A and C takes trumpet B. We need also to know how many ways A, B and C can be chosen from the five. This is the same as the number of ways in which the two with the right trumpets

Enumeration5 1. Enumeration 5

can be chosen, this being ten (e.g. if the students are called A, B, C, D and E these are A and B, A and C, A and D, A and E, B and C, B and D, B and E, C and D, C and E, and D and E. Thus the answer is 10 × 2 = 20, hence (D). Generalisation to this question: In the following discussion n! n = is the number of combinations possible from choosing r r!(n − r)! r objects from a set of n (without regard to order), and D(r) is the number of derangements of the numbers 1, 2, . . ., r, as discussed in Example 1, i.e. 1 1 1 1 D(r) = r! 1 − + − + . . . + (−1)r . 1! 2! 3! r! It is now easily established that if there are n children, each owning their own trumpets, the number of ways in which exactly r take home the wrong trumpet is nr × D(r). For n = 5 the solutions can be tabulated as 5 5 r D(r) × D(r) r r 0 1 2 3 4 5

1 5 10 10 5 1

1 0 1 2 9 44

1 0 10 20 45 44

(the solution to our problem)

Note: The sum of the solutions, 1 + 0 + 10 + 20 + 45 + 44 = 120 or 5!, is the total number of ways of allocating 5 trumpets to 5 children. INCLUSION-EXCLUSION PRINCIPLE Let us first look at the problem of counting numbers between 1 and 1000 which are divisible, say, by the factors 5, 7 and 11. Let us adopt the notation that numbers divisible by 5 (and less than or equal to 1000) are in set A, those divisible by 7 are in set B and those divisible by 11 are in set C. If we adopt the notation |X| for the number of members in set X

6Enumeration 6 1. Enumeration

then |A| = 200 (since 5 × 200 = 1000), |B| = 142 (since 7 × 142 = 994) and |C| = 90 (since 11 × 90 = 990). Now if we wish to count the number of numbers divisible by 5, 7 or 11 we can’t just add these numbers to give 200 + 142 + 90 = 432 as some numbers have been counted more than once. We also need to know the values of the intersections, that is, the number of numbers in both A and B, etc. The first number divisible by 5 and 7 is 35, and then they repeat in sequence as 35, 70, 105, . . .. Since 35×28 = 980 we have |A ∩ B| = 28. Similarly 55 × 18 = 990 so |A ∩ C| = 18 while 77 × 12 = 924, giving |B ∩ C| = 12. So we can subtract all of these double counts, giving 432 − 28 − 18 − 12 = 374. But we may have repeated something here also, that is, subtracted numbers divisible by 5, 7 and 11 more than once. In our case the smallest number which is divisible by all three is 5 × 7 × 11 = 385. Thus only this number and 770 are divisible by all three. These numbers were originally counted 3 times instead of once, but then we have subtracted them all three times. They need to be reinstated by adding them back. Thus the final answer is 374 + 2 = 376. This can all be represented by the Venn diagram: .................................... .................................... .......... ....... ....... .......... .... .... .......... ....... .... .... ... ..... .... ..... ... . . . . ... . .. .. .... . ... . . . ... .. .. .. . . .. ... .. .. . . . .. . .. .. . . . . .. .. ..... .... .. ... .. .. ... ... . .. . . . . 156 104 B ..... .... A .... 26 ... . . .. .. . . . . ... ... . . . . . . .. . .............................................. .. .. ......... .... ...... ... .. ..... .... ... 2 .... ... ... .... .... ... .... ... ... . .... . . . ... . . . . . . . ..... 16 .................. 10 ...... .... ... ..... .... ....... .. ........ ... ... ........... ....... ............................................. ................................... . ... .. .. .. .. .. .... . 62 .. ... ... .. . . .. .. .. . .. .. .. .. ... .. ... ... ... ... . .... . .. ..... .... ...... ..... ......... ....... ..............C .....................

624

The Principle of Inclusion-Exclusion for the case of three subsets A, B and C is |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|. There will be variations on this when the number of subsets of the original set is different from three. We now look at a problem which is quite different and lends itself in the first instance to a systematic counting process.

Enumeration7 1. Enumeration 7

Example 3 (Australian Mathematics Competition, 1983) I have four pairs of socks to be hung out side by side on a straight clothes line. The socks in each pair are identical but the pairs themselves have different colours. How many different colour patterns can be made if no sock is allowed to be next to its mate? (A) 792 (B) 630 Solution 3 Alternative 1 Call the socks aa, bb, cc, dd.

(C) 2520

(D) 864

(E) 720

There are 4 × 3 × 2 = 24 ways of selecting the first three as abc (i.e. the first three different). These can be arranged abc

a

b da

db dc

bdcd cdbd dbcd dbdc dcbd dcdb symmetric with a bcd bdc cbd cdb dbc dcb symmetric with da symmetric with da

6 6

6 6 6

30

So there are 24 × 30 = 720 patterns commencing with abc. There are 4 × 3 = 12 ways of selecting the first three as aba.

8Enumeration 8 1. Enumeration

These can be arranged aba

b c

d

cdcd dcdc bdcd dbcd dbdc dcbd dcdb symmetric with c

2

5 5

12

So there are 12 × 12 = 144 patterns commencing with aba. Finally, 720 + 144 = 864,

hence (D). Alternative 2 (Uses Inclusion-Exclusion Principle) This solution has the advantage of generalising the problem to n pairs of socks when n = 4.

Consider the problem for 2 pairs of socks, as illustrated in the Venn diagram.

............................................. .............................................. ......... ........ ....... ....... ....... ....... ..... ..... .... .... .... . . . . . . . . . .... ... .... 2 1...... ...... .... . .. . . . . ... . .. .. .... ... . . . . ... .. ... . .... ... .. .. ... . .. .. ... ... .. .. ... ... .. .. .. .. ... ... . ... .. .... . ... ... .. .. . ... . .. .. . . ... .. .. . . . . . .. ... .. .. .. .. . . . . .. .. .. .. .. .. ... .. ... ... ... .... .. ... .. ... ... ... . . ..... .... ... .... ... ....... ..... .... .... ......... ...... .... .... ...... ........ ...... ...... ........ . . . . . . . . . . . . ............ . . . .............. ................................... .............................

A

A

The set A1 , for instance, is the set of arrangements in which pair 1 is together. We wish to compute N (A1 ∪ A2 ) = N () −

2 i=1

N (Ai ) + N (A1 ∩ A2 ).

Enumeration9 1. Enumeration 9

In this problem

4! . 22 (We divide by 22 because the socks within each pair can be changed without changing the arrangement.) Now 3! N (A1 ) = N (A2 ) = 1 2 and 2! N (A1 ∩ A2 ) = 0 , 2 so the solution is N () =

3! 4! − 2 + 2 = 6 − 6 + 2 = 2. 4 2 The Principle of Inclusion and Exclusion generalises to give, in the case of n pairs n n N = N () − Ai N (Ai ) + N (Ai ∩ Aj ) i=1

i=1

−

i=j=k

i=j

N (Ai ∩ Aj ∩ Ak )

+ . . . + (−1) N n

n

i=1

Ai

.

In the case of 3 pairs, this gives 5! 4! 6! − 3 2 + 3 1 − 3! = 90 − 90 + 36 − 6 = 30. 23 2 2 In the case of 4 pairs, this gives 7! 6! 5! 8! − 4 3 + 6 2 − 4 + 4! = 4 2 2 2 2 =

2520 − 2520 + 1080 − 240 + 24 864,

hence (D). In the case of n pairs this gives n (2n)! n (2n − 1)! n (2n − 2)! n n n! − + + . . . + (−1) . 0 2n 2n−1 2n−2 n 2n 1 2 Note that the first two terms always cancel each other in this particular problem.

10Enumeration 10 1. Enumeration

THE NECKLACE PROBLEM As the final enumeration method we discuss in this chapter, we will address the necklace problem. This method was developed by George Polya in the 1930s, while working on a problem in chemistry. However, like the above two methods, we will be looking at small problems where the counting can be done by intuitive, systematic methods, and then find the generalisation (due to Polya) which enables large problems to be solved. Example 4 In how many ways can I make a necklace with 5 beads if the beads can each be of one of 3 different colours and there is an inexhaustible supply of each colour? Solution 4 We first count the necklaces with all beads having one and the same colour. There exist 3 such necklaces which we denote by type AAAAA. When only two of the colours are used denote them by A and B. Without loss of generality assume there are less beads of colour A than the beads of colour B. We may choose colour A in three ways and then to choose colour B in two ways. Therefore the pair of colours (A, B) can be chosen by 6 ways. For each such pair there exists only one necklace with one bead of colour A and 4 beads of colour B, namely ABBBB; two necklaces with two beads A and three beads B, namely AABBB and ABABB. Therefore there are 6 × 3 = 18 necklaces with beads of only two colours. Finally, suppose all three colours occur and denote them by A, B and C. The triple of colours (A, B, C) can be chosen by 6 ways. Again, without loss of generality we assume that the beads of colour A are not more than the beads of colour B and the beads of colour B are not more than the beads of colour C. Since we have 5 beads we conclude that there is only one bead of colour A. Now we have two possibilities: 1. One bead of colour B and three beads of colour C. We have two such necklaces, namely ABBBC and ABBCB. 2. Two beads of each colour B and C. We have three such necklaces, namely ABCBC, ABCCB and ABBCC. Hence, we found 5 necklaces for each triple (A, B, C) and therefore when all three colours occur, there are 5 × 6 = 30 necklaces altogether. By adding the above results we find that there are 3 + 18 + 30 = 51 necklaces.

Enumeration11 1. Enumeration 11

We now generalize the necklace problem when the number of beads is a prime number. Example 5 Let p be a prime number. In how many ways can I make a necklace with p beads if the beads each have one of n different colours and there is an inexhaustible supply of each colour? Solution 5 As in Example 4 we first count the number of necklaces with all beads having one and the same colour. Clearly, there are n such necklaces. In order to count the number of necklaces with at least two colours, consider a regular p-gon A1A2 . . . Ap . Place a bead in every vertex and colour each bead in one of the given n colours. Since each bead allows one of the given n colours, we have np distinct colourings of this “numbered” necklace. It is clear that some of the colourings give one and the same necklace. For example, for p = 5 and n = 3 (as in Example 4) the two colourings ABABB and ABBAB of the “numbered” necklace actually give one and the same necklace. Thus, every necklace may be obtained by p “numbered” necklaces (some of which may coincide) depending on which bead is number 1. Note that there exist np −n necklaces with at least two colours. Consider one such necklace. We show that this necklace is counted exactly p times as a “numbered” necklace. For suppose there exist two “numbered” necklaces that give the same necklace. That means that for some positive integer t every two beads that are t beads apart always have the same colour. But p is a prime which implies that there exists positive integer k such that kt ≡ 1 (mod p). Consider bead number 1 and denote its colour by A. All beads numbered t + 1, 2t + 1, . . . , kt + 1 have colour A. Note that since kt+1 ≡ 2 (mod p) bead number kt+1 is actually bead number 2. We conclude that beads numbered 1 and 2 have the same colour A. By the same way we prove that any two neighbouring beads have the same colour, i.e. only one colour occurs, a contradiction. p Hence, there exist n p−n necklaces with at least two colours. Therefore the number of all necklaces is given by n+

np − n . p

Note: The above formula proves that for prime p and for any integer n, np − n is divisible by p (known as Fermat’s Little Theorem).

12Enumeration 12 1. Enumeration

Generalisation The case when p is not prime is much more complicated. The results for this can be found on the website of the Australian Mathematics Trust at www.amt.edu.au/polyanecklace.pdf.

Enumeration13 1. Enumeration 13

PROBLEMS Enumeration 1. Consider all 5-digit integers whose digits are among 1, 2 or 3. Find the number of such integers in which the sum of their digits is at least 12. Enumeration 2. Find the number of triples (x1 , x2, x3) of positive integers such that x1 + x2 + x3 = 15. Enumeration 3. Let m and n be positive integers. Find the number of n-tuples (x1 , . . . , xn) of positive integers such that x1 + x2 + · · · + xn = m. Enumeration 4. Consider 8 cubes with edge lengths 1, 2, . . ., 8 respectively. One builds a tower using all eight cubes following the rules: 1. Any of the cubes could be the base of the tower. 2. On top of a cube of edge k another cube would have a maximum edge length of k + 2. Find the number of all such towers. Enumeration 5. Fifteen points are given on a circle. Find the number of nonintersecting broken lines passing through all fifteen points. Enumeration 6. Consider a 2 × n table. Find the number of distinct ways the table can be covered by dominoes. Enumeration 7. There are 2n seats in a row in a cinema. In how many ways could n girls and n boys occupy these seats such that no two girls are neighbours? Enumeration 8. In how many distinct ways can 8 distinguishable checkers be put in the cells of an 8 × 8 table such that there is exactly one checker in every row and in every column?

14Enumeration 14 1. Enumeration

Enumeration 9. The passenger seats in a plane are arranged in a rows of 7 seats each. The seats in every row are labeled by 1, 2, . . ., 7. If there are 449 passengers in the plane, prove that there are two rows having the same seats occupied. Enumeration 10. Consider n points on a circle such that no three chords connecting such points intersect in a common point. All possible chords are drawn. Determine the number of regions of the disc as divided by the chords. Enumeration 11. Consider a square grid with 100 points. Find the number of squares having all their vertices among the given points. Enumeration 12. Let A be alphabet having three letters a, b and c. Find the number of n-letter words over A in which the letter a appears an even number of times. Enumeration 13. Let n be a positive integer. Find the number of sequences a1a2 . . . a2n of 1s and −1s satisfying the inequality 2l ai ≤ 2 i=2k−1

for all k and l, such that 1 ≤ k ≤ l ≤ n.

Enumeration15 1. Enumeration 15

SOLUTIONS Enumeration 1. Since 4 × 2 + 3 = 11 < 12 we have that at least two digits equal 3. The sum of the remaining three digits is at least 6. Any such number is obtained by a permutation of the digits of one of the integers 12333, 13333, 22233, 22333, 23333 and 33333. Let us first count the number of 5-digit numbers with digits 1,2,3,3,3. The digit 1 could be any one of the digits, so there are 5 possible places. After placing the digit 1 there are 4 places for the digit 2. The remaining three positions are occupied by the digit 3. Therefore there exist exactly 5 × 4 = 20 5-digit numbers with digits 1, 2, 3, 3, 3. By the same reasoning one obtains that there exist 5, 10, 10, 5 and 1 integers having digits 13333, 22233, 22333, 23333 and 33333, respectively. Thus the desired number equals 20 + 5 + 10 + 10 + 5 + 1 = 51. Enumeration 2. Direct counting of the triples (x1, x2, x3) for which x1 + x2 + x3 = 15 is possible, but long and tedious. Instead, we present a general approach for solving problems of this type. Consider 15 points in a row an equal distance apart. For any triple (x1 , x2, x3) for which x1 + x2 + x3 = 15 place a star after the first x1 points and another star after the first x1 + x2 points. For example, the triple (4, 3, 8) implies the following distribution of stars.

. . . .. . .. . . . . . . . Conversely, consider two stars placed on some of the fourteen empty spaces between the given fifteen points. Take x1 to be equal to the number of points from the beginning to the first star, x2 to be the number of points between the two stars and x3 to be the number of points after second star. It is clear that x1 + x2 + x3 = 15. Therefore the number of triples (x1 , x2, x3) of positive integers such that x1 + x2 + x3 = 15 equals the number of ways one can place two stars in fourteen positions. Since there are exactly 14.13 14 = = 91 2 2 ways to put two stars in fourteen positions we conclude that the answer is 91.

16Enumeration 16 1. Enumeration

Enumeration 3. We follow the approach from the previous problem. Consider m points in a row an equal distance apart. Every n-tuple (x1 , x2, . . . , xn) of positive integers such that x1 + x2 + · · · + xn = m corresponds to n − 1 stars each placed in one of m − 1 empty spaces in between the given m points. We may choose the stars in m−1 n−1 ways, hence the answer is

m−1 . n−1

Enumeration 4. Denote by S(n) the number of towers satisfying the two rules using n cubes with edges 1, 2, . . ., n. We shall show that if n ≥ 2 then S(n + 1) = 3S(n). First consider a tower of height n. We may add a cube of edge length n + 1 in three positions: below all cubes, above the cube of edge n, or above the cube of edge n − 1 (which exists since n ≥ 2). The new tower satisfies the two conditions. Thus, any tower of height n can be expanded to three towers of height n + 1. Now consider an arbitrary tower of height n + 1. The cube below the cube of edge length n + 1 (if any) according to rule 2 has edge length at least n − 1. Therefore removing the cube of edge n + 1 and keeping all the remaining cubes in the same order will result in a tower of height n satisfying the conditions of the problem. The above observation implies that all towers of height n + 1 are obtained from a tower of height n by adding the cube of edge n + 1 in one of the three possible positions. Therefore S(n + 1) = 3S(n) for all n ≥ 2 and since S(2) = 2 we have S(n) = 3S(n − 1) = 32 Sn−2 = · · · = 3n−2S2 = 2 × 3n−2. Hence S(8) = 2 × 36 = 1458.

Enumeration 5. Denote the points in a clockwise direction by A1, A2 , . . ., A15. We count first the number of nonintersecting broken lines having one end at A1. Let the first segment be A1Ai for some i = 2, 3, . . ., 15. Suppose i = 2 and i = 15. Since the broken line is not intersecting it is passing through all points only from one side of the chord A1 Ai and therefore not all points lie on it. Therefore the first segment is A1 A2 or A1 A15. By the same reasoning there exist two options for the next segment of the broken line. This is

Enumeration17 1. Enumeration 17

true for all consecutive segments but the last one for which there is only one option. Thus, the broken lines having one end at A1 are 213. The same is true for the broken lines having one end at Ai for i = 2, 3, . . ., 15. Since any broken line has two ends and there are 15 points we conclude that all nonintersecting broken lines are exactly 15.213 = 15.212. 2 Enumeration 6. Denote by Tn the answer to the problem, i.e. the number of distinct ways the table 2 × n can be covered by dominoes. It is easy to find that T1 = 1 and T2 = 2. The idea of the following observations is to express Tn in terms of Tk for smaller ks. Consider a 2 × n, n ≥ 3 table and observe that the cell marked by can be covered by a domino in two ways – by a horizontal domino or by a vertical one. ... ...

In the first case the only way to cover the cell below the marked one is to put a horizontal domino below the first one. The remaining part of the table is of dimension 2 × (n − 2) and there exist T (n − 2) ways to cover it. In the second case the remaining part of the table is of dimension 2 × (n − 1) and there are exactly T (n − 1) ways to cover it. We conclude that T (n) = T (n − 1) + T (n − 2). Thus, we find that T1 = 1, T2 = 2 and

Tn = Tn−1 + Tn−2 for n ≥ 3. This is exactly the well known Fibonacci sequence (defined by F0 = F1 = 1 and Fn+1 = Fn +Fn−1). Therefore the answer is T (n) = Fn. Enumeration 7. Number the seats consequently by 1, 2, . . ., 2n. It follows from the condition of the problem that seats 1 and 2 are not both occupied by girls. The same is true for seats 3 and 4, 5 and 6 and so on, for seats 2n − 1 and 2n. Since there are n girls it follows that there is exactly one girl on seats 1 and 2, one girl on seats 3 and 4, and so on, one girl on seats 2n − 1 and 2n.

18Enumeration 18 1. Enumeration

Therefore there is exactly one girl on any pair of seats (2i − 1, 2i) for each i = 1, 2, . . ., n. Call a pair (2i − 1, 2i) left pair if the girl sits on seat 2i − 1 and right pair if the girl sits on the seat 2i. Since no two girls sit next to each other any right pair is followed by another right pair. If there are no right pairs then all pairs are left ones and there is one such distribution. Otherwise consider the leftmost right pair and denote it by (2j − 1, 2j). All pairs on the right of (2j − 1, 2j) are right pairs and all pairs on the left of it are left pairs. Therefore a distribution of all boys and girls is uniquely determined by the leftmost right pair (2j − 1, 2j). Since the possible values for j are j = 1, 2, . . ., n we have n such distributions. Therefore there exist n + 1 ways to fix the seats for the girls and for the boys. The girls now can occupy their seats in n! ways and the same is true for the boys. Finally, we obtain that there exist (n+1)n!n! = n!(n+1)! ways for n girls and n boys to occupy the seats such that no two girls are neighbours. Enumeration 8. Denote by (i, j) for i, j ∈ {1, 2, . . ., 8} the cell in the i-th row and the j-th column of the table. Let us first choose the cells for the checkers. By the condition of the problem there are no two cells in a row or in a column. Therefore the cells for the checkers are (1, j1), (2, j2), . . . , (8, j8) where (j1 , j2, . . . , j8) is a permutation of 1, 2, . . ., 8. Since there are exactly 8! permutations of 1, 2, . . ., 8 we conclude that we may choose the cells for the checkers in exactly 8! ways. The checker for the first cell is any of the 8 checkers, thus there are 8 possible choices. There are 7 choices for the checker in the second cell and so on, and a unique choice for the last checker. Therefore there are 8! distinct ways to place the checkers in the cells. Finally, we conclude that there are (8!)2 distinct ways to place 8 distinguishable checkers in the cells of 8 × 8 table such that there is exactly one checker in every row and in every column. Enumeration 9. Suppose the assertion of the problem is not true. Then we have only one empty row. The rows having only one passenger are at most 71 = 7 and they have at most 7 × 1 = 7 passengers altogether. The rows with two passengers are at most 72 = 21 and they have at most 21 × 2 = 42 passengers altogether. By the same way we conclude that the rows with k, k ≤ 7 passengers are at most k7 and they have at most k k7 passengers altogether. Therefore the number of passengers is at most

Enumeration19 1. Enumeration 19

7 7 7 1. + 2. + · · · + 7. 1 2 7

= 7 + 42 + 105 + 140 + 105 + 42 + 7 = 448 < 449,

a contradiction. Therefore there exist two rows having the same seats occupied. Enumeration 10. Let’s draw the chords one by one and keep record of the number of regions created. We start with no chords and one region (the whole disc). The first chord divides the disc into two regions. With the second chord we have two cases: 1. If it intersects the first one then the number of regions increases by 2. 2. If the second chord does not intersect the first one then the number of regions increases by 1. Observe that in both cases the number of regions increases by t+1 where t is the number of intersecting points of the two chords. The same is true for any chord, namely: the number of regions added by this chord equals t + 1 where t is the number of points of this chord intersecting all those previously drawn. Therefore after the last chord is drawn the number of regions equals 1 (for the initial region) plus the number of all chords (all ones in the sum t + 1 when drawing chords one by one) plus the number of intersecting points. It remains to count the number of chords and the number of intersecting points. Since a chord is uniquely determined by two points we have that the number of chords is n2 . To find the number of intersecting points observe that any 4 points determine exactly one intersecting point (they form a convex quadrilateral and its diagonals meet at a point) and according to the conditions of the problem nopoint is repeated twice. Thus, the number of intersecting points is n4 and the answer to the problem is n n 1+ + . 2 4 Remark. If we try to guess the answer by counting the number of regions for small values of n we see that for n = 1, 2, 3, 4, 5 and the answer is 1, 2, 4, 8, 16 respectively. This leads to the wrong conclusion that the

20Enumeration 20 1. Enumeration

answer of the problem might be 2n−1. The reason for this is that for all n ≤ 5 we have n n n−1 2 = 1+ + . 2 4 Enumeration 11. Note that there exist two types of squares – upright squares, i.e. squares with vertical and horizontal sides and leaning squares. Observe that any leaning square is inscribed in an upright square, meaning that its vertices lie on the sides of this upright square.

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

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

For convenience assign to each point a pair of positive integers (x, y), where x is the number of the row of this point and y is the number of its column. Observe that there are exactly 8 leaning squares inscribed in the big square of 100 points (two of them are given on the figure).

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

Consider a upright square of 81 points. Its lower left vertex could be any of the points (1, 1), (1, 2), (2, 1) or (2, 2), i.e. there exist 4 such squares. For any such square there exist exactly 7 leaning squares inscribed in it. The lower left vertex of a right square of 64 could be any of the points (a, b) for 1 ≤ a ≤ 3 and 1 ≤ b ≤ 3. Therefore there exist exactly 9 such squares. In any of them there are 6 leaning squares. Continuing this way we find that the number of all squares equals 1.9 + 4.8 + 9.7 + 16.6 + 25.5 + 36.4 + 49.3 + 64.2 + 81.1 = 825.

Enumeration21 1. Enumeration 21

Enumeration 12. Alternative 1 n Denote by An 0 (A1 ) the number of words over A of length n in which the letter a appears an even (odd) number of times. There are 3n words of n n length n, hence An 0 + A1 = 3 . A word of length n + 1 and even number of as can be obtained from a word w of length n, by adding one extra letter, in two ways: 1. by writing b or c if w has an even number of as; 2. by writing a if w has an odd number of as. n n n Therefore An+1 = 2An 0 + A1 = 3 + A0 which implies 0

An+1 0

=

3n + An 0

=

3n + 3n−1 + An−1 0 ... 3n + 3n−1 + · · · + 31 + A10 .

= Since A10 = 2 we compute An+1 0

= = =

3n + 3n−1 + . . . + 31 + 2 3n+1 − 1 +1 2 3n+1 + 1 . 2

Alternative 2 First we count the number of words of length n in which the letter a appears n 2k times where 2k ∈ [0, n]. We can fix the places of the letter a in 2k ways. For each of the remaining n − 2k positions we have two n n−2k .2 words of length n having choices – b or c. Therefore there are 2k 2k letters a. It remains to compute the sum n n n n−0 n−2 S= .2 + .2 +...+ .2n−2k + . . . 0 2 2k It follows from Newton’s binomial formula that n n n n−2 n x + x +...+ .2n−2k + . . . . (x − 1)n + (x + 1)n = 2 0 2 2k By setting x = 2 we obtain that n n n .2n−0 + .2n−2 + . . . + .2n−2k + . . . = 2S, 3n + 1 = 2 0 2 2k

22Enumeration 22 1. Enumeration

3n + 1 . 2 Enumeration 13. A sequence for which a2k−1 +a2k = 0 for 1 ≤ k ≤ n satisfies the condition 2l since all the sums of the kind i=2k−1 ai equal 0. There are 2n such sequences. Now, count the sequences for which there exists k such that i.e. S =

a2k−1 + a2k = 0. Let k1, k2, . . . , ks be all ks having the above property. It is clear now that if a2ki −1 +a2ki = 2 (or −2) then a2ki+1−1 +a2ki+1 = −2 (or 2). Therefore all sums a2ki−1 +a2ki (and thus all pairs a2ki−1 a2ki ) are uniquely defined from a2k1−1 + a2k1 (there are two choices for a2k1−1 a2k1 ). There are two choices for any of the remaining n − s pairs (for which a2t−1 + a2t = 0). Therefore the desired number equals: n n−1 n n−2 n 2 + 2.2 + 2.2 + 1 2 n n n−k n + · · · + 2.2 + 2. · · · + 2.2 k n−1 n By adding and subtracting 2n we obtain: n n n 2. 2n + 2n−1 + 2n−2 + 0 1 2 n n ···+ 2 + − 2n n−1 n = 2.3n − 2n . Finally, the answer is 2.3n − 2n .

COMBINATORIAL GAMES

2. COMBINATORIAL GAMES Combinatorial Games

Combinatorial games We consider problems of the following type: We are given a certain structure – piles of stones, candies, numbers, playing cards, table etc. Two players in turn change the structure using only a number of allowed operations. Each of them is trying to reach a certain winning position. The problem is to find (if it exists) a winning strategy for one of the players. One basic and powerful tool for solving such problems is to consider small values of some of the given entries. This gives a reasonable amount of information to make correct conclusions and good insights about the problem. Example 1 Two piles having m and n cards are given. Two players play the following game. Each of them in turn is allowed to perform one of the following moves: 1. remove one card from a pile of his choice; 2. remove one card from each of the two piles; 3. shift a card from one pile to the other. A player who has no move loses the game. Determine in terms of m and n who has a winning strategy. Solution 1 We begin by inspection for small values of m and n. To do so consider a 11×11 table and label the rows and the columns of the table consecutively by the numbers 0, 1, 2, . . ., 10, starting from the bottom row and the leftmost column (Diagram 1).

26 26

CombinatorialGames Games 2. Combinatorial

10 9 8 7 6 5 4

3

2 1 0 0

1

2

3

4

5

6

7

8

9

10

Diagram 1 In this notation, (0, 0) is the left cell in the bottom of the table. The cells (5, 2) and (3, 3) are marked. Choose one of the given piles to be the first, then the other pile is the second. Suppose that at a certain moment there are x cards in the first pile and y cards in the second pile. We will denote the situation of the game at this moment by the pair (x, y) of the numbers of cards in the piles; and, for a better visual representation, we associate with it the respective cell (x, y) in Diagram 1. The player who has the move can perform one of the allowed moves described in the statement of the problem. In terms of the above interpretation he can change the situation in the game from the cell (x, y) to one of the five cells (x−1, y), (x, y −1), (x − 1, y − 1), (x − 1, y + 1), (x + 1, y − 1) (see Diagram 2), but only if x and y are different from 0. (x − 1, y + 1) (x − 1, y)

(x − 1, y − 1)

(x, y) (x, y − 1)

(x + 1, y)

Diagram 2 In the case when x = 0 and y is different than 0 there are two possible moves – to (0, y − 1) and (1, y − 1); in the case when y = 0 and x is

Combinatorial Games27 2. Combinatorial Games 27

different than 0 there are also two possible moves, and in the case when x = y = 0 no move is possible. The last observation gives us a key to the solution of the problem, because the player, who has the move when the game is in the position (0, 0), loses. We label the cell (0, 0) by L, indicating, that it is “losing” for the player who has the move when the game is at that point. Now we can label the neighboring three cells by W as “winning” (see Diagram 3), because the player, who has the move when the situation in the game is represented by one of these cells, clearly has a winning strategy: he can transform the game to cell (0, 0), which is losing for his opponent. 3

3

2

2

L

1

W

W

1

W

W

0

L

W

0

L

W

L

0

1

0

1

2

2

Diagram 3

3

3

Diagram 4

But then clearly the cells (0, 2) and (2, 0) are losing (Diagram 4) for the player who has the move, since all possible moves from these cells lead to cells labelled W , i.e. to positions which are winning for the opponent.

28 28

CombinatorialGames Games 2. Combinatorial

10 9 8 7 6 5 4 3

W

W

2

L

W

1

W

W

W

W

0

L

W

L

W

0

1

2

3

4

5

6

7

8

9

10

Diagram 5 As it was explained above, each of the last two “losing” cells has three “winning” neighbours (Diagram 5). Continuing this construction, we see, that the “L-cells” and the “W -cells” form a specific pattern. This pattern can be described also in the following way: all cells of the form (2p, 2q), where p and q are non-negative integers, are “losing”, and all remaining cells are “winning” (Diagram 6) .

Combinatorial Games29 2. Combinatorial Games 29

10

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

L

9 8 7 6 5 4 3 2 1 0

0

1

2

3

4

5

6

7

8

9

10

Diagram 6 Or, in other words, the second player wins if and only if both numbers m and n are even. We will prove that this assertion is true. Suppose m and n are even numbers, i.e. m = 2m0 and n = 2n0. With respect to the move of the first player and up to equivalence the following distributions may occur: (a) m = 2m0 − 1 and n = 2n0 cards. The second player takes a card from the first pile and it remains m = 2m0 − 2 and n = 2n0 cards respectively. (b) m = 2m0 − 1 and n = 2n0 − 1 cards. The second player takes one card from the two piles and it remains m = 2m0 −2 and n = 2n0 −2 cards. (c) m = 2m0 − 1 and n = 2n0 + 1 cards. The second player takes one card from the two piles and it remains m = 2m0 − 2 and n = 2n0 cards. By the above strategy after two moves, one by each player, the number of cards decreases and the second player always has move. It follows that the first player loses. Assume m and n are not both even. With his first move the first player makes both m and n even and then follows the above strategy.

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CombinatorialGames Games 2. Combinatorial

Example 2 Consider a 2 × 2011 table. Two players in turn place dominoes on it, the first one places only horizontal dominoes and the second one places only vertical dominoes. The dominoes may not overlap. The player who has no legal move loses the game. Solution 2 Denote the player who places horizontal dominoes by A and the player who places vertical ones by B. We prove that A has a winning strategy. We show first that if it is player B’s move on an empty 2 × 4 table then he loses the game. Indeed, observe that no matter where B places his vertical domino then A always has a move. It is easy to see now that irrelevant to the move of A, B has a unique move after which A also has a move. Now the table is covered and player A wins. Consider now a 2 × 2011 table. Let A split the table into 502 tables 2 × 4 and one table 2 × 3. His first move is to put a domino anywhere in the 2 × 3 table. Note that the vertical dominoes of B lie entirely inside one of the small tables. Player A now follows the strategy: Each time B puts a domino in one of the 2 × 4 tables then A places a domino in the same 2 × 4 table. According to the above observation this is always possible. If B puts his vertical domino in the 2 × 3 table then A puts another horizontal domino in this table. Thus the 2 × 3 table is completely covered and no other moves in this table are possible. The described strategy shows that no matter what moves B makes, A always has a move. Thus the last move will belong to A. Therefore the first player has a winning strategy. Example 3 Two piles of a and b stones are given. Two players in turn take a certain number of stones from one pile and move the same number of stones from this pile to the other one. A player who can not play according to that rule loses. Determine in terms of a and b who has a winning strategy. Solution 3 We present two solutions. The first one is based on our main approach, that is, to consider particular values of a and b and draw conclusions for the general case. Alternative 1 Without loss of generality assume that b ≥ a. Denote the pile with a stones by A and that with b stones by B. If at a certain moment there

Combinatorial Games31 2. Combinatorial Games 31

are x stones in A and y stones in B, we will say that the game is in position (x, y), or simply in (x, y). This terminology and notation help to make a suitable visual representation of the rules of the game and the moves of the players. The diagram shows a Cartesian system of coordinates and points with integer coordinates in it. We will assume that the point P = (12, 7) represents the position (12, 7). From this position, the following moves are possible (for the player who has the move): 1) Take 1 stone from A and move 1 stone from A to B: the game passes to position A1(10, 8) with 10 stones in A and 8 stones in B. 2) Take 2 stones from A and move 2 stones from A to B: the game passes to position A2(8, 9) with 8 stones in A and 9 stones in B. 3) Take 3 stones from A and move 3 stones from A to B: the game passes to position A3(6, 10) with 6 stones in A and 10 stones in B. 4) Take 4 stones from A and move 4 stones from A to B: the game passes to position A4(4, 11) with 4 stones in A and 11 stones in B. 5) Take 5 stones from A and move 5 stones from A to B: the game passes to position A5(2, 12) with 2 stones in A and 12 stones in B. 6) Take 6 stones from A and move 6 stones from A to B: the game passes to position A6(0, 13) with 0 stones in A and 13 stones in B. 7) Take 1 stone from B and move 1 stone from B to A: the game passes to position B1 (13, 5) with 13 stones in A and 5 stones in B. 8) Take 2 stones from B and move 2 stones from B to A: the game passes to position B2 (14, 3) with 14 stones in A and 3 stones in B. 9) Take 3 stones from B and move 3 stones from B to A: the game passes to position B3 (15, 1) with 15 stones in A and 1 stone in B. .... ... .. ... ... .... ..A6 (0,13) ............. ... ........... A5 (2,12) ........ ... ....... ........ ... ....... A4 (4,11) ... ....... ........ ... ....... A (6,10) ... ........ 3 ... ....... ........ ... ........ A2 (8,9) ... ....... ........ ... ....... ... ........ A1 (10,8) ........ ... ....... ... ........ ...... P .... .. .. .. ... .. ... .. .. ... ... B (13,5) ... .. 1 .. ... .. ... .. ... .. .. ... ... ... ..B2 (14,3) .... .. .. .. .. ... .. .. B (15,1) ... ... 3 ... ... ... ..........................................................................................................................................................................................................

• • • • • • •

•

•

•

•

•

•

•

•

• • •

•

• • • • • • • • Diagram 1 In other words, the player, who has the move, can transform the game from position/point P to one of the positions/points A1 , A2, A3 , A4 , A5 , •

•

32 32

CombinatorialGames Games 2. Combinatorial

A6 , B1 , B2 , B3 . Of these points the first 6 lie on the line 1 y = − (x − 12) + 7, 2 and their abscissas and ordinates are in the intervals [0, 10] and [8, 13], respectively; the last 3 lie on the line y = −2(x − 12) + 7, and their abscissas and ordinates are in the intervals [13, 16] and [0, 5], respectively. Note also that the abscissas of any two consecutive points from A1 , A2, A3 , A4 , A5, A6 differ by 2 whereas their ordinates differ by 1. For B1 , B2 , B3 the abscissas of any two consecutive points differ by 1 and the ordinates differ by 2. Summarizing the above observations we come to the following conclusion. From a position point P we “move” along one of the two semi-lines passing through P and having slopes −2 and − 12 respectively. Moreover we may reach any point with non-negative coordinates on these two lines. Note that if (i) a = 0, b = 1, i.e. if the game is in position (0, 1); (ii) a = 1, b = 0, i.e. if the game is in position (1, 0); (iii) a = 1, b = 1, i.e. if the game is in position (1, 1); then the first player has no move and he loses. These three positions determine a “losing set” L, consisting of the positions on the three lines with equations y = x + 1, passing through (0, 1); y = x, passing through (1, 1); and y = x − 1, passing through (1, 0). ... .... y=x+1 .. ... .... .... .... y=x .... . .. . .. .... .... ... ... .... . . y=x−1 . . .... . . . .. ... .... .... .... .... .... ... ... ... ... . . . . . . . .... . . . ... ... .. .... ... ... .... ... .... .... ... .... .... .... .... ... . .... . . . . .. . . . .. ... .... .... .... ... ... .. .... .... .... ... ... .... ... .... .... ... . . . . . . .... . . . ... ... ... .. ... ... ... ... .... .... .... ... .... .... ... ... .... ... . . . . . . . . .... . .. ... ... .. .... ... ... .... .... .... .... .... .... .... .. ... .... ... . . . . . ... . . . . ... .... .... ...... ... ... .. ..... .... .... .... .. ... .... .... ..... ... . . . .... ..... .... .. ... ... .... ...................................................................................................................................................................................... .

•

• •

Diagram 2

Combinatorial Games33 2. Combinatorial Games 33

We will show that if a player has the move when the game is in one of the positions of this losing set, then the other player has a winning strategy. This strategy is based on two observations: (a) If the game is in a position of L, then any possible move leads to position outside L. (b) If the game is in a position outside L, then there exists a possible move that leads to position of L.

The two observations follow from the fact that any line of slope −2 or − 12 intersects the set L in exactly one point. It remains to note that the number of stones decreases and eventually we will reach one of the losing positions (0, 1), (1, 0) or (1, 1). Alternative 2 Without loss of generality assume that b ≥ a. We prove that the second player wins if and only if b − a ≤ 1.

Indeed, suppose b = a + i where i = 0 or i = 1. If a = 0 or a = 1 then the first player has no move and he loses, so let a > 1. Assume the first player takes x > 0 stones from one of the piles and moves the same number of x stones to the other one. The stones then become a − 2x and a + i + x or a + x and a + i − 2x. In the first case since a ≥ 2x we have a + i + x ≥ 2x and the second player makes the same move using the other pile and he gets piles with a − x and a + i − x stones. In the second case since a + i ≥ 2x and x ≥ i we have a ≥ 2x − i ≥ x

and the second player again makes the same move using the other pile and he gets a − x and a + i − x stones. In both cases the differences between the stones remains i ≤ 1. Eventually, the number of stones will become 1, 1 or 0, 1 and the first player loses the game. It remains to consider the case when b − a > 1. The first player takes y stones from the second pile and moves y stones to the first pile. He gets piles with a + y and b − 2y stones respectively. We show that it is always possible to choose y such that S = |(a + y) − (b − 2y)| ≤ 1.

Indeed, if b − a = 3k + j for j = −1, 0, 1, then by taking y = k we obtain S = |j| ≤ 1.

Therefore with his first move the first player makes the difference between the stones less than or equal to 1 and, according to the first part of the proof, he wins the game.

34 34

CombinatorialGames Games 2. Combinatorial

PROBLEMS Combinatorial Games 1. A pile of a stones is given. Two players play the following game: each of them in turn takes 1, 2 or 3 stones. The player who takes the last stone wins. Determine, in terms of a, who has a winning strategy. Combinatorial Games 2. Two piles comprising a and b candies are given. Each of two players in turn eats all the candies in one of the piles and splits the other pile into two (not necessarily equal in size) non-empty piles. The player who has no move loses the game. Who has a winning strategy? Combinatorial Games 3. Two piles of M and N stones are given. Two players in turn take an arbitrary number of stones from one of the piles. The player who takes the last stone wins. Who has a winning strategy? Combinatorial Games 4. All cells of a 4 × 4 table are painted white. In turn two players paint a white cell in black. If a player makes a move, after which there is a black square 2 × 2, the player loses. Who has a winning strategy? Combinatorial Games 5. Two piles of a and b stones are given. Two players in turn are allowed to remove all stones from one of the piles and if there is one stone in the second pile the player wins; if not the player splits the stones from this pile into two (not necessarily distinct) piles. Who has a winning strategy? Combinatorial Games 6. A chess rook is placed on the leftmost bottom cell of a rectangular table of m rows and n columns. Two players in turn are allowed to make a move up or right any number of cells. The player who has no legal move loses. Who has a winning strategy? Combinatorial Games 7. There are k candies on the table. Each of two players in turn eats at least one but not more than half of the candies. The player who can not play by this rule loses. Who has a winning strategy?

Combinatorial Games35 2. Combinatorial Games 35

Combinatorial Games 8. There are 2005 candies on the table. Each of two players in turn eats either half of the candies (if the number of candies is 2k + 1 he eats k) or one candy (if it is less than half the number of candies). The first player who can not play by this rule loses. Who has a winning strategy? Combinatorial Games 9. Two players in turn take stones from a pile. The first player takes 1 or 10 stones, while the second one takes m or n stones. The player who can not play according the rules loses. Find all pairs (m, n) for which the first player has a winning strategy whatever the initial number of stones. Combinatorial Games 10. Three piles of a, b and c stones are given. Each of two players in turn takes an arbitrary number of stones from one of the piles. The player who takes the last stone wins. Who has a winning strategy? Combinatorial Games 11. Two players in turn write S or O in an empty cell of 1 × 2000 table. The player who first achieves S O S wins. Prove that the player who plays second has a winning strategy. Combinatorial Games 12. Number 2 is written on the blackboard. Two players in turn choose two numbers from the blackboard (not necessarily distinct) and write on the blackboard their sum or their product. Moreover a player is not allowed to write a number that is already on the blackboard and all numbers should be less than 2007. A player who first writes 2006 is declared a winner. Determine who has a winning strategy – the player having the first move or his opponent. Combinatorial Games 13. A checker is placed on the number 2. Two players in turn move the checker rightwards at most 1000 units at their discretion each time. The player who places the checker on a composite number loses the game. Who has a winning strategy?

36 36

CombinatorialGames Games 2. Combinatorial

SOLUTIONS Combinatorial Games 1. Suppose a is divisible by 4, i.e. a = 4k. The second player has the following symmetric winning strategy. Whenever the first player takes x, x ∈ {1, 2, 3} stones he takes 4 − x stones. Since x ∈ {1, 2, 3} implies 4 − x ∈ {1, 2, 3} his move is possible. After one move of the first player and one move of the second one the number of stones is reduced by 4 and therefore after k moves there are no stones left, so the second player wins. When a is not divisible by 4, i.e. a = 4k + i for 0 < i ≤ 3 then the first player wins. He takes i stones and then follows the above strategy. Combinatorial Games 2. We start by considering small values of the number of candies. It follows from the condition of the problem that a player who tries to split a pile of 1 candy loses. Therefore a player who splits a pile of two candies wins. There is only one way to split a pile of 3 candies – one pile of 1 and one pile of 2. The next player now will choose to eat 1 candy and to split the pile of 2 and will win. Thus, a player who splits a pile of 3 candies loses. A pile of 4 can be divided into two piles of 1 and 3 candies which implies that a player who divides a pile of 4 wins. We come to an assertion that if a player splits a pile with an odd number of candies he loses and when he splits a pile of an even number of candies he has a strategy to win. We prove the above assertion by induction on the number of candies. 1. For 1 or 2 candies the assertion is obvious. 2. Suppose it is true for any pile of k candies for k ≤ 2n. 3. We prove the assertion for piles with 2n + 1 and 2n + 2 candies. Assume that a player splits a pile of 2n + 1 candies. As a result he gets a pile of an even number of candies and this number is not greater than 2n. The next player may choose to divide exactly this pile and by the induction hypothesis he can win. So, the first player loses. Suppose a player divides a pile of 2n + 2 candies into two piles having 1 and 2n − 1 candies each. No matter the choice of the next player he has to divide a pile of odd number of candies and this number is less that 2n. By the induction hypothesis he loses.

Combinatorial Games37 2. Combinatorial Games 37

Return to the main problem. If both a and b are odd then the first player has to divide one of them. According to our assertion he loses. If at least one of a and b is even then the first player may choose to split the pile with an even number of candies and therefore he has a winning strategy. Combinatorial Games 3. Suppose first that M = N = 1. The first player takes one of the stones and loses. If M = 1, N = 2, the first player takes one of the stones from the second pile, this puts the second player in the above unpleasant position and now clearly the second player loses. The first player could repeat the same idea also in the case M = 1 and where N is an arbitrary integer greater than 1: he can take N − 1 stones from the second pile and this puts the second player in a losing position. The above reasoning generalizes in the following way: Suppose first that M = N . If the first player takes x stones from one of the piles then the second player takes the same number of stones from the other pile. The resulting piles will have again an equal number of stones. Using this symmetric strategy the second player wins. If M = N then the first player with his first move makes the stones in the two piles equal and then follows the above symmetric strategy. Combinatorial Games 4. Write a number in every cell of the table as shown in the figure. 5

6

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8

1

2

3

4

5

6

7

8

1

2

3

4

After any move by the first player the second one paints the cell having the same number. It is easy to see that if a black 2 × 2 square occurs after a move by the second player then such a square exists before the move. Therefore the second player always wins. Combinatorial Games 5. We try first to solve the problem for small values of a and b, say a ≤ 10 and b ≤ 10 and then to infer a proper assumption.

Consider a 10 × 10 table. The cell in the ith column and jth row represents a game with two piles with i and j stones. Call such a cell winning if the player who has the turn wins, and losing otherwise. Write W in

38 38

CombinatorialGames Games 2. Combinatorial

a cell that is winning and L in a cell that is losing. According to the condition of the problem all cells (1, j) and (i, 1) are winning cells. So we begin with Diagram 1. 10

W

9

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1

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1

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5

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10

Diagram 1 In order to fill the table further we need to know which cells are accessible from the cell (i, j). It follows from the condition of the problem that from (i, j) the game “jumps” to a cell (x, y) if and only if x+y = i or x+y = j. Therefore we fill the table using the following rules. If all cells (x, y) for which x + y = i or x + y = j are winning cells then the cell (i, j) is losing and we write L in it. If there is at least one losing cell (x, y) with x + y = i or x + y = j then the cell (i, j) is winning. Note also that the table is symmetric with respect to the main diagonal. According to the above observations we easily find that: 1. (2, 2), (2, 3), (3, 2) and (3, 3) are all losing; 2. All cells (x, y) for which x = 4, 5 or 6, or y = 4, 5 or 6 are winning. We obtain Diagram 2.

Combinatorial Games39 2. Combinatorial Games 39

10

W

W

W

W

9

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8

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6

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4

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3

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L

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2

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L

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1

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1

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Diagram 2 It is easy to see now that all cells (x, y) for which x, y ∈ {2, 3, 7, 8} are losing. Finally, all remaining cells become winning. We now have Diagram 3. 10

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7

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L

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L

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6

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5

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4

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3

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L

L

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L

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2

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1

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Diagram 3 Using Diagram 3 we have to describe all losing cells. We conclude that

40 40

CombinatorialGames Games 2. Combinatorial

a cell (i, j) is losing if and only if i = 2 or i = 3 modulo 5 and j = 2 or j = 3 modulo 5. We prove that this is true in general. Call all numbers of the form 5k ±2 good and let all remaining numbers be bad. The basic observation is that the sum of two good numbers is bad, and any bad number can be expressed as a sum of two good numbers. We prove that the second player wins if and only if both a and b are good. Suppose the number of stones in one of the piles is bad. Without loss of generality let this be a. Then a = 5k, a = 5k + 1 or a = 5k + 4. The first player removes the stones from the other pile and splits a into two piles having 2 and 5(k − 1) + 3 stones in the first case; 3 and 5(k − 1) + 3 stones in the second case (if k = 0 the player wins); 2 and 5k + 2 stones in the third case. Thus the number of stones in both piles is good. Now, assume both a and b are good. Since the sum of two good numbers is bad the player having the move leaves at least one bad number. Therefore if a and b are good after the move of the first player there is at least one bad number and the second player can make the number of stones in both piles good. Following this line eventually the number of stones becomes 2 or 3 and then the first player leaves a pile of 1 stone and he loses. When a or b is not good the first player splits the pile having a bad number of stones into two piles each with a good number of stones and wins. Combinatorial Games 6. Alternative 1 To analyse the game, we label the rows and the columns of the table consecutively with the numbers 1, 2, 3, . . . starting from the top row and the right column, as shown in Diagram 1. In accordance with this notation, (1, 1) is the right cell at the top of the table; in Diagram 1, the cells (2, 5) and (3, 3) are marked by X.

Combinatorial Games41 2. Combinatorial Games 41

7

6

5

4

3

2

1 1

X

2

X

3 4 5 6 7

Diagram 1 If at some moment of the game the rook is in, say, (2, 5), then the player who has the move can place it in one of the cells to the right (in the same row) – (2, 4), (2, 3), (2, 2), (2, 1), or up (in the same column) – (1, 5). Suppose that after a certain number of moves the rook is in the top-right corner (1, 1): this is the only cell in the table from which there is no move, and hence the player, who has the move, loses. We label this cell by the letter L (see Diagram 2). 7

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7

Diagram 2 Now it is easy to see, that if the rook is in any cell from the first row except for (1, 1), the player who has the move, has a winning strategy: he can move the rook to (1, 1), and his opponent loses. Therefore we label the cells in the first row to the left of (1, 1) by W (Diagram 2). The same reasoning can be applied to the cells of the first column: the result is shown in Diagram 3. Suppose that the rook is in (2, 2). The possible moves from this cell are to (2, 1) or to (1, 2); both these two positions

42 42

CombinatorialGames Games 2. Combinatorial

are winning for the next move; therefore (2, 2) is a losing position for the player who has the move and we label it by L (Diagram 3). 7

6

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1

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Diagram 3 As above, the cells below and to the left of (2, 2) are winning, because from any of them the rook can be moved to the losing cell (2, 2). We label these cells by W (Diagram 3). Continuing in the same manner, we obtain a visual picture of the distribution of the losing and the winning cells in the table; it is shown in Diagram 4. The conclusion from Diagram 4 is that if the rook is in any cell of the form (k, k) for a certain positive integer k, then there is no winning strategy for the player who has the move. All these cells form a diagonal of the given table, starting from the top-right cell and going to the left and down. In particular, if the dimensions m and n of the given table are equal (i.e. if m = n), then the starting position of the rook lies on this diagonal, and hence there is no winning strategy for the first player (who has the first move). In other words, there is a winning strategy for the second player.

Combinatorial Games43 2. Combinatorial Games 43

7

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7

Diagram 4 But if the initial position of the rook is not on the diagonal described above, i.e. when m and n are different, then there is a winning strategy for the first player (who has the first move): he can simply move the rook to a cell of the diagonal, thus putting it in a losing cell. Alternative 2 Note that the only cell from which there is no move is the upper-right corner of the table. First let m = n. At the beginning the rook is placed on the diagonal of the n × n square. After the move of the first player the rook is not on the diagonal. With the next move the second player can return the rook to the diagonal. Eventually the rook will reach the upper-right corner after a move of the second player. Thus when m = n the second player wins. Let m = n and assume for clarity that m > n. The first player can move the rook m − n cells up. Thus, he reduces the game to the case of a square n × n table when he plays second. Therefore in this case the first player has a winning strategy. Combinatorial Games 7. It is natural to start with cases of the game for small values of k. Denote by A and B the players, and let X be either of them. If k = 1 and X has the move, then clearly X loses, since he cannot make a “legal” move, satisfying the rules of the game. If k = 2 and X has the move, this move is uniquely determined, i.e. X eats one candy. Then the other player (say, Y ) has his move when there is one candy on the table and loses (see above); hence X wins. If k = 3 and X has the move, this move again is uniquely determined: X eats one candy. The other player Y is in the above situation – with two candies on the table, as it is described above, he wins; hence X loses.

44 44

CombinatorialGames Games 2. Combinatorial

For k > 3 the player with the first move – X – has a choice: to eat 1, or 2, or 3, . . . or [n/2] candies (here [x] denotes the greatest integer not exceeding x). In particular for k = 4, X has a choice: to eat one candy or to eat two candies. So he can leave either 3 or 2 candies on the table after his move. But if for the next move – Y has 3 candies on the table, he loses, as it is established above, and if for the next move – Y has 2 candies on the table, he wins. Therefore in the case with k = 4 there is a winning strategy for the first player X: the first move of this strategy is to eat one candy. If k = 5, the player with the first move – X – has a choice: to eat 1 or 2 candies, thus leaving 4 or 3 candies on the table for Y ’s move. But, as it is established above, if for the next move Y has – 3 candies on the table, he loses, – 4 candies on the table, he has a winning strategy. Therefore in the case with k = 5 there is a winning strategy for the first player X: the first move of this strategy is to eat two candies. Continuing this analysis of the strategies of the players, it is easy to find that the player X (who has the first move) has a winning strategy for k = 2, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, and that the player Y (who has the second move) has a winning strategy for k = 1, 3, 7, 15. These results lead to the hypothesis that Y has a winning strategy if k is of the form 2n − 1, where n is a positive integer, and that X has a winning strategy in all other cases. We prove by induction on n that if the number of candies on the table equals 2n − 1 then the second player has a winning strategy. For n = 1 the assertion is clear. Suppose it is true for some positive integer n and consider 2n+1 − 1 candies on the table. Playing by the rules of the game the first player may leave p candies for any p ∈ [2n; 2n+1 − 2]. In all cases the second player may leave exactly 2n − 1 candies and by the induction hypothesis will win. Thus, the assertion is proved. Suppose k = 2n − 1. Then there exists a positive integer m such that k ∈ [2m ; 2m+1 − 2]. As above the first player may leave on the table 2m − 1 candies and according to the above assertion win. Combinatorial Games 8. We shall prove first that if there is an even number of candies on the table, the player who has the move wins. Suppose there are 2n candies and denote the player who has the move by A and the other one by B.

Combinatorial Games45 2. Combinatorial Games 45

If the player who has the move when n candies are on the table loses, then A eats n candies and wins. Consider the opposite case – the player who has the move when n candies are on the table, wins. Player A eats 1 candy and leaves 2n − 1 candies. If B is to eat 1 candy he loses by induction. If B eats n − 1 candies he leaves n candies and it is A’s move. According to the case considered A wins. For 2005 candies the first player wins. First he eats 1002 candies leaving 1003 on the table. The second player can eat 1 or 501 candies in both cases leaving an even number of candies. Therefore the first player wins. Comment: Note that if the number of candies is 4k + 3 the first player loses the game since he is allowed to eat 1 or 2k + 1 and in both cases there is even number of candies left. Now consider the general case for number of candies 2 k + 1 where k is odd. The answer depends on the parity of . If is odd then the first player loses the game; otherwise (for even) the first player wins. To prove the above assertion note that for 2 k + 1 candies the first player is forced to eat 2(−1) k candies (otherwise he eats one candy and loses the game because of even arguments). Then there are 2(−1)k +1 candies left and so the second player is also forced to eat 2(−2)k candies. And so on, they eventually reach 2k + 1 candies and the player having the turn loses since he has to eat 1 or k in both cases leaving even number of candies (note that k is odd). Combinatorial Games 9. A standard observation for small values of m and n leads to the following conclusion. The first player has a winning strategy for a pair (m, n) if and only if m ≥ 9, n ≥ 9, |m − n| = 9. We prove that this is indeed true. Consider a pair (m, n) for which the first player always wins and assume m < 9. For a pile with m + 1 stones the first player is forced (since m + 1 < 10) to take 1 stone. The second player takes m stones and wins: contradiction. Therefore m ≥ 9 and in the same way n ≥ 9. If m − n = 9 consider a pile of m + 1 = n + 10 stones. If the first player takes 1 or 10 stones then the second one takes m or n stones respectively and wins: contradiction. Therefore m−n = 9 and by analogy n−m = 9, i.e. |m − n| = 9. It remains to prove that for all pairs (m, n) for which m ≥ 9, n ≥ 9 and |m − n| = 9 the first player wins. It suffices to show that for any number of stones the first player can make a move according to the rules such that the remaining stones are not m or n.

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CombinatorialGames Games 2. Combinatorial

Denote the number of stones by k. If k ≤ 10, the first player wins by taking 10 stones when k = 10, and by taking 1 stone when k ≤ 9.

If k > 10, then at least one of the numbers k −1 and k −10 is not equal to m or n, otherwise |m − n| = 9, a contradiction. Therefore the first player can make a move such that the second player can not win with his next move. Since the number of stones decreases the first player eventually wins. Combinatorial Games 10. Denote the first player by A and his opponent by B. Without loss of generality assume a ≤ b ≤ c. The approach used so far is limited since we need a three-dimensional grid to represent the game. First we analyze the game by some general observations. When a = 0 we are left with two piles and according to the solution of Combinatorial games 3 the first player wins only if b = c. Suppose now that a = 0. If a = b or b = c then A wins by converting the game to a two-pile game with an equal number of stones. Therefore 1. if a = 0 and b = c the winner is A;

2. if a = 0 and b = c the winner is B; 3. if a = 0 and a = b or b = c the winner is A.

Thus we may assume that a < b < c. By standard reasoning we complete the following table. a

b

c

winner

1

2

3

B

1

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>3

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3

>3

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1

4

5

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4

>5

A

1

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>5

A

2

3

>3

A

2

4

6

B

2

4

>6

A

2

5

7

B

2

5

= 7

A

Combinatorial Games47 2. Combinatorial Games 47

This table may easily lead to the conclusion that each time when B wins we have a + b = c. What about a = 3, b = 5 and c = 6? Using the results from the table above it is easy to see that whatever move A makes player B wins. It turns out that the answer to the problem depends on the binary representation of a, b and c. Represent a, b and c in base 2. If necessary we may write 0s in front of the two smallest numbers to match the number of digits of the largest one. Therefore we may assume that a = anan−1 . . . a1 , b = bnbn−1 . . . b1 , c = cn cn−1 . . . c1 where each of ai , bi and ci equals 0 or 1. We shall prove that if ai + bi + ci is an even number for all i = 1, 2, . . ., n then B has a winning strategy. Without loss of generality assume the first player takes stones from the pile of a stones. Let the number of remaining stones be d. Since d = a there exists at least one i for which ai = di. For this i we have that di + bi + ci is odd. Denote by X the set of all i, i ∈ {1, 2, . . ., n} for which di + bi + ci is odd and let j = max{i}. i∈X

Since dj +bj +cj is odd at least one of dj , bj and cj equals 1. Without loss of generality assume bj = 1. The aim of the second player is to restore the parity condition. Consider the number e = en en−1 . . . e1 obtained from b by changing all bi for all i ∈ X from 1 to 0 and vice versa. Since the first binary digit in which e and b differ is in position j and ej = 0, bj = 1, we have that e < b. On the other hand, ei + bi + ci has the same parity as di + bi + ci for i ∈ X and different parity from di + bi + ci for i ∈ X. Thus, ei + bi + ci is even for all i. Therefore the second player takes b − e > 0 stones from the pile of b stones and restores the parity condition. Following the described strategy the second player always has a move and therefore he wins. It is clear from the above that when there exists i for which ai + bi + ci is odd then the first player wins by restoring the parity condition. Combinatorial Games 11. During the game denote the player who has the move by A. If A is able to make S

O

S

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CombinatorialGames Games 2. Combinatorial

he wins. Suppose this is not the case and consider the empty cells. Assume there exist three empty adjacent cells.

Now A puts O in the middle cell to obtain O and the other player can not win in his next move. Assume there is an isolated empty cell, i.e. a cell that is either one of the end cells or a cell both neighbours of which are filled by S or O. Then A writes O in this cell and again the other player can not win in his next move. Note that there are 2000 empty cells at the beginning of the game. Therefore when the second player has a turn there is an odd number of empty cells. Consider all blocks of consecutive empty cells. Since the total number of empty cells is odd, there exists a block of odd length. Therefore one of the two described above situations must occur and the second player can not lose. On the other hand with his first two moves the second player can always S S make . Indeed, after the first move of the first player there are 11 (actually many more) adjacent empty cells. The second player writes S in the middle cell of these 11 empty cells. With his next move he either wins or makes S S such that the first player can not win with his next move. After that the second player makes moves in a way not to lose. Thus, the first player either loses the game or has to fill one of the empty cells in S S . It is easy to see that after that the second player always wins. Combinatorial Games 12. We prove that the second player has a winning strategy. First note that only even numbers appear on the blackboard. Moreover a number of the form 4k + 2 is obtained only as a sum of two numbers. Hence 2006 can appear on the blackboard only as a sum of two numbers. Consider the pairs (4k, 4k + 2) for k = 1, 2, . . ., 500, i.e. the pairs (4, 6), (8, 10), . . ., (2000, 2002). Note that if a + b = 2006 then a and b are not in one and the same pair. Moreover since 2006 is of the form 4k + 2 then a and b are not congruent modulo 4. Therefore if a is paired with c and b is paired with d then c + d = a + b = 2006.

Combinatorial Games49 2. Combinatorial Games 49

Suppose the first player writes number x. If the second player can get 2006 he does so and wins. Assume the second player can not get 2006. We show that he can write the number that is paired with x. If x = 4m then the second player simply adds 2 to it and gets 4m + 2 which is in parity with x. If x = 4m + 2 then x = a + b and without loss of generality a = 4p + 2. According to the strategy 4p is also on the blackboard and therefore the second player may write 4m = 4p + b. According to the above observation the first player can not win using the number paired with x. Combinatorial Games 13. The first question that needs answering is whether the game always has a winner. To show that this is the case we find 1000 consecutive composite numbers. Consider the following 1000 consecutive numbers 1001! + 2, 1001! + 3, . . . , 1001! + 1001. Note that each of these numbers is composite, since the first one is divisible by 2, the second one by 3, and so on. Therefore one of the players will not be able to skip these 1000 composite numbers. Hence the game will have a winner. We show that the first player has a winning strategy. Call a number lucky if the player who has the move when the checker is on this number wins the game. We have to prove that 2 is a lucky number. Assume first that 3 is a lucky number. Therefore there exists a prime number p ≤ 1003 such that when the checker is placed on p then the player who has the move loses the game. Note that 1003 = 17. × 59 is not a prime number, so p ≤ 1002. This implies that 2 is also a lucky number since the checker can be moved from 2 to p. Assume now that 3 is not a lucky number. Then obviously 2 is a lucky one since 3 is a prime and it is enough to move the checker from 2 to 3. We proved that no matter whether 3 is lucky or not 2 is always a lucky number. Therefore the first player has a winning strategy. Remark The solution does not give the exact strategy for winning but rather shows that it exists.

SUM OF SEGMENTS

3. SUM OF SEGMENTS Sum of Segments

When we say that the sum of two segments AB and CD is equal to a segment EF , we usually mean that the sum of the length of AB and CD equals the length of EF . If we wish to avoid the notion of length (and thus to remain close to the basic notions of geometry), we can also mean that on EF there exists a point G, such that EG is equal to AB and GF is equal to CD. From the “sum of two segments” it is easy to move on to the “sum of n segments”, where n is an arbitrary positive integer. But even the simplest case, when n = 2, various interesting problems demonstrate the essence of these ideas. Here we expose to the readers attention some of them in combination with some of the most useful methods for attacking them. Example 1 A point M lies on the circumcircle of the equilateral triangle ABC. Prove that one of the segments M A, M B and M C equals the sum of the other two. Solution 1

Suppose M is a point on a smaller arc BC. We show that M A = M B + M C. C

•

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

M

•

•

P

•

A

•

B

Let P be a point on AM such that M P = M C. Since CM P = CBA = 60◦ and M C = M P we have that CM P is equilateral. Therefore CP = CM . Further, it follows from ACB = P CM that M CB = P CA. Now CM B ∼ CP A (CB = CA, CM = CP and M CB = P CA) and thus, M B = P A. Hence M A = M P + P A = M C + M B.

54 54

SumofofSegments Segments 3. Sum

Alternative Solution Since ABM C is cyclic it follows by Ptolemy’s Theorem that M A.BC = M B.AC + M C.AB, which (since BC = AC = AB) implies M A = M B + M C. Example 2 Let AA1 and BB1 be angular bisectors of ABC. Let M be any point on the segment A1B1 and let X, Y and Z be the feet of the perpendiculars from M to AB, BC and CA, respectively. Prove that M X = M Y + M Z. Solution 2 Denote the feet of the perpendiculars from A1 to AB and AC by A2 and A3 , respectively and the feet of the perpendiculars from B1 to AB and BC by B2 and B3 , respectively. C

.... ... .. .... ... ... .. . . . .. .. .. ... ... .. .. .... . . .. .. . . .. .. A3........... .. .... . . .. .... .. . . .. . .... .. . . . . . .... . ........B3 . .. . . . . . . . . . . . . . .. .............. . . . . ... . . . . . . . . . . . Z............ ............................ .........................................Y .... .. .... .. B1............................................................................................. . .. . ........ . . .. ... ... ......... .... M .......................................................... ....... ... .... A .... .... ............... ..... 1 ......... . . . ... . . . . . . . .... .. ..... .............. ... .... ........ .... .... ... .. P .... ............... .............. ........ . .... . . ... ........ ... ... ... .. ... ... . . ............................ ........ ... . . .... ... . . .... ........ .... ........ . ... . . ... .. . . . . . . . . . . . .......... .. ............. .. . ... ... . . . . . . . . . .... .... ... .. ........ . . ... . ... .. . . . . . . . . . . . . . . .... .... . ... ..... .. .. . . . . . . . . . . . . . . . . . . .... ..... .... ... ... ..... ... ..... .. . . . . . . . . . . . . . . .... .... ... .. .... . Q....... .. . . . . . . . . . . . . .... ........ ... ... .. ... . .. ............. . . . . . . . . . . .... .. ..... .. . ... .... ........ .... ... .. .. .. ............ ................................................................................................................................................................................................... A2 B A B2 X

Since AA1 and BB1 are angular bisectors we have that A1A2 = A1 A3 and B1 B2 = B1 B3 . Let P = M X ∩ B1 A2 and Q = M X ∩ A1 B2 . Since M P A1A2 and M Z A1 A3 we obtain MP B1 M MZ = = , A1 A2 B1 A2 A1 A3 and thus M P = M Z. Further, A2 P A1 M MY PX = = = , B1 B2 A2 B1 A1 B1 B1 B3 and thus P X = M Y . Therefore M X = M P + P X = M Z + M Y .

Sum of Segments55 3. Sum of Segments 55

Example 3 A square sheet of paper ABCD is folded in such a way that D is moved to point D on BC (see diagram). A denotes the position of A after the folding, E is the intersecting point of AB and AD , and r is the inradius of triangle BD E. Prove that ED + r = AB. Solution 3 Let AB = a. The symmetry s with respect to the line of folding (denoted by l in the diagram) is defined by the conditions s : A −→ A and s : D −→ D . Let A B C D be the image of ABCD after applying s; its side also equals a. ........................................................ ............ ......... ......... ........ ....... . ...... . . . . . .... .... .... .... . .... .... . . .... .. ... . . ..... .. . .... . .. . . ... ... .. .. ..... . .. ... . .. . . ... . .. .. .. .. .. ... .. .. .. ... . .. . . . . .. . C.. . . .. . . ... .. . .. ......... . ... . . .. ..... . . .. .. . ... . . ..... .. .. .. ..... . .. . . ...... ... . .... l . D .............................................................................................................. .. . . . . C ... ...... .. ......... . . ... ...... ... .. ... ... .... ..... .. ... ...... ... ... ...... .. .. ..... ...... ... .. ... . . . ..... .. .. ... .. ......... ... ... .. . .. .... .. ... ... .. .. .... .. D .. .. ... ... ...... ..... .. . . . . . . . . . . .. ... .. .. . . . . .. .. . . . . . ... .. ... ... .... ..... ... ... ... ... ... ... ... ... ... ... ... ... ... B ............ .. ... . . . . . . . . . .... ...... ... . ........................ .... . . . .... . ..... . . . . ... .. ........ .. .... ..... ... . . . . . ..... ..... ... .... .. ... ...... . . . ...... . ..... . ...... .. .... .. .. .. .......... .. .. ........ ...... ... .... . .. .... ... ......... ... ................ .......... .. .. . . . .. ............. .. ................................................................................................................................................................. . ... ... ......... .... E A B ..... .. ... .. ... ...

A

Since D is at a distance a from AD, then by symmetry D is also at a distance a from A D . This means that D is at equal distances from the lines D E, EB and BD , i.e. D is the centre of the exscribed circle of triangle BD E, touching the side D E, and the lines BD and BE at C and A, respectively. Hence the perimeter of triangle BD E is equal to AB + BC = 2a, and therefore BE + BD + ED = 2a, BE + BD = 2a − ED .

(1)

Since triangle BD E is right angled ( B = 90◦), r = 12 (BE+BD −ED ), which together with (1) leads to r = 12 (2a − 2ED ), i.e. ED + r = a = AB.

56 56

SumofofSegments Segments 3. Sum

PROBLEMS Sum of Segments 1. Let ABC be an isosceles triangle, ACB = 90◦ . A point M lies on the circumcircle of ABC. A point P from the segment AM is such that CP ⊥ M A. Prove that: (a) If M and C lie in one and the same semiplane with respect to AB, then AP = BM + M P . (b) If M and C lie in different semiplanes with respect to AB, then M P = AP + M B. Sum of Segments 2. Let ABCD be a square and M and N be points on BC and CD respectively, such that M AN = 45◦ . Prove that M N = BM + DN . Sum of Segments 3. Let AA1 , A1 ∈ BC and BB1 , B1 ∈ AC be interior and external bisectors, respectively, of triangle ABC with C > A. Let M be an arbitrary point of the segment A1 B1 and let X, Y and Z be the feet of the perpendiculars from M to AB, BC and CA, respectively. Prove that M X = M Y + M Z. Is this result true if C < A? Sum of Segments 4. Let AA1 and BB1 be angular bisectors of ABC. Let M ∈ A1 B1 be an arbitrary point and let X ∈ AB, Y ∈ BC and Z ∈ AC be such that AXM = BY M = CZM . Prove that M X = M Y + M Z. Sum of Segments 5. Let ABCD be a rectangle and E and F be points on the segments BC and DC respectively, such that DAF = F AE. Show that ABCD is a square if and only if DF + BE = AE. Sum of Segments 6. (Journal The Mathematical Visitor which existed from 1877 to 1896) Let ABCD be a convex quadrilateral and let M be the midpoint of the side CD. Lines BM and AM are perpendicular. Prove that lines BC and AD are parallel if and only if AB = BC + AD.

Sum of Segments57 3. Sum of Segments 57

Sum of Segments 7. Consider an isosceles triangle ABC, AC = BC. Find the number of points M on the circumcircle of ABC such that one of the segments M A, M B and M C equals the sum of the other two. Sum of Segments 8. (Romanian Olympiad) Denote the lengths of the sides of ABC by BC = a, AC = b and AB = c and suppose a > b > c. Determine the number of points M on the circumcircle of ABC such that one of the segments M A, M B and M C equals the sum of the other two. Sum of Segments 9. (Romanian Olympiad) Let AA1 , BB1 and CC1 be angular bisectors of ABC. Let A2 , B2 and C2 be the second intersecting points of BC, AC and AB with the circumcircle of A1B1 C1 . Prove that one of the segments A1 A2 , B1 B2 and C1C2 equals the sum of the other two. Sum of Segments 10. (Romanian Olympiad) Let ABCD be a rhombus and M , N and P be interior points on the sides AB, BC and CD respectively. Show that the centroid of the triangle M N P belongs to the line AC if and only if AM + DP = BN . Sum of Segments 11. (Bulgarian competition (Kazanlyk 2007)) Consider a triangle ABC, such that BC > AB > AC and cos A + cos B + cos C =

11 . 8

Points X ∈ BC and Y ∈ AC → are such that BX = AY = c. (a) Prove that XY =

AB . 2

(b) Point Z from the arc AB of the circumcircle of ABC not containing C is such that ZC = ZA + ZB. Find the ratio

ZC . XC + Y C

58 58

SumofofSegments Segments 3. Sum

Sum of Segments 12. (Urquhart’s Theorem; given also in the Iranian Olympiad) Let ABCD be a convex quadrilateral. The lines AB and CD intersect at P (C lies between D and P ) and the lines AD and BC intersect at Q (C lies between B and Q). Prove that if AB + BC = AD + DC then AP + P C = AQ + QC. Sum of Segments 13. (Chinese Olympiad) Let I be the incentre of triangle ABC and let AI meet the circumcircle of ABC at D. Denote the feet of the perpendiculars from I to BD and CD by E and F , respectively. AD If IE + IF = , calculate BAC. 2

Sum of Segments59 3. Sum of Segments 59

SOLUTIONS Sum of Segments 1. (a) Let N ∈ AP be such that AN = BM . C

•

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

•

.•

•M

P

N

•

A

•

B

Now CAN ≡ CBM (CA = CB, AN = BM and CAN = CBM ). Therefore CN = CM and since CP ⊥ N M we conclude that N P = M P . Thus AP = AN + N P = BM + M P . (b) Hint: Take point Q ∈ AM , M Q = M B and consider M BC and M QC. Sum of Segments 2. Alternative 1 Let R be the symmetric point of B with respect to AM . D N C ...........................................................................................................................................

... ..... .... ... .. ... .. ... .. ..... ... ... .. ... . . .... ... ... . . .. . ... ... . . ... . ... ... .. ... . . ... ... .. ... . ... ... . .. . ... . . ... ... . . . ... ... ... . . . . ... ... ... .. ... ... . ... .. ... . .. . . ............ R ... .. ... . . ............... ... .. . . . . . .... ... ... ... . . .. . . . . ... .... ... .. ...... ... ... ... .. ...... . . . . . . . . ... .. ... ... ... ...... .. ...... ...... ... ... . ...... . .......... M .. ... . . . . . . . . . . . . .... ......... . ... . .. . . . . . . . . . . . . .... . . .... ... ........ . . . . . . . . . . . . . ... . . . . . . .. ... ... ............ ...... ... ... .. ............ ..... . . . . . . . . . . . . . . ... . . . ... .. ...... ....... . . . . . . . . . . ... . . . . . ..................................... . ...................................................................................................................................................

..

A

B

Then ABM ∼ = ARM and therefore

and

AR = RM = RAM =

ARM

=

AB, BM, BAM 90◦ .

60 60

SumofofSegments Segments 3. Sum

Since BAM + DAN = 45◦ we have

RAN = M AN − M AR = 45◦ − M AB = DAN.

Thus RAN ∼ = DAN (AR = AD, AN is a common side and RAN = DAN ) gives RN = DN and ARN = ADN = 90◦ . Since M RN = M RA + N RA = 180◦ we obtain that M , R and N are collinear. Therefore M N = RM + RN = BM + DN, which completes the proof. Alternative 2 Let AB = 1 and BAM = α. Since M AN = 45◦ we obtain

DAN = 45◦ − α.

Therefore BM = tan α, DN = tan(45◦ − α). Using the identity tan α + tan(45◦ − α) + tan α tan(45◦ − α) = 1, which is easily verified by tan(45◦ − α) =

1 − tan α , 1 + tan α

we obtain MN2

= = = = =

CM 2 + CN 2 = (1 − tan α)2 + (1 − tan(45◦ − α))2 2 − 2 tan α − 2 tan(45◦ − α) + tan2 α + tan2(45◦ − α) 2 tan α tan(45◦ − α) + tan2 α + tan2 (45◦ − α) (tan α + tan(45◦ − α))2 (BM + DN )2 .

Therefore M N = BM + DN . Sum of Segments 3. First we have to determine the position of point B1 on the line AC. It is easy to see that 180◦ − C + (90◦ −

B ) < 180◦ 2

is equivalent to C > A. We conclude that point C lies between points A and B1 . Therefore points A1 and B1 lie in one and the same semiplane with respect to line AB.

Sum of Segments61 3. Sum of Segments 61

The proof now is the same as the proof of Example 2. When C < A the point A lies between C and B1 . Thus, A1 and B1 lie in different semiplanes with respect to the line AB. For point M = AB ∩ A1 B1 we have M X = 0 and M X = M Y + M Z is not true. Sum of Segments 4. See the solution of Example 2. In this case B1 B2 and M X are no longer perpendicular to AB but they are parallel to each other. Sum of Segments 5. Alternative 1 Denote DAF = F AE = α. D F C ..........................................................................................................................................

... ... .... ... ... ... ... .. ... .... .. . ... .. . .. ... . ... ... .. . ... . ... . ... . . ... . . ... ... . . . ... ... . . ... . ... . . . ... . . ...... E . . . ... . . . . ............. . . .. ... . . . . .... . ... . .. .. . . . . .. ... 2α ...... ....... ... ... ..... .. .. ... ...... .. . . . . . . . ... ... ...... .. ... ... ...... .. . . . . . . ... . . ... ... . . .. . . ... . . ... ... . . .. . . ... . ... . .... . . .. ... . . ... . . .... . ... α ... . . . . .... ......... ... α .......... .. ... ........ ........ ... ....... .......... ... ....................................................................................................................................................... ...... . .. B A .......... . ...... .... ....... ...... .... ...... ... ....... ... ....... ....... ... ....... ...... ..... ...... ... ....... ... ....... ....... .... ....... ... ...... ... ...... ....... ...... .... ......... .

G

Then AEB = 2α, DF = AD tan α and AE − BE =

AB cos 2α AB − = AB sin 2α sin 2α

2 sin2 α sin 2α

= AB tan α.

It follows that DF + BE = AE if and only if AD = AB, i.e. ABCD is a square.

62 62

SumofofSegments Segments 3. Sum

Alternative 2 Let G be a point on line BC (B lies between C and G) such that GE = AE. If DAF = F AE = α then AEG = 2α and therefore 90◦ − α. Thus ADF ∼ ABG.

AN E =

If AE = DF + BE then GB = GE − BE = AE − BE = DF and hence ADF ∼ = ABG, i.e. AD = AB. If AD = AB then ADF ∼ = ABG giving GB = DF , i.e. AE = GE = GB + BE = DF + BE. Sum of Segments 6. Alternative 1 Let DP ⊥ AM , P ∈ AM , CQ ⊥ BM , Q ∈ BM and R = DP ∩ CQ. C

.. ............... ........... .... .. ........... ........ .... ............ .... .. .......... . . . . . . . . . . . . . . . . M .............. . ... .... ... .............. .. ..... ............ ..... ... .... .. ........... .... ............... .. .... ........... ........ . . . . . . . . . . . . . . . . . . . . . ...... .... ... ..... ..... ........... D .... ............. ... .......... .... ... ..... .... .. ..... ...... .. ... ..... .. .... ... . . . . . . . . ... ...... . Q .... . .. .... ... . .... ... .. .. ... ......... .... ... .. .. ... .... ...... .. .. ... .. .. ..... ..... ........ . . . . . . ... . . . ... ... ... ... ... .... .. .... P ............. ....... ... .. .. .... .... ...... .. .. .. ..... . . ...................................................................................................................................................................

.

.

.

.

A

B

R

The quadrilateral P RQM is a rectangle and M P and M Q are middle segments in DRC. Hence CQ = QR, DP = P R and we conclude that ADR and BCR are both isosceles. Therefore AR = AD, BR = BC

(1)

ARB = ARD + DRC + CRB = ADR + 90◦ + BCR.

(2)

and

Suppose AB = AD + BC. It follows from (1) that AD + BC = AR + RB ≥ AB = AD + BC and we conclude that R ∈ AB. Thus, ADR + BCR = 90◦. Therefore

ARB = 180◦ and (2) implies

ADC + BCD = ADR + BCR + RDC + RCD = 180◦

giving that AD BC.

Sum of Segments63 3. Sum of Segments 63

Suppose AD BC. We have ADC + BCD = 180◦ and since RDC + RCD = 90◦ we conclude that ARD + BRC = 90◦ . It follows by (2) that ARC = 180◦ and therefore AB = AR + RB = AD + BC. Alternative 2 Let N be a point on AM (M lies between A and N ) such that AM = M N . Since DM = CM we have that ACN D is a parallelogram, i.e. CN = AD. On the other hand, BM ⊥ AN and AM = M N , implying that ABN is isosceles. Hence AB = BN . If AB = BC + AD then BN = AB = BC + AD = BC + CN , which is possible only if point C lies on the segment BN . Therefore BN AD. If BC AD then since CN AD we have that BN = BC + CN . Therefore AB = BN = BC + CN = BC + AD, which completes the proof. Sum of Segments 7.

Let M be a point on the arc BC not containing A. C

............................................................ ......... ........ .. ... ....... ...... .. ...... ..... .... ..... .. ........ . .... . . .. .... .. ... ... . . . . .. ... ... . .. . . . ... . . . .. .. .. ... .. . . . . . .. .. . . .. . . . . . . .. .. ... .. . . .. . . . .. .... .. . .. . . .. .. . .. .. .. ... ... .. . ... ... ... . .. .. ... ... .. .. . . ... .... ... .... ... ... .. .. .. ... .. .. . .. . . .... ... . .. ... . . . . . . ... ... . .. .. . . . . ... ... . .. . .. .. ... .... .. .. .. .. ... .. .. . .. ... .. ... ................ .. ... . ..... ... ... .......................... ...... M ... .. ........................ .. .... ...... ............................................. . ............................................................................................................... . ..... . . A ........ .... B ....... ......... . . . . . . . . ............. ....................................

Denote BAC = α and BCM = ψ. Ptolemy’s Theorem for quadrilateral ABM C implies M A.a = M B.b + M C.c.

(1)

If M A = M B+M C or M B = M A+M C then (1) implies (a−c)M C = 0. Since a = c we have M ≡ C. If M C = M A + M B then (1) implies MC 2a = . MB a−c

We conclude that if a < c then such a point does not exist. Otherwise, observe that sin(α − ψ) MC = = sin α cot ψ − cos α. MB sin ψ

64 64

SumofofSegments Segments 3. Sum

Since cot ψ is a decreasing function there exists exactly one point M for which 2a MC = . MB a−c

Let M be a point on the arc AB not containing C. Ptolemy’s Theorem for quadrilateral ABM C implies M C.c = M A.a + M B.b.

(2)

If M C = M A + M B then a = c, a contradiction. If M A = M B + M C then (2) implies 2a MC = . MB c−a

Therefore if a > c then such a point does not exist. Otherwise, one can see as above that such a point exists and it is unique. Analogously, there exists a unique point M for which M B = M A + M C. Therefore: 1. If a > c then there exist three points with the desired property –

point C and two points, one on each of the two arcs BC and AC. 2. If a < c then there exist three points with the desired property –

point C and two points both on the arc AB. Sum of Segments 8.

Let M be a point on the arc BC not containing A. Ptolemy’s Theorem for quadrilateral ABM C implies M A.a = M B.b + M C.c

(1)

With respect to the largest segment from M A, M B and M C we have three cases: 1. If M A = M B + M C, then (1) is equivalent to MB c−a = < 0. MC a−b 2. If M B = M A + M C, then (1) is equivalent to a−b MB = > 0. MC a+c

Sum of Segments65 3. Sum of Segments 65

3. If M C = M A + M B, then (1) is equivalent to a−c MB = > 0. MC a+b

Note that when M moves from B to C on the arc BC not containing A, MB takes all values in the interval (0, ∞). Moreover, if then the ratio MC CBM = ϕ then sin(α − ϕ) MB = = sin α cot ϕ − cos α MC sin ϕ MB is a strictly increasing function. Therefore in this implying that MC case there exist two points with the desired property.

Similar arguments show that when M lies on the arc CA not containing B, there exist two points with the desired property, and when M lies

on the arc AB not containing C, there exists one point with the desired property. Thus there exist 5 such points M . Sum of Segments 9. We consider the case (other cases are treated similarly) when A2 lies between A1 and C, B2 lies between B1 and A, and C2 lies between C1 and A. C

... .......... .. ... .... .. .. ..... ... .... ....... .. .... . ... .... ... ... .. . .. ........................................... A ... .......... .... ..... 2 .. ..... ......... .. .. ... .... ........A1 . . ... ........... . . . .. B1 ............ ... .. ..... . . . . . . ... .... B2 ..... ...................... .... ........ .. ....... ............ ... .... .... ................... . . ....... .... . ......... .. .... ... ... . ... . . . . . . . . . ............ .... .. .. . . . . ... .... .... . . . . ......... . .. . .... . . . . . ... ... ........ . . .......... ... . . . . .. . . . . . . . . . ......... ....... ... . .. .......... .... ... . ... ....... ...... .. . ............ ..... .. ......... ....... . ................................................................................................................................................................................ A B C2 C1

•

• •

•

•

•

Set A1 A2 = x, B1 B2 = y and C1 C2 = z. It follows from the Power of a Point Theorem that AC1.(AC1 − z) = AB1 .(AB1 − y) then

cb a+b

cb −z a+b

bc = c+a

bc −y , c+a

66 66

SumofofSegments Segments 3. Sum

and after simplification we obtain z 1 1 y − = bc . − c+a a+b (c + a)2 (a + b)2

(1)

By analogy, it follows from BC1 .(BC1 + z) = BA1 .(BA1 + x) and CA1.(CA1 − x) = CB1 .(CB1 + y) that x 1 1 z − = ac (2) − a+b b+c (b + c)2 (a + b)2 x y 1 1 + = ab . (3) − c+a b+c (b + c)2 (a + c)2

Now (1) + (2) + (3), −(1) + (2) + (3) and −(1) − (2) + (3) give −c a b(c − a) 2y = + + , c+a a + b b + c (c + a)2 c(b − a) −b a 2z = + , + 2 a+b (a + b) b+c c+a c a(b − c) 2x −b = + . + 2 b+c a + b (b + c) c+a Thus,

c2 + cb ab − ac −b2 − bc + + , a+b b+c c+a −c2 − ca a2 + ac bc − ba 2y = + + , a+b b+c c+a cb − ca a2 + ab −b2 − ba 2z = + + , a+b b+c c+a and it is clear that x + y = z. 2x =

Sum of Segments 10. Without loss of generality suppose that AB = 1. Let AM = x, DP = y and BN = z. Denote the centroid of M N P by G and let G ∈ AC. y P D C ............................................................................................................................................

.... ........ .. .... . ... ......... ... .... .. ..... .. .... .... ... .. ..... .... . . . .... .... ..... . . ... ..... ... .. ..... .. ..... .. ... ..... ..... ... ... .. ..... ....... ... . . ......... .. ... .. .. .. ... .. ... ..... ......... .. ..... .... ... ... ... ..... ..... . . . ... . ..... . ..... .. ... ....... . ... .... ... ..... N ..... .... .... ..... .. ... .. .... G .... .... . . . . . . .... . .... . .... .... . . ... . . ... . ... . .... ... .. ..... .. .... ... ..... ..... ... ... .... ......... .. . . ... . . . ... .... ... z .. .... ...... .. .. .... .. ...... .. .. .... .... ..... .. ... . .... ..... . . . . . ... . . . . ..... ..... ... ... .. .... ..... .. ... .. ..... ..... .. .. .... .... ........ ... . . . ..............................x . . . . .................... ............................................................................................

•

A

M

B

Sum of Segments67 3. Sum of Segments 67

−→ −→ − −→ −−→ We have 3AG = λAC = λ(AB + BC) and −→ −−→ −−→ −→ −−→ −−→ 3AG = AM + AN + AP = (x + y)AB + z BC −−→ −−→ − −→ −−→ and therefore (x + y − λ)AB = (z − λ)CB. Since AB and CB are not collinear we have that x + y − λ = 0 and z − λ = 0, i.e. x + y = z. When x + y = z then −→ −−→ −−→ −→ −−→ −−→ AG = AM + AN + AP = (x + y)AB + z BC −−→ −−→ = (x + y)(AB + BC) −→ = (x + y)AC, which implies that G lies on AC. Sum of Segments 11. (a) Denote AB = c, BC = a and CA = b. Y.

. ......... ..... .... .......................................... ... .. .......... ....... ..... .... .... ........... ..... ... ......... .... ... ....... .... ... C ........ .. ... . . . . . . . ... .......... .............. X . .. ...... ... ... ... ..... .. ...... .. ... ... .. ...... ... ... ... .. ...... ..... .. .... ... .. ...... ... ... ... ... ...... ..... .. ... ... .. ...... ... .. . . ...... .. .. .. ... . . . . . ..... .. .. .. .. ...... .. ... . . .. ...... .. . . . .. ... .................................................................................................................................. .... . ........... ............ .... B A ................ .... ............ . . . . . . . . . . . . ....... .. ... ....... . . . . . . . ....... . . . . . . . . . ......... ................. .... ............... ..... . . . . ........ . . .. Z .................................................

It follows from the condition of the problem that XC = a − c and Y C = c − b. The Cosine Law for XY C implies: XY 2

=

XC 2 + Y C 2 − 2XC.Y C cos XCY

=

(a − c)2 + (c − b)2 + 2(a − c)(c − b) cos ACB

=

a2 + 2c2 + b2 − 2ac − 2cb +(ac − ab − c2 + cb)

=

=

a2 + b2 − c2 2ab

1 3 c(a + b3 + c3 − a2b − a2 c − b2 a − b2 c ab −c2 a − c2b + 3abc)

1 c(3abc − a(b2 + c2 − a2) − b(a2 + c2 − b2) ab

68 68

Sum Segments 3. Sum ofofSegments

−c(a2 + b2 − c2)) = = = and so XY =

c(3abc − 2abc(cos A + cos B + cos C)) ab c(3abc − 11 4 abc) ab c2 , 4 c . 2

(b) Ptolemy’s Theorem for the cyclic quadrilateral AZBC implies ZA.a + ZB.b = ZC.c. Y.

...... ........ .... .... ............................................... ... ... ......... ....... . ...... ..... .... ........... .... . .... ..... ........ ... ... C ....... ... ... ........... ... . .. . ....... ..........X .. . ...... .. ... .... .... ...... ... ...... ... .... .... ... ...... .. .. .. .. ...... ... ... .. .. ...... ..... ... .. . ... ...... ... ... .. . .. . ...... .. .. .... ... . . . . . ...... . ... . ... ... . . . ...... . .. .. .. . . . . ...... .. .. .. .. ... ..... ............................................................................................................................... .... . ............. ........... ..... B . . . . A ................ .... . . . . . . .. ....... . ............ ... ....... .. ............ ......... ...................... .... ................ ..... ........ ..... . . . . . . . ............. . ................................

Z

Since ZC = ZA + ZB we have that a−c ZA = . ZB c−b

The latter equality and AZB = XCY give ZAB ∼ CXY . c = 2 and therefore It follows from (a) that XY ZA = 2(a − c) and ZB = 2(c − b). Thus ZC = ZA + ZB = 2(a − b), and therefore 2(a − b) ZC = = 2. XC + Y C (a − c) + (c − b)

Sum of Segments69 3. Sum of Segments 69

Sum of Segments 12. Let M be a point on the ray BP → such that BM = BC and N be a point on the ray DQ→ such that DN = DC. ... .... .... ... . . . .. .... .... .... ... . . . .... .... ... .... N ......... .. ........ .... .. ... D ..................... ..... ..... . ........ .... .......... ........................................ ................... .... ........ .... .. .................. O .. .. .. .... .. .. .. .... .... . .... . .. ............ .. . . . . ......... .... . . . . C .. ........ ........ . . ... .. ...... . . . . . . . . . . ........... .. . ... . ... . . . . . . . . . . . . ... ....... .. .. ... ... .. ........ ......... .... ... ... ........... ...... .... ... ............ .......... .... ...... .. ............ . . ... . . . . . . . . . . ... .. . ... ......... ... ... ..... ...... .... ......... .. .... .... .... .. ........ ... .... ... ......... ... ... .. ....... .... ................ . ... .. . . . . . . .. ..... ... ........ .............. ... ............. ...........................................................................................................................................................................................................................

A

M

B

Since AM N , BM C and DN C are isosceles the angular bisectors of M AN , M BC and CDN are perpendicular bisectors of M N , M C and N C. Therefore the angular bisectors intersect at the circumcentre O of M N C. . .... .... .... ... . . . ... ... ....................................... ......... .... ............ ........... ...... .... ........... . . . . .... .... H.......... .... . . ... ... . . . . ... Q............ ... . .... . .. . .. .... . .. . . .. .. ... . . .. . D......... .... T ... . .. .. . . . . O .. .... ... ... . . . .... .... .. . . .... . . ... ..... ... . . . . . . . .... ... ... ... C . . . . . . .. . ...... . . . . . . . . . ...... .. . . . . . . . . .. ... ... .... ... .. ...S ... .. .... .. .. ... .... .. ... .. ..... . . . . . . . ..... . ... ... ...... ... ... .. ... .... .... ... .. ..... .... .... ... .. ..... .... . . .... .. . . . . . . . . . . . ....... ... .. . .... ......... .... ... ......................................................................................................................................................................................................................................

•

•

•

•

A

B

P

•

R

Thus, the distances from O to AP , AQ, DP and BQ are equal, i.e. there exists a circle k centered at O which is tangent to AP , AQ, DP and BQ. Denote by R, S, T and H the tangent points of k with AP , DP , BQ and AQ. Using that AR = AH, P R = P S, SC = T C and QH = QT we obtain AP + P C

= = =

AP + P S + SC = AP + P R + CT = AR + CT AH + CT = AQ + QH + CT = AQ + QT + CT AQ + QC.

70 70

SumofofSegments Segments 3. Sum

Sum of Segments 13. We use the well known fact that DB = DI = DC. ............................................................... ........... ......... ......... ....... ...... ...... ...... ..... . . . . .... ... . . .... . . .... .... . . ... .. ... . . ... C ... . . .... ... ....... ...... . ..... ............ .. . . . . . . . .. ...... .... . . . . . . . .. .. .. ... .. .. .. .. ...... .. .. .. ... ...... .. ...... .. .. .. . . . ... . . .. ..... .... . . . ... . . .. . .. .... . . . . .. .......F ... . ... . .. ......... . . . .... . ..... . . ......... ... . .. . . . . . . .... ........................... D . .... . ... . . . . . . . . . . .. .. ....................... .. . . . . . . . . . .... . . . . . .. ........ .. I............................................ . ....... ..................................... .... ...... ..... ............ .. .. ..E ...... ............. . . . . . . . . . . . . . . .... . . . .. .. ... ........ ..... .. .. .. ... ............. ...... .. .. .. ............. ...... ... .. ... ... ............ . ...... . . . . . . . . . . ... . . . . . .. .. .. ....... ... . . . . . . . . . .. . . . . . ...... . . . .. ......... .................... ....... .. ................... ................................................................................................................................................................................................................... . . . A ... .. B . . .. ... ... .. ... . . ... ... ... ... .... ... ... ... .... . . . .... .... ..... .... ..... .... ...... ...... ....... ...... . . . . ......... . . . ..... ........... ...............................................................

Set DI = x. We have IE = x sin γ and IF = x sin β. Also α sin β + AD 2 . = α x sin 2 Now IE + IF =

AD is equivalent to 2 α sin β + 2 sin γ + sin β = α 2 sin 2

which in turn implies sin(β + γ) = 12 . Therefore BAC = α = 30◦ or 150◦.

CEVA'S THEOREM

4. CEVA’S THEOREM Ceva's Theorem

It is well known that in any triangle • the medians meet at a point; • the altitudes meet at a point; • the angle bisectors meet at a point. When do three lines, joining each of the vertices of an arbitrary triangle with a point on the respective opposite side, have a common intersecting point? A suitable and useful answer to this question is given by the famous Ceva’s Theorem, which in its simplest form states that: Theorem 1 Let A1 , B1 and C1 be points on the sides BC, AC and AB of ABC. The line segments AA1 , BB1 and CC1 intersect in a point if and only if AC1 BA1 CB1 · · = 1. C1 B A1 C B1 A Proof Suppose AA1 , BB1 and CC1 intersect in a point and denote the intersecting point by F . C

.. ...... .... .. ....... .... . . .. . . ... ... .... .. ... .. .. ... . .. .. .... . . .. .. ... ... . .. . .. . . .. ... .... .. . . .. . . .... . . .. .. ........ . . . . .. ...... .. .. ... ....... .... ... . . ...... . . .. . . . . . . . . . .. .. . .. ... . ......... .. .. .. .. 1 ........... .. ... ........ .. .. ... ....... ... .. . . .. . . ... . . . . . . . .. . ..... ... .. .. .. ....... .. .. ... .. ....... .. .. ...... .. .. .. ............... .. ....... ... .. ... .... .................. .............. . ... .............. . .. .. .. .. .... ..... .. .. .. .. 1........ .. ...................... .. ... .. .... .. ... ....... .... .......... .. .. .......... ....... ............ . . . .......... .. .... ..... .......... ....... .. ............. . . ................................................................................................................................................

Q

•

A•

P•• F B •

A

•

B

C1

Let the lines through A and B parallel to CC1 intersect the lines BB1 and AA1 at points P and Q respectively. Since BQ CF AP we have BQA1 ∼ CF A1 and CB1 F ∼ AB1 P which implies that BQ BA1 = A1 C CF

and

CB1 CF = . B1A AP

(1)

74 74

Ceva'sTheorem Theorem 4. Ceva’s

Further, since AC1F ∼ ABQ and AF P ∼ QF B we obtain AF AP AC1 = = . C1B FQ BQ

(2)

Using (1) and (2) we have AP BQ CF AC1 BA1 CB1 · · = · · = 1. C1 B A1 C B1 A BQ CF AP Alternative proof We repeatedly use the following observation: If a point M lies on the side P Q of a triangle P QR then SP M R PM = . MQ SQM R (SXY Z is the area of the triangle XY Z; we will use this notation further.) We have SAC1 F SAC1 C AC1 = = C1B SBC1 F SBC1 C and therefore By analogy

Thus,

AC1 C1B which completes

SAC1 C − SAC1 F SAF C AC1 = = . C1B SBC1 C − SBC1 F SBF C SBF A BA1 = A1 C SCF A

and

SCF B CB1 = . B1A SAF B

BA1 CB1 SAF C SBF A SCF B · = · · = 1, A1 C B1A SBF C SCF A SAF B the proof. ·

Further Development Suppose that

AC1 BA1 CB1 · · =1 (3) C1B A1 C B1A and denote the intersecting point of AA1 and BB1 by F . If X is the intersecting point of CF and AB then according to the first part of the proof we have AX BA1 CB1 · · = 1. (4) XB A1 C B1A Now (3) and (4) give AX AC1 = C1 B XB

Ceva's Theorem75 4. Ceva’s Theorem 75

which implies

i.e.

AX AC1 +1= + 1, C1B XB AB AB = . C1B XB

Therefore C1B = XB giving C1 ≡ X, so AA1 , BB1 and CC1 intersect in a point.

It is natural to generalize the condition of Theorem 1 and to consider the case when the points A1 , B1 and C1 lie on the lines BC, AC and AB respectively. In order to state and prove, similar to Theorem 1, necessary and sufficient conditions for the lines AA1, BB1 and CC1 to intersect in a point, we need some preliminary notation. Consider a line and three points X, Y and Z on . Define XY if Y lies between X and Z XY YZ = YZ − XY , otherwise. YZ Theorem 2

Consider ABC and let A1, B1 and C1 be points on the lines BC, AC and AB. (a) If lines AA1 , BB1 and CC1 intersect in a point then AC1 BA1 CB1 · · = 1. C1B A1 C B1 A

(5)

(b) If (5) holds true and two of the lines AA1 , BB1 and CC1 intersect in a point, then all three intersect in a point. The proof of Theorem 2 is analogous to the proof of Theorem 1 and is left to the reader. Example 1 The excircle to the side BC of ABC is tangent to BC, CA and AB at points A1 , B1 and C1 . Prove that the lines AA1 , BB1 and CC1 intersect in a point.

76 76

Ceva'sTheorem Theorem 4. Ceva’s

Solution 1 Note that AC1 = AB1 , BA1 = BC1 and CA1 = CB1 . .. .. ........ ....... ........ . . . . .... ....... ....... ..... .. ....... .... . . .. . . ... ... .. ...... .. ... ... ....... .. ... ............ .... . . ..... ............. .. .... ......... ... ... ......... .. .... ... .. . .. ... ........... .... ..... . . . . . .... .. .... ... .. . . . . . ... .... .. .. .. . . . ... . . .. . . . ... ... .... 1... ......... ... ... ...... . . ... . . . .. ..... ... .. ...... . . . . . ... .. ...... .. ... ..... . ... . .. ... .... .. ...... ... . . ..... ... .. .. .. ..... . ... . . . ..... .... ........ ... . . . . . . . ........ . ..... ........ . . . . . . . .... .... .............................................................................................................................................................................................

B1

C• A

A

•

•

•

•

•

C1

B

We have

AC1 BA1 CB1 · · =1 C1B A1 C B1A and according to Ceva’s Theorem the lines AA1, BB1 and CC1 intersect in a point. Example 2 Consider ABC and let point C1 be the symmetric point of the vertex A with respect to the vertex B, and A1 be the symmetric point of the vertex B with respect to the vertex C. If B1 is a point on the line CB1 = 14 prove that the lines AA1 , BB1 and segment AC such that B1 A CC1 intersect in a point. Solution 2 It follows from the condition of the problem that AC1 = −2, C1 B

BA1 = −2 A1 C

and

1 CB1 = . 4 B1 A

A•1

.... .. .. .. .. .. ... .. .. . .. .. .. .. .. .. .. ............. .. . ... ..................... .... . . . . . .......... . . ... .. . . .... ....................... .. ......... .. .......... .. .. ........ .... .. ........ . .... ..... .. . ........ . . ... .... ... . ........ . . ... ......... 1 ....... .... .... . . . ....... . . . . .... .. ........ ... ...... ... .. ........ . ..... . . . ........ .... .. . .... ........ . . . ...... ...... ... . . . . . .........................................................................................................................................................................

•

•

•C

B

•

A

Therefore

•

B

•

C1

1 AC1 BA1 CB1 · · = (−2).(−2). = 1. 4 C1B A1 C B1 A

Ceva's Theorem77 4. Ceva’s Theorem 77

Since AA1 and BB1 obviously intersect it follows by Ceva’s Theorem that the lines AA1 , BB1 and CC1 intersect in a point. We say that three lines are concurrent when they either intersect in a point or are parallel. Theorem 3 (Ceva’s Theorem) Consider ABC and let A1, B1 and C1 be points on the lines BC, AC and AB. Lines AA1 , BB1 and CC1 are concurrent if and only if AC1 BA1 CB1 · · = 1. C1 B A1 C B1 A When two of the lines AA1 , BB1 and CC1 intersect the assertion of the theorem follows from Theorem 2. Suppose AA1 (B between C and A1), BB1 (B1 between A and C) and CC1 (B between A and C) are parallel. Then C1BC ∼ ABA1 , giving BC BC1 = AB BA1 and therefore BC1 BC A1 C AC1 = 1+ = 1+ = . AB AB BA1 BA1 Thus

AC1 BA1 · = 1. A1 C AB

Since CC1 BB1 we have C1 B CB1 = . B1 A AB Therefore AC1 BA1 CB1 · · C1B A1 C B1A

=

AC1 BA1 CB1 · · C1B A1 C B1 A

=

AC1 BA1 C1B · · C1B A1 C AB

=

AC1 BA1 · A1C AB

=

1.

Suppose that AC1 BA1 CB1 · · =1 C1B A1 C B1 A

78 78

Ceva'sTheorem Theorem 4. Ceva’s

and two of the lines, say AA1 and CC1 are parallel (the other cases are treated in a similar way). As above we have A1 C AC1 = AB BA1 and since

AC1 BA1 CB1 AC1 BA1 CB1 · · · · = C1B A1 C B1 A C1B A1 C B1 A

we obtain

C1 B CB1 = , B1 A AB

which implies that BB1 CC1.

The alternative proof of Theorem 1 prompts us to make a conceptual connection between Ceva’s Theorem and barycentric (areal) coordinates, as discussed in Book 1 of this series. Indeed, for triangle ABC of unit area, the key areas used in this proof, SCF B , SAF C , SBF A , are actually the barycentric coordinates of F with respect to the triangle ABC. Ceva’s Theorem could be expressed in terms of barycentric coordinates in the following way: Given the barycentric coordinates of the points A(αA , αB , αC ), B (βA , βB , βC ), C (γA , γB , γC ) with respect to ABC. Lines only if αB αC

AA , BB and CC are concurrent if and ·

γA β C · = 1. γB βA

Theorem 4 (Trigonometric Form of Ceva’s Theorem.) Let A1 , B1 and C1 be points on the lines BC, AC and AB respectively. The lines AA1 , BB1 and CC1 are concurrent if and only if sin A1 AB sin B1 BC sin C1 CA · · = 1. sin A1 AC sin B1 BA sin C1 CB Proof Since

and

SA1 AB

=

SA1 AC

=

SA1 AB SA1 AC

=

1 A1 A.BA sin A1AB, 2 1 A1 A.CA sin A1AC 2 BA1 A1 C

Ceva's Theorem79 4. Ceva’s Theorem 79

we have SA1 AB BA sin A1AB A1 A.BA sin A1 AB BA1 = = . = A1 C SA1 AC A1 A.CA sin A1 AC CA sin A1 AC Therefore

BA1 .CA sin A1 AB = . sin A1AC A1 C.BA

Thus, sin A1 AB sin B1 BC sin C1 CA BA1 CB1 AC1 · · = · · . sin A1 AC sin B1 BA sin C1 CB A1 C B1 A C1B The assertion of the theorem now follows from Theorem 3. Remark. It follows from trigonometric form of Ceva’s Theorem that if P is an interior point of ABC then sin P BA sin P CB sin P AC · · = 1. sin P BC sin P CA sin P AB

80 80

Ceva's Theorem 4. Ceva’s Theorem

PROBLEMS Ceva’s Theorem 1. Let A1 be a point on side BC of ABC. Points B1 and C1 on CA and AB respectively are such that A1 B1 and A1 C1 are angular bisectors of AA1 C and AA1 B. Prove that the lines AA1 , BB1 and CC1 intersect in a point. Ceva’s Theorem 2. For ABC points A1 , A2, B1 , B2 and C1, C2 on the sides BC, AC and AB, respectively are such that BAA1 = CAA2, CBB1 = ABB2 and BCC1 = ACC2. Prove that AA1 , BB1 and CC1 intersect in a point if and only if AA2 , BB2 and CC2 intersect in a point. Ceva’s Theorem 3. In a hexagon ABCDEF it is given that AB = BC, CD = DE, EF = F A and ABC = CDE = EF A. Prove that the lines AD, CF and EB intersect in a point. Ceva’s Theorem 4. Points P , Q and R lie on the sides AB, BC and CA of an acute triangle ABC, respectively. If BAQ = CAQ, QP ⊥ AB, QR ⊥ AC and CP and BR intersect at S prove that AS ⊥ BC. Ceva’s Theorem 5. The opposite sides of a hexagon are parallel. Prove that the lines through the midpoints of the opposite sides intersect in a point. Ceva’s Theorem 6. Points A1 ∈ BC, B1 ∈ AC and C1 ∈ AB are such that AA1 , BB1 and CC1 intersect in a point and AA1 ⊥ BC. Prove that C1A1 A = B1 A1 A. Ceva’s Theorem 7. Let C1 be a point on the base of an isosceles ABC, AC = BC. Points P and Q on CC1 are such that P AB = QBC. Prove that QAC = P BA. Ceva’s Theorem 8. Point P in the interior of ABC is such that P AB = 30◦ , 20◦ , P BA = 10◦ and P BC = 40◦ . Find P CB.

P AC =

Ceva's Theorem81 4. Ceva’s Theorem 81

Ceva’s Theorem 9. The points A and D lie on a circle k. The tangent lines to k at A and D intersect at S. The points B and C lie on the smaller arc AD. If AC and BD intersect at P and AB and CD intersect at Q prove that points P , Q and S are collinear. Ceva’s Theorem 10. Points A1 , B1 and C1 on the sides BC, CA and AB of triangle ABC are such that AA1 , BB1 and CC1 are angular bisectors of the corresponding angles. If AA1 and CC1 intersect C1B1 and B1 A1 at M and N respectively, prove that M BB1 = N BB1 .

82 82

Ceva'sTheorem Theorem 4. Ceva’s

SOLUTIONS Ceva’s Theorem 1. Using the bisector property we obtain AA1 AC1 = C1 B BA1

and

A1 C CB1 = . B1 A AA1

•C

.. ....... ... ....... ... ........ ... ... .. . ... ... ... ... ... ... ... ... .. . ... .... .. . ... .. .. . .. . ... . . . ... 1............................................. ..... ... ...... ............... . ....... ................ .... . . . ....... .. ................ 1 .. . . . . ...... .. ............ . ......... ...... .. .. .. .... . . ... ........... .. .. .. ..... .... .............. ..... . ...... .. . .. ... ....... ... . . . . ...... .. ... . ... . . . . . . . ....... . .. ... ... ... ...... .... ....... ...... .... ........................ ... ...... . . . . .. . . . . . . . . . . ...... . . . .... . . . . . . . . . . . . . . . . . . ..... ............ .. ...... ............ 1 ... ....... ........................ .. .. ..... ..................................... . . ..............

B •

•A

•B

•

C

•

A

Therefore

AA1 BA1 A1 C AC1 BA1 CB1 · · = · · = 1, C1B A1 C B1 A BA1 A1 C AA1 and according to Ceva’s Theorem we conclude that AA1 , BB1 and CC1 intersect in a point. Ceva’s Theorem 2. Using the trigonometric form of Ceva’s Theorem we have that AA1 , BB1 and CC1 intersect in a point if and only if sin BAA1 sin ACC1 sin CBB1 · · = 1. sin CAA1 sin BCC1 sin ABB1 C •

..... ....... .. ....... .... .......... ....... . . . ... ... .. . .... ... . ..... .. ..... ... ... ... .. ... 2........... ...... ... .. . . . . ........ ... ........ . . . . . . .... ........ .. 1 .... ... ...... ............. ... ... .... ... .. ... ........... .... ... ..... ... .. .. .... ... ............. ...... . ... . ... . . . . 1........... . ........ ... . ... . . . . . . . . . ....... .......... ... .... ... .. . . . . . . . . . . . ... ... .. . ................ . . . . . . . . . . . . ...... ............ ........ ... ..... .... . . . . . . . . . . . . . . . . . .. . . . . ........ ............. .... .... . . . . .... . . . . . . . .. . . . . . . . . . . . . . . . . . ... ... ................ .. ........... . .. . . . . . . . . . . . . . . . . . . . . ....... .. ... ... . ..................... ... ........... . . . . . . . . . . . . .......... ... ... .............. ... .......... . . . . . . . . . . . . . . . . . . . . . . ... .......... ........ .............. ........................ .... . . . ........... .. ....................... .. .........................................................................................................................................................................................

B•

B•

•

A

•

C1

•

•A

• • A2

•

C2

•

B

By the same reasoning AA2 , BB2 and CC2 intersect in a point if and only if sin CAA2 sin ABB2 sin BCC2 · · = 1. sin BAA2 sin CBB2 sin ACC2

Ceva's Theorem83 4. Ceva’s Theorem 83

It follows from the condition of the problem that

BAA2 = BAC − CAA2 = BAC − A1 AB = CAA1 ,

and in the same way,

CBB2 = ABB1

and

BCC2 = ACC1.

Therefore sin CAA2 sin ABB2 sin BCC2 sin BAA1 sin ACC1 sin CBB1 · · = · · , sin CAA1 sin BCC1 sin ABB1 sin BAA2 sin CBB2 sin ACC2 which completes the proof. Remark. Suppose that in the notation of the last problem AA1 , BB1 and CC1 intersect in a point X1 ; then the result proved in the problem implies that AA2 , BB2 and CC2 also meet in a point, say X2 . The point X2 is called the isogonal conjugate to X1 . Obviously, if X2 is the isogonal conjugate to X1 , then X1 is also the isogonal conjugate to X2 . Ceva’s Theorem 3. Note that since ABC, CDE and EF A are isosceles and ABC = CDE = EF A

we have ABC ∼ CDE ∼ EF A. D

•

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

•C

E• F•

•

•

A

B

Therefore BCE = DCA, DEA = F EC and F AC = BAE. It follows from the sine rule for ACD and ADE that sin DCA AD = CD sin CAD

and

AD sin DEA = . ED sin EAD

Since CD = ED we have sin AED sin ACD = sin CAD sin EAD

84 84

Ceva'sTheorem Theorem 4. Ceva’s

which implies

sin DCA sin CAD = . sin EAD sin DEA

By the same reasoning, sin BCE sin CEB = sin AEB sin BAE and

sin F EC sin ECF = . sin ACF sin F AC

Therefore sin CAD sin CEB sin ECF . . sin EAD sin AEB sin ACF = =

sin DCA sin BCE sin F EC . . sin DEA sin BAE sin F AC 1,

i.e. the lines AD, CF and EB intersect in a point. Ceva’s Theorem 4. We use standard notation for the elements of ABC. By the bisector property we have BQ c = . QC b •C R• •T •

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

S

•

A

•Q

•

P

•

B

It follows from the right triangles BP Q and CRQ that P B = BQ cos β and CR = CQ cos γ. Let AS intersect BC at T . Using that AP = AR and Ceva’s Theorem we have 1= giving

AP BT CR BT.CR BT.CQ. cos γ b.BT. cos γ · · = = = P B T C RA P B.CT BQ.CT. cos β c.CT. cos β c cos β BT = . CT b cos γ

Ceva's Theorem85 4. Ceva’s Theorem 85

On the other hand, if AH ⊥ BC, H ∈ BC, then H is a point on the side BC and BH = c cos β and CH = b cos γ. Thus BH c cos β BT = = , CH b cos γ CT which implies that H ≡ T . Therefore AS ⊥ BC. Ceva’s Theorem 5. Let diagonals AD and BE intersect at P . Denote by K and L the midpoints of AB and DE, respectively. Since ABDE is a trapezoid we have that K, P and L are collinear. L

E

•

•D

•

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

•

•

M

F•• •

•

•P

•

A

•C

•

N

•

•

•

B

K

It follows from AK = BK that SAP K = SBP K which in turn implies that AP.P K sin AP K = BP.P K sin BP K, i.e. BP sin AP K = . sin BP K AP Since ABP ∼ DEP we have

BP AP

=

EP DP

and therefore

BP + EP BE BP = = . AP AP + DP AD Let CF intersect AD and BE at points M and N , respectively. If M ≡ N ≡ P then all three lines through the midpoints of opposite sides pass through this point. If the three points are distinct then the results follows from the trigonometric form of Ceva’s Theorem for triangle M N P .

86 86

Ceva'sTheorem Theorem 4. Ceva’s

Ceva’s Theorem 6. Denote C1A1 A = ϕ and B1 A1A = ψ. C

•

..... ..... ...... .... ........ .. .... .. . .............. 1 .. . .. .. ............ ....... .. .. ...... .... ... ..... .. ........... ..... ... .... . ..... . ... .. .. .. ..... 1................. ............ .... .... ............ .. ..... .. . . . . . . . .... .. ......... .................. ..... . ......... .. .... .. ... ..... . ......... .... .. ...... .... ... . . . . . . ........ ..... . ... .. ..... . . . . . ..... . . ......... . .... ... ... . .... . . . . ........ .. .. . .... . . . . . . . . ......... ........ . .. ......... .... ... .... ......... ......... ..... ... .... . . ............. ..... ..... ... .. ...............................................................................................................................................................................

•A

B•

•

•

A

•

B

C1

Compute SAC1 A1 AA1 sin ϕ AA1 sin ϕ AC1 = = = C1B SBC1 A1 BA1 sin(90◦ − ϕ) BA1 cos ϕ and analogously

SCB1 A1 CA1 cos ψ CB1 = . = B1 A SAB1 A1 AA1 sin ψ

Ceva’s Theorem implies sin ϕ cos ψ AC1 BA1 CB1 · · = · = 1, C1B A1 C B1 A cos ϕ sin ψ i.e. tan ϕ = tan ψ. Since 0 < ϕ < 90◦ and 0 < ψ < 90◦ we obtain that ϕ = ψ. Ceva’s Theorem 7. First we apply the trigonometric form of Ceva’s Theorem for the points P and Q. C .. ...... ..• ..... ... . . ... ...... ... .... .... .. ... ... . ... ... .... . ... . .. ... ... . . ... ... ... . .... . .. ... ... . . . .... .. . . . . . ... ................. .. . . . . . ... ....... .. .... ... ... . . . . . . . ....... ... .. ...... ... . . . . . . ........ ... .. ....... ... . . . .... . . . ....... . . ... ... ....... . . . . . . ........ . .. ...... ............................................ . . . . . ........ ..... .................. ....... ........... .. . . . . . . .................. ........ .... . .................. . . . . . . . . . . . . . . .................................. . .............. . . . . . . ...............................................................................................................................................................

•Q

•

A

and

•

•P

C1

•

B

sin P BA sin P CB sin P AC · · sin P BC sin P CA sin P AB

(1)

sin QBA sin QCB sin QAC · · . sin QBC sin QCA sin QAB

(2)

Ceva's Theorem87 4. Ceva’s Theorem 87

Since QCB = P CB = C1CB and obtain from (1) and (2) that

QCA =

P CA =

C1CA we

sin C1CB sin P BC sin P AB sin QBC sin QAB = . = . . sin C1 CA sin P BA sin P AC sin QBA sin QAC

(3)

Observe that by the condition of the problem P AB = QBC and since P AC + P AB = QBA + QBC we conclude that P AC = QBA. The above equalities and (3) imply sin(α − QAC) sin(α − P BA) = , sin P BA sin QAC

(4)

where α = BAC = ABC. Using the formula sin(α−x) = sin α cos x− cos α sin x we obtain from (4) that sin α cot P BA − cos α = sin α cot QAC − cos α, and since sin α = 0 we have cot P BA = cot QAC. Therefore 0 = cot P BA − cot QAC =

sin( QAC − P BA) , sin QAC sin P BA

i.e. QAC − P BA = kπ, k ∈ Z, and since QAC, follows that QAC = P BA.

P BA ∈ (0, 90◦) it

Ceva’s Theorem 8. Using the trigonometric form of Ceva’s Theorem we obtain sin 20◦ sin 10◦ sin x = 1, sin 30◦ sin 40◦ sin(80◦ − x) where

P CB = x and x ∈ (0, 80◦). C •

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

x

•

A

• P

Since sin 30◦ = and

•

B

1 2

sin 40◦ = 2 sin 20◦ cos 20◦

88 88

Ceva'sTheorem Theorem 4. Ceva’s

the above equality implies sin 10◦ sin x = sin 70◦ sin(80◦ − x). Further, using that sin α sin β = 2(cos(α + β) − cos(α − β)) we have cos(x + 10◦ ) − cos(150◦ − x) = 0

which in turn implies

sin 80◦ sin(x − 70◦) = 0. Thus sin(x − 70◦) = 0, i.e. x = 70◦. Ceva’s Theorem 9. It follows from the trigonometric form of Ceva’s Theorem that

and

sin ASP sin SDP sin DAP · · =1 sin P SA sin ADP sin SAP

(1)

sin ASQ sin SDQ sin DAQ · · = 1. sin QSA sin ADQ sin SAQ

(2)

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

D...................

•

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

•C

•

Q

•P •

B

•

S

•

A

But DAP = SDQ, SDP = DAQ, SAP = DAQ and ADP = SAQ, and (1) and (2) imply sin ASQ sin ASP = . sin DSP sin DSQ Denote that

ASD = ϕ,

DSP = α and

(3)

DSQ = β. It follows from (3)

sin(ϕ − β) sin(ϕ − α) = , sin α sin β

Ceva's Theorem89 4. Ceva’s Theorem 89

which π implies that cot α = cot β. Since cot x is a monotone function in 0, 2 we conclude that α = β. Therefore DSP = DSQ, i.e. S, P and Q are collinear. Ceva’s Theorem 10. The sine rule for triangles AC1 B1 and BC1B1 implies BC1 B1 C1 B1 C1 AC1 and = = . sin AB1 C1 sin BAC sin BB1 C1 sin ABB1 C

•

...... ..... .... .. ... .... .. .... ...... . ... .. ... .... ... .. ... .. .. ... .. .. .... .. ... . ... .. .. ... . .. .... ... ... . . . . ................ 1 . . . . .................................................................... ........ ...... . . ... .... ..... 1................. ... . . . .. . ... . . . .... ... ..... ............... .... .................. ............. ..... ... ... .. ..... ... ..... ... .. ...... .... ... ......... .... ................ .. . . . ... . . . . . . . . ........ ...... . ...................... ... . .. . . . . . . . . . . . ......... ..... ...... ... .... .................... . .. . . . . . . . . . . ........ ..... .... ............ ... .. ... . . . .. . . . . . . . . . . . . . . . ............ ............... ..... ... .. .. ............ ............. ... ... ......... ... ... ............................ . ... .. .................. .. ...... .. ...........................................................................................................................................................................

B•

•A

N•

•

•

M

•

A

•

C1

•

B

Dividing the above equalities gives sin ABB1 sin BB1 C1 AC1 · = sin AB1 C1 BC1 sin BAC and using the bisector property sin ABC AC AC1 = = BC1 BC sin BAC we have

sin ABB1 sin BB1 C1 . (1) = sin AB1 C1 sin ABC On the other hand, the trigonometric form of Ceva’s Theorem for triangle ABB1 and point M implies sin BB1 C1 sin B1 AM sin ABM · · = 1. sin AB1 C1 sin C1AM sin M BB1 Finally, since AA1 is the angular bisector of BAC, (1) and (2) give sin M BB1 sin ABB1 sin BB1 C1 = . = sin AB1 C1 sin ABC sin ABM In the same way, we obtain sin CBB1 sin N BB1 = sin ABC sin CBN

(2)

FUNCTIONAL EQUATIONS

5. FUNCTIONAL EQUATIONS

Functional Equations

Functional equations play a very important role in olympiad mathematics. Their difficulty may vary from trivial problems to extremely difficult ones. As such they appear in many mathematical competitions up to International Mathematical Olympiad. Functional equations are often related to some famous ones such as f(xy) = f(x)f(y) or

f(x + y) = f(x) + f(y).

These are called the Cauchy multiplicative and additive functional equations (some only define the additive version and call it the Cauchy functional equation). These equations have been extensively studied in various domains and they appear at some point in the solution of many problems. There are numerous books dealing with such equations and presenting all basic approaches and tricks for solving them. We will take a slightly different approach. We will focus on equations for which one needs some common sense, and good choice of substitutions and algebraic manipulations. In most cases the way to the solution of a functional equation involves the calculation of the value(s) of the unknown function at a certain point(s). For instance, the standard procedure for solving the additive functional equation f(x + y) = f(x) + f(y) starts with putting x = y = 0 in the equation, which leads to the algebraic equality f(0) = f(0) + f(0) and from it to the conclusion that f(0) = 0. Knowing that f(0) = 0, we continue the process of solving by reasoning of the following type. In the given equation f(x + y) = f(x) + f(y) we take y = x = 1, the result is f(2) = f(1) + f(1) = 2f(1); and further, we continue by induction to establish that f(n) = nf(1) for any positive integer n, and so on. Similar algebraic manipulation for finding suitable values of the unknown function are the key element in the algorithms suggested below.

94 94

FunctionalEquations Equations 5. Functional

In what follows we will use the standard notation: N – the set of all positive integers, i.e. N = {1, 2, 3, . . .}; Z – the set of all integers, i.e. Z = {. . . , −2, −1, 0, 1, 2, . . .}; R – the set of all real numbers. Example 1 Given f : N → N such that: 1. f(ab) = f(a)f(b) for any relatively prime positive integers a and b. 2. f(p + q) = f(p) + f(q) for any prime numbers p and q, prove that f(2) = 2, f(3) = 3 and f(1999) = 1999. Note: In problems such as this the domain and conditions are important. Whereas the domain is N the conditions only apply on subsets of N and the broader results must be built about this. Solution 1 For a = b = 1 the first equality implies f(1) = f(1).f(1) and since f(1) = 0 we conclude that f(1) = 1. Our next goal is to find values of f(x) at some prime values of x. In our case it is reasonable to try to find f(2), f(3), f(5), and possibly (if necessary) some other small prime x. For f(2) and f(3) we can obtain a system of simultaneous algebraic equations by setting a = 2, b = 3 in the first equality and p = q = 3 in the second one: f(6) = f(2)f(3) and f(6) = 2f(3). Therefore f(2)f(3) = 2f(3), giving (since f(3) = 0) that f(2) = 2. Now we can find some values of f or express them in terms of f(3). 1) Setting p = q = 2 we obtain f(4) = 2f(2) = 4. 2) Similarly p = 2, q = 3 gives f(5)

=

f(2) + f(3) = 2 + f(3).

Functional Equations95 5. Functional Equations 95

3) Then p = 2, q = 5 gives = f(2) + f(5) = 4 + f(3).

f(7) 4) Finally, p = 5, q = 7 gives f(12)

= =

f(5) + f(7) 6 + 2f(3).

5) But a = 3, b = 4 implies f(12)

=

f(4)f(3)

=

4f(3).

Hence 6) 6 + 2f(3) = 4f(3), i.e. f(3) = 3. It follows now that 7) f(5) = 5 and f(7) = 7. And further, we may find easily now that 8) f(15) = f(3).f(5) = 15, and also that 9) f(13) = f(15) − f(2) = 13, and that 10) f(11) = f(13) − f(2) = 11. Now it is clear that we could continue our calculations in the same manner to find consecutively the values of f(x) for larger and larger positive integers, since according to the first equality in the condition of the problem, if we know the values of f on a set of prime numbers we may compute the value of f for any product of some of these prime numbers; and according to the first equality in the condition of the problem, if we know the values of f on a set of prime numbers we may compute the value of f for any sum of some of these prime numbers. However, such a procedure could be rather long. It could be shorter, if we could express 1999 as a product of prime numbers. But 1999 is a prime number. Therefore, in order to find f(1999) we could first try to come closer to 1999. To achieve this, we look at the numbers around 1999 that are product of small prime numbers. One such number is 2002 = 2.7.11.13. We have f(2002) = = =

f(2)f(7)f(11)f(13) 2.7.11.13 2002

and therefore f(1999) = f(2002) − f(3) = 1999. Note: The reader might form the hypothesis that f(n) = n for all n ∈ N and indeed this satisfies the conditions, but the reader might as an exercise try to prove this.

96 96

FunctionalEquations Equations 5. Functional

Example 2 Let a be a real number. Consider a function f : R → R satisfying the following conditions: 1. f(x + y) = f(x)f(a − y) + f(y)f(a − x) for any x, y ∈ R. 2. f(0) = 12 . Prove that f is constant. Solution 2 Here the statement of the problem gives us that f(0) = 12 . To move further, we can observe, that by setting x = y = 0 we obtain f(0) = 2f(0)f(a). Since f(0) = 12 , this implies f(a) = 12 , and we have another value of the unknown function f(x). Now setting x = 0 we have f(y)

= =

f(0).f(a − y) + f(y).f(a) 1 (f(a − y) + f(y)) , 2

i.e. f(y) = f(a − y) for any y. Therefore f(x + y)

= = = = =

Setting x = y =

f(x)f(a − y) + f(y)f(a − x)

f(x)f(y) + f(y)f(x) 2f(x)f(y) 2f(a − x)f(y) f(a − x + y).

1 z in the above equality we obtain f(z) = f(a) = . 2 2

Example 3 Find all functions f : R → R for which f(x + y) − 2f(x − y) + f(x) − 2f(y) = y − 2 for all x, y ∈ R.

Solution 3 Although the left-hand side of the given equation looks complicated, the “standard” substitution y = 0 reduces the equation to −2f(0) = −2

Functional Equations97 5. Functional Equations 97

and therefore f(0) = 1. It is reasonable now to take x = 0 in the initial equation. We obtain f(y) + 2f(−y) = 3 − y.

(1)

By replacing y by −y in (1) we have f(−y) + 2f(y) = 3 + y.

(2)

Consider (1) and (2) as a system of simultaneous equations with unknowns f(y) and f(−y). By standard manipulations we obtain f(y) = y + 1. It is easy to verify that f(x) = x+1 satisfies the condition of the problem, so it is the only solution. Example 4 Prove that a function f : Z → Z such that f(f(n)) = n + 1 does not exist. Solution 4 The basic observation here is that the value of f(f(f(m))) can be computed in two ways. First, using the condition of the problem for n = m, i.e. f(f(m)) = m + 1 we have f(f(f(m))) = f(m + 1).

(1)

On the other hand, for n = f(m) we obtain f(f(f(m))) = f(m) + 1.

(2)

Equations (1) and (2) now imply f(m + 1) = f(m) + 1

(3)

for all m ∈ Z. Equation (3) allows us to find f(n) for any n in terms of a = f(0), where a is an integer. Indeed, for n > 0 we have f(n)

= = =

f(n − 1) + 1 f(n − 2) + 2 ··· f(0) + n

=

a + n,

98 98

FunctionalEquations Equations 5. Functional

and for n < 0 we obtain f(n)

= = = =

f(n + 1) − 1 f(n + 2) − 2 ··· f(0) − |n| a + n.

Therefore f(n) = n + a for any n. Now f(f(n)) = f(n + a) = n + 2a and the condition of the problem gives n + 2a = n + 1, which is a contradiction since a is an integer. Therefore a function satisfying f(f(n)) = n + 1 does not exist. Example 5 Find all functions f : R → R such that (x − 2)f(y) + f(y + 2f(x)) = f(x + yf(x)) for all x, y ∈ R.

Solution 5 The equation looks rather complicated. What substitution simplifies it as much as possible? Careful thought prompts us to consider x0 (if such x0 exists) for which f(x0 ) = 0. Thus for x = x0 the equation becomes (x0 − 1)f(y) = 0.

(1)

If x0 = 1 then f(y) = 0 for all y, so f is identically zero. This function satisfies the equation. Suppose now that f is not identically zero. Equation (1) implies that x0 = 1. Therefore if there exists x0 for which f(x0 ) = 0 then x0 = 1. Other possible simplification is to try to find x and y for which y + 2f(x) = x + yf(x), i.e.

y(f(x) − 1) = 2f(x) − x.

(2)

If f(x) = 1 for some x then (2) implies x = 2. Therefore for x = 2 we have f(x) = 1 and then we may set y=

2f(x) − x . f(x) − 1

(3)

Functional Equations99 5. Functional Equations 99

The equation now becomes (x − 2)f(y) = 0 and since x = 2 we have f(y) = 0. According to the above observation this is possible only if y = 1. Therefore (3) gives f(x) − 1 = 2f(x) − x, i.e. f(x) = x − 1.

Direct verification shows that this function is a solution of the problem. Thus we have two solutions f(x) = 0 and f(x) = x − 1.

100 100

FunctionalEquations Equations 5. Functional

PROBLEMS Functional Equations 1. Find all functions f : R → R for which f(x − f(y)) = 1 − x − y for all x, y ∈ R. Functional Equations 2. Find all functions f : R\{−1, 1} → R for which 3+x x−3 +f =x f x+1 1−x for all x ∈ R\{−1, 1}. Functional Equations 3. Find all functions f : Z → Z that obey f(m + n) + f(mn − 1) = f(m)f(n) + 2. Functional Equations 4. Find all functions f : N0 → N0 such that f(0) = 0 and 1. f(2n + 1) = f(2n) + 1 for any n ∈ N0 ; 2. f(2n) = f(n) for any n ∈ N0 . Functional Equations 5. Find all functions f : R → R such that f(x)f(y) = f(x + y) − f(x − y). Functional Equations 6. Show that if f : R → R satisfies f(xy) = xf(x) + yf(y) then f is identically zero.

Functional Equations101 5. Functional Equations 101

Functional Equations 7. Determine all functions f : Z+ → Z+ such that xf(y) + yf(x) = (x + y)f(x2 + y2 ) for all positive integers x, y. Functional Equations 8. Find all functions f : Z → Z such that f(x + y) + f(xy) = f(x)f(y) + 1 for all x, y ∈ Z. Functional equations 9. Let a, b and c be non-negative integers. Prove that a function f : N → N such that f(x + y) + f(x) + f(y) = xy + ax + by + c for all x, y ∈ N does not exist. Functional equations 10. Find all functions f : R → R for which f(f(x + y)) = f(x + y) + f(x)f(y) − xy for all x, y ∈ R. Functional equations 11. A function f : Z → N is such that f(m) − f(n) is divisible by f(m − n) for all m, n ∈ Z. Prove that for f(n) ≤ f(m) it is true that f(m) is divisible by f(n). Functional equations 12. Find all functions f : N → N for which f(f(n)) = n + 2 for all n ∈ N. Functional equations 13. Find all functions f : R → R for which f(x + y) − f(x − y) = 4xy for all x, y ∈ R.

102 102

FunctionalEquations Equations 5. Functional

SOLUTIONS Functional Equations 1. Alternative 1 Note that functional values appear only in the left-hand side of the given equation. The best way to simplify it is to find y0 ∈ R for which f(y0 ) = 0. We show first that such a y0 exists. Indeed, since f(x − f(y)) = 1 − x − y it follows that for all x, y ∈ R the expression 1 − x − y is a functional value for f. Thus for x = 1, y = 0 (or any values for x and y for which x + y = 1) we have f(1 + f(0)) = 0 (or f(x − f(1 − x)) = 0). Therefore y0 = 1 + f(0) is the desired value. By setting y = y0 we have f(x) = 1 − x − y0 . This implies that f is a linear function of the form f(x) = c − x where c = 1 − y0 is a constant. The condition of the problem now gives 1−x−y

= = = =

Therefore 2c = 1, i.e. c =

1 . 2

f(x − f(y)) f(x − (c − y))

c − (x − c + y) 2c − x − y.

1 It is straightforward to check that the function f(x) = − x satisfies the 2 condition of the problem, so it is the only solution. Alternative 2 As in Alternative 1 we obtain that f(x − f(1 − x)) = 0 for any x ∈ R and f(x) = c − x for some constant c. The latter means that the only value of f for which f(x) = 0 is x = c. Therefore x − f(1 − x) = c for any x ∈ R. Since f(1 − x) = c − 1 + x we conclude that x − c + 1 − x = c, i.e. c = 12 .

Functional Equations103 5. Functional Equations 103

Functional Equations 2. First note that when x = ±1 then

x−3 = ±1 x+1

and

3+x = ±1. 1−x Let t be an arbitrary real number distinct from ±1. Then t=

x−3 x+1

x=

3+t . 1−t

is equivalent to

Since

3+t 3+ 3+x 1−t = t−3 = 3+t 1−x t+1 1− 1−t we may write the given equation in terms of t as 3+t t−3 = . f(t) + f t+1 1−t

(1)

Repeat the above by setting t=

3+x . 1−x

x=

t−3 t+1

We have that and

3+t x−3 = . x+1 1−t Again, writing the given equation in terms of t we get t−3 3+t f + f(t) = . 1−t t+1

Adding (1) and (2) gives 3+t t−3 t−3 3+t 2f(t) + f +f = + t+1 1−t 1−t t+1

(2)

104 104

Functional Equations Equations 5. Functional

which since f implies

t−3 t+1

+f

2f(t) + t = Thus the function is f(t) =

3+t 1−t

=t

8t . 1 − t2

4t t − . 2 1−t 2

Functional Equations 3. The functional equation from the condition of the problem is easily simplified by setting m = 0. We obtain f(n) + f(−1) = f(0)f(n) + 2 which leads to f(n)(f(0) − 1) = f(−1) − 2.

(1)

If f(0) = 1 then for any n ∈ Z we have f(n) =

f(−1) − 2 (f(0) − 1)

and hence f is a constant. Let f(n) = k for some k ∈ Z. The condition of the problem yields 2k = k2 + 2 so (k − 1)2 + 1 = 0 which is impossible. Therefore f(0) = 1 and (1) implies f(−1) = 2. It is reasonable now to set m = −1 in the initial equation. We obtain f(n − 1) + f(−n − 1) = 2f(n) + 2.

(2)

Note that the left-hand side of (2) is invariant with respect to replacing n by −n. Therefore the right-hand side should also be invariant under the same transformation. Thus 2f(n) + 2 = = =

f(n − 1) + f(−n − 1) f((−n) − 1) + f(−(−n) − 1) 2f(−n) + 2

which implies that f is an even function, i.e. f(n) = f(−n) for all n ∈ Z. Hence it suffices to find f(n) for all positive integers n. Replacing f(−n − 1) with f(n + 1) in (2) yields f(n + 1) + f(n − 1) = 2f(n) + 2.

Functional Equations105 5. Functional Equations 105

We conclude that f(n + 1) = 2f(n) − f(n − 1) + 2.

(3)

Note that if we know the functional values for two consecutive integers n − 1 and n then by (3) we may find f(n + 1). But we already know that f(−1) = 2, f(0) = 1. So by setting n = 0 in (3) we may find f(1). We have f(1)

= =

2f(0) − f(−1) + 2 2.1 − 2 + 2 = 2.

Further, n = 1 gives f(2)

= = =

2f(1) − f(0) + 2 4−1+2 5.

Continuing this way we fill the following table n f(n)

0 1 2 3 4 5 6 7 1 2 5 10 17 26 37 50

We see from the table that one reasonable assumption for the function is f(n) = n2 +1. We shall prove by induction on n that indeed f(n) = n2 +1 for all n ≥ 0. For n = 0 and n = 1 this is true since f(0) = 1 and f(1) = 2. Suppose that f(n) = n2 + 1 and

f(n − 1) = (n − 1)2 + 1.

Using (3) we find f(n + 1) = = =

2f(n) + 2 − f(n − 1)

2(n2 + 1) + 2 − (n − 1)2 − 1 n2 + 2n + 2 = (n + 1)2 + 1,

which completes the inductive step. Since f is even we conclude that f(n) = n2 + 1 for all n. Functional Equations 4. By setting n = 0 in the first condition we obtain f(1) = f(0) + 1 = 1. Now the second condition implies that f(2) = f(1) = 1. Turning to the first condition for n = 1 yields f(3) = 2 and so on. Therefore if we know f(0), f(1), . . . , f(m − 1) we may find f(m). Indeed, if m = 2n+1 is odd then f(m) = f(m−1)+1, and if m = 2n is even then

106 106

FunctionalEquations Equations 5. Functional

f(m) = f(n). It follows that f is uniquely determined by the condition of the problem, i.e. there exists only one function that is a solution of the problem. To find this function we start by finding the first few values. Thus we fill the following table n f(n)

0 1 0 1

2 3 4 5 1 2 1 2

6 7 8 2 3 1

9 10 2 2

The above values and the two conditions suggest that f(n) equals the number of ones in the binary expansion of n. In order to establish this we need two lemmas. Lemma 1. If g(n) represents the number of ones in the binary representation of n then g(2n) = g(n). Proof. Let ak ak−1 . . . a1 a0, where ai ∈ {0, 1} for i = 0, 1, . . ., k be the binary representation of n, i.e. n = ak 2k + ak−12k−1 + · · · + a121 + a020 . Thus 2n = ak 2k+1 + ak−12k + · · · + a122 + a021

and therefore ak ak−1 . . . a1a0 0 is the binary representation of 2n. Hence g(2n) = g(n). Lemma 2. If g(n) represents the number of ones in the binary representation of n then g(2n + 1) = g(2n) + 1. Proof. As in the proof of Lemma 1 we see that if ak ak−1 . . . a1 a0 is the binary representation of n then ak ak−1 . . . a1 a00 is the binary representation of 2n. Moreover the binary representation of 2n + 1 is given by ak ak−1 . . . a1 a0 1. Therefore g(2n + 1) = g(2n) + 1. Since obviously g(0) = 0, it follows from Lemma 1 and Lemma 2 that g(n) satisfies the two conditions for f in the statement of the problem. In view of uniqueness of f this means that g coincides with f. Functional Equations 5. First assume that f is a constant, i.e. for all x ∈ R we have f(x) = k for a constant k. The condition of the problem implies that k2 = 0 so k = 0. Thus f(x) = 0 for all x is a solution.

Functional Equations107 5. Functional Equations 107

We show that this is the only solution of the problem. Suppose there exists a function f that satisfies the condition of the problem and for which f(x) = 0 for at least one value of x ∈ R. By setting x = 0, y = 0 we obtain f 2 (0) = 0 so f(0) = 0. By setting x = 0 in the equation we have f(0)f(y) = f(y) − f(−y), which implies

f(y) = f(−y)

for all y ∈ R, i.e. f is an even function. Replace y with −y in the equation. We have f(x)f(−y)

= = =

f(x − y) − f(x + y) −(f(x + y) − f(x − y)) −f(x)f(y).

Since f(x) = 0 for at least one value of x the above equation implies that f(y) = −f(−y) for any y, i.e. f is an odd function. Finally, since f is both an even and odd function we have f(y) = f(−y) = −f(−(−y)) = −f(y) and therefore f(y) = 0 for all y ∈ R. Remark. After finding that f is an even function another way of showing that f is identically zero is the following. When y = x we have f(x)f(x) = f(2x) and when y = −x we obtain f(x)f(−x) = −f(2x). Since f(x) = f(−x) we conclude that f(2x) = −f(2x) and therefore f(2x) = 0 so f is identically zero. Functional Equations 6. First we show that f is a constant, i.e. for all x ∈ R we have f(x) = k for a constant k. By setting x = 0 we obtain f(0) = yf(y) for all y ∈ R. Therefore f(xy) = xf(x) + yf(y) = 2f(0). It is clear that when x and y vary then xy may take any real value. Thus f(x) = 2f(0) for any x so f is a constant. If f(x) = k then the condition of the problem implies k = (x + y)k. If k = 0 we conclude that x + y = 1 for all x and y, which is not true. Therefore k = 0 and f is identically zero.

108 108

FunctionalEquations Equations 5. Functional

Functional Equations 7. Note first that the constant function f(x) = k, where k is any positive integer, is a solution of the problem. We show that this is the only solution. Suppose that a function f satisfies the given condition and f is not constant. Then there exist some two positive integers x and y such that f(x) < f(y). This implies that (x + y)f(x) < xf(y) + yf(x) < (x + y)f(y) and using the condition of the problem we obtain (x + y)f(x) < (x + y)f(x2 + y2 ) < (x + y)f(y). We conclude that f(x) < f(x2 + y2 ) < f(y) because x + y is positive. Therefore for two different values f(x) and f(y), we find another value of the function strictly between them. We may repeat this step an arbitrary number of times. However, the function takes only integer values, so there is a finite number of positive integers between any two values of the function, which is a contradiction. Thus the function must be constant Functional Equations 8. By setting x = 0 and y = 0 into the functional equation, we find that f(0)2 − 2f(0) + 1 = 0 and hence f(0) = 1. By taking y = 1 we obtain f(x + 1) + f(x) = f(x)f(1) + 1, i.e. f(x + 1) = f(x)(f(1) − 1) + 1.

(1)

This equation shows that if we know f(1) and at least one value of f (note that we already know f(0) = 1) we may easily find by induction all values of f. Thus it is crucial to find f(1). Setting x = 1 and y = −1 implies f(0) + f(−1) = f(1)f(−1) + 1 and therefore f(−1)(f(1) − 1) = 0.

This yields that one possible value of f(1) is f(1) = 1. In this case (1) reduces to f(x + 1) = 1 for any x, so f is the constant 1. When f(1) = 1 we have f(−1) = 0.

Functional Equations109 5. Functional Equations 109

Now setting x = 2, y = −1 and x = −2, y = 1 yields f(1) + f(−2) = 1 and f(−2) = f(−2)f(1) + 1. Substituting f(−2) = 1 − f(1) into the second equation, we find that h(1)2 − 2h(1) = 0, i.e. h(1)(h(1) − 2) = 0,

implying that f(1) = 0 or f(1) = 2.

When f(1) = 0 equation (1) becomes f(x + 1) + f(x) = 1. Since f(0) = 1 by setting x = 0 and x = −1 we obtain f(1) = f(−1) = 0. Now x = 1 and x = −2 give f(2) = f(−2) = 1 and so on. The function becomes f(x) = 0 for x odd and f(x) = 1 for x even. When f(1) = 2 equation (1) implies f(x + 1) − f(x) = 1 and it follows that f(x) = x + 1. Functional Equations 9. Alternative 1 Suppose such a function exists. Note first that the left-hand side of the equation from the condition of the problem is symmetric with respect to x and y. This prompts that the right-hand side should be also symmetric with respect to x and y. The latter is possible only if a = b. We prove now that this is indeed true. Let x = y be positive integers. Using the condition of the problem we have xy + ax + by + c = = =

f(x + y) + f(x) + f(y) f(y + x) + f(y) + f(x) yx + ay + bx + c

which implies (a − b)(x − y) = 0. Since x = y we conclude that a = b. Hence we find

f(x + y) + f(x) + f(y) = xy + a(x + y) + c.

110 110

FunctionalEquations Equations 5. Functional

Our main goal now is to try to find the value of the function for some small values of x, say x = 1, 2, 3, 4. To begin with we use the substitution x = y = 1. We have f(2) + 2f(1) = 1 + 2a + c which is a relation between f(1) and f(2). The next useful substitution could be x = 1 and y = 2. It implies f(3) + f(1) + f(2) = 2 + 3a + c. At this stage we have two equations for three functional values f(1), f(2) and f(3). Note that the two substitutions x = 1, y = 3 and x = y = 2 will give two more equations for f(1), f(2), f(3) and f(4). Therefore we will have a system of 4 simultaneous equations with 4 unknowns f(1), f(2), f(3) and f(4). Hence the system becomes f(2) + 2f(1) = 1 + 2a + c f(3) + f(1) + f(2) = 2 + 3a + c f(4) + f(1) + f(3) = 3 + 4a + c f(4) + 2f(2) = 4 + 4a + c

Straightforward computations show that

3a + 2c 3a + c + 3 , f(2) = , 6 3 9a + 2c + 6 6a + c + 6 f(3) = , f(4) = . 6 3

f(1) =

In order to find the next functional value f(5) we have two options – setting x = 1, y = 4 or x = 2, y = 3. The corresponding equations become f(5)+f(1)+ f (4) = 4+5a+c and f(5)+f(2)+f (3) = 6+5a+c which imply f(2) + f(3) − f(1) − f(4) = 2.

But

f(2) + f(3) =

3a + c + 3 9a + 2c + 6 15a + 4c + 12 + = 3 6 6

and

15a + 4c + 12 3a + 2c 6a + c + 6 + = 6 6 6 which contradicts f(2) + f(3) − f(1) − f(4) = 2. Therefore a function having the desired property does not exist. f(1) + f(4) =

Functional Equations111 5. Functional Equations 111

Alternative 2 We show first that a function f that satisfies the condition of the problem is linear, i.e. f(x) = px + q for constants p, q ∈ N. Set y = 1 in the given equation. We obtain

f(x + 1) + f(x) = x(a + 1) + b + c − f(1).

(1)

Replace x by x + 1 in (1). We have f(x + 2) + f(x + 1) = (x + 1)(a + 1) + b + c − f(1).

(2)

By subtracting equation (1) from equation (2) we find f(x + 2) − f(x) = a + 1.

(3)

On the other hand, y = 2 gives f(x + 2) + f(x) = x(a + 2) + 2b + c − f(2).

(4)

By subtracting equation (3) from equation (4) we obtain f(x) = x

a + 2 2b + c − f(2) − a − 1 + , 2 2

which implies that f(x) is a linear function. Let f(x) = px + q for constants p ∈ N and q ∈ Z. The condition becomes xy = x(2p − a) + y(2p − b) + 3q − c. For x = y we find that x2 = x(2p + 2q − a − b) + 3q − c

(5)

for any x ∈ N. The latter is clearly impossible since there exist at most two values of x for which (5) is true. We conclude that a function having the desired property does not exist. Functional Equations 10. To simplify the equation we first set y = 0. We obtain f(f(x)) = f(x)(1 + f(0)). Replacing x by x + y in (1) we have f(f(x + y)) = f(x + y)(1 + f(0))

(1)

112 112

FunctionalEquations Equations 5. Functional

which together with the initial equation implies f(x)f(y) − xy = f(x + y)f(0).

(2)

Our next goal is to find f(0). Observe that if there exists t such that f(t) = 0 then by setting x = t in (1) we have f(0) = 0. Set x = −f(0) and y = f(0) in (2). We obtain f(−f(0))f(f(0)) = 0 which implies that there exists t (t = −f(0) or t = f(0)) for which f(t) = 0. According to our observation we conclude that f(0) = 0. Now (3) reduces to f(x)f(y) = xy. For x = 0 and y = 0 rewrite this equation in the form f(x) f(y) . = 1. x y

(3)

f(x) = c is a constant. Now (3) implies x 2 c = 1 and therefore c = 1 or c = −1, i.e. f(x) = −x or f(x) = x. Direct verification shows that f(x) = −x does not satisfy the initial equation, whereas f(x) = x is indeed a solution. Therefore the only solution is f(x) = x. When y is fixed we have that

Functional Equations 11. Note that all functional values are positive integers. Let m and n be such that f(n) ≤ f(m). (1)

We shall prove that f(n) is a factor of f(m). The condition of the problem implies that f(n − m) is a factor of f(n) − f(m). Therefore f(n − m) ≤ |f(n) − f(m)| ≤ f(m).

(2)

Further, since m = n − (n − m) it follows from the condition of the problem that f(m) is a factor of f(n) − f(n − m). If f(n) = f(n − m) we have f(m) ≤ |f(n) − f(n − m)| < max{f(n), f(n − m)}.

(3)

Now (3) contradicts either (1) (if f(n) > f(n − m)) or (2) (if f(n) < f(n − m)). Therefore f(n) = f(n − m).

Functional Equations113 5. Functional Equations 113

We already know that f(n − m) is a factor of f(n) − f(m), i.e. f(n) is a factor of f(n) − f(m). Hence f(n) is a factor of f(m), which completes the proof. Remark. This is IMO 2011 Problem 5. Functional Equations 12. We apply the approach from Example 4, i.e. we compute f(f(f(m))) in two ways. Since f(f(m)) = m + 2 we have first that f(f(f(m))) = f(m + 2).

(1)

On the other hand, for n = f(m) we obtain f(f(f(m))) = f(m) + 2.

(2)

Equations (1) and (2) imply f(m + 2) = f(m) + 2.

(3)

Equation (3) allows us to find the values of f(n) for odd n in terms of f(1). Indeed, for n = 2t + 1 we have f(2t + 1) = = =

f(2t − 1) + 2 f(2t − 3) + 4 ... f(1) + 2t.

Therefore, for n odd we have f(n) = f(1) + n − 1.

(4)

Similarly, for n = 2t we have f(2t)

= f(2t − 2) + 2

= f(2t − 4) + 4 ... = f(2) + 2t − 2,

hence for even n we obtain f(n) = f(2) + n − 2.

(5)

Set a = f(1) and b = f(2). Note that we may compute f(a) in two ways. First, directly from the condition of the problem we have f(a) = f(f(1)) = 3.

114 114

FunctionalEquations Equations 5. Functional

Secondly, if we know the parity of a then we may use equation (4) or equation (5). Suppose first that a is odd. According to (4) we write f(a) = a + a − 1 = 2a − 1. Therefore 2a − 1 = 3, so a = 2, a contradiction with the parity of a. Hence a is even and according to (5) we have f(a) = b + a − 2. Thus b + a − 2 = 3, i.e. a + b = 5. Since a is even we have a = 2, b = 3 or a = 4, b = 1. In the first case f1 (n) = n + 1 for all n and in the second f2 (n) = n + 3 for n odd and f2 (n) = n − 1 for n even. It remains to check whether the two functions satisfy the condition of the problem. We have f1 (f1 (n))

= =

f1 (n + 1) n+2

which means that f1 is a solution. For n odd we obtain f2 (f2 (n))

= = =

f2 (n + 3) (n + 3) − 1 n+2

and for n even we have f2 (f2 (n))

= = =

f2 (n − 1) (n − 1) + 3 n + 2.

Therefore both f1 and f2 are solutions of the problem. Functional Equations 13. By taking y = x we have f(2x) − f(0) = 4x2 which implies that f(x) = x2 + a for a = f(0). Therefore a function that obeys the equation from the problem is of the form f(x) = x2 + a for a constant a.

Functional Equations115 5. Functional Equations 115

We show that any function f(x) = x2 + a satisfies the condition of the problem. Indeed, f(x + y) − f(x − y)

= =

(x + y)2 + a − ((x − y)2 + a) 4xy,

and therefore the solution is f(x) = x2 + a.

GENERAL REFERENCES

General References The following is a general list of books of interest to the problem solver.

1. Andreescu T and Feng Z, 101 Problems in Algebra: From the Training of the USA IMO Team, AMT Publishing, Canberra, 2001. 2. Andronache M, Bˇ alunˇ a M, Gologan R, Popescu G, Schwarz D, S¸erbˇ anescu D, Romanian Mathematical Competitions, Romanian Mathematical Society, 2007. 3. Andronache M, Bˇ alunˇ a M, Becheanu M, Enescu B, Gologan R, Popescu G, Schwarz D, S ¸ erbˇ anescu D, Romanian Mathematical Competitions, Romanian Mathematical Society, 2011. 4. Artino RA, Gaglione AM and Shell N, The Contest Problem Book IV, Mathematical Association of America, 1983. 5. Atkins WJ, Problem Solving via the AMC, AMT Publishing, Canberra, 1992. 6. Atkins WJ, Edwards JD, King DJ, O’Halloran PJ and Taylor PJ, Australian Mathematics Competition, Book 1, 1978–1984, AMT Publishing, Canberra, 1986 and 2000. 7. Atkins WJ, Munro JEM and Taylor PJ, Australian Mathematics Competition, Book 3, 1992–1998, AMT Publishing, Canberra, 1998. 8. Atkins WJ and Taylor PJ, Australian Mathematics Competition, Book 4, 1999–2005, AMT Publishing, Canberra, 2006. 9. Barbeau EJ, Polynomials, Springer-Verlag, New York, 1989. 10. Beckenbach EF and Bellman R, An Introduction to Inequalities, Mathematical Association of America, 1961. 11. Berzsenyi G, International Mathematical Talent Search, Part 1, AMT Publishing, Canberra, 2010. 12. Berzsenyi G, International Mathematical Talent Search, Part 2, AMT Publishing, Canberra, 2011.

118 118

General References General References

13. Berzsenyi G and Maurer SB, The Contest Problem Book V, The Mathematical Association of America, 1996. 14. Burns JC, Seeking Solutions, AMT Publishing, Canberra, 2000. 15. Chinn WG and Steenrod NE, First Concepts of Topology, Mathematical Association of America, 1966. 16. Coxeter HSM, Introduction to Geometry, John Wiley & Sons, Inc., 1961. 17. Coxeter HSM and Greitzer SL, Geometry Revisited, Mathematical Association of America, 1967. 18. Fomin D and Kirichenko A, Leningrad Mathematical Olympiads, Mathpro Press, Westford, Massachusetts, 1994. 19. Greitzer SL, International Mathematical Olympiads 1959–1977, Mathematical Association of America, 1978. 20. Grossman I and Magnus W, Groups and Their Graphs, Mathematical Association of America, 1964. 21. Gueron S, Hungary-Israeli Mathematics Competition: The First Twelve Years, AMT Publishing, Canberra, 2004. 22. Henry JB, Dowsey J, Edwards AR, Mottershead LJ, Nakos A, Vardaro G, Taylor PJ, Challenge! Book 1, 1991–1998, AMT Publishing, Canberra, 2008. 23. Henry JB and Taylor PJ, Challenge! Book 2, 1999–2006, AMT Publishing, Canberra, 2009. 24. Holton D, Let’s Solve Some Math Problems, Canadian Mathematics Competition, Waterloo, Ontario, 1993. 25. Honsberger R, From Erd¨ os to Kiev, Mathematical Association of America, 1996. 26. Honsberger R, Ingenuity in Mathematics, Mathematical Association of America, 1970. 27. Honsberger R, Mathematical Gems I, Mathematical Association of America, 1973. 28. Honsberger R, Mathematical Gems II, Mathematical Association of America, 1976. 29. Honsberger R, Mathematical Gems III, Mathematical Association of America, 1985.

General References119 References 119 30. Honsberger R, Mathematical Morsels, Mathematical Association of America, 1978. 31. Honsberger R, Mathematical Plums, Mathematical Association of America, 1989. 32. Honsberger R, Mathematical Chestnuts from Around the World, Mathematical Association of America, 2001. 33. Kazarinoff ND, Geometric Inequalities, New Mathematical Association of America, 1961. 34. Klamkin MS, International Mathematical Olympiads 1978–1985, Mathematical Association of America, 1986. 35. Klamkin MS, USA Mathematical Olympiads, 1972–1986, Mathematical Association of America, 1988. 36. Kolev E, Boyvalenkov P, Mushkarov O, Nikolov N, Bulgarian Mathematical Competitions 2003-2006, GIL Publishing house, Zalau, 2007. 37. Kuczma ME, 144 Problems of the Austrian-Polish Mathematics Competition 1978–1993, The Academic Distribution Center, Freeland, Maryland, 1994. 38. Kuczma ME and Windisbacher E, Polish and Austrian Mathematical Olympiads 1981–1995: Selected Problems with Multiple Solutions, AMT Publishing, Canberra, 1998. 39. Larson LC, Problem Solving Through Problems, Springer-Verlag, New York, 1983. 40. Lausch H and Bosch Giral C, Asian Pacific Mathematics Olympiad, 1989–2000, AMT Publishing, Canberra, 2000. 41. Lausch H and Taylor PJ, Australian Mathematical Olympiads Book 1 1979–1995, AMT Publishing, Canberra, 1997. 42. Lausch H, Di Pasquale A, Hunt DC and Taylor PJ, Australian Mathematical Olympiads Book 2 1995–2011, AMT Publishing, Canberra, 2011. 43. Lazarov B, Tabov JB, Taylor PJ and Storozhev AM, Bulgarian Mathematics Competition, 1992–2001, AMT Publishing, Canberra, 2004. 44. Liu A, Chinese Mathematics Competitions and Olympiads Book 1, 1981–1993, AMT Publishing, Canberra, 1997.

120 120

General References General References

45. Liu A, Chinese Mathematics Competitions and Olympiads, Book 2, 1993–2001, AMT Publishing, Canberra, 2005. 46. Liu A, Hungarian Problem Book III: Based on the E¨ otvos Competitions, 1929–1943, Mathematical Association of America, 2001. 47. Liu A and Taylor PJ, International Mathematics Tournament of Towns, Problems and Solutions, Book 6, 2002–2007, AMT Publishing, Canberra, 2009. 48. Niven Ivan, Mathematics of Choice - How to count without counting, Mathematical Association of America, 1965. 49. O’Halloran PJ, Pollard GH and Taylor PJ, Australian Mathematics Competition, Book 2, 1985–1991, AMT Publishing, Canberra, 1992 and 1998. 50. Ore O, Graphs and Their Uses, Mathematical Association of America, 1963, revised and updated by R Wilson, 1990. 51. Ore O, Invitation to Number Theory, Mathematical Association of America, 1967. 52. Plank AW and Williams NH, Mathematical Toolchest, AMT Publishing, Canberra, 1992 and 1996. 53. Rabinowitz S, Index to Mathematical Problems 1980–1984, Mathpro Press, Westford, Massachusetts, 1992. 54. Rapaport E, Hungarian Problem Book I, Mathematical Association of America, 1963. 55. Rapaport E, Hungarian Problem Book II, Mathematical Association of America, 1963. 56. Salkind CT, The Contest Problem Book, Random House, 1961. 57. Salkind CT, The Contest Problem Book II, Random House, 1966. 58. Salkind CT and Earl JM, The Contest Problem Book III, Random House, 1973. 59. Schneider LJ, The Contest Problem Book VI, Mathematical Association of America, 1998. 60. Sharygin IF, Problems in Plane Geometry, Mir, Moscow, 1988. 61. Sharygin IF, Problems in Solid Geometry, Mir, Moscow, 1986.

General References121 References 121 62. Slinko AM, USSR Mathematical Olympiads 1989–1992, AMT Publishing, Canberra, 1997. 63. Storozhev A, International Mathematics Tournament of Towns, Problems and Solutions, Book 5, 1997–2002, AMT Publishing, Canberra, 2005. 64. Tabov JB and Taylor PJ, Methods of Problem Solving, Book 1, AMT Publishing, Canberra, 1996. 65. Tabov JB and Taylor PJ, Methods of Problem Solving, Book 2, AMT Publishing, Canberra, 2002. 66. Taylor PJ, International Mathematics Tournament of Towns, Problems and Solutions, Book 1, 1980–1984, AMT Publishing, Canberra, 1993 and 2006. 67. Taylor PJ, International Mathematics Tournament of Towns, Problems and Solutions, Book 2, 1984–1989, AMT Publishing, Canberra, 1992 and 2003. 68. Taylor PJ, International Mathematics Tournament of Towns, Problems and Solutions, Book 3, 1989–1993, AMT Publishing, Canberra, 1994. 69. Taylor PJ and Storozhev A, International Mathematics Tournament of Towns, Problems and Solutions, Book 4, 1993–1997, AMT Publishing, Canberra, 1998. 70. Yaglom IM, Geometric Transformations, Mathematical Association of America, 1962. 71. Yaglom IM, Geometric Transformations II, Mathematical Association of America, 1968. 72. Yaglom IM, Geometric Transformations III, Mathematical Association of America, 1973.

ABOUT THE AUTHORS Jordan Tabov

Emil Kolev Emil Kolev is an Associate Professor at the Institute of Mathematics and Informatics at the Bulgarian Academy of Sciences, a recent Bulgarian Team Leader at IMO and a former IMO Bronze Medallist as a participant.

Peter Taylor Peter Taylor is Executive Director of the Australian Mathematics Trust. He was Chairman of the Problems Committee of the Australian Mathematics Competition from 1979 to 1994.

AUSTRALIAN MATHS TRUST

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Copyright © 2019 Australian Mathematics Trust

METHODS OF PROBLEM SOLVING — BOOK 3

Jordan Tabov is a Professor at the Institute of Mathematics and Informatics at the Bulgarian Academy of Sciences, a former Bulgarian Team Leader at IMO and a former IMO Bronze Medallist as a participant.