Shortlisted Problems and Solutions - International Mathematical ...

51 st International Mathematical Olympiad
Astana, Kazakhstan 2010
Shortlisted Problems with Solutions

Contents
Note of Confidentiality 5
Contributing Countries & Problem Selection Committee 5
Algebra 7
Problem A1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Problem A2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Problem A3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Problem A4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Problem A5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Problem A6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Problem A7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
Problem A8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Combinatorics 23
Problem C1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Problem C2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
Problem C3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Problem C4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
Problem C4½. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
Problem C5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Problem C6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Problem C7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
Geometry 44
Problem G1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Problem G2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Problem G3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Problem G4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
Problem G5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
Problem G6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Problem G6½. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Problem G7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Number Theory 64
Problem N1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
Problem N1½. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
Problem N2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
Problem N3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Problem N4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
Problem N5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
Problem N6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

Note of Confidentiality
The Shortlisted Problems should be kept
strictly confidential until IMO 2011.
Contributing Countries
The Organizing Committee and the Problem Selection Committee of IMO 2010 thank the
following 42 countries for contributing 158 problem proposals.
Armenia, Australia, Austria, Bulgaria, Canada, Columbia, Croatia,
Cyprus, Estonia, Finland, France, Georgia, Germany, Greece,
Hong Kong, Hungary, India, Indonesia, Iran, Ireland, Japan,
Korea (North), Korea (South), Luxembourg, Mongolia, Netherlands,
Pakistan, Panama, Poland, Romania, Russia, Saudi Arabia, Serbia,
Slovakia, Slovenia, Switzerland, Thailand, Turkey, Ukraine,
United Kingdom, United States of America, Uzbekistan
Problem Selection Committee
Yerzhan Baissalov
Ilya Bogdanov
Géza Kós
Nairi Sedrakyan
Damir Yeliussizov
Kuat Yessenov

Algebra
A1. Determine all functions f : RR such that the equality
fÔÖx×yÕfÔxÕÖfÔyÕ×. (1)
holds for all x, yÈR. Here, byÖx×we denote the greatest integer not exceeding x.
Answer. fÔxÕconstC, where C0 or 1C2.
fÔ0ÕfÔ0ÕÖfÔyÕ× (2)
(France)
Solution 1. First, setting x0 in (1) we get
for all yÈR. Now, two cases are possible.
Case 1. Assume that fÔ0Õ0. Then from (2) we conclude thatÖfÔyÕ×1 for all
yÈR. Therefore, equation (1) becomes fÔÖx×yÕfÔxÕ, and substituting y0we have
fÔxÕfÔ0ÕC0. Finally, fromÖfÔyÕ×1ÖC×we obtain that 1C2.
Case 2. Now we have fÔ0Õ0. Here we consider two subcases.
NÖz×
Subcase 2a. Suppose that there exists 0α1 such that fÔαÕ0. Then setting xα
in (1) we obtain 0fÔ0ÕfÔαÕÖfÔyÕ×for all yÈR. Hence,ÖfÔyÕ×0 for all yÈR. Finally,
substituting x1in (1) provides fÔyÕ0for all yÈR, thus contradicting the condition
fÔαÕ0.
Subcase 2b. Conversely, we have fÔαÕ0for all 0α1. Consider any real z; there
exists an integer N such that αz
1Õ(one may set 1 if z0and
NÈÖ0, NÖzס1
otherwise). Now, from (1) we get fÔzÕfÔÖN×αÕfÔNÕÖfÔαÕ×0 for all zÈR.
Finally, a straightforward check shows that all the
fÔxÕfÔÖx×aÕ
obtained functions satisfy (1).
Solution 2. Assume thatÖfÔyÕ×0 for some y; then the substitution x1 provides
fÔyÕfÔ1ÕÖfÔyÕ×0. Hence, ifÖfÔyÕ×0for all y, then fÔyÕ0for all y. This function
obviously satisfies the problem conditions.
So we are left to consider the case whenÖfÔaÕ×0 for some a. Then we have
fÔÖx×aÕfÔxÕÖfÔaÕ×, or
This means that fÔx 1ÕfÔx 2ÕwheneverÖx fÔxÕfÔ0Õ 1×Öx 2×, hence fÔxÕfÔÖx×Õ, and we may assume
that a is an integer.
Now we have
fÔaÕf 2a¤1
2¨fÔ2aÕf
1
2¨fÔ2aÕÖfÔ0Õ×;
this impliesÖfÔ0Õ×0, so we may even assume that a0. Therefore equation (3) provides
ÖfÔ0Õ×C0
for each x. Now, condition (1) becomes equivalent to the equation CCÖC×which holds
exactly whenÖC×1.
ÖfÔaÕ×. (3)

8
A2. Let the real numbers a, b, c, d satisfy the relations a b c d6 and a 2 b 2 c 2 d 212.
Prove that
364Ôa 3 b 3 c 3 d 3Õ¡Ôa 4 b 4 c 4 d 4Õ48.
¡ 4Õ¡ Ôd¡1Õ4¨ dÕ
(Ukraine)
Solution 1. Observe that
4Ôa 3 b 3 c 3 d 3Õ¡Ôa 4 b 4 c 4 d Ôa¡1Õ4 6Ôa 2 b 2 c 2 d 2Õ¡4Ôa b c 4
Ôa¡1Õ4 Ôb¡1Õ4 Ôc¡1Õ4 52.
dÕ
Now, introducing xa¡1, yb¡1, zc¡1, td¡1, we need to prove the inequalities
16x 4 y 4 z 4 t 44,
under the constraint
x 2 y 2 z 2 t 2Ôa 2 b 2 c 2 d 2Õ¡2Ôa b c 44 (1)
(we will not use the value of x y z t though it can be found).
Now the rightmost inequality in (1) follows from the power 4Õ mean inequality:
x 4 y 4 z 4 t 4Ôx2 y 2 z 2 t 2Õ2
4.
4
For the other one, expanding the brackets we note that
Ôx 2 y 2 z 2 t 2Õ2Ôx 4 y 4 z 4 t q,
where q is a nonnegative number, so
and we are done.
x 4 y 4 z 4 t 4Ôx 2
Ôb¡1Õ4 Ôc¡1Õ4 Ôd¡1Õ4¨
y 2 z 2 t 2Õ216,
Comment 1. The estimates are sharp; the lower and upper bounds are attained atÔ3,1,1,1Õand
Ô0,2,2,2Õ, respectively.
Comment 2. After the change of variables, one can finish the solution in several different ways.
The latter estimate, for instance, it can be performed by moving the variables – since we need only
the second of the two shifted conditions.
Solution 2. First, we claim that 0a, b, c, d3. Actually, we have
a b c6¡d, a 2 b 2 c 212¡d 2 ,
hence the power mean inequality
a 2 b 2 c 2Ôa b cÕ2
3
rewrites as
2dÔd¡3Õ0,
12¡d 2Ô6¡dÕ2
3

which implies the desired inequalities for d; since the conditions are symmetric, we also have
the same estimate for the other variables.
Now, to prove the rightmost inequality, we use the obvious inequality x 2Ôx¡2Õ20for
each real 4Õ 4Õ x; this inequality 4Õ rewrites 4Õ as 4x 3¡x44x2 . It follows 2Õ that
Ô4a 3¡a 2 b 2 c 2 d 2Õ48,
as desired.
Now we prove the leftmost inequality in an analogous way. For each xÈÖ0, 3×, we have
Ôx 1ÕÔx¡1Õ2Ôx¡3Õ0 which is equivalent to 4x 3¡x42x2 4x¡3. This implies that
Ô4a 3¡a Ô4b 3¡b Ô4c 3¡c Ô4d 3¡d 4Õ2Ôa 2 b 2 c 2 d 4Ôa b c dÕ¡1236,
as desired.
4Õ Ô4b 3¡b
4Õ Ô4c 3¡c
Ô4d 3¡d 4Õ4Ôa
Comment. It is easy to guess the extremal pointsÔ0,2,2,2ÕandÔ3,1,1,1Õfor this inequality. This
provides a method of finding the polynomials used in Solution 2. Namely, these polynomials should
have the form x 4¡4x3 ax 2 bx c; moreover, the former polynomial should have roots at 2 (with
an even multiplicity) and 0, while the latter should have roots at 1 (with an even multiplicity) and 3.
These conditions determine the polynomials uniquely.
Solution 3. First, expanding 484Ôa 3 2Õ 2 b 3 2 c 3 2Õ 2Õ 2© 2Õ
2 d 2Õand applying the AM–GM inequality,
we have
a 4 b 4 c 4 d 4 48Ôa 4 4a Ôb 4 4b Ôc 4 4c Ôd 4 4d
2¡a 4¤4a b 2 4¤4b c 2 4¤4c d 2 4¤4d
4Ôa b c 3 b 3 c 3 d 3Õ,
c
which establishes the rightmost inequality.
To prove the leftmost inequality, we first show that a, b, c, dÈÖ0, 3×as in the previous
solution. Moreover, we can assume that 0abcd. Then we have a bb 2
3Ôb c dÕ2
3¤64.
Next, we show 2Õ that 4b¡b24c¡c2 . Actually, this inequality rewrites asÔc¡bÕÔb c¡4Õ0,
which follows from the previous estimate. The inequality 4a¡a24b¡b2 can be proved
analogously.
Further, the inequalities abctogether with 4a¡a24b¡b24c¡c2 allow us to
apply the Chebyshev inequality obtaining
a 2Ô4a¡a 4Õ 2Õ b 2Ô4b¡b 4Õ c
2Ô4c¡c 4Õ 2Õ 2ÕÕ
2Õ¨
2Õ1
3Ôa 2 b 2 c 4Ôa b cÕ¡Ôa 2 b 2 c
Ô12¡d 2ÕÔ4Ô6¡dÕ¡Ô12¡d .
3
This implies that
12Õ
Ô4a 3¡a Ô4b 3¡b Ô4c 3¡c Ô4d 3¡d 4ÕÔ12¡d 2ÕÔd2¡4d 4d 3¡d 4
3
144¡48d 16d 3¡4d4
4
36
3Ô3¡dÕÔd¡1ÕÔd 2¡3Õ. (2)
3
Finally, we have d 21
4Ôa2 b 2 c 2 d 2Õ3(which implies d1); so, the expression
3Ô3¡dÕÔd¡1ÕÔd 4
2¡3Õin the right-hand part of (2) is nonnegative, and the desired inequality
is proved.
Comment. The rightmost inequality is easier than the leftmost one. In particular, Solutions 2 and 3
show that only the condition a 2 b 2 c 2 d 212 is needed for the former one.
d 3Õ4Ôa
9

10
A3. Let x 1 , . . .,x 100 be nonnegative real numbers such that x i x i 1 x i 21 for all
i1, . . .,100 (we put x 101x 1 , x 102x 2 ). Find the maximal possible value of the sum
S100
x i x i 2.
ôi1
Answer. 25
2 .
1Õ 2i x 2i
(Russia)
Solution 1. Let x 2i0, x 2i¡11
for all i1, . . ., 50. Then we have S50¤
1Õ
1
2 2¨225
. So, 2
we are left to show that S25
for all values of x 2 i’s satisfying the problem conditions.
Consider any 1i50. By the problem condition, we get x 2i¡11¡x 2i¡x 2i 1 and
x 2i 21¡x 2i¡x 1Õ
2i 1. Hence by the AM–GM inequality we get
x 2i¡1x 2i 1 x 2i x 2i 2Ô1¡x 2i¡x 2i 1Õx 2i 1 x 2iÔ1¡x 2i¡x 2i
1Õ¢Ôx Ô1¡x 2i¡x Ôx 2i x 2i 1ÕÔ1¡x 2i
Å21
2i¡x 2i
2 4 .
Summing up these inequalities for i1, 2, . . ., 50, we get the desired inequality
50ôi1Ôx 2i¡1x 2i 1 x 2i x 2i 2Õ50¤1
425
2 .
Comment. This solution shows that a bit more general fact holds. Namely, consider 2n nonnegative
numbers x 1 ,... ,x 2n in a row (with no cyclic notation) and suppose that x i x i 1 x i 21 for all
i1,2,... ,2n¡2. Then
2n¡2 ôi1
x i x i 2n¡1
.
4
The proof is the same as above, though if might be easier to find it (for instance, applying
induction). The original estimate can be obtained from this version by considering the sequence
x 1 ,x 2 ,... ,x 100 ,x 1 ,x 2 .
Solution 2. We present another proof of the estimate. From the problem condition, we get
S100
x i x i 2100 ôi1x iÔ1¡x i¡x i 1Õ100 ôi1x i¡100
xi¡100
ôi1 ôi1 2 x i x i 1
ôi1
100 100 ôi1x i¡1 ôi1Ôx i x 1Õ2 i .
2
By the AM–QM inequality, we haveÔx i x i 1Õ21
100
Ôx
S100 ôi1x i¡1 ôi1Ôx i x i 1Õ«2100
200£100
And finally, by the AM–GM inequality
i x 1Õ¨2 i
x i«2
ôi1
ôi1 x i¡2
100£100
2 ôi1x i«£100
100£100
ôi1x i««22
, so
2¡100 ôi1
x i«.
100
S2
x i
100¤£1
2£100
2¡100 ôi1
100¤¢100
4Å225
2 .

11
Comment. These solutions are not as easy as they may seem at the first sight. There are two
different optimal configurations in which the variables have different values, and not all of sums of
three consecutive numbers equal 1. Although it is easy to find the value 25 2
, the estimates must be
done with care to preserve equality in the optimal configurations.

12
x
¤¤¤
A4. A sequence x 1 , x 2 , . . . is defined by x 11and x 2k¡x k , x 2k¡1Ô¡1Õk k1. Prove that x 1 x 2
n0 for all n1.
1
x k for all
(Austria)
Solution 1. We start with some observations. First, from the definition of x i it follows that
for each positive integer k we have
x 4k¡3x 2k¡1¡x 4k¡2 2kx k . (1)
Hence, denoting S nn
i1 x i, we have
S Ôx 4k¡3 x 4kkôi1 4k¡2Õ Ôx 4k¡1 x 4kÕ¨kôi1Ô0 2x kÕ2S k , (2)
S 4k 2S 4k Ôx 4k 1 x 4k 2ÕS 4k . (3)
Observe also that S nn
i1 x in
i1 1nÔmod 2Õ.
Now we prove by induction on k that S i0for all i4k. The base case is valid since
x 1x 3x 41, x 2¡1. For the induction step, assume that S i0for all i4k. Using
the relations (1)–(3), we obtain
4k 2 S 4k
S 4k 42S k 10, S 4k 2S 4k0, S 4k 3S 4k 2 x 4k
1
4
3S 0.
2
So, we are left to prove that S 4k 10. If k is odd, then S 4k2S k0; since k is odd, S k
is odd as well, so we have S 4k2 and hence S 4k 1S 4k x 4k 11.
Conversely, if k is even, then we have x 4k 1x 2k 1x k 1, hence S 4k 1S 4k x 4k
2S k x k 1S k S k 10. The step is proved.
and x 4k¡1x 4k¡x
Solution 2. We will use the notation of S n and the relations (1)–(3) from the previous
solution.
Assume the contrary and consider the minimal n such that S n 10; surely n1, and
from S n0 we get S n0, x n 1¡1. Hence, we are especially interested in the set
MØn : S n0Ù; our aim is to prove that x n 11whenever nÈM thus coming to a
contradiction.
For this purpose, we first describe the set M inductively. We claim that (i) M consists only
of even numbers, (ii) 2ÈM, and (iii) for every even n4we have nÈMÖnß4×ÈM.
Actually, (i) holds since S nnÔmod 2Õ, (ii) is straightforward, while (iii) follows from the
relations S 4k 2S 4k2S k .
Now, we are left to prove that x n 11if nÈM. We use the induction on n. The base
case is n2, that is, the minimal element of M; here we have x 31, as desired.
For the induction step, consider some 4nÈM and let mÖnß4×ÈM; then m is even,
and x m 11by the induction hypothesis. We prove that x n 1x m 11. If n4m then we
have x n 1x 2m 1x m 1 since m is even; otherwise, n4m 2, and x n 1¡x 2x m 1,
as desired. The proof is complete.
2m
Comment. Using the inductive definition of set M, one can describe it explicitly. Namely, M consists
exactly of all positive integers not containing digits 1 and 3 in their 4-base representation.

13
A5. Denote by Q the set of all positive rational numbers. Determine all functions f : Q Q
which satisfy the following equation for all x, yÈQ :
f fÔxÕ2 y¨x 3
fÔxyÕ. (1)
(Switzerland)
fÔxÕ1
Answer. The only such function is
x .
Solution. By substituting y1, we get
fÔxÕ2¨
fÔyÕ2¨
f fÔxÕ2¨x 3 fÔxÕ. (2)
Then, whenever fÔxÕfÔyÕ, we have
x 3f f
y 3
which implies xy, so the function f is injective.
Now replace x by xy in (2), and apply (1) twice, second time to y, fÔxÕ2¨instead ofÔx, yÕ:
Since f is injective, we get
f fÔxyÕ2¨ÔxyÕ3
fÔxyÕy 3 f fÔxÕ2 y¨f
fÔxyÕfÔxÕfÔyÕ.
Therefore, f is multiplicative. This also implies fÔ1Õ1
fÔxyÕ2fÔxÕ2 fÔyÕ2 ,
fÔxÕ2
fÔyÕ2¨.
and fÔxnÕfÔxÕn for all integers n.
Then the function equation (1) can be re-written as
f fÔxÕ¨2 fÔxÕ
fÔyÕx xfÔxÕ¨5ß2 fÔxÕ¨
3 fÔxÕfÔyÕ,
f fÔxÕ¨x 3 fÔxÕ. (3)
Let gÔxÕxfÔxÕ. Then, by (3), we have
ÐÓÓÓÓÑÓÓÓÓÒÔxÕ...¨© g gÔxÕ¨g xfÔxÕ¨xfÔxÕ¤f xfÔxÕ¨xfÔxÕ2 f
xfÔxÕ2x 3 gÔxÕ¨5ß2 ,
and, by induction,
gÔxÕ¨Ô5ß2Õn
g¡g ...g
(4)
n 1
for every positive integer n.
Consider (4) for
ÐÓÓÓÓÑÓÓÓÓÒ
gÔxÕ¨Ô5ß2Õn a fixed x. The left-hand side is always rational, so must be
rational for every n. We show that this is possible only if gÔxÕ1. Suppose that gÔxÕ1,
and let the prime factorization of gÔxÕbe gÔxÕp α 1
1 . . .p α k
k
where p 1 , . . .,p k are distinct primes
and α 1 , . . .,α k are nonzero integers. Then the unique prime factorization of (4) is
g¡g ...g gÔxÕ¨Ô5ß2ÕnpÔ5ß2Õn α ¤¤¤pÔ5ß2Õn 1 α k
1 k
n 1
ÔxÕ...¨©

14
where the exponents should be integers. But this is not true for large values of n, for example
. Therefore, gÔxÕ1 is impossible.
yÕ y
Hence, gÔxÕ1 and for all x.
fÔxÕ1
The function satisfies the equation (1):
x fÔfÔxÕ2 1 1
fÔxÕ2 x¨2 1
yx3
xyx 3 fÔxyÕ.
fÔxÕ1
Comment. Among R R functions, is not the only solution. Another solution is
x
f 1ÔxÕx . Using transfinite tools, infinitely many other solutions can be constructed.
Ô5
2Õn α 1 cannot be a integer number when 2 nЧα
3ß2
fÔxÕ1
thus
x
1

gÔnÕ fÔgÔnÕÕfÔnÕ gÔfÔnÕÕ
15
A6. Suppose that f and g are two functions defined on the set of positive integers and taking
positive integer values. Suppose also that the equations 1 and
1 hold for all positive integers. Prove that fÔnÕgÔnÕfor all positive integer n.
(Germany)
kÔxÕ¨gÔxÕ k¡1ÔxÕ¨ 1¤¤¤fÔxÕ ÐÓÓÓÓÑÓÓÓÓÒ
Solution 1. Throughout the solution, by N we denote the set of all positive integers. For
any function h: NNand for any positive integer k, define h kÔxÕh h . . .hÔxÕ...¨¨(in
k
particular, h 0ÔxÕx).
Observe that f g kÔxÕ¨f g k for any positive integer k, and
similarly g f k. Now let a and b are the minimal values attained by f and g,
respectively; say fÔn fÕa, gÔn gÕb. Then we have f g kÔn fÕ¨a k, g f kÔn gÕ¨b k, so
the function f attains all values from the set N fØa, a 1, . . .Ù, while g attains all the values
from the set N gØb, b 1, . . .Ù.
Next, note that fÔxÕfÔyÕimplies gÔxÕg fÔxÕ¨¡1g fÔyÕ¨¡1gÔyÕ; surely, the
converse implication also holds. Now, we say that x and y are similar (and write xy) if
fÔxÕfÔyÕ(equivalently, gÔxÕgÔyÕ). For every xÈN, we defineÖx×ØyÈN:xyÙ;
surely, y 1y 2 for all y 1 , y 2ÈÖx×, soÖx×Öy×whenever yÈÖx×.
Ð
Now we investigate the structure of the setsÖx×.
Claim 1. Suppose that fÔxÕfÔyÕ; then xy, that is, fÔxÕfÔyÕ. Consequently, each
classÖx×contains at most one element from N f , as well as at most one element from N g .
Proof. If fÔxÕfÔyÕ, then we have gÔxÕg fÔxÕ¨¡1g fÔyÕ¨¡1gÔyÕ, so xy. The
second statement follows now from the sets of values of f and g.
Next, we clarify which classes do not contain large elements.
fÔyÕ
Claim 2. For any xÈN, we haveÖx×Ø1,
gÔyÕ¨fÔyÕ gÔyÕ¨
2, . . .,b¡1Ùif and only if fÔxÕa. Analogously,
Ð
Öx×Ø1, 2, . . ., a¡1Ùif and only if gÔxÕb.
Proof. We will prove thatÖx×Ø1, 2, . . .,b¡1ÙfÔxÕa; the proof of the second
statement is similar.
Note that fÔxÕaimplies that there exists some y satisfying fÔyÕfÔxÕ¡1, so f
1fÔxÕ, and hence xgÔyÕb. Conversely, if bcxthen cgÔyÕfor some yÈN,
which in turn follows fÔxÕf 1a 1, and hence fÔxÕa.
Claim 2 implies that there exists exactly one class contained inØ1, . . .,a¡1Ù(that is, the
classÖn g×), as well as exactly one class contained inØ1, . . .,b¡1Ù(the classÖn f×). Assume for a
moment that ab; thenÖn g×is contained inØ1, . . .,b¡1Ùas well, hence it coincides withÖn g×.
So, we get that
Ð
Claim 3. ab.
Proof. By Claim 2, we haveÖa×Ön f×, soÖa×should contain some element a½b by Claim 2
again. If aa½, thenÖa×contains two elementsawhich is impossible by Claim 1. Therefore,
aa½b. Similarly, ba.
Now we are ready to prove the problem statement. First, we establish the following
fÕ Ð
d.
Proof. Induction on d. For d0, the statement follows from (1) and Claim 3. Next, for d1
f g dÔn fÕ¨fÔn da d.
d is analogous.
fÔxÕagÔxÕbxn fn g . (1)
Claim 4. For every integer d0, f d 1Ôn fÕgd 1Ôn fÕa from the induction hypothesis we have f d 1Ôn fÕf dÔn fÕ¨f The equality g d 1Ôn fÕa

16
Finally, for each xÈN, we have fÔxÕa d for some d0, so fÔxÕf g dÔn fÕ¨and
hence xgdÔn fÕ. It follows that gÔxÕg g dÔn fÕ¨gd 1Ôn fÕa fÔaÕ¨gÔaÕ
dfÔxÕby Claim 4.
Solution 2. We start with the same observations, introducing the relationand proving
Claim
fÔaÕfÔxÕ
1 from the previous solution.
Note that fÔaÕa since otherwise we have fÔaÕaand hence gÔaÕg Ð1,
which is false.
Claim 2½. ab.
Proof. We can assume that ab. Since fÔaÕa 1, there exists some xÈNsuch that
1, which is equivalent to fÔaÕf gÔxÕ¨and agÔxÕ. fÔaÕfÔxÕ
Since gÔxÕba, by
Claim 1 we have agÔxÕb, which together with ab proves the Claim.
Now, almost the same method allows to find the values fÔaÕand gÔaÕ.
Claim 3½. fÔaÕgÔaÕa 1.
Proof. Assume the contrary; then fÔaÕa 2, hence there exist some x, yÈNsuch that
fÔxÕfÔaÕ¡2 and fÔyÕgÔxÕ(as gÔxÕab). Now we get 2f g 2ÔxÕ¨,
so ag2ÔxÕa, and by Claim
gÔxÕ¨fÔxÕ
1 we get ag2ÔxÕg 1Õ
fÔyÕ¨1 gÔyÕ1 a; this is
impossible. The equality gÔaÕa 1 is similar.
Now, we are prepared for the proof of the problem statement. First, we prove it for na.
Claim 4½. For each integer xa, we have fÔxÕgÔxÕx 1.
Proof. Induction on x. The base case xais provided by Claim 3½, while the induction
step
fÔnÕ gÔnÕ¨gÔnÕ fÔnÕ fÔfÔnÕÕ 1
follows from fÔx 1Õf 1Ôx 1 and the similar computation
for gÔx 1Õ.
Finally, for an arbitrary nÈNwe have gÔnÕa, so by Claim 4½we have
f 1, hence fÔnÕgÔnÕ.
Comment. It is not hard now to describe all the functions f : NN satisfying the property
1. For each such function, there exists n 0ÈN such that fÔnÕn 1 for all nn 0 , while for
each nn 0 , fÔnÕis an arbitrary number greater than of equal to n 0 (these numbers may be different
for different nn 0 ).

17
A7. Let a 1 , . . .,a r be positive real numbers. For nr, we inductively define
a nmax
1kn¡1Ôa k a n¡kÕ. (1)
Prove that there exist positive integers lr and N such that a na n¡l a l for all nN.
¤¤¤
(Iran)
Solution 1. First, from the problem conditions we have that each a n (nr) can be expressed
as a na j1 a j2 with j 1 , j 2n, j 1 j 2n. If, say, j 1r then we can proceed in the same
way with a j1 , and so on. Finally, we represent a n in a form
a na i1 a ik , (2)
i kn. (3)
Moreover, if a i1
¤¤¤
and a i2 are the numbers in (2) obtained on the last step, then i 1 i 2r.
Hence we can adjust (3) as
i kn, i 1 i 2r.
¤¤¤
(4)
On the other hand, suppose that the indices i 1 , . . .,i k satisfy the conditions (4). Then,
denoting s ji 1 i j , from (1) we have
a na ska sk¡1 ¤¤¤
a ika sk¡2 a ik¡1 a ik¤¤¤a i1 a ik
Ð
.
Summarizing these observations we get the following
Claim. For every nr, we have
a nmaxØa i1 a ik : the collectionÔi 1 , . . .,i kÕsatisfies (4)Ù.
¤¤¤
Now we denote
a i
smax
1ir i
l
and fix some index lr such that sa l .
Consider some nr2 l 2r and choose an expansion of a n in the form (2), (4). Then we have
ni 1 i krk,
¤¤¤ ¤¤¤
so knßrrl 2. Suppose that none of the numbers i 3 , . . ., i k equals l.
Then by the pigeonhole principle there is an index 1jr which appears among i 3 , . . .,i k
at least l times, and surely jl. Let us delete these l occurrences of j fromÔi 1 , . . ., i kÕ, and
add j occurrences of l instead, obtaining a sequenceÔi 1 , i 2 , i½3 , . . .,i½k½Õalso satisfying (4). By
Claim, we have
a i1 a ika na i1 a i2 a i½3
a i½k½,
or, after removing the coinciding terms, la jja¤¤¤ l , so a l j
. By the definition of l, this
la j
means that la jja l , hence
a na i1 a i2
¤¤¤
a i½3
a i½k½.
Thus, for every nr2 l 2r we have found a representation of the form (2), (4) with i jl
for some j3. Rearranging the indices we may assume
ik¡1Õ
that i kl.
Finally, observe that in this representation, the indicesÔi 1 , . . ., i k¡1Õsatisfy the conditions
(4) with n replaced by n¡l. Thus, from the Claim we get
a n¡l a lÔa i1 a a la n ,
which by (1) implies
a na n¡l a l for each nr 2 l 2r,
as desired.
1i jr, i 1 ¤¤¤
1i jr, i 1 ¤¤¤

18
Solution 2. As in the previous solution, we involve the expansion (2), (3), and we fix some
index 1lr such that
a l a i
lsmax
1ir i .
Now, we introduce the sequenceÔb nÕas b na n¡sn; then b l0.
We prove by induction on n that b n0, andÔb nÕsatisfies the same recurrence relation
asÔa nÕ. The base cases nrfollow from the definition of s. Now, for nrfrom the
induction hypothesis we have
b nmax
1kn¡1Ôa k a n¡kÕ¡nsmax
1kn¡1Ôb k b n¡k nsÕ¡nsmax
1kn¡1Ôb k b n¡kÕ0,
as required.
Now, if b k0 for all 1kr, then b n0 for all n, hence a nsn, and the statement is
trivial. Otherwise, define
Mmax
1irb i,
εminØb i:1ir, b i0Ù.
Then for nr we obtain
b nmax
1kn¡1Ôb k b n¡kÕb
l
b n¡lb n¡l,
so
¤¤¤
contained in a set
b
We claim that this set is finite. Actually, for any xÈT, let xb i1 b ik (i 1 , . . .,i kr).
Then among b ij ’s there are at most M nonzero terms (otherwise xM
Thus
ε ε¤Ô¡εÕ¡M).
x can be expressed in the same way with kM
, and there is only a finite number of such
ε
sums.
0b nb ¤¤¤
n¡lb n¡2l¤¤¤¡M.
Thus, in view of the expansion (2), (3) applied to the sequenceÔb nÕ, 0×
we get that each b n is
TØb i1 b i2 ik : i 1 , . . .,i krÙÖ¡M,
Finally, for every t1, 2, . . ., l we get that the sequence
b r t, b r t l, b r t 2l, . . .
is non-decreasing and attains the finite number of values; therefore it is constant from some
index. Thus, the sequenceÔb nÕis periodic with period l from some index N, which means that
b nb Ôn¡lÕsÕ
n¡lb n¡l b l for all nN l,
and hence
a nb n nsÔb n¡l Ôb l lsÕa n¡l a l for all nN l,
as desired.

TÕ
19
A8. Given six positive numbers a, b, c, d, e, f such that abcdef. Let a c eS
and b d fT. Prove that
2ST3ÔS SÔbd bf dfÕ TÔac ae ceÕ¨. (1)
fÕÔx¡aÕÔx¡cÕÔx¡eÕ eÕÔx¡bÕÔx¡dÕÔx¡fÕ
(South Korea)
Solution 1. We define also σac ce ae, τbd bf df. The idea of the solution is to
interpret (1) as a natural inequality on the roots of an appropriate polynomial.
Actually, consider the polynomial
PÔxÕÔb d Ôa c
σx¡aceÕ TÔx 3¡Sx 2 SÔx 3¡Tx 2 τx¡bdfÕ. (2)
Surely, P is cubic with leading coefficient S T0. Moreover, we have
PÔaÕSÔa¡bÕÔa¡dÕÔa¡fÕ0,
PÔeÕSÔe¡bÕÔe¡dÕÔe¡fÕ0,
PÔcÕSÔc¡bÕÔc¡dÕÔc¡fÕ0,
PÔfÕTÔf¡aÕÔf¡cÕÔf¡eÕ0.
Hence, each of the intervalsÔa, cÕ,Ôc, eÕ,Ôe, fÕcontains at least one root of PÔxÕ. Since there
are at most three roots at all, we obtain that there is exactly one root in each interval (denote
them by αÈÔa, cÕ, βÈÔc, eÕ, γÈÔe, fÕ). Moreover, the polynomial P can be factorized as
PÔxÕÔT SÕÔx¡αÕÔx¡βÕÔx¡γÕ. (3)
Equating the coefficients in the two representations (2) and (3) of PÔxÕprovides
α β γ2TS
T S ,
αβ αγ βγSτ Tσ
T S .
Now, since the numbers α, β, γ are distinct, we have
0Ôα¡βÕ2 which implies
β
or
4S 2 T 23ÔT SÕÔTσ SτÕ,
4S 2 T 2
ÔT SÕ2Ôα
Ôα¡γÕ2 Ôβ¡γÕ22Ôα β γÕ2¡6Ôαβ αγ βγÕ,
TσÕ
γÕ23Ôαβ αγ βγÕ3ÔSτ
,
T S
which is exactly what we need.
Comment 1. In fact, one can locate the roots of PÔxÕmore narrowly: they should lie in the intervals
Ôa,bÕ,Ôc,dÕ,Ôe,fÕ.
Surely, if we change all inequality signs
fÕ
in the problem statement to non-strict ones, the (non-strict)
inequality will also hold by continuity. One can also find when the equality is achieved. This happens
in that case when PÔxÕis a perfect cube, which immediately implies that bcdeÔαβγÕ,
together with the additional condition that P¾ÔbÕ0. Algebraically,
6ÔT SÕb¡4TS0
3bÔa fÕ2Ôa 2bÕÔ2b
fbÔ4b¡aÕ 3Ôb¡aÕ
2a bb¢1
2a bÅb.
This means that for every pair of numbers a, b such that 0ab, there exists fb such that the
pointÔa,b,b,b,b,fÕis a point of equality.

20
Solution 2. Let
U1
Ôe¡aÕ2 Ôc¡aÕ2 Ôe¡cÕ2¨S 2¡3Ôac ae ceÕ
2
TÕ
and
2¡VÕ TÕ
V1
Ôf¡bÕ2 Ôf¡dÕ2 Ôd¡bÕ2¨T 2¡3Ôbd bf dfÕ.
2
ThenÔL.H.S.Õ2¡ÔR.H.S.Õ2Ô2STÕ2¡ÔS S¤3Ôbd bf dfÕ ceÕ¨
T¤3Ôac ae
4S 2 T 2¡ÔS SÔT SVÕ
TÔS 2¡UÕ¨ÔS TÕÔSV TUÕ¡STÔT¡SÕ2 ,
and the statement is equivalent with
ÔS TÕÔSV TUÕSTÔT¡SÕ2 . (4)
By the Cauchy-Schwarz inequality,
ÔS TÕÔTU
S¤TU
T¤SV¨2ST
U V¨2
Ôe¡aÕ Ôc¡aÕ Ôf¡bÕ Ôf¡dÕ 2Å
. (5)
Estimate the quantitiesU andV by the QM–AM inequality with the positive termsÔe¡cÕ2
andÔd¡bÕ2 being omitted:
ÔT¡SÕ 2Ôe¡dÕ 2¡a© Ôf¡bÕ2 U
VÔe¡aÕ2 Ôc¡aÕ2
Ôf¡dÕ2
2 2
¡e c
2 2¡b
2
Ôc¡bÕÔc¡dÕ
3
Ôe¡fÕÔe¡dÕ
3
(6)
2Ôc¡bÕT¡S.
The estimates (5) and (6) prove (4) and hence the statement.
Solution 3. We keep using the notations σ and τ from Solution 1. Moreover, let sc e.
Note that
Ôe¡fÕÔc¡bÕ0,
since each summand is negative. This rewrites as
Ôbd bf dfÕ¡Ôac ce aeÕÔc eÕÔb d f¡a¡c¡eÕ, or
τ¡σsÔT¡SÕ. (7)
Then we have
Sτ TσSÔτ¡σÕ
SsÔT¡SÕ
TÕσSsÔT¡SÕ ÔS ÔS TÕÔce asÕ
TÕsÅ
TÕ¢s 2
ÔS¡sÕsÅs¢2ST¡3
ÔS TÕsÅ.
4 4ÔS Using this inequality together with the AM–GM inequality we get
3
TσÕ3
TÕÔSτ TÕs¢2ST¡3
4ÔS TÕ
4ÔS 4ÔS 3
4ÔS TÕs 2ST¡3
4ÔS TÕs
ST.
2
Hence,
2ST3ÔS SÔbd bf dfÕ TÔac ae ceÕ¨.
2
¢f¡d

Comment 2. The expression (7) can be found by considering the sum of the roots of the quadratic
polynomial qÔxÕÔx¡bÕÔx¡dÕÔx¡fÕ¡Ôx¡aÕÔx¡cÕÔx¡eÕ.
Solution 4. We introduce the expressions σ and τ as in the previous solutions. The idea of
the solution is to change the values of variables a, . . .,f keeping the left-hand side unchanged
and increasing the right-hand side; it will lead to a simpler inequality which can be proved in
a direct way.
abcd
Namely, we change the variables (i) keeping the (non-strict) inequalities
x½
ef; (ii) keeping the values of sums S and T unchanged; and finally (iii) increasing
zÕ
the
values of σ and τ. Then the left-hand side of (1) remains unchanged, while the right-hand
side increases. Hence, the inequality (1) (and even a non-strict version of (1)) for the changed
values would imply the same (strict) inequality for the original values.
First, we find
zÕzÔx½
the sufficient conditions for (ii) and (iii) to be satisfied.
y½Õ x½y½zÔx½
Lemma. Let x, y, z0; denote UÔx, y, zÕx y z, υÔx, y, zÕxy xz yz. Suppose that
y½x y butx¡yx½¡y½; then we
yÕ y½Õ Ôx½
have UÔx½, y½, zÕUÔx, y, zÕand υÔx½, y½,
υÔx, y, zÕwith equality achieved only whenx¡yx½¡y½.
Proof. The first equality is obvious. For the second, we have
y½Õ2¡Ôx½¡y½Õ2
υÔx½, y½,
4
Ôx yÕ2¡Ôx¡yÕ2
zÔx υÔx, y, zÕ,
4
with the equality achieved only forÔx½¡y½Õ2Ôx¡yÕ2x½¡y½x¡y, as desired.
ÐNow, we apply Lemma several times making the following changes. For each change, we
denote the new values by the same letters to avoid cumbersome notations.
1. Let kd¡c
2 . ReplaceÔb, c, d, eÕbyÔb k, c k, d¡k, e¡kÕ. After the change we have
abcdef, the values of S, T remain unchanged, but σ, τ strictly increase by
Lemma.
2. Let le¡d
2 . ReplaceÔc, d, e, fÕbyÔc l, d l, e¡l, f¡lÕ. After the change we
dÕ
have
abcdef, the values of S, T remain unchanged, but σ, τ strictly increase by the
Lemma.
3. Finally, let mc¡b
3 . ReplaceÔa, b, c, d, e, fÕbyÔa 2m, b 2m, c¡m, d¡m, e¡m, f¡mÕ.
After the change, we have abcdef and S, T are unchanged. To check (iii),
we observe that our change can be considered as a composition of two changes:Ôa, b, c,
Ôa m, b m, c¡m, d¡mÕandÔa, b, e, fÕÔa fÕ
m, b m, e¡m, f¡mÕ. It is easy to see that
each of these two consecutive changes satisfy the conditions of the Lemma, hence the values
of σ and τ increase.
Finally, we come to the situation when abcdef, and we need to prove the
inequality
2Ôa 2bÕÔ2b 2bfÕ 2Õ¨
fÕ3Ôa Ô2b fÕÔ2ab b
Now, observe that
fÕ¤
4b
3bÔa 4b
2¤2Ôa 2bÕÔ2b fÕ3bÔa 4b fÕ
Ôa 2bÕÔb
2 2bÕÔb
Ôa
Ôa 2bÕÔb
2fÕ
2fÕ
21
Ô2b fÕÔ2a bÕ¨. (8)
Ô2a bÕÔ2b fÕ¨.

22
HenceÔ4Õrewrites as
3bÔa 4b fÕ
2fÕ
fÕ¤ 23bÔa 4b
Ôa 2bÕÔb
which is simply the AM–GM inequality.
fÕ¨
Ô2a bÕÔ2b
2fÕ
Ôa 2bÕÔb
Ô2b fÕÔ2a bÕ¨,
Comment 3. Here, we also can find all the cases of equality. Actually, it is easy to see that if
some two numbers among b,c,d,e are distinct then one can use Lemma to increase the right-hand side
of (1). Further, if bcde, then we need equality inÔ4Õ; this means that we apply AM–GM to
equal numbers, that is,
3bÔa 4b 2fÕ fÕÔa 2bÕÔb Ô2a bÕÔ2b fÕ,
which leads to the same equality as in Comment 1.

Combinatorics
C1. In a concert, 20 singers will perform. For each singer, there is a (possibly empty) set of
other singers such that he wishes to perform later than all the singers from that set. Can it
happen that there are exactly 2010 orders of the singers such that all their wishes are satisfied
Answer. Yes, such an example exists.
(Austria)
Solution. We say that an order of singers is good if it satisfied all their wishes. Next, we
say that a number N is realizable by k singers (or k-realizable) if for some set of wishes of
these singers there are exactly N good orders. Thus, we have to prove that a number 2010 is
20-realizable.
We start with the following simple
Lemma. Suppose that numbers n 1 , n 2 are realizable by respectively k 1 and k 2 singers. Then
the number n 1 n 2 isÔk 1 k 2Õ-realizable.
Proof. Let the singers A 1 , . . ., A k1 (with some wishes among them) realize n 1 , and the singers B 1 ,
..., B k2 (with some wishes among them) realize n 2 . Add to each singer B i the wish to perform
later than all the singers A j . Then, each good order of the obtained set of singers has the form
ÔA i1 , . . .,A ik1 , B j1 , . . .,B jk2Õ, whereÔA i1 , . . .,A ik1Õis a good order for A i ’s andÔB j1 , . . .,B jk2Õ
Ð
is a good order for B j ’s. Conversely, each order of this form is obviously good. Hence, the
number of good orders is n 1 n 2 .
In view of Lemma, we show how to construct sets of singers containing 4, 3 and 13 singers
and realizing the numbers 5, 6 and 67, respectively. Thus the number 20106¤5¤67 will be
realizable by 4 3 1320 singers. These companies of singers are shown in Figs. 1–3; the
wishes are denoted by arrows, and the number of good orders for each Figure stands below in
the brackets.
a 3
a 2
a 1
a
b
a 4
c
(5)
Fig. 1
d
(3)
Fig. 2
y
a 5
a 6
a 7
a 8
a 9
a 10
x
a 11
(67)
Fig. 3
For Fig. 1, there are exactly 5 good ordersÔa, b, c, dÕ,Ôa, b, d, cÕ,Ôb, a, c, dÕ,Ôb, a, d, cÕ,
Ôb, d, a, cÕ. For Fig. 2, each of 6 orders is good since there are no wishes.

24
Finally, for Fig. 3, the order of a 1 , . . .,a 11 is fixed; in this line, singer x can stand before
each of a i (i9), and singer y can stand after each of a j (j5), thus resulting in 9¤763
cases. Further, the positions of x and y in this line determine the whole order uniquely unless
both of them come between the same pairÔa i , a i 1Õ(thus 5i8); in the latter cases, there
are two orders instead of 1 due to the order of x and y. Hence, the total number of good orders
is 63 467, as desired.
Comment. The number 20 in the problem statement is not sharp and is put there to respect the
original formulation. So, if necessary, the difficulty level of this problem may be adjusted by replacing
20 by a smaller number. Here we present some improvements of the example leading to a smaller
number of singers.
Surely, each example with20 singers can be filled with some “super-stars” who should perform
at the very end in a fixed order. Hence each of these improvements provides a different solution for
the problem. Moreover, the large variety of ideas standing behind these examples allows to suggest
that there are many other examples.
1. Instead of building the examples realizing 5 and 6, it is more economic to make an example
realizing 30; it may seem even simpler. Two possible examples consisting of 5 and 6 singers are shown
in Fig. 4; hence the number 20 can be decreased to 19 or 18.
For Fig. 4a, the order of a 1 ,... ,a 4 is fixed, there are 5 ways to add x into this order, and there
are 6 ways to add y into the resulting order of a 1 ,...,a 4 ,x. Hence there are 5¤630 good orders.
On Fig. 4b, for 5 singers a,b 1 ,b 2 ,c 1 ,c 2 there are 5!120 orders at all. Obviously, exactly one half
of them satisfies the wish b 1b 2 , and exactly one half of these orders satisfies another wish c 1c 2 ;
hence, there are exactly 5!ß430 good orders.
a 1
a 2
a 3
x
y
b 1
a
c 2
b 2
c 1
y
b 2 b
b 3 1
b 4 b 5
a 6
a7
a8
a 9
x
b 1 b 2
b 3 b 4
a 5
y a 6
a7 x
a8
c 9 c 10
a 4
(30)
a)
(30)
b)
a 11
(2010)
a 10
(2010)
Fig. 4 Fig. 5 Fig. 6
c 11
2. One can merge the examples for 30 and 67 shown in Figs. 4b and 3 in a smarter way, obtaining
a set of 13 singers representing 2010. This example is shown in Fig. 5; an arrow from/to group
Øb 1 ,...,b 5Ùmeans that there exists such arrow from each member of this group.
Here, as in Fig. 4b, one can see that there are exactly 30 orders of b 1 ,... ,b 5 ,a 6 ,...,a 11 satisfying
all their wishes among themselves. Moreover, one can prove in the same way as for Fig. 3 that each
of these orders can be complemented by x and y in exactly 67 ways, hence obtaining 30¤672010
good orders at all.
Analogously, one can merge the examples in Figs. 1–3 to represent 2010 by 13 singers as is shown
in Fig. 6.

25
a 5
a 1
a 2
a 3
b 5
b 6
(67)
b 4
a 4
b 1
b 2
b 3
a 4
b 3
b 2
b 4
a 3
a
a 6
2
a 1
b 1
(2010)
Fig. 7 Fig. 8
3. Finally, we will present two other improvements; the proofs are left to the reader. The graph in
Fig. 7 shows how 10 singers can represent 67. Moreover, even a company of 10 singers representing 2010
can be found; this company is shown in Fig. 8.

26
C2. On some planet, there are 2 N countries (N4). Each country has a flag N units wide
and one unit high composed of N fields of size 1¢1, each field being either yellow or blue. No
two countries have the same flag.
We say that a set of N flags is diverse if these flags can be arranged into an N¢N square so
that all N fields on its main diagonal will have the same color. Determine the smallest positive
integer M such that among any M distinct flags, there exist N flags forming a diverse set.
Answer. M2 N¡2 1.
(Croatia)
Solution. When speaking about the diagonal of a square, we will always mean the main
diagonal.
Let M N be the smallest positive integer satisfying the problem condition. First, we show
that M N2N¡2 . Consider the collection of all 2 N¡2 flags having yellow first squares and blue
second ones. Obviously, both colors appear on the diagonal of each N¢N square formed by
these flags.
We are left to show that M N2N¡2 1, thus obtaining the desired answer. We start with
establishing this statement for N4.
Suppose that we have 5 flags of length 4. We decompose each flag into two parts of 2 squares
each; thereby, we denote each flag as LR, where the 2¢1 flags L, RÈSØBB, BY, YB, YYÙ
are its left and right parts, respectively. First, we make two easy observations on the flags 2¢1
which can be checked manually.
(i) For each AÈS, there exists only one 2¢1 flag CÈS (possibly CA) such that A
and C cannot form a 2¢2 square with monochrome diagonal (for part BB, that is YY, and
for BY that is YB).
(ii) Let A 1 , A 2 , A 3ÈS be three distinct elements; then two of them can form a 2¢2 square
with yellow diagonal, and two of them can form a 2¢2 square with blue diagonal (for all parts
but BB, a pair (BY, YB) fits for both statements, while for all parts but BY, these pairs are
(YB, YY) and (BB, YB)).
Now, let l and r be the numbers of distinct left and right parts of our 5 flags, respectively.
The total number of flags is 5rl, hence one of the factors (say, r) should be at least 3. On
the other hand, l, r4, so there are two flags with coinciding right part; let them be L 1 R 1
and L 2 R 1 (L 1L 2 ).
Next, since r3, there exist some flags L 3 R 3 and L 4 R 4 such that R 1 , R 3 , R 4 are distinct.
Let L½R½be the remaining flag. By (i), one of the pairsÔL½, L 1ÕandÔL½, L 2Õcan form a
2¢2 square with monochrome diagonal; we can assume that L½, L 2 form a square with a blue
diagonal. Finally, the right parts of two of the flags L 1 R 1 , L 3 R 3 , L 4 R 4 can also form a 2¢2
square with a blue diagonal by (ii). Putting these 2¢2 squares on the diagonal of a 4¢4
square, we find a desired arrangement of four flags.
We are ready to prove the problem statement by induction on N; actually, above we have
proved the base case N4. For the induction step, assume that N4, consider any 2 N¡2 1
flags of length N, and arrange them into a large flag of sizeÔ2N¡2 1Õ¢N. This flag contains
a non-monochrome column since the flags are distinct; we may assume that thisÎ2 column is the
first one. By the pigeonhole principle, this column contains at leastÊ2 N¡2 1 N¡3 1
2
squares of one color (say, blue). We call the flags with a blue first square good.
Consider all the good flags and remove the first square from each of them. We obtain at
least 2 N¡3 1M N¡1 flags of length N¡1; by the induction hypothesis, N¡1 of them

27
can form a square Q with the monochrome diagonal. Now, returning the removed squares, we
obtain a rectangleÔN¡1Õ¢N, and our aim is to supplement it on the top by one more flag.
If Q has a yellow diagonal, then we can take each flag with a yellow first square (it exists
by a choice of the first column; moreover, it is not used in Q). Conversely, if the diagonal of Q
is blue then we can take any of the2 N¡3 1¡ÔN¡1Õ0remaining good flags. So, in both
cases we get a desired N¢N square.
Solution 2. We present a different proof of the estimate M N2N¡2 1. We do not use the
induction, involving Hall’s lemma on matchings instead.
Consider arbitrary 2 N¡2 1 distinct flags and arrange them into a largeÔ2N¡2 1Õ¢N
flag. Construct two bipartite graphs G yÔVV½, E yÕand G bÔVV½, E bÕwith the
common set of vertices as follows. Let V and V½be the set of columns and the set of flags
under consideration, respectively. Next, let the edgeÔc, fÕappear in E y if the intersection of
column c and flag f is yellow, andÔc, fÕÈE otherwise. Then we have to prove exactly that
one of the graphs G y and G b contains a matching with all the vertices of V involved.
b
two sets of columns S y , S
yÕ
bV such thatE y
yÔS yÕS flag is connected to c either in G y or in G b , hence E yÔS yÕE 2 N¡2 1V½E yÔS E bÔS bÕS S
So, we have S yS b∅. Let yS bS yÔS yÕE
Assume that these matchings do not exist. By Hall’s lemma, it means that there exist
y¡1 andE bÔS bÕS b¡1 (in the
left-hand sides, E yÔS yÕand E bÔS bÕdenote respectively the sets of all vertices connected to S y
and S b in the corresponding graphs). Our aim is to prove that this is impossible. Note that
S y , S bV since N2N¡2 1.
First, suppose that S yS b∅,
V¾V½Þ
so there exists some cÈS yS b . Note that each
bÔS bÕV½. Hence we have
b¡22N¡4; this is impossible for N4.
y, b. From the construction of our graph,
bÕ
we have that all the flags in the set E bÔS bÕ¨have blue squares in the
columns of S y and yellow squares in the columns of S b . Hence the only undetermined positions
in these flags are the remaining N¡y¡b ones, so 2 N¡y¡bV¾V½¡E yÔS yÕ¡E bÔS 2 N¡2 1¡Ôy¡1Õ¡Ôb¡1Õ, or, denoting cy b, 2 N¡c c2N¡2 2. This is impossible
since Nc2.

28
C3. 2500 chess kings have to be placed on a 100¢100 chessboard so that
(i) no king can capture any other one (i.e. no two kings are placed in two squares sharing
a common vertex);
(ii) each row and each column contains exactly 25 kings.
Find the number of such arrangements. (Two arrangements differing by rotation or symmetry
are supposed to be different.)
Answer. There are two such arrangements.
(Russia)
Solution. Suppose that we have an arrangement satisfying the problem conditions. Divide the
board into 2¢2 pieces; we call these pieces blocks. Each block can contain not more than one
king (otherwise these two kings would attack each other); hence, by the pigeonhole principle
each block must contain exactly one king.
Now assign to each block a letter T or B if a king is placed in its top or bottom half,
respectively. Similarly, assign to each block a letter L or R if a king stands in its left or right
half. So we define T-blocks, B-blocks, L-blocks, and R-blocks. We also combine the letters; we call
a block a TL-block if it is simultaneously T-block and L-block. Similarly we define TR-blocks,
BL-blocks, and BR-blocks. The arrangement of blocks determines uniquely the arrangement of
kings; so in the rest of the solution we consider the 50¢50 system of blocks (see Fig. 1). We
identify the blocks by their coordinate pairs; the pairÔi, jÕ, where 1i, j50, refers to the
jth block in the ith row (or the ith block in the jth column). The upper-left block isÔ1, 1Õ.
The system of blocks has the following properties..
(i½) IfÔi, jÕis a B-block thenÔi 1, jÕis a B-block: otherwise the kings in these two blocks
can take each other. Similarly: ifÔi, jÕis a T-block thenÔi¡1, jÕis a T-block; ifÔi, jÕis an
L-block thenÔi, j¡1Õis an L-block; ifÔi, jÕis an R-block thenÔi, j 1Õis an R-block.
(ii½) Each column contains exactly 25 L-blocks and 25 R-blocks, and each row contains
exactly 25 T-blocks and 25 B-blocks. In particular, the total number of L-blocks (or R-blocks,
or T-blocks, or B-blocks) is equal to 25¤501250.
Consider any B-block of the formÔ1, jÕ. By (i½), all blocks in the jth column are B-blocks;
so we call such a column B-column. By (ii½), we have 25 B-blocks in the first row, so we obtain
25 B-columns. These 25 B-columns contain 1250 B-blocks, hence all blocks in the remaining
columns are T-blocks,
and we obtain 25 T-columns. Similarly, there are exactly 25 L-rows and
exactly 25 R-rows.
Now consider an arbitrary pair of a T-column and a neighboring B-column (columns with
numbers j and j 1).
1 2 3
j j+1
1 TL BL TL
i
L
2 TL BR TR
i+1
3 BL BL TR
T B
Fig. 1 Fig. 2
Case 1. Suppose that the jth column is a T-column, and theÔj 1Õth column is a B-
column. Consider some index i such that the ith row is an L-row; thenÔi, j 1Õis a BL-block.
Therefore,Ôi 1, jÕcannot be a TR-block (see Fig. 2), henceÔi 1, jÕis a TL-block, thus the

29
Ôi 1Õth row is an L-row. Now, choosing the ith row to be the topmost L-row, we successively
obtain that all rows from the ith to the 50th are L-rows. Since we have exactly 25 L-rows, it
follows that the rows from the 1st to the 25th are R-rows, and the rows from the 26th to the
50th are L-rows.
Now consider the neighboring R-row and L-row (that are the rows with numbers 25 and
26). Replacing in the previous reasoning rows by columns and vice versa, the columns from the
1st to the 25th are T-columns, and the columns from the 26th to the 50th are B-columns. So
we have a unique arrangement of blocks that leads to the arrangement of kings satisfying the
condition of the problem (see Fig. 3).
TR TR BR BR
TR TR BR BR
TL TL BL BL
TL TL BL BL
Fig. 3 Fig. 4
Case 2. Suppose that the jth column is a B-column, and theÔj 1Õth column is a T-column.
Repeating the arguments from Case 1, we obtain that the rows from the 1st to the 25th are
L-rows (and all other rows are R-rows), the columns from the 1st to the 25th are B-columns
(and all other columns are T-columns), so we find exactly one more arrangement of kings (see
Fig. 4).

30
C4. Six stacks S 1 , . . ., S 6 of coins are standing in a row. In the beginning every stack contains
a single coin. There are two types of allowed moves:
Move 1: If stack S k with 1k5contains at least one coin, you may remove one coin
from S k and add two coins to S k 1.
Move 2: If stack S k with 1k4contains at least one coin, then you may remove
one coin from S k and exchange stacks S k 1 and S k 2.
Decide whether it is possible to achieve by a sequence of such moves that the first five stacks
are empty, whereas the sixth stack S 6 contains exactly 2010 20102010 coins.
C4½. Same as Problem C4, but the constant 2010 20102010 is replaced by 2010 2010 .
2, 0ÕÔa¡1,
(Netherlands)
Answer. Yes (in both variants of the problem). There exists such a sequence of moves.
Solution. Denote byÔa 1 , a 2 , . . .,a nÕÔa½1, a½2, . . .,a½nÕthe following: if some consecutive stacks
contain a 1 , . . .,a n coins, then it is possible to perform several allowed moves such that the stacks
contain a½1 , . . .,a½n coins respectively, whereas the contents of the other stacks remain unchanged.
Let A2010 2010 or A2010 20102010 , respectively. Our goal is to show that
Ô1, 1, 1, 1, 1, 1ÕÔ0, 0, 0, 0, 0, AÕ.
First we prove two auxiliary observations.
Lemma 1.Ôa, 0, 0ÕÔ0, 2 a , 0Õfor every a1.
Proof. We prove by induction thatÔa, 0, 0ÕÔa¡k, 2 k , 0Õfor every 1ka. For k1,
apply Move 1 to the first stack:
Ôa, 0, 0ÕÔa¡1, 2 1 , 0Õ.
Now assume that kaand the statement holds for some ka. Starting fromÔa¡k, 2 k , 0Õ,
apply Move 1 to the middle stack 2 k times, until it becomes empty. Then apply Move 2 to the
first stack:
ÐÓÑÓÒ Ð
Ôa¡k, 2 k , 0ÕÔa¡k, 2 k 1 , 0Õ.
Hence,
Ôa, 0, 0ÕÔa¡k, 2 k 1 , 0Õ. Lemma 2. For every positive integer n, let P n2 2...2
(e.g. P 322216). Then
n
2 k¡1, 2Õ¤¤¤Ôa¡k,
2 k , 0ÕÔa¡k¡1,
0, 2 k 1ÕÔa¡k¡1,
Ôa, 0, 0, 0ÕÔ0, P a , 0, 0Õfor every a1.
Proof. Similarly to Lemma 1, we prove thatÔa, 0, 0, 0ÕÔa¡k,
For k1, apply Move 1 to the first stack:
Ôa, 0, 0, 0ÕÔa¡1,
2, 0, 0ÕÔa¡1,
P k , 0, 0Õfor every 1ka.
P 1 , 0, 0Õ.
Ð
P k 1, 0, 0Õ.
k 1, 0, 0Õ.
Now assume that the lemma holds for some ka. Starting fromÔa¡k, P k , 0, 0Õ, apply
Lemma 1, then apply Move 1 to the first stack:
Therefore,
Ôa¡k, P k , 0, 0ÕÔa¡k,
0, 2 P k
, 0ÕÔa¡k,
Ôa, 0, 0, 0ÕÔa¡k,
P k , 0, 0ÕÔa¡k¡1,
0, P k 1, 0ÕÔa¡k¡1,
P

31
0Õ
Now we prove the statement of the problem.
First apply Move 1 to stack 5, then apply Move 2 to stacks S 4 , S 3 , S 2 and S 1 in this order.
Then apply Lemma 2 twice:
Ô1, 1, 1, 1, 1, 1ÕÔ1, 1, 1, 1, 0, 3ÕÔ1, 1, 1, 0, 3, 0ÕÔ1, 1, 0, 3, 0, 0ÕÔ1, 0, 3, 0, 0,
Ô0, 3, 0, 0, 0, 0ÕÔ0, 0, P 3 , 0, 0, 0ÕÔ0, 0, 16, 0, 0, 0ÕÔ0, 0, 0, P 16 , 0, 0Õ.
We already have more than A coins in stack S 4 , since
A2010 20102010Ô2 11Õ201020102 11¤201020102 201020112Ô211Õ20112 211¤20112 2215P 16 .
To decrease the number of coins in stack S 4
0Õ
, apply Move 2 to this stack repeatedly until its
size decreases to Aß4. (In every step, we remove a coin from S 4 and exchange the empty stacks
0, 0, P 16¡1, 0, 0ÕÔ0, 0, 0, P 16¡2, 0,
¤¤¤Ô0, 0, 0, Aß4, 0, 0Õ.
Finally, apply Move 1 repeatedly to empty stacks S 4 and S 5 :
Ô0, 0, 0, Aß4, 0, 0Õ¤¤¤Ô0, 0, 0, 0, Aß2, 0Õ¤¤¤Ô0, 0, 0, 0, 0, AÕ.
Comment 1. Starting with only 4 stack, it is not hard to check manually that we can achieve at
most 28 coins in the last position. However, around 5 and 6 stacks the maximal number of coins
Ô1,1,1,1,1,1ÕÔ0,3,1,1,1,1ÕÔ0,1,5,1,1,1ÕÔ0,1,1,9,1,1Õ
explodes. With 5 stacks it is possible to achieve more than 2 214 coins. With 6 stacks the maximum is
greater than P . P2 14
It is not hard to show that the numbers 2010 2010 and 2010 20102010 in the problem can be replaced
by any nonnegative integer up to P . P2 14
Comment 2. The simpler variant C4½of the problem can be solved without Lemma 2:
Ô0,1,1,1,17,1ÕÔ0,1,1,1,0,35ÕÔ0,1,1,0,35,0ÕÔ0,1,0,35,0,0Õ
Ô0,0,35,0,0,0ÕÔ0,0,1,2 34 ,0,0ÕÔ0,0,1,0,2 234 ,0ÕÔ0,0,0,2 ,0,0Õ
234
S 5 and S 6 .)Ô0, 0, 0, P 16 , 0, 0ÕÔ0,
Ô0,0,0,2 234¡1,0,0Õ...Ô0,0,0,Aß4,0,0ÕÔ0,0,0,0,Aß2,0ÕÔ0,0,0,0,0,AÕ.
For this reason, the PSC suggests to consider the problem C4 as well. Problem C4 requires more
invention and technical care. On the other hand, the problem statement in C4½hides the fact that the
resulting amount of coins can be such incredibly huge and leaves this discovery to the students.

32
C5. n4 players participated in a tennis tournament. Any two players have played exactly
one game, and there was no tie game. We call a company of four players bad if one player
was defeated by the other three players, and each of these three players won a game and lost
another game among themselves. Suppose that there is no bad company in this tournament.
Let w i and l i be respectively the number of wins and losses of the ith player. Prove that
nôi1Ôw i¡l iÕ30. (1)
(South Korea)
Solution. For any tournament T satisfying the problem condition, denote by SÔTÕsum under
consideration, namely
SÔTÕnôi1Ôw i¡l iÕ3 .
4
4
First, we show that the statement holds if a tournament T has only 4 players. Actually, let
AÔa 1 , a 2 , a 3 , a 4Õbe the number of wins of the players; we may assume that a 1a 2a 3a 4 .
We have a 1 a 2 a 3 a 2¨6, hence a 41. If a 40, then we cannot have
a 1a 2a 32, otherwise the company of all players is bad. Hence we should have
AÔ3, 2, 1, 0Õ, and SÔTÕ33 1 3 Ô¡1Õ3 Ô¡3Õ30. On the other hand, if a 41, then
only two possibilities, AÔ3, 1, 1, 1Õand AÔ2, 2, 1, 1Õcan take place. In the former case we
have SÔTÕ33 3¤Ô¡2Õ30, while in the latter one SÔTÕ13 1 3 Ô¡1Õ3 SÔTÕô
Ô¡1Õ30, as
desired.
Now we turn to the general problem. Consider a tournament T with no bad companies and
enumerate the players by the numbers from 1 to n. For every 4 players i 1 , i 2 , i 3 , i 4 consider a
“sub-tournament” T i1 i 2 i 3 i 4
consisting of only these players and the games which they performed
with each other. By the abovementioned, we have SÔT i1 i 2 i 3 i 4Õ0. Our aim is to prove that
SÔT i1 i 2 i 3 i 4Õ, (2)
i 1 ,i 2 ,i 3 ,i 4
ε ij«3
ôε ij1 ε ij2 ε ij3 .
j 1 ,j 2 ,j 3i
where the sum is taken over all 4-tuples of distinct numbers from the setØ1, . . .,nÙ. This way
the problem statement will be established.
We interpret the numberÔw i¡l iÕ3 as following. For ij, let ε ij1 if the ith player wins
against the jth one, and ε ij¡1 otherwise. Then
Ôw i¡l iÕ3£ôji SÔTÕ ô
Hence,
ij1 ε ij2 ε ij3 .
,j 2 ,j 3Ùε
SÔTÕ ô
iÊØj 1
To simplify this expression, consider all the terms in this sum where two indices are equal.
Øi,j 1 ,j 2 ,j 3Ù4
If, for instance, j 1j 2 , then the term contains ε 2 ij 11, so we can replace this term by ε ij3 .
Make such replacements for each such term; obviously, after this change each term of the form
ε ij3 will appear PÔTÕtimes, hence
PÔTÕS 2ÔTÕ.
ε
1ÔTÕ ij1 ε ij2 ε ij3 ε PÔTÕôij
ijS

33
We show that S 2ÔTÕ0and hence SÔTÕS 1ÔTÕfor each tournament. Actually, note that
ε ij¡ε ji , and the whole sum can
1ÔTÕô
be split into such pairs. Since the sum in each pair is 0, so
is S 2ÔTÕ.
Thus the desired equality (2) rewrites as
S S 1ÔT i1 i 2 i 3 i 4Õ. (3)
i 1 ,i 2 ,i 3 ,i 4
Now, if all the numbers j 1 , j 2 , j 3 are distinct, then the setØi, j 1 , j 2 , j 3Ùis contained in exactly
one 4-tuple, hence the term ε ij1 ε ij2 ε ij3 appears in the right-hand part of (3) exactly once, as
well as in the left-hand part. Clearly, there are no other terms in both parts, so the equality is
established.
Solution 2. Similarly to the first solution, we call the subsets of players as companies, and
the k-element subsets will be called as k-companies.
In any company of the players, call a player the local champion of the company if he defeated
all other members of the company. Similarly, if a player lost all his games against the others
in the company then call him the local loser of the company. Obviously every company has
at most one local champion and at most one local loser. By the condition of the problem,
whenever a 4-company has a local loser, then this company has a local champion as well.
Suppose that k is some positive integer, and let us count all cases when a player is the local
champion of some k-company. The ith player won against w i other player. To be the local
champion of a k-company, he must be a member of the company, and the other k¡1 members
must be chosen from those whom he defeated. Therefore, the ith player is the local champion
of¢w i
k¡1Åk-companies. Hence, the total number of local champions of all k-companies is
nôi1¢w i
k¡1Å.
Similarly, the total number of local losers of the k-companies is
Now apply this for k2, 3 and 4.
Since every game has a winner and a loser, we have
w
nôi1
inôi1
nôi1¢l i
k¡1Å.
i¢n
l and hence
2Å,
nôi1 w i¡l i¨0. (4)
In every 3-company, either the players defeated one another in a cycle or the company has
both a local champion and a local loser. Therefore, the total number of local champions and
nôi1¢w i
local losers in the 3-companies is the same,
2Ånôi1¢l i
2Å. So we have
nôi1£¢w 2Å¡¢l i
2Å«0. i
(5)
In every 4-company, by the problem’s condition, the number of local losers is less than or
equal to the number of local champions. Then the same holds for the total numbers of local

34
champions and local losers in all 4-companies, so
nôi1¢w i
3Ånôi1¢l i
3Å. Hence,
3Å«
nôi1£¢w 3Å¡¢l i
3Å«0. i
(6)
2Å«¡
3Å«
Ôx¡yÕ324£¢x 3Å¡¢y 24£¢x 2Å¡¢y
2Å«¡
3Ôx yÕ2¡4¨Ôx¡yÕ.
iÕ324£¢w 3Å¡¢l
Ôw i
24£¢w 2Å¡¢l i i
i¡l ÐÓÓÓÓÓÓÓÓÓÑÓÓÓÓÓÓÓÓÓÒ 3Å« ÐÓÓÓÓÓÓÓÓÓÑÓÓÓÓÓÓÓÓÓÒ 2Å«
3Ôn¡1Õ2¡4¨¡w
ÐÓÓÓÓÓÑÓÓÓÓÓÒ
i¡l i©.
nôi1£¢w nôi1£¢w 3Å¡¢l i i
2Å¡¢l i i
Now we establish the problem statement (1) as a linear combination of (4), (5) and (6). It
is easy check that
Apply this identity to xw 1 and yl i . Since every player played n¡1 games, we have
w i l in¡1, and thus
i
i©
Then
nôi1Ôw i¡l iÕ324
3Ôn¡1Õ2¡4¨nôi1¡w i¡l 0.
0
24
0
¡
0

35
C6. Given a positive integer k and other two integers bw1. There are two strings of
pearls, a string of b black pearls and a string of w white pearls. The length of a string is the
number of pearls on it.
One cuts these strings in some steps by the following rules. In each step:
(i) The strings are ordered by their lengths in a non-increasing order. If there are some
strings of equal lengths, then the white ones precede the black ones. Then k first ones (if they
consist of more than one pearl) are chosen; if there are less than k strings longer than 1, then
one chooses all of them.
(ii) Next, one cuts each chosen string into two parts differing in length by at most one.
(For instance, if there are strings of 5, 4, 4, 2 black pearls, strings of 8, 4, 3 white pearls and
k4, then the strings of 8 white, 5 black, 4 white and 4 black pearls are cut into the parts
Ô4, 4Õ,Ô3, 2Õ,Ô2, 2ÕandÔ2, 2Õ, respectively.)
The process stops immediately after the step when a first isolated white pearl appears.
Prove that at this stage, there will still exist a string of at least two black pearls.
(Canada)
Solution 1. Denote the situation after the ith step by A i ; hence A 0 is the initial situation, and
A i¡1A i is the ith step. We call a string containing m pearls an m-string; it is an m-w-string
or a m-b-string if it is white or black, respectively.
We continue the process until every string consists of a single pearl. We will focus on three
moments of the process: (a) the first stage A s when the first 1-string (no matter black or
white) appears; (b) the first stage A t where the total number of strings is greater than k (if
such moment does not appear then we put t); and (c) the first stage A f when all black
pearls are isolated. It is sufficient to prove that in A f¡1 (or earlier), a 1-w-string appears.
We start with some easy properties of the situations under consideration. Obviously, we
have sf. Moreover, all b-strings from A f¡1 become single pearls in the fth step, hence all
of them are 1- or 2-b-strings.
Next, observe that in each step A iA i 1 with it¡1, allÔ1Õ-strings were cut since
there are not more than k strings at all; if, in addition, is, then there were no 1-string, so
all the strings were cut in this step.
Now, let B i and b i be the lengths of the longest and the shortest b-strings in A i , and
let W i and w i be the same for w-strings. We show by induction on iminØs, tÙthat (i) the
situation A i contains exactly 2 i black and 2 i white strings, (ii) B iW i , and (iii) b iw i .
The base case i0is obvious. For the induction step, if iminØs, tÙthen in the ith step,
each string is cut, thus the claim (i) follows from the induction hypothesis; next, we have
B iÖB i¡1ß2×ÖW i¡1ß2×W i and b iØb i¡1ß2ÙØw i¡1ß2Ùw i , thus establishing (ii)
and (iii).
For the numbers s, t, f, two cases are possible.
Case 1. Suppose that stor ft 1 (and hence st 1); in particular, this is true
when t. Then in A s¡1 we have B s¡1W s¡1, b s¡1w s¡11as s¡1minØs, tÙ.
Now, if sf, then in A s¡1, there is no 1-w-string as well as noÔ2Õ-b-string. That is,
2B s¡1W s¡1b s¡1w s¡11, hence all these numbers equal 2. This means that
in A s¡1, all strings contain 2 pearls, and there are 2 s¡1 black and 2 s¡1 white strings, which
means b2¤2s¡1w. This contradicts the problem conditions.
Hence we have sf¡1 and thus st. Therefore, in the sth step each string is cut
into two parts. Now, if a 1-b-string appears in this step, then from w s¡1b s¡1 we see that a

36
1-w-string appears as well; so, in each case in the sth step a 1-w-string appears, while not all
black pearls become single, as desired.
Case 2. Now assume that t 1sand t 2f. Then in A t we have exactly 2 t white
and 2 t black strings, all being larger than 1, and 2 t 1k2t (the latter holds since 2 t is the
total number of strings in A t¡1). Now, in theÔt 1Õst step, exactly k strings are cut, not more
than 2 t of them being black; so the number of w-strings in A t 1 is at least 2 t Ôk¡2tÕk.
Since the number of w-strings does not decrease in our process, in A f¡1 we have at least k
white strings as well.
Finally, in A f¡1, all b-strings are not larger than 2, and at least one 2-b-string is cut in
the fth step. Therefore, at most k¡1 white strings are cut in this step, hence there exists a
w-string W which is not cut in the fth step. On the other hand, since a 2-b-string is cut, all
Ô2Õ-w-strings should also be cut in the fth step; hence W should be a single pearl. This is
exactly what we needed.
Comment. In this solution, we used the condition bw only to avoid the case bw2 t . Hence,
if a number bw is not a power of 2, then the problem statement is also valid.
Solution 2. We use the same notations as introduced in the first paragraph of the previous
solution. We claim that at every stage, there exist a u-b-string and a v-w-string such that
either
(i) uv1, or
(ii) 2uv2u, and there also exist k¡1 ofÔvß2Õ-strings other than considered
above.
First, we notice that this statement implies the problem statement. Actually, in both
cases (i) and (ii) we have u1, so at each stage there exists aÔ2Õ-b-string, and for the last
stage it is exactly what we need.
Now, we prove the claim by induction on the number of the stage. Obviously, for A 0 the
condition (i) holds since bw. Further, we suppose that the statement holds for A i , and prove
it for A i 1. Two cases are possible.
Case 1. Assume that in A i , there are a u-b-string and a v-w-string with uv. We can
assume that v is the length of the shortest w-string in A i ; since we are not at the final stage,
we have v2. Now, in theÔi 1Õst step, two subcases may occur.
Subcase 1a. Suppose that either no u-b-string is cut, or both some u-b-string and some
v-w-string are cut. Then in A i 1, we have either a u-b-string and aÔvÕ-w-string (and (i) is
valid), or we have aÖuß2×-b-string and aØvß2Ù-w-string. In the latter case, from uv we get
Öuß2×Øvß2Ù, and (i) is valid again.
Subcase 1b. Now, some u-b-string is cut, and no v-w-string is cut (and hence all the strings
which are cut are longer than v). If u½Öuß2×v, then the condition (i) is satisfied since we
have a u½-b-string and a v-w-string in A i 1. Otherwise, notice that the inequality uv2
implies u½2. Furthermore, besides a fixed u-b-string, other k¡1 ofÔv 1Õ-strings should
be cut in theÔi 1Õst step, hence providing at least k¡1 ofÔÖÔv 1Õß2×Õ-strings, and
ÖÔv 1Õß2×vß2. So, we can put v½v, and we have u½vu2u½, so the condition (ii)
holds for A i 1.
Case 2. Conversely, assume that in A i there exist a u-b-string, a v-w-string (2uv2u)
and a set S of k¡1 other strings larger than vß2 (and hence larger than 1). In theÔi 1Õst
step, three subcases may occur.
Subcase 2a. Suppose that some u-b-string is not cut, and some v-w-string is cut. The latter
one results in aØvß2Ù-w-string, we have v½Øvß2Ùu, and the condition (i) is valid.

37
Subcase 2b. Next, suppose that no v-w-string is cut (and therefore no u-b-string is cut as
uv). Then all k strings which are cut have the lengthv, so each one results in aÔvß2Õstring.
Hence in A i 1, there exist kk¡1 ofÔvß2Õ-strings other than the considered u- and
v-strings, and the condition (ii) is satisfied.
Subcase 2c. In the remaining case, all u-b-strings are cut. This means that allÔuÕ-strings
are cut as well, hence our v-w-string is cut. Therefore in A i 1 there exists aÖuß2×-b-string
together with aØvß2Ù-w-string. Now, if u½Öuß2×Øvß2Ùv½then the condition (i) is
fulfilled. Otherwise, we have u½v½u2u½. In this case, we show that u½2. If, to the
contrary, u½1 (and hence u2), then all black and whiteÔ2Õ-strings should be cut in the
Ôi 1Õst step, and among these strings there are at least a u-b-string, a v-w-string, and k¡1
strings in S (k 1 strings altogether). This is impossible.
Hence, we get 2u½v½2u½. To reach (ii), it remains to check that in A i 1, there exists
a set S½of k¡1 other strings larger than v½ß2. These will be exactly the strings obtained from
the elements of S. Namely, each sÈS was either cut in theÔi 1Õst step, or not. In the former
case, let us include into S½the largest of the strings obtained from s; otherwise we include s
itself into S½. All k¡1 strings in S½are greater than vß2v½, as desired.

38
C7. Let P 1 , . . .,P s be arithmetic progressions of integers, the following conditions being
satisfied:
(i) each integer belongs to at least one of them;
(ii) each progression contains a number which does not belong to other progressions.
Denote by n the least common multiple of steps of these progressions; let np α 1
1 . . .p α k
k
be
its prime factorization. Prove that
s1 kôi1α iÔp i¡1Õ.
N
(Germany)
Solution 1. First, we prove the key lemma, and then we show how to apply it to finish the
k grid we mean the set
ØÔa 1 , . . ., a kÕ:a iÈZ, 0a in i¡1Ù; the elements of N will be referred to as points. In this
grid, we define a subgrid as a subset of the form
: b i1x i1 , . . .,b itx itÙ, (1)
where IØi 1 , . . .,i tÙis an arbitrary nonempty set of indices, and x ijÈÖ0, n ij¡1×(1jt)
are fixed integer numbers. Further, we say that a subgrid (1) is orthogonal to the ith coordinate
axis if iÈI, and that it is parallel to the ith coordinate axis otherwise.
Lemma. Assume that the grid N is covered by subgrids L 1 , L 2 , . . .,L s (this means Näs
i1 L i)
so that
(ii½) each subgrid contains a point which is not covered by other subgrids;
(iii) for each coordinate axis, there exists a subgrid L i orthogonal to this axis.
Then
solution.
Let n 1 , . . .,n k be positive integers. By an n 1¢n 2¢¤¤¤¢n
LØÔb 1 , . . .,b kÕÈN
s1 kôi1Ôn i¡1Õ.
Proof. Assume to the contrary that siÔn i¡1Õs½. Our aim is to find a point that is not
covered by L 1 , . . ., L s .
The idea of the proof is the following. Imagine that we expand each subgrid to some maximal
subgrid so that for the ith axis, there will be at most n i¡1 maximal subgrids orthogonal to
this axis. Then the desired point can be found easily: its ith coordinate should be that not
covered by the maximal subgrids orthogonal to the ith axis. Surely, the conditions for existence
of such expansion are provided by Hall’s lemma on matchings. So, we will follow this direction,
although we will apply Hall’s lemma to some subgraph instead of the whole graph.
Construct a bipartite graph GÔVV½, EÕas follows. Let VØL 1 , . . .,L sÙ, and let
V½Øv ij : 1is, 1jn i¡1Ùbe some set of s½elements. Further, let the edgeÔL m , v
appear iff L m is orthogonal to the ith axis.
For each subset WV , denote
ijÕ
fÔWÕØvÈV½:ÔL, vÕÈE for some LÈWÙ.
Notice that fÔVÕV½by (iii).
Now, consider the set WV containing the maximal number of elements such thatW
fÔWÕ; if there is no such set then we set W∅. Denote W½fÔWÕ, UVÞW, U½V½ÞW½.

39
By our
WXW
assumption and
XfÔWÕ
the Lemma condition,fÔVÕV½V, hence WV and U∅.
Permuting the coordinates, we can assume that U½Øv ij : 1ilÙ, W½Øv ij : l 1ikÙ.
Consider the induced subgraph G½of G on the vertices UU½. We claim that for every
XU, we getfÔXÕU½X(so G½satisfies the conditions of Hall’s lemma). Actually, we
haveWfÔWÕ, so ifXfÔXÕU½for some XU, then we have
fÔXÕU½fÔWÕÔfÔXÕU½ÕfÔWXÕ.
This contradicts the maximality ofW.
Thus, applying Hall’s lemma, we can assign to each LÈU some vertex v ijÈU½so that to
distinct elements of U, distinct vertices of U½are assigned. In this situation, we say that LÈU
corresponds to the ith axis, and write gÔLÕi. Since there are n i¡1 vertices of the form v ij ,
we get that for each 1il, not more than n i¡1 subgrids correspond to the ith axis.
Finally, we are ready to present the desired point. Since WV , there exists a point
bÔb 1 , b 2 , . . .,b kÕÈNÞÔLÈWLÕ. On the other hand, for every 1il, consider any subgrid
LÈU with gÔLÕi. This means exactly that L is orthogonal to the ith axis, and hence all
its elements have the same ith coordinate c L . Since there are at most n i¡1 such subgrids,
there exists a number 0a in i¡1 which is not contained in a setØc L : gÔLÕiÙ. Choose
such number for every 1il. Now we claim that point aÔa 1 , . . .,a l , b l 1, . . .,b kÕis not
covered, hence contradicting the Lemma condition.
Surely, point a cannot lie in some LÈU, since all the points in L have gÔLÕth coordinate
c La gÔLÕ. Ð
On the other hand, suppose that aÈL for some LÈW; recall that bÊL. But the
points a and b differ only at first l coordinates, so L should be orthogonal to at least one of
the first l axes, and hence our graph contains some edgeÔL, v ijÕfor il. It contradicts the
definition of W½. The Lemma is proved.
Now we turn to the problem. Let d j be the step of the progression P j . Note that since
nl.c.m.Ôd ÐÓÓÓÓÓÓÑÓÓÓÓÓÓÒ
1 , . . .,d sÕ, for each
ÐÓÓÓÓÓÓÑÓÓÓÓÓÓÒ
1ik there exists an index jÔiÕsuch that p α i
i§d jÔiÕ. rÔmÕ
We
assume that n1; otherwise the problem statement is trivial.
For each 0mn¡1 and 1ik, let m i be the residue of m modulo p α i
i , and let
m ir iαi . . .r i1 be the base p i representation of m i (possibly, with some leading zeroes). Now,
we put into correspondence to m the sequence rÔmÕÔr 11 , . . .,r 1α1 , r 21 , . . .,r kαkÕ. Hence
lies in a p 1¢¤¤¤¢p 1 ¢¤¤¤¢p k¢¤¤¤¢p k grid N.
α 1 times
α k times
Surely, if rÔmÕrÔm½Õthen p α i
i§m i¡m½i , which follows pα i
i§m¡m½for all 1ik;
consequently, n§m¡m½. So, when m runs over the setØ0, . . .,n¡1Ù, the sequences rÔmÕdo
not repeat; sinceNn, this means that r is a bijection betweenØ0, . . ., n¡1Ùand N. Now
we will show that for each 1is, the set L iØrÔmÕ:mÈP iÙis a subgrid, and that for
each axis there exists
a subgrid orthogonal to this axis. Obviously, these subgrids cover N, and
the condition (ii½) follows directly from (ii). Hence the Lemma provides exactly the estimate
we need.
Consider some 1jsand let d jp rÔqÕ
γ 1
1 . . .pγ k
k . Consider some qÈP j and let
Ôr 11 , . . ., r kαkÕ. Then for an arbitrary q½, setting rÔq½ÕÔr½11, . . .,r½kα kÕwe have
p γ i
i§q¡q½for each 1ik r i,tr½i,t for all tγ i .
Hence L jØÔr½11 , . . .,r½kα kÕÈN : r i,tr½i,t for all tγ iÙwhich means that L j is a subgrid
containing rÔqÕ. Moreover, in L jÔiÕ, all the coordinates corresponding to p i are fixed, so it is
orthogonal to all of their axes, as desired.
q½ÈP j

40
Comment 1. The estimate in the problem is sharp for every n. One of the possible examples is the
following one. For each 1ik, 0jα i¡1, 1kp¡1, let
P i,j,kkp j i
p j 1
i
Z,
and add the progression P 0nZ. One can easily check that this set satisfies all the problem conditions.
There also exist other examples.
On the other hand, the estimate can be adjusted in the following sense. For every 1ik, let
0α i0 ,α i1 ,... ,α ihi be all the numbers of the form ord piÔd jÕin an increasing order (we delete the
repeating occurences of a number, and add a number 0α i0 if it does not occur). Then, repeating
the arguments from the solution one can obtain that
s1 kôi1
h i ôj1Ôp α j¡α j¡1¡1Õ.
Note that p α¡1αÔp¡1Õ, and the equality is achieved only for α1. Hence, for reaching the
minimal number of the progressions, one should have α i,jj for all i,j. In other words, for each
1jα i , there should be an index t such that ord piÔd tÕj.
Solution 2. We start with introducing some notation. For positive integer r, we denote
Ör×Ø1, 2, . . .,rÙ. Next, we say that a set of progressions PØP 1 , . . .,P sÙcover Z if each
integer belongs to some of them; we say that this covering is minimal if no proper subset of P
covers Z. Obviously, each covering contains a minimal subcovering.
Next, for a minimal coveringØP 1 , . . .,P sÙand for every 1is, let d i be the step of
progression P i , and h i be some number which is contained in P i but in none of the other
progressions. We assume that n1, otherwise the problem is trivial. This implies d i1,
otherwise the progression P i covers all the numbers, and n1.
tÙÖk×
We will prove a more general statement, namely the following
Claim. Assume that the progressions P 1 , . . .,P s and number np α 1
1 . . .pα k
k1 are chosen as
in the problem statement. Moreover, choose some nonempty set of indices IØi 1 , . . .,i
and some positive integer β iα i for every iÈI. Consider the set of indices
Tj : 1js, and p α i¡β i 1
i
§d j for some iÈIµ.
Then
ôiÈI
T1 β iÔp i¡1Õ. (2)
Observe that the Claim for IÖk×and β iα i implies the problem statement, since the
left-hand side in (2) is not greater than s. Hence, it suffices to prove the Claim.
1. First, we prove the Claim assuming that all d j ’s are prime numbers. If for some 1ik
we have at least p i progressions with the step p i , then they do not intersect and hence cover all
the integers; it means that there are no other progressions, and np i ; the Claim is trivial in
this case.
Now assume that for every 1ik, there are not more than p i¡1 progressions with
step p i ; each such progression covers the numbers with a fixed residue modulo p i , therefore
there exists a residue q i mod p i which is not touched by these progressions. By the Chinese
Remainder Theorem, there exists a number q such that qq iÔmod p iÕfor all 1ik; this
number cannot be covered by any progression with step p i , hence it is not covered at all. A
contradiction.

41
2. Now, we assume that the general Claim is not valid, and hence we consider a counterexampleØP
1 , . . .,P sÙfor the Claim; we can choose it to be minimal in the following sense:
­the number n is minimal possible among all the counterexamples;
­the sumi d i is minimal possible among all the counterexamples having the chosen value
of n.
As was mentioned above, not all numbers d i are primes; hence we can assume that d 1 is
composite, say p 1§d 1 and d½1d 1
p 11. Consider a progression P½1 having the step d½1, and
containing P 1 . We will focus on two coverings constructed as follows.
(i) Surely, the progressions P½1, P 2 , . . .,P s cover Z, though this covering in not necessarily
minimal. So, choose some minimal subcovering P½in it; surely P½1ÈP½since h 1 is not covered
by P 2 , . . .,P s , so we may assume that P½ØP½1 P 2, . . ., P s½Ùfor some s½s. Furthermore, the
,
period of the covering P½can appear to be less than n; so we denote this period by
l.c.m. d½1, d 2 , . . ., d s½¨.
n½p α 1¡σ 1
1 . . .p α k¡σ k
k
Observe that for each P jÊP½, we have h jÈP½1, otherwise h j would not be covered by P.
(ii) On the other hand, each nonempty set of the form R iP iP½1 (1is) is also a
progression with a step r il.c.m.Ôd i , d½1Õ, and such sets cover P½1. Scaling these progressions
with the ratio 1ßd½1 , we obtain the progressions Q i with steps q ir ißd½1 which cover Z. Now we
choose a minimal subcovering Q of this covering; again we should have Q 1ÈQ by the reasons
of h 1 . Now, denote the period of Q by
n¾l.c.m.Øq i : Q iÈQÙl.c.m.Ør i : Q iÈQÙ
d½1
p γ 1
1 . . .p γ k
k
.
d½1
Note that if h jÈP½1 , then the image of h j under the scaling can be covered by Q j only; so, in
this case we have Q jÈQ.
Our aim is to find the desired number of progressions in coverings P and Q. First, we have
nn½, and the sum of the steps in P½is less than that in P; hence the Claim is valid for P½.
We apply it to the set of indices I½ØiÈI : β iσ iÙand the exponents β½iβ i¡σ i ; hence
the set under consideration is
T½j : 1js½, and i¡σ pÔα iÕ¡β½i
1p α i¡β i 1
i i
§d j for some iÈI½µTÖs½×,
and we obtain that
ôiÈIÔβ whereÔxÕ
iÕ
xÔx¡yÕ i¡1Õ1 i¡σ Ôp i¡1Õ,
maxØx, 0Ù; the latter equality holds as for
i¡1Õ
iÊI½we have β iσ i .
TTÖs½× TØs½
Observe that
TØs½
minØx, yÙfor all x, y. So, if we find at least
minØβ GôiÈI i , σ iÙÔp iÕ
indices in 1, . . ., sÙ, then we would have
1, . . .,
ôiÈI sÙ1 Ôβ i¡σ minØβ i , σ
ôiÈI
iÙ¨Ôp i¡1Õ1 β iÔp i¡1Õ,
ôiÈI½Ôβ TÖs½×T½1 i¡σ iÕÔp
thus leading to a contradiction with the choice of P. We will find those indices among the
indices of progressions in Q.

42
3. Now denote I¾ØiÈI : σ i0Ùand consider some iÈI¾; then p α i
the
other hand, there exists an index jÔiÕsuch that p α i
i§d jÔiÕ; this means that d jÔiÕЧn½and hence
P jÔiÕcannot appear in P½, so jÔiÕs½. Moreover, we have observed before that in this case
h jÔiÕÈP½1 , hence Q jÔiÕÈQ. This means that q jÔiÕ§n¾, therefore γ iα i for each iÈI¾(recall
here that q ir ißd½1 and hence d jÔiÕ§r jÔiÕ§d½1n¾).
Let d½1p τ 1
1 . . .p τ k
k
. Then n¾p γ 1¡τ 1
1 . . .p γ i¡τ i
k
. Now, if iÈI¾, then for every β the condition
i¡τ pÔγ iÕ¡β 1
i
§q j is equivalent to p α i¡β 1
i
§r j .
Note that n¾nßd½1n, hence we can apply the Claim to the covering Q. We perform
this with the set of indices I¾and the exponents β¾iminØβ i , σ iÙ0. So, the set under
consideration is
iÙ iÙ iÈI¾µ
T¾j : Q jÈQ, and i¡τ iÕ¡minØβ i
T¾T
,σ 1
pÔγ i
§q j
Ø1ÙØs½
for some
j : Q jÈQ, and p α i¡minØβ i ,σ 1
i
§r j for some iÈI¾µ,
and we obtainT¾1 G. Finally, we claim that 1, . . .,sÙ¨; then we
will obtainTØs½
iÙ iÙ
1, . . .,sÙG, which is exactly what we need.
To prove this, consider any jÈT¾. Observe first that α i¡minØβ i , σ 1α i¡σ iτ i ,
hence from p α i¡minØβ i ,σ 1
1
i
§d j , which means that
jÈT. Next, the exponent of p i in d j is greater than that in n½, which means that P jÊP½. This
may appear only if j1 or js½, as desired. This completes the proof.
i
§r jl.c.m.Ôd½1, d jÕwe have p α i¡minØβ i ,σ iÙ
Comment 2. A grid analogue of the Claim is also valid. It reads as following.
Claim. Assume that the grid N is covered by subgrids L 1 ,L 2 ,... ,L s so that
(ii½) each subgrid contains a point which is not covered by other subgrids;
(iii) for each coordinate axis, there exists a subgrid L i orthogonal to this axis.
Choose some set of indices IØi 1 ,... ,i tÙÖk×, and consider the set of indices
TØj : 1js,and L j is orthogonal to the ith axis for some iÈIÙ.
Then ôiÈIÔn T1 i¡1Õ.
This Claim may be proved almost in the same way as in Solution 1.
iЧn½. On

Geometry
G1. Let ABC be an acute triangle with D, E, F the feet of the altitudes lying on BC, CA, AB
respectively. One of the intersection points of the line EF and the circumcircle is P. The
lines BP and DF meet at point Q. Prove that APAQ.
(United Kingdom)
Solution 1. The line EF intersects the circumcircle at two points. Depending on the choice
of P, there are two different cases to consider.
Case 1: The point P lies on the ray EF (see Fig. 1).
LetCABα,ABCβ andBCAγ. The quadrilaterals BCEF and CAFD are
cyclic due to the right angles at D, E and F. So,
BDF180¥¡FDCCAFα,
AFE180¥¡EFBBCEγ,
DFB180¥¡AFDDCAγ.
Since P lies on the arc AB of the circumcircle,PBABCAγ. Hence, we have
PBD BDFPBA ABD BDFγ β α180¥,
and the point Q must lie on the extensions of BP and DF beyond the points P and F,
respectively.
From the cyclic quadrilateral APBC we get
QPA180¥¡APBBCAγDFBQFA.
Hence, the quadrilateral AQPF is cyclic. ThenAQP180¥¡PFAAFEγ.
We obtained thatAQPQPAγ, so the triangle AQP is isosceles, APAQ.
Q
γ
A
A
α
P
γ
γ
F γ
γ
E
F
γ
γ
γ
E
γ
P
B
γ
β
α
D
γ
C
B
Q
D
γ
C
Fig. 1 Fig. 2

45
Case 2: The point P lies on the ray FE (see Fig. 2). In this case the point Q lies inside
the segment FD.
Similarly to the first case, we have
QPABCAγDFB180¥¡AFQ.
Hence, the quadrilateral AFQP is cyclic.
ThenAQPAFPAFEγQPA. The triangle AQP is isosceles again,
AQPQPA and thus APAQ.
Comment. Using signed angles, the two possible configurations can be handled simultaneously, without
investigating the possible locations of P and Q.
Solution 2. For arbitrary points X, Y on the circumcircle, denote byXY the central angle
of the arc XY .
Let P and P½be the two points where the line EF meets the circumcircle; let P lie on
the arc AB and let P½lie on the arc CA. Let BP and BP½meet the line DF and Q and Q½,
respectively (see Fig. 3). We will prove that APAP½AQAQ½.
Q
A
P
F
γ
γ
γ
E
P ′
Q ′
γ
B D
C
Fig. 3
Like in the first solution, we haveAFEBFPDFBBCAγ from the
cyclic quadrilaterals BCEF and CAFD.
ByPB P½A2AFP½2γ2BCAAP PB, we have
APP½A,PBAABP½
Ô1Õ
and APAP½.
Ô2Õ
Due toAPP½A, the lines BP and BQ½are symmetrical about line AB.
Similarly, byBFPQ½FB, the lines FP and FQ½are symmetrical about AB. It
follows that also the points P and P½are symmetrical to Q½and Q, respectively. Therefore,
APAQ½and AP½AQ.
The relations (1) and (2) together prove APAP½AQAQ½.

46
G2. Point P lies inside triangle ABC. Lines AP, BP, CP meet the circumcircle of ABC
again at points K, L, M, respectively. The tangent to the circumcircle at C meets line AB
at S. Prove that SCSP if and only if MKML.
(Poland)
Solution 1. We assume that CACB, so point S lies on the ray AB.
From the similar triangles △PKM△PCA and △PLM△PCB we get
KMPA PM
CA
and
PMCB LM . Multiplying these two equalities, we get
PB
LM
KMCB
CA¤PA
PB .
Hence, the relation MKML is equivalent to CB
CAPB
PA .
Denote by E the foot of the bisector of angle B in triangle ABC. Recall that the locus of
points X for which
XBCA XA is the Apollonius circle Ω with the center Q on the line AB,
CB
and this circle passes through C and E. Hence, we have MKML if and only if P lies on Ω,
that is QPQC.
L
C
Ω
K
P
A
E
M
Fig. 1
Now we prove that SQ, thus establishing the problem statement. We haveCES
CAE ACEBCS ECBECS, so SCSE. Hence, the point S lies on AB
as well as on the perpendicular bisector of CE and therefore coincides with Q.
Solution 2. As in the previous solution, we assume that S lies on the ray AB.
1. Let P be an arbitrary point inside both the circumcircle ω of the triangle ABC and the
angle ASC, the points K, L, M defined as in the problem. We claim that SPSC implies
MKML.
Let E and F be the points of intersection of the line SP with ω, point E lying on the
segment SP (see Fig. 2).
B
S

47
F
L
ω
P
K
C
M
E
A
B
Fig. 2
S
We have SP 2SC2SA¤SB, so
SBSA SP
SP , and hence △PSA△BSP. Then
BPSSAP. Since 2BPSBE LF and 2SAPBE EK we have
LFEK. (1)
On the other hand, fromSPCSCP we haveEC MFEC EM, or
MFEM. (2)
From (1) and (2) we getǑMFLMF FLME EKǑMEK and hence MKML.
The claim is proved.
2. We are left to prove the converse. So, assume that MKML, and introduce the
points E and F as above. We have SC 2SE¤SF; hence, there exists a point P½lying on the
segment EF such that SP½SC (see Fig. 3).
L ′
L
C
F
ω
P ′
P
K ′
K
E
A
B
S
M ′
M
Fig. 3

M½
48
Assume that PP½. Let the lines AP½, BP½, CP½meet ω again at points K½, L½,
respectively. Now, if P½lies on the segment PF then by the first part of the solution we have
ǑM½FL½ǑM½EK½. On the other hand, we haveǑMFLǑM½FL½ǑM½EK½ǑMEK, therefore
ǑMFLǑMEK which contradicts MKML.
Similarly, if point P½lies on the segment EP then we getǑMFLǑMEK which is impossible.
Therefore, the points P and P½coincide and hence SPSP½SC.
Solution 3. We present a different proof of the converse direction, that is, MKML
SPSC. As in the previous solutions we assume that CACB, and line meets ω
at E and F.
From MLMK we getǑMEKǑMFL. Now we claim thatMEMF andEKFL.
To the contrary, suppose first thatMEMF; thenEKǑMEK¡MEǑMFL¡MF
FL. Now, the inequalityMEMF implies 2SCMEC MEEC MF2SPC
and hence SPSC. On the other hand, the inequalityEKFL implies 2SPK
EK AFFL AF2ABL, hence
SPA180¥¡SPK180¥¡ABLSBP.
L
C
F
K
P
E
A
A ′
B
S
M
ω
Fig. 4
Consider the point A½on the ray SA for whichSPA½SBP; in our case, this point lies
on the segment SA (see Fig. 4). Then △SBP△SPA½and SP 2SB¤SA½SB¤SASC 2 .
Therefore, SPSC which contradicts SPSC.
Similarly, one can prove that the inequalityMEMF is also impossible. So, we get
MEMF and therefore 2SCMEC MEEC MF2SPC, which implies
SCSP.

50
G3. Let A 1 A 2 . . .A n be a convex polygon. Point P inside this polygon is chosen so that its
projections P 1 , ..., P n onto lines A 1 A 2 , ..., A n A 1 respectively lie on the sides of the polygon.
Prove that for arbitrary points X 1 , ..., X n on sides A 1 A 2 , ..., A n A 1 respectively,
maxX 1 X 2
P 1 P 2
, . . ., X nX 1
P n P 11.
(Armenia)
Solution 1. Denote P n 1P 1 , X n 1X 1 , A n 1A 1 .
Lemma. Let point Q lies inside A 1 A 2 . . .A n . Then it is contained in at least one of the circumcircles
of triangles X 1 A 2 X 2 , ..., X n A 1 X 1 .
Proof. If Q lies in one
2Õ ¤¤¤
of the triangles X 1 A 2 X 2 , ..., X n A 1 X 1 , the claim is obvious. Otherwise
Q lies inside the polygon X 1 X 2 . . .X n
1Õ 1Õ ¤¤¤
(see Fig. 1). Then we have
X 1 QX ÔX n A 1 X 1 X n QX
X n A 1 X ÔX 1 QX 2
Ð
2πnπ,
hence there exists an index i such thatX i A i 1X i X 1 i QX i 1πn
nπ. Since the
quadrilateral QX i A i 1X i 1 is convex, this means exactly that Q is contained the circumcircle
of △X i A i 1X i 1, as desired.
ÔX 1 A 2 X 2
¤¤¤ ÔX 1 A 1 X 2
X n QX 1ÕÔn¡2Õπ
Now we turn to the solution. Applying lemma, we get that P lies inside the circumcircle of
triangle X i A i 1X i 1 for some i. Consider the circumcircles ω and Ω of triangles P i A i 1P i 1 and
X i A i 1X i 1 respectively (see Fig. 2); let r and R be their radii. Then we get 2rA i 1P2R
(since P lies inside Ω), hence
QED.
P i P i 12r sinP i A i 1P i 12R sinX i A i 1X i 1X i X i 1,
X 4
A 4
X 3
Ω
A 3
P
A 5 X i+1
Q
X 2 ω
P i+1
X 5
A 1 X 1 A P i 2
X i A i+1
Fig. 1 Fig. 2

51
Solution 2. As in Solution 1, we assume that all indices of points are considered modulo n.
We will prove a bit stronger inequality, namely
1 X 2
maxX
cosα 1 , . . ., X nX 1
cosα n1,
P 1 P 2 P n P 1
where α i (1in) is the angle between lines X i X i 1 and P i P i 1. We denote β iA i P i P i¡1
and γ iA i 1P i P i 1 for all 1in.
i
Suppose that for some 1in, point X i lies on the segment A i P i , while point X i 1 lies on
the segment P i 1A i 2. Then the projection of the segment X i X i 1 onto the line P i P i 1 contains
segment P i P i 1, since γ i and β i 1 are acute angles (see Fig. 3). Therefore, X i X i 1 cosα
P i P i 1, and in this case the statement is proved.
So, the only case left is when point X i lies on segment P i A i 1 for all 1in(the case
when each X i lies on segment A i P i
Y½ Y½
1Y½
is completely analogous).
Now, assume to the contrary that the inequality
X i X i 1 cosα iP i P i 1 (1)
holds for every 1in. Let Y i and i 1 be the projections of X i and X i 1 onto P i P i 1. Then
inequality (1) means exactly that Y i i 1P i P i 1, or P i Y iP i i 1 (again since γ i and β i 1
are acute; see Fig. 4). Hence, we have
X i P i cosγ iX i 1P i
1¨
1 cosβ i 1, 1in.
Multiplying these inequalities, we get
cos γ 1 cos γ 2¤¤¤cosγ ncosβ 1 cosβ 2¤¤¤cosβ n . (2)
On the other hand, the sines theorem applied to triangle PP i P i 1 provides
π
PP i 2¡β i i 1
.
PP
π
i 1sin
sin 2¡γ i¨cosβ γ i
Multiplying these equalities we get
which contradicts (2).
1cos β 2
cos γ 1¤cos β 3
cos γ 2¤¤¤cosβ 1
cosγ n
α i
X i+1
P
P i+1
β i+1
γ i
X i P i A i+1
P i+1
γ i
β i+1
α i
A i+1
X i
P
X i−1
X i+1
Fig. 3 Fig. 4
Y i
Y ′
i+1

52
G4. Let I be the incenter of a triangle ABC and Γ be its circumcircle. Let the line AI
intersect Γ at a point DA. Let F and E be points on side BC and arc BDC respectively
such thatBAFCAE1
2BAC. Finally, let G be the midpoint of the segment IF.
Prove that the lines DG and EI intersect on Γ.
(Hong Kong)
Solution 1. Let X be the second point of intersection of line EI with Γ, and L be the foot
of the bisector of angle BAC. Let G½and T be the points of intersection of segment DX with
lines IF and AF, respectively. We are to prove that GG½, or IG½G½F. By the Menelaus
theorem applied to triangle AIF and line DX, it means that we need the relation
1G½F
IG½TF
AT¤AD
ID , or TF
ATID
AD .
Let the line AF intersect Γ at point KA(see Fig. 1); Furthermore,DCLDCB
sinceBAKCAE we have
BKCE, hence KE ‖ BC. Notice thatIATDAKEADEXDIXT, so
the points I, A, X, T are concyclic. Hence we haveITAIXAEXAEKA, so
IT ‖ KE ‖ BC. Therefore we obtain TF
ATIL
AI .
Since CI is the bisector ofACL, we get
AICL IL
AC .
DABCAD1
2BAC, hence the triangles DCL and DAC are similar; therefore we get
CL
. Finally, it is known that the midpoint D of arc BC is equidistant from points I,
ACDC
AD
B, C, hence DC
ADID
AD .
Summarizing all these equalities, we get
as desired.
TF
ATIL
AICL
ACDC
ADID
AD ,
X
A
A
I
T
I
B
C
B
F
G ′
L
C
D
K
E
D
L
Fig. 1 Fig. 2

53
Comment. The equality AI is known and can be obtained in many different ways. For
ILAD
DI
instance, one can consider the inversion with center D and radius DCDI. This inversion takes
ǑBAC to the segment BC, so point A goes to L. Hence
DIAI IL
, which is the desired equality.
AD
Solution 2. As in the previous solution, we introduce the points X, T and K and note that
it suffice to prove the equality
TF
TF AT
ATDI DI AD AT AF
AD AT AD ADDI AD .
SinceFADEAI andTDAXDAXEAIEA, we get that the triangles
ATD and AIE are similar, therefore
ADAI AT
AE .
Next, we also use the relation DBDCDI. Let J be the point on the extension
of segment AD over point D such that DJDIDC (see Fig. 2). ThenDJC
JCD1
2Ôπ¡JDCÕ1
2ADC1
2ABCABI. Moreover,BAIJAC, hence
triangles ABI and AJC are similar, so AB
AJAI
AC , or AB¤ACAJ¤AIÔDI ADÕ¤AI.
On the other hand, we getABFABCAEC andBAFCAE, so triangles
ABF and AEC are also similar, which implies AF , or AB¤ACAF¤AE.
ACAB
AE
Summarizing we get
AI AF
AE
AT AF
AD
ÔDI ADÕ¤AIAB¤ACAF¤AE
as desired.
Comment. In fact, point J is an excenter of triangle ABC.
AD DI
AD DI ,

54
G5. Let ABCDE be a convex pentagon such that BC ‖ AE, ABBC AE, andABC
CDE. Let M be the midpoint of CE, and let O be the circumcenter of triangle BCD. Given
thatDMO90¥, prove that 2BDACDE.
(Ukraine)
Solution 1. Choose point T on ray AE such that ATAB; then from AE ‖ BC we have
CBTATBABT, so BT is the bisector ofABC. On the other hand, we have
ETAT¡AEAB¡AEBC, hence quadrilateral
BDCCDEABC180¥¡
BCTE is a parallelogram, and the
midpoint M of its diagonal CE is also the midpoint of the other diagonal BT.
Next, let point K be symmetrical to D with respect to M. Then OM is the perpendicular
bisector of segment DK, and hence ODOK, which means that point K lies on the circumcircle
of triangle BCD. Hence we haveBDCBKC. On the other hand, the angles
BKC and TDE are symmetrical with respect to M, soTDEBKCBDC.
Therefore,BDTBDE EDTBDE
BAT. This means that the points A, B, D, T are concyclic, and henceADBATB
1
2ABC1
2CDE, as desired.
B
C
B
C
O 2ϕ α
D
β
α + β
K
M
2ϕ − β − γ
D
γ
A E
T
A
E
Solution 2. LetCBDα,BDCβ,ADEγ, andABCCDE2ϕ. Then
we haveADB2ϕ¡β¡γ,BCD180¥¡α¡β,AED360¥¡BCD¡CDE
180¥¡2ϕ α β, and finallyDAE180¥¡ADE¡AED2ϕ¡α¡β¡γ.
2ϕ − α − β − γ
2ϕ − α − β
B
C
B
C
O
M
N
D
M
N
O
D
E
E
Let N be the midpoint of CD; thenDNO90¥DMO, hence points M, N lie on
the circle with diameter OD. Now, if points O and M lie on the same side of CD, we have
DMNDON1
2DOCα; in the other case, we haveDMN180¥¡DONα;

55
so, in both casesDMNα(see Figures). Next, since MN is a midline triangle CDE,
we haveMDEDMNα andNDM2ϕ¡α.
Now we apply the sine rule to the triangles ABD, ADE (twice), BCD and MND obtaining
AB
AE
γ DE
ADsinÔ2ϕ¡β¡γÕ
sinÔ2ϕ¡αÕ,
ADsinÔ2ϕ¡α¡βÕ,
ADsinÔ2ϕ¡α¡β¡γÕ
sinÔ2ϕ¡α¡βÕ,
BC β
CDsin α ,
α
DECDß2
DEß2ND
NMsinÔ2ϕ¡αÕ,
which implies
β¤sinÔ2ϕ¡α¡β¡γÕ
ADBC
CD¤CD
DE¤DE
ADsin
sinÔ2ϕ¡αÕ¤sinÔ2ϕ¡α¡βÕ.
Hence, the condition ABAE BC, or equivalently
ADAE AB BC
, after multiplying
AD
by the common denominator rewrites as
sinÔ2ϕ¡α¡βÕ¤sinÔ2ϕ¡β¡γÕsin γ¤sinÔ2ϕ¡αÕ sin β¤sinÔ2ϕ¡α¡β¡γÕ
γ¡αÕ cosÔγ¡αÕ¡cosÔ4ϕ¡2β¡α¡γÕcosÔ2ϕ¡α¡2β¡γÕ¡cosÔ2ϕ
cosÔγ¡αÕ cosÔ2ϕ γ¡αÕcosÔ2ϕ¡α¡2β¡γÕ cosÔ4ϕ¡2β¡α¡γÕ
cosϕ¤cosÔϕ γ¡αÕcosϕ¤cosÔ3ϕ¡2β¡α¡γÕ
cosÔϕ γ¡αÕ¡cosÔ3ϕ¡2β¡α¡γÕ¨0
cosϕ¤sinÔ2ϕ¡β¡αÕ¤sinÔϕ¡β¡γÕ0.
Since 2ϕ¡β¡α180¥¡AED180¥and ϕ1
2ABC90¥, it follows that ϕβ γ,
henceBDA2ϕ¡β¡γϕ1
2CDE, as desired.

56
G6. The vertices X, Y , Z of an equilateral triangle XY Z lie respectively on the sides BC,
CA, AB of an acute-angled triangle ABC. Prove that the incenter of triangle ABC lies inside
triangle XY Z.
G6½.
The vertices X, Y , Z of an equilateral triangle XY Z lie respectively on the sides
BC, CA, AB of a triangle ABC. Prove that if the incenter of triangle ABC lies outside
triangle XY Z, then one of the angles of triangle ABC is greater than 120¥.
(Bulgaria)
Solution 1 for G6. We will prove a stronger fact; namely, we will show that the incenter I of
triangle ABC lies inside the incircle of triangle XY Z (and hence surely inside triangle XY Z
itself). We denote by dÔU, V WÕthe distance between point U and line V W.
Denote by O the incenter of △XY Z and by r, r½and R½the inradii of triangles ABC, XY Z
and the circumradius of XY Z, respectively. Then we have R½2r½, and the desired inequality
is OIr½. We assume that OI; otherwise the claim is trivial.
Let the incircle of △ABC touch its sides BC, AC, AB at points A 1 , B 1 , C 1 respectively.
The lines IA 1 , IB 1 , IC 1 cut the plane into 6 acute angles, each one containing one of the
points A 1 , B 1 , C 1
BCÕ
on its border. We may assume that O lies in an angle defined by lines IA 1 ,
IC 1 and containing point C 1 (see Fig. 1). Let A½and C½be the projections of O onto lines IA 1
and IC 1 , respectively.
Since OXR½, we have dÔO, BCÕR½. Since OA½‖ BC, it follows that dÔA½,
A½I rR½, or A½IR½¡r. On the other hand, the incircle of △XY Z lies inside △ABC,
hence dÔO, ABÕr½, and analogously we get dÔO, ABÕC½C 1r¡IC½r½, or IC½r¡r½.
B
X
C 1
C ′
Z
O C′
I
A ′
O
I
A
B 1 Y C
A ′
ÔR½¡rÕ Fig. 1 Fig. 2
Finally, the quadrilateral IA½OC½is circumscribed due to the right angles at A½and
(see Fig. 2). On its circumcircle, we haveǑA½OC½2A½IC½180¥OC½I, hence 180¥
IC½A½O. This means that IC½A½O. Finally, we have OIIA½ A½OIA½ IC½ Ôr¡r½ÕR½¡r½r½, as desired.
Solution 2 for G6. Assume the contrary. Then the incenter I should lie in one of triangles
AY Z, BXZ, CXY — assume that it lies in △AY Z. Let the incircle ω of △ABC touch
sides BC, AC at point A 1 , B 1 respectively. Without loss of generality, assume that point A 1
lies on segment CX. In this case we will show thatC90¥thus leading to a contradiction.
Note that ω intersects each of the segments XY and Y Z at two points; let U, U½and V ,
V½be the points of intersection of ω with XY and Y Z, respectively (UYU½Y , V YV½Y ;
see Figs. 3 and 4). Note that 60¥XY Z1
2ÔUV¡U½V½Õ1
2UV , henceUV120¥.
A 1
C ′

On the other hand, since I lies in △AY Z, we getǑV UV½180¥, henceǑUA 1 U½ǑUA 180¥¡UV60¥.
Now, two cases are possible due to the order of points Y , B 1 on segment AC.
1
V½
57
A
ω C 1
B 1
C 1
Y
V ′
Y
I
V ′
I
U ′
V
U ′
B
Z
1 V
Z
U
ω
U
C A 1 X
B C A 1 X B
Fig. 3 Fig. 4
Case 1. Let point Y lie on the segment AB 1 (see Fig. 3). Then we haveY XC
1A 2 1 U½¡A U¨1
1 2ǑUA 1 U½30¥; analogously, we getXY C1
2ǑUA 1
X
U½30¥. Therefore,
Y CX180¥¡Y XC¡XY C120¥, as desired.
Case 2. Now let point Y lie on the segment CB 1 (see Fig. 4). Analogously, we obtain
Y XC30¥. Next,IY XZY X60¥, butIY XIY B 1 , since Y B 1 is a tangent
and Y X is a secant line to circle ω from point Y . Hence, we get 120¥IY B 1 IY
B 1 Y XY XC CX30¥ Y Y CX, henceY CX120¥¡30¥90¥, as desired.
Comment. In the same way, one can prove a more general
Claim. Let the vertices X, Y , Z of a triangle XY Z lie respectively on the sides BC, CA, AB of a
triangle ABC. Suppose that the incenter of triangle ABC lies outside triangle XY Z, and α is the
least angle of △XY Z. Then one of the angles of triangle ABC is greater than 3α¡90¥.
A
Solution for G6½. Assume the contrary. As in Solution 2, we assume that the incenter I of
△ABC lies in △AY Z, and the tangency point A 1 of ω and BC lies on segment CX. Surely,
Y ZA180¥¡Y ZX120¥, hence points I and Y lie on one side of the perpendicular
bisector to XY ; therefore IXIY . Moreover, ω intersects segment XY at two points, and
therefore the projection M of I onto XY lies on the segment XY . In this case, we will prove
thatC120¥.
Let Y K, Y L be two tangents from point Y to ω (points K and A 1 lie on one side of XY ;
if Y lies on ω, we say KLY ); one of the points K and L is in fact a tangency point B 1
of ω and AC. From symmetry, we haveY IKY IL. On the other hand, since IXIY ,
we get XMXY which impliesA 1
MIKÕ
XYKY X.
Next, we haveMIY90¥¡IY X90¥¡ZY X30¥. Since IA 1ÃA 1 X, IMÃXY ,
IKÃY K we getMIA 1A 1 XYKY XMIK. Finally, we get
A 1 IKA 1 IM ÔKIY ILÕ Y
2MIK 2KIY2MIY60¥.
1 ILÔA
Hence,A 1 IB 160¥, and thereforeACB180¥¡A 1 IB 1120¥, as desired.

58
X A 1 K(= B 1 )
Y
M
C
A 1 M Y
L(= B 1 )
X
I
B
Z
A
I
Fig. 5 Fig. 6
Comment 1. The estimate claimed in G6½is sharp. Actually, ifBAC120¥, one can consider an
equilateral triangle XY Z with ZA, YÈAC, XÈBC (such triangle exists sinceACB60¥). It
intersects with the angle bisector ofBAC only at point A, hence it does not contain I.
Comment 2. As in the previous solution, there is a generalization for an arbitrary triangle XY Z,
but here we need some additional condition. The statement reads as follows.
Claim. Let the vertices X, Y , Z of a triangle XY Z lie respectively on the sides BC, CA, AB of a
triangle ABC. Suppose that the incenter of triangle ABC lies outside triangle XY Z, α is the least
angle of △XY Z, and all sides of triangle XY Z are greater than 2r cot α, where r is the inradius
of △ABC. Then one of the angles of triangle ABC is greater than 2α.
The additional condition is needed to verify that XMY M since it cannot be shown in the
original way. Actually, we haveMY Iα, IMr, hence Y Mr cot α. Now, if we have
XYXM YM2r cot α, then surely XMY M.
On the other hand, this additional condition follows easily from the conditions of the original
problem. Actually, if IÈ△AY Z, then the diameter of ω parallel to Y Z is contained in △AY Z and
is thus shorter than Y Z. Hence Y Z2r2r cot 60¥.

60
G7. Three circular arcs γ 1 , γ 2 , and γ 3 connect the points A and C. These arcs lie in the same
half-plane defined by line AC in such a way that arc γ 2 lies between the arcs γ 1 and γ 3 . Point
B lies on the segment AC. Let h 1 , h 2 , and h 3 be three rays starting at B, lying in the same
half-plane, h 2 being between h 1 and h 3 . For i, j1, 2, 3, denote by V ij the point of intersection
of h i and γ j (see the Figure below).
Denote byǑV ij V kjǑV kl V il the curved quadrilateral, whose sides are the segments V ij V il , V kj V kl
and arcs V ij V kj and V il V kl . We say that this quadrilateral is circumscribed if there exists a circle
touching these two segments and two arcs.
Prove that if the curved quadrilateralsǑV 11 V 21ǑV 22 V 12 ,ǑV 12 V 22ǑV 23 V 13 ,ǑV 21 V 31ǑV 32 V 22 are circumscribed,
then the curved quadrilateralǑV 22 V 32ǑV 33 V 23 is circumscribed, too.
h
V 2 23
h 1
V 22
γ 3
V 13
V 33
V 12
V 11
V 32
h 3
A
γ 2
V 21 V 31
γ 1
B
Fig. 1
C
(Hungary)
Solution. Denote by O i and R i the center and the radius of γ i , respectively. Denote also by H
the half-plane defined by AC which contains the whole configuration. For every point P in
the half-plane H, denote by dÔPÕthe distance between P and line AC. Furthermore, for any
r0, denote by ΩÔP, rÕthe circle with center P and radius r.
Lemma 1. For every 1ij3, consider those circles ΩÔP, rÕin the half-plane H which
are tangent to h i and h j .
(a) The locus of the centers of these circles is the angle bisector β ij between h i and h j .
(b) There is a constant u ij such that ru ij¤dÔPÕfor
Ð
all such circles.
Proof. Part (a) is obvious. To prove part (b), notice that the circles which are tangent to h i
and h j are homothetic with the common homothety center B (see Fig. 2). Then part (b) also
becomes trivial.
Lemma 2. For every 1ij3, consider those circles ΩÔP, rÕin the half-plane H which
are externally tangent to γ i and internally tangent to γ j .
(a) The locus of the centers of these circles is an ellipse arc ε ij with end-points A and C.
(b) There is a constant v ij such that rv ij¤dÔPÕfor all such circles.
Proof. (a) Notice that the circle ΩÔP, rÕis externally tangent to γ i and internally tangent to γ j
if and only if O i PR i r and O jR j¡r. Therefore, for each such circle we have
O i P O j PO i A O j AO i C O j CR i R j .
Such points lie on an ellipse with foci O i and O j ; the diameter of this ellipse is R i R j , and it
passes through the points A and C. Let ε ij be that arc AC of the ellipse which runs inside the
half plane H (see Fig. 3.)
This ellipse arc lies between the arcs γ i and γ j . Therefore, if some point P lies on ε ij ,
then O i PR i and O j PR j . Now, we choose rO i P¡R iR j¡O j P0; then the

61
Ω(P, r)
γ j
h i
β ij
r
P
R j
ε ij
P
r
P ′
r ′
d(P ′ )
h j
A
d(P)
B
⃗ρ
Fig. 2 Fig. 3
r, i j
circle ΩÔP, rÕtouches γ i externally and touches γ j
,P
internally, so P belongs to the locus under
investigation.
(b) Let AP, ⃗ρ AO i , and ⃗ρ AO j ; let d ijO i O j , and let ⃗v be a unit vector
orthogonal to AC and directed toward H. Then we have⃗ρ iR i ,⃗ρ jR j O i
⃗ρ¡⃗ρ iR i O j
jÕ P⃗ρ¡⃗ρ jR j¡r,
jÕ
hence
Ô⃗ρ¡⃗ρ iÕ2¡Ô⃗ρ¡⃗ρ jÕ2ÔR i rÕ2¡ÔR j¡rÕ2 ,
Ô⃗ρ i¡⃗ρ 2 2 2⃗ρ¤Ô⃗ρ j¡⃗ρ iÕÔR i¡R 2 2 2rÔR i R jÕ,
d ij¤dÔPÕd ij
r
⃗v¤⃗ρÔ⃗ρ j¡⃗ρ iÕ¤⃗ρrÔR i R jÕ.
Therefore,
d R i R j¤dÔPÕ,
and the value v ij
d
ijǑ Ǒ Ð
ij
does not depend on P.
R i R j
Lemma 3. The curved quadrilateral Q V i,j V i 1,jV i 1,j 1V i,j 1 is circumscribed if and only
if u i,i 1v j,j 1.
Proof. First suppose that the curved quadrilateral Q ij is circumscribed and ΩÔP, rÕis its inscribed
circle. By Lemma 1 and Lemma 2 we have ru i,i 1¤dÔPÕand rv j,j 1¤dÔPÕas
well. Hence, u i,i 1v j,j 1.
To prove the opposite direction, suppose u i,i 1v j,j
Ð
1. Let P be the intersection of the
angle bisector β i,i 1 and the ellipse arc ε j,j 1. Choose ru i,i 1¤dÔPÕv j,j 1¤dÔPÕ. Then
the circle ΩÔP, rÕis tangent to the half lines h i and h i 1 by Lemma 1, and it is tangent to the
arcs γ j and γ j 1 by Lemma 2. Hence, the curved quadrilateral Q ij is circumscribed.
By Lemma 3, the statement of the problem can be reformulated to an obvious fact: If the
equalities u 12v 12 , u 12v 23 , and u 23v 12 hold, then u 23v 23 holds as well.
⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ⃗ρ
r
⃗ρ j
⃗ρ i
O j
O i
R i
γ i
⃗v
C

62
Comment 1. Lemma 2(b) (together with the easy Lemma 1(b)) is the key tool in this solution.
If one finds this fact, then the solution can be finished in many ways. That is, one can find a circle
touching three of h 2 , h 3 , γ 2 , and γ 3 , and then prove that it is tangent to the fourth one in either
synthetic or analytical way. Both approaches can be successful.
Here we present some discussion about this key Lemma.
1. In the solution above we chose an analytic proof for Lemma 2(b) because we expect that most
students will use coordinates or vectors to examine the locus of the centers, and these approaches are
less case-sensitive.
Here we outline a synthetic proof. We consider only the case when P does not lie in the line O i O j .
The other case can be obtained as a limit case, or computed in a direct way.
Let S be the internal homothety center between the circles of γ i and γ j , lying on O i O j ; this point
does not depend on P. Let U and V be the points of tangency of circle σΩÔP,rÕwith γ i and γ j ,
respectively (then rPUPV ); in other words, points U and V are the intersection points of
rays O i P, O j P with arcs γ i , γ j respectively (see Fig. 4).
Due to the theorem on three homothety centers (or just to the Menelaus theorem applied to
triangle O i O j
P
P), the points U, V and S are collinear. Let T be the intersection point of line AC and
the common tangent to σ and γ i at U; then T is the radical center of σ, γ i and γ j , hence TV is the
common tangent to σ and γ j
Ð
.
Let Q be the projection of P onto the line AC. By the right angles, the points U, V and Q lie on
the circle with diameter PT. From this fact and the equality PUPV we getUQPUV
V UPSUO i . Since O i S ‖ PQ, we haveSO i UQPU. Hence, the triangles SO i U and UPQ
r
i S i S
are similar and thus
; the last expression is constant since S is a constant
dÔPÕPU
PQO O i UO R i
point.
l
V
P
σ
γ j
d l (A)
P
d l (P)
O j
ε ij
U
γ i
d(P)
O j
S
A
C
T
A
Q
C
O i
O i
Fig. 4 Fig. 5
2. Using some known facts about conics, the same statement can be proved in a very short way.
Denote by l the directrix of ellipse of ε ij related to the focus O j ; since ε ij is symmetrical about O i O j ,
we have l ‖ AC. Recall that for each point PÈε ij , we have PO jǫ¤d lÔPÕ, where d lÔPÕis the
distance from P to l, and ǫ is the eccentricity of ε ij
Ð
(see Fig. 5).
Now we have
rR j¡ÔR j¡rÕAO j¡PO jǫ d lÔAÕ¡d lÔPÕ¨ǫ dÔPÕ¡dÔAÕ¨ǫ¤dÔPÕ,
and ǫ does not depend on P.

63
Comment 2. One can find a spatial interpretations of the problem and the solution.
For every pointÔx,yÕand radius r0, represent the circle Ω Ôx,yÕ,r¨by the pointÔx,y,rÕ
in space. This point is the apex of the cone with base circle Ω Ôx,yÕ,r¨and height r. According to
Lemma 1, the circles which are tangent to h i and h j correspond to the points of a half line β½ij , starting
at B.
Now we translate Lemma 2. Take some 1ij3, and consider those circles which are
internally tangent to γ j . It is easy to see that the locus of the points which represent these circles is
a subset of a cone, containing γ j . Similarly, the circles which are externally tangent to γ i correspond
to the points on the extension of another cone, which has its apex on the opposite side of the base
plane Π. (See Fig. 6; for this illustration, the z-coordinates were multiplied by 2.)
The two cones are symmetric to each other (they have the same aperture, and their axes are
parallel). As is well-known, it follows that the common points of the two cones are co-planar. So the
intersection of the two cones is a a conic section — which is an ellipse, according to Lemma 2(a). The
points which represent the circles touching γ i and γ j is an ellipse arc ε½ij with end-points A and C.
β 12
′
ε ′ ij
γ i
γ j
Π
Σ
ε ′ 23
ε ′ 12
β ′ 23
Fig. 6 Fig. 7
Thus, the curved quadrilateral Q ij is circumscribed if and only if β½i,i 1 and ε½j,j 1 intersect, i.e. if
they are coplanar. If three of the four curved quadrilaterals are circumscribed, it means that ε½12 , ε½23 ,
and β½23 lie in the same plane Σ, and the fourth intersection comes to existence, too (see Fig. 7).
β½12
A connection between mathematics and real life:
the Palace of Creativity “Shabyt” (“Inspiration”) in Astana

Number Theory
N1. Find the least positive integer n for which there exists a setØs 1 , s 2 , . . .,s nÙconsisting of
n distinct positive integers such that
¢1¡1
1Å¢1¡1
s s 2Å...¢1¡1
s nÅ51
2010 .
N1½. Same as Problem N1, but the constant
51
2010
is replaced by
42
2010 . (Canada)
Answer for Problem N1. n39.
Solution for Problem N1. Suppose that for some n there exist the desired numbers; we
may assume that s 1s 2¤¤¤s n . Surely s 11since
nÅ
otherwise 1¡1
So we have
s 10.
2s 1s 2¡1¤¤¤s n¡Ôn¡1Õ, hence s ii 1 for each i1, . . .,n. Therefore
51
2010¢1¡1
1Å¢1¡1
s s 2Å...¢1¡1
1
s
¢1¡1 2Å¢1¡1
3Å...¢1¡1
1Å1
n 2¤2
3¤¤¤n 1
n n 1 ,
which implies
n 12010
51670
1739,
so n39.
Now we are left to show that n39 fits. Consider the setØ2, 3, . . ., 33, 35, 36, . . ., 40, 67Ù
Ô1Õ
which contains exactly 39 numbers. We have
1
2¤2
3¤¤¤32
33¤34
35¤¤¤39
40¤66
671
33¤34
40¤66
6717
67051
2010 ,
hence for n39 there exists a desired example.
Comment. One can show that the exampleÔ1Õis unique.
Answer for Problem N1½. n48.
Solution for Problem N1½. Suppose that for some n there exist the desired numbers. In
the same way we obtain that s ii 1. Moreover, since the denominator of the fraction
42
20107
335 is divisible by 67, some of s i’s should be divisible by 67, so s ns i67. This
3¤¤¤n¡1
means that
42
20101
2¤2
n¤¢1¡1
67Å66
67n ,

65
which implies
n2010¤66
42¤67330
747,
so n48.
Now we are left to show that n48 fits. Consider the setØ2, 3, . . ., 33, 36, 37, . . ., 50, 67Ù
which contains exactly 48 numbers. We have
1
2¤2
3¤¤¤32
33¤35
36¤¤¤49
50¤66
671
33¤35
50¤66
677
33542
2010 ,
hence for n48 there exists a desired example.
Comment 1. In this version of the problem, the estimate needs one more step, hence it is a bit
harder. On the other hand, the example in this version is not unique. Another example is
1
2¤2
3¤¤¤46
47¤66
67¤329
3301
67¤66
330¤329
477
67¤542
2010 .
Comment 2. N1½was the Proposer’s formulation of the problem. We propose N1 according to the
number of current IMO.

66
N2. Find all pairsÔm, nÕof nonnegative integers for which
m 2 2¤3 nm 2 n 1¡1¨. (1)
Answer.Ô6, 3Õ,Ô9, 3Õ,Ô9, 5Õ,Ô54, 5Õ.
(Australia)
Solution. For fixed values of n, the equation (1) is a simple quadratic equation in m. For
n5 the solutions are listed in the following table.
case equation discriminant integer roots
n0 m 2¡m 20 ¡7 none
n1 m 2¡3m 60 ¡15 none
n2 m 2¡7m 180 ¡23 none
n3 m 2¡15m 540 9 m6 and m9
n4 m 2¡31m 1620 313 none
n5 m 2¡63m 4860 202545 2 m9 and m54
We prove that there is no solution for n6.
Suppose thatÔm, nÕsatisfies (1) and n6. Since m§2¤3nm 2 n 1¡1¨¡m
m3p with some 0pn or m2¤3q with some 0qn.
In the first case, let qn¡p; then
In the second case let pn¡q. Then
2 n 1¡1m 2¤3 n
m3 p 2¤3 q .
2 , we have
2 n 2¤3 1¡1m n
m2¤3 q 3 p .
1Õ
Hence, in both cases we need to find the nonnegative integer solutions of
3 p 2¤3 q2 n 1¡1, p qn. (2)
Next, we prove
1Õ
1Õ
bounds for p, q. From (2) we get
3 p2 n 18 n 1
39 n 1
33 1Õ
2Ôn 3
and
2¤3 q2 n 12¤8 n 32¤9 n 32¤3 2n 32¤3 2Ôn 3 ,
so p, q2Ôn . Combining these inequalities with p qn, we obtain
3
n¡2
q2Ôn p, . (3)
3 3
Now let hminÔp, qÕ. By (3) we have hn¡2
3 ; in particular, we have h1. On the
left-hand side of (2), both terms are divisible by 3 h , therefore 9§3 h§2 n 1¡1. It is easy check
that ord 9Ô2Õ6, so 9§2 n 1¡1 if and only if 6§n 1. Therefore, n 16r for some positive
integer r, and we can write
2 n 1¡14 3r¡1Ô4 2r 4 r 1ÕÔ2
r¡1ÕÔ2 r 1Õ. (4)

Notice that the factor 4 2r 4 r 1Ô4 divisible by 9. The other two factors in (4), 2 r¡1 and 2 r 1 are coprime: both are odd and
67
r¡1Õ2 3¤4r is divisible by 3, but it is never
their difference is 2. Since the whole product is divisible by 3 h , we have either 3 h¡1§2 r¡1 or
3 h¡1§2 r 1. In any case, we have 3 h¡12 r 1. Then
3 h¡12 r 13 r3 n 1
6 ,
n¡2
1
,
3¡1h¡1n 6
n11.
But this is impossible since we assumed n6, and we proved 6§n 1.

¤¤¤
68
N3. Find the smallest number n such that there exist polynomials f 1 , f 2 , ..., f n with rational
coefficients satisfying
x 2 7f 1ÔxÕ2 f 2ÔxÕ2 f nÔxÕ2 .
(Poland)
Answer. The smallest n is 5.
Solution 1. The equality x 2 7x2 2 2 1 2 1 2 1 2 shows that n5. It remains to
show that x 2 7 is not a sum of four (or less) squares of polynomials with rational coefficients.
Suppose by way of contradiction that x 2 7f 1ÔxÕ2 f 2ÔxÕ2 f 3ÔxÕ2 f 4ÔxÕ2 , where the
coefficients of polynomials f 1 , f 2 , f 3 and f 4 are rational (some of these polynomials may be
zero).
Clearly, the degrees of f 1 , f 2 , f 3 and f 4 are at most 1. Thus f iÔxÕa i x b i for i1, 2, 3, 4
and some rationals a 1 , b 1 , a 2 , b 2 , a 3 , b 3 , a 4 , b 4 . It follows that x 2 74
i1Ôa i x b iÕ2 and
hence
4ôi1
a 2 i1,
4ôi1
a i b i0,
Let p ia i b i and q ia i¡b i for i1, 2, 3, 4. Then
4ôi1
b 2 i7. (1)
4ôi1
p 2 i4ôi1 a 2 i 2
4ôi1
a i b i
4ôi1b 2 i8,
4ôi1
q 2 i4ôi1 a 2 i¡2
4ôi1
a i b i
4ôi1b 2 i8
and
4ôi1
p i q i4ôi1
a 2 i¡4ôi1 b 2 i¡6,
which means that there exist a solution in integers x 1 , y 1 , x 2 , y 2 , x 3 , y 3 , x 4 , y 4 and m0of
the system of equations
4ôi1
4ôi1
4ôi1
(i)
x 2 i8m 2 ,
(ii)
y 2 i8m 2 ,
(iii)
x i y i¡6m 2 .
We will show that such a solution does not exist.
Assume the contrary and consider a solution with minimal m. Note that if an integer x is
odd then x 21Ômod 8Õ. Otherwise (i.e., if x is even) we have x 20Ômod 8Õor x 24
Ômod 8Õ. Hence, by (i), we get that x 1 , x 2 , x 3 and x 4 are even. Similarly, by (ii), we get that
y 1 , y 2 , y 3 and y 4 are even. Thus the LHS of (iii) is divisible by 4 and m is also even. It follows
thatÔx 1
2
, y 1
, x 2
2 2
, y 2
, x 3
2 2
, y 3
, x 4
2 2
, y 4
, m 2 2Õis a solution of the system of equations (i), (ii) and (iii),
which contradicts the minimality of m.
Solution 2. We prove that n4 is impossible. Define the numbers a i , b i for i1, 2, 3, 4 as
in the previous solution.
By Euler’s identity we have
Ôa 2 1 a 2 2 a 2 3 a 2 4ÕÔb 2 1 b 2 2 b 2 3 b 2 4ÕÔa
1
b 1 a 2 b 2 a 3 b 3 a 4 b 4Õ2 Ôa 1 b 3¡a 3 b 1 a 4 b 2¡a 2 b 4Õ2
Ôa 1 b 2¡a 2 b 1 a 3 b 4¡a 4 b 3Õ2
Ôa 1 b 4¡a 4 b 1 a 2 b 3¡a 3 b 2Õ2 .

69
So, using the relations (1) from the Solution 1 we get that
where
7¡m 1
m©2 ¡m 2
m©2 ¡m 3
m©2
, (2)
m 1
1 b 2¡a 2 b 1 a 3 b 4¡a 4 b 3 ,
ma m 2
1 b 3¡a 3 b 1 a 4 b 2¡a 2 b 4 ,
ma m 3
1 b 4¡a 4 b 1 a 2 b 3¡a 3 b 2
ma and m 1 , m 2 , m 3ÈZ,
2¨
mÈN.
Let m be a minimum positive integer number for which (2) holds. Then
8m 2m 2 1 m 2 2 m 2 3 m 2 .
m
As in the previous solution, we get that m 1 , m 2 , m 3 , m are all even numbers. Then 1
, m 2
, m 3
, m
2 2 2
is also a solution of (2) which contradicts the minimality of m. So, we have n5. The example
with n5 is already shown in Solution 1.

70
N4. Let a, b be integers, and let PÔxÕax 3 bx. For any positive integer n we say that the
pairÔa, bÕis n-good if n§PÔmÕ¡PÔkÕimplies n§m¡k for all integers m, k. We say that
Ôa, bÕis very good ifÔa, bÕis n-good for infinitely many positive integers n.
(a) Find a pairÔa, bÕwhich is 51-good, but not very good.
(b) Show that all 2010-good pairs are very good.
(Turkey)
Solution. (a) We show that the pairÔ1,¡51 2Õis good but not very good. Let PÔxÕx3¡512 x.
Since PÔ51ÕPÔ0Õ, the pairÔ1,¡51 2Õis not n-good for any positive integer that does not
divide 51. Therefore,Ô1,¡51 2Õis not very good.
On the other hand, if PÔmÕPÔkÕÔmod 51Õ, then m 3k3Ômod 51Õ. By Fermat’s
theorem, from this we obtain
mm 3k 3kÔmod 3Õand mm 33k 33kÔmod 17Õ.
Hence we have mkÔmod 51Õ. ThereforeÔ1,¡51 2Õis 51-good.
PÔm½ÕPÔmÕPÔkÕPÔk½Õ
(b) We will show that if a pairÔa, bÕis 2010-good thenÔa, bÕis 67 i -good for all positive
integer i.
Ð
Claim 1. IfÔa, bÕis 2010-good thenÔa, bÕis 67-good.
Proof. Assume that PÔmÕPÔkÕÔmod 67Õ. Since 67 and 30 are coprime, there exist integers
m½and k½such that k½kÔmod 67Õ, k½0Ômod 30Õ, and m½mÔmod 67Õ, m½0
Ômod 30Õ. Then we have PÔm½ÕPÔ0ÕPÔk½ÕÔmod 30Õand
Ômod 67Õ, hence PÔm½ÕPÔk½ÕÔmod 2010Õ. This implies m½k½Ômod 2010ÕasÔa, bÕis
2010-good. It follows that mm½k½kÔmod 67Õ. Therefore,Ôa, bÕis 67-good.
Claim 2. IfÔa, bÕis 67-good then 3Õ 67§a.
bÔm¡kÕÔm¡kÕ 2Õ b¨
Proof. Suppose that 67Чa. Consider the setsØat 2Ômod 67Õ:0t33ÙandØ¡3as 2¡b
mod 67 : 0s33Ù. Since a0Ômod 67Õ, each of these sets has 34 elements. Hence they
have at least one element in common. If at 2¡3as2¡bÔmod 67Õthen for mt¨s, k©2s
we have
PÔmÕ¡PÔkÕaÔm 3¡k aÔm 2 mk k
Ôt¨3sÕÔat 2 3as 2 bÕ0Ômod 67Õ.
SinceÔa, bÕis 67-good, we must have mkÔmod 67Õin both cases, that is, t3sÔmod 67Õ
and t¡3sÔmod i§PÔmÕ¡PÔkÕÔm¡kÕ 67Õ. This means ts0Ômod 67Õand b¡3as 2¡at20Ômod 67Õ.
But then 67§PÔ7Õ¡PÔ2Õ67¤5a 2Õ 2Õ 5b and 67Ч7¡2, contradicting thatÔa, bÕis 67-good.Ð
bÕ Ð
Claim 3. IfÔa, bÕis 2010-good thenÔa, bÕis 67 i -good all i1.
Proof. By Claim 2, we have 67§a. If 67§b, then PÔxÕPÔ0ÕÔmod 67Õfor all x, contradicting
thatÔa, bÕis 67-good. Hence, 67Чb.
Suppose that 67 aÔm2 mk k b¨. Since 67§a and 67Чb,
the second factor aÔm2 mk k b is coprime to 67 and hence 67 i§m¡k. Therefore,Ôa,
is 67 i -good.
Comment 1. In the proof of Claim 2, the following reasoning can also be used. Since 3 is not
a quadratic residue modulo 67, either au 2¡bÔmod 67Õor 3av 2¡bÔmod 67Õhas a solution.
The settingsÔm,kÕÔu,0Õin the first case andÔm,kÕÔv,¡2vÕin the second case lead to b0
Ômod 67Õ.
Comment 2. The pairÔ67,30Õis n-good if and only if nd¤67 i , where d§30 and i0. It shows
that in part (b), one should deal with the large powers of 67 to reach the solution. The key property
of number 67 is that it has the form 3k 1, so there exists a nontrivial cubic root of unity modulo 67.

fÔmÕ n¨
71
N5. Let N be the set of all positive integers. Find all functions f : NNsuch that the
number m fÔnÕ¨is a square for all m,
fÔmÕ n¨
nÈN.
fÔnÕ
(U.S.A.)
Answer. All functions of the form fÔnÕn c, where cÈNØ0Ù.
Solution. First, it is clear that all functions of
fÔlÕfÔkÕ
the form fÔnÕn c with a constant nonnegative
integer c satisfy the problem conditions since m¨Ôn m cÕ2 is a
square.
We are left to prove that there are no other functions. We start with the following
Lemma. Suppose that p§fÔkÕ¡fÔlÕfor some prime p and positive integers k, l. Then p§k¡l.
p§
Proof.
fÔnÕ fÔkÕ
Suppose
k¨¡ n¨
first
fÔnÕ fÔnÕ fÔlÕ
that p 2§fÔkÕ¡fÔlÕ, fÔnÕ n¨ fÔnÕ fÔnÕ
so p 2 a for some integer a. Take some
positive integer DmaxØfÔkÕ, fÔlÕÙwhich is not divisible by p and set npD¡fÔkÕ. Then
the positive numbers n fÔkÕpD and n fÔlÕpD fÔlÕ¡fÔkÕ¨pÔD paÕare
both divisible by p but not by p 2 . Now, applying the problem conditions, we get that both the
numbers k¨and l¨are squares divisible by p (and thus
by p 2 ); this
p§
means
fÔnÕ
that
k¨¡
the
fÔnÕ fÔnÕ fÔkÕ fÔnÕ fÔlÕ n
multipliers k and l are also divisible by p, therefore
Ð
l¨k¡l as well.
On the other hand, if fÔkÕ¡fÔlÕis divisible by p but not by p 2 , then choose the same
number D and set np3 D¡fÔkÕ. Then the positive numbers np3 D and
p 3 D fÔlÕ¡fÔkÕ¨are respectively divisible by p 3 (but not by p 4 ) and by p (but not by p 2 ).
Hence in analogous way we obtain that the numbers k and l are divisible by p,
therefore l¨k¡l.
We turn to the problem. First, suppose that fÔkÕfÔlÕfor some k, lÈN. Then by Lemma
we
fÔnÕ¨qfÔ1Õ fÔnÕfÔ1Õ
have that k¡l is divisible by every prime number, so k¡l0, or kl. Therefore, the
function f is injective.
Next, consider the
fÔnÕfÔ1Õ fÔnÕfÔn¡2ÕfÔ1Õ 1Õ
numbers fÔkÕand fÔk 1Õ. Since the numberÔk 1Õ¡k1has no
prime divisors, by Lemma the same holds for fÔk 1Õ¡fÔkÕ; fÔnÕÔfÔ1Õ¡1Õ fÔnÕfÔ1Õ nfÔ1Õ
thusfÔk 1Õ¡fÔkÕ1.
Now, let fÔ2Õ¡fÔ1Õq,q1. Then we prove by induction that qÔn¡1Õ.
The base for n1, 2 holds by the definition of q. For the step, if n1we have qÔn¡1Õ¨q. Since qÔn¡2Õ, we get qn,
as desired.
Finally, we have qÔn¡1Õ. Then q cannot be¡1 since otherwise for 1
we have fÔnÕ0 which is impossible. Hence q1 and n for each nÈN,
and fÔ1Õ¡10, as desired.

72
N6. The rows and columns of a 2 n¢2 n table are numbered from 0 to 2 n¡1. The cells of the
table have been colored with the following property being satisfied: for each 0i, j2 n¡1,
the jth cell in the ith row and theÔi jÕth cell in the jth row have the same color. (The
indices of the cells in a row are considered modulo 2 n .)
Prove that the maximal possible number of colors is 2 n .
jÕ
(Iran)
Solution. Throughout the solution we denote the cells of the table by coordinate pairs;Ôi,
refers to the jth cell in the ith row.
Consider the directed graph, whose vertices are the cells of the board, and the edges are
the arrowsÔi, jÕÔj, i jÕfor all 0i, j2n¡1. From each vertexÔi, jÕ, exactly one edge
passes (toÔj, i j mod 2 nÕ); conversely, to each cellÔj, kÕexactly one edge is directed (from
the cellÔk¡j mod 2 n , jÕÕ. Hence, the graph splits into cycles.
Now, in any coloring considered, the vertices of each cycle should have the same color by
the problem condition. On the other hand, if each cycle has its own color, the obtained coloring
obviously satisfies the problem conditions. Thus, the maximal possible number of colors is the
same as the number of cycles, and we have to prove that this number is 2 n .
Next, consider any cycleÔi 1 , j 1Õ,Ôi 2 , j 2Õ, . . .; we will describe it in other terms. Define a
sequenceÔa 0 , a 1 , . . .Õby the relations a 0i 1 , a 1j 1 , a n 1a n a n¡1 2
for all n1(we
say that such a sequence is a Fibonacci-type sequence). Then an obvious induction shows
that i ka k¡1Ômod 2 nÕ, j ka kÔmod 2 nÕ. Hence we need to investigate the behavior of
Fibonacci-type sequences modulo 2 n .
Denote by F 0 , F 1 , . . . the Fibonacci numbers defined by F 00, F 11, and F n
F n 1 F n for n0. We also set F¡11 according to the recurrence relation.
νÔmÕ k
For every positive integer m, denote by νÔmÕthe exponent of 2 in the prime factorization
of m, i.e. for which 2 νÔmÕ§m but 2 1Чm.
Lemma 1. For every Fibonacci-type sequence a 0 , a 1 , a 2 , . . ., and every k0, we have a
F k¡1a 0 F k a 1 .
Proof. Apply induction on k. The base cases
1Õ Ð
k0, 1 are trivial. For the step, from the
induction hypothesis we get
a k 1a k a k¡1ÔF k¡1a 0 F k a ÔF k¡2a 0 F k¡1a 1ÕF k a 0 F k 1a 1 .
Lemma 2. For every m3,
(a) we have νÔF m¡2Õm;
3¤2 (b) d3¤2m¡2 is the least positive index for which 2 m§F d ;
(c) F 3¤2m¡2 11 2 m¡1Ômod 2 mÕ.
Proof. Apply induction on m. In the base case m3we have νÔF m¡2ÕF 3¤2 68,
so
νÔF m¡2ÕνÔ8Õ3, 3¤2 the preceding Fibonacci-numbers are not divisible by 8, and indeed
F 3¤2m¡2 1F 7131 4Ômod 8Õ.
Now suppose that m3 and let k3¤2m¡3 . By applying Lemma 1 to the Fibonacci-type
2Ôm¡1ÕÔm¡1Õ
sequence F k , F k 1, . . . we get
F 2kF k¡1F k F k F k 1ÔF k 1¡F kÕF k F k 1F k2F k 1F k¡F kÕ
k 2 ,
F 2k 1F k¤F k F k 1¤F k 1F k 2 Fk 2 1.
By the induction hypothesis, νÔF kÕm¡1, and F k 1 is odd. Therefore we get νÔF 2
1νÔ2F k F k 1Õ, which implies νÔF 2kÕm, establishing statement (a).

73
Moreover, since F k 11 2 m¡2 a2 m¡1 for some integer a, we get
F 2k 1F k 2 Fk 2 10 Ô1 2 m¡2 a2 m¡1Õ21 2 m¡1Ômod 2 mÕ,
as desired in statement (c).
We are left to prove that 2 mЧF l for l2k. Assume the contrary. Since 2 m¡1§F l
Ð
, from
the induction hypothesis it follows that lk. But then we have F lF k¡1F l¡k F k F l¡k 1,
where the second summand is divisible by 2 m¡1 but the first one is not (since F k¡1 is odd and
l¡kk). Hence the sum is not divisible even by 2 m¡1 . A contradiction.
Now, for every pair of integersÔa, bÕÔ0, 0Õ, let µÔa, bÕminØνÔaÕ, νÔbÕÙ. By an obvious induction,
for every Fibonacci-type sequence AÔa 0 , a 1 , . . .Õwe have µÔa 0 , a 1ÕµÔa 1 , a 2Õ. . .;
denote this common value by µÔAÕ. Also denote by p nÔAÕthe period of this sequence modulo
2 n , that is, the least p0 such that a k pa kÔmod 2 nÕfor all k0.
Lemma 3. Let AÔa 0 , a 1 , . . .Õbe a Fibonacci-type sequence such that µÔAÕkn. Then
p nÔAÕ3¤2n¡1¡k .
Proof. First, we note that the sequenceÔa 0 , a 1 , . . .Õhas period p modulo 2 n if and only if the
sequenceÔa 0ß2k , a 1ß2k , . . .Õhas period p modulo 2 n¡k . Hence, passing to this sequence we can
.Õ
assume that k0.
We prove the statement by induction on n. It is easy to see that for n1, 2 the claim
is true; actually, each Fibonacci-type sequence A with µÔAÕ0behaves as 0, 1, 1, 0, 1, 1, . . .
modulo 2, and as 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, ... modulo 4 (all pairs of residues from which at
least one is odd appear as a pair of consecutive terms in this sequence).
Now suppose that n3and consider an arbitrary Fibonacci-type sequence AÔa 0 , a 1 , . .
with µÔAÕ0. Obviously we should have p n¡1ÔAÕ§p nÔAÕ, or, using the induction hypothesis,
0
s3¤2n¡2§p nÔAÕ. Next, we may suppose that a 0 is even; hence a 1 is odd, and a 02b 0 ,
a 12b 1 1 for some integers b 0 , b 1 .
Consider the Fibonacci-type sequence BÔb 0 , b 1 , . . .Õstarting withÔb 0 , b 1Õ. Since a
2b 0 F 0 , a 12b 1 F 1 , by an easy induction we get a k2b k F k for all k0. By
the induction hypothesis, we have p n¡1ÔBÕ§s, hence the sequenceÔ2b 0 , 2b 1 , . . .Õis s-periodic
modulo 2 n . On the other hand, by Lemma 2 we have F s 11 2 n¡1Ômod 2 nÕ, F 2s0
Ômod 2 nÕ, F 2s 11Ômod 2 nÕ, hence
a s 12b s 1 F s 12b 1 1 2 n¡12b 1 1a 1Ômod 2 nÕ,
a 2s2b 2s F 2s2b 0 0a 0Ômod 2 nÕ,
a 2s 12b 2s 1 F 2s 12b 1 1a 1Ômod
Ð
2 nÕ.
The first line means that A is not s-periodic, while the other two provide that a 2sa 0 ,
a 2s 1a 1 and hence a 2s ta t for all t0. Hence s§p nÔAÕ§2s and p nÔAÕs, which means
that p nÔAÕ2s, as desired.
0Õ
Finally, Lemma 3 provides a straightforward method of counting the number of cycles.
Actually, take any number 0kn¡1 and consider all the cellsÔi, jÕwith µÔi, jÕk. The
total number of such cells is 22Ôn¡kÕ¡22Ôn¡k¡1Õ3¤22n¡2k¡2 . On the other hand, they are split
into cycles, and by Lemma 3 the length of each cycle is
¤¤¤
3¤2n¡1¡k . Hence the number of cycles
2n¡2k¡2
3¤2 consisting of these cells is exactly
3¤2n¡1¡k2 n¡k¡1 . Finally, there is only one cellÔ0,
which is not mentioned in the previous computation, and it forms a separate cycle. So the total
number of cycles is
n¡1
1 2
ôk0
n¡1¡k1 Ô1 2 4 2 n¡1Õ2 n .

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Comment. We outline a different proof for the essential part of Lemma 3. That is, we assume that
k0and show that in this case the period ofÔa iÕmodulo 2 n coincides with the period of the Fibonacci
numbers modulo 2 n ; then the proof can be finished by the arguments from Lemma 2..
Note that p is a (not necessarily minimal) period of the
0Õ
sequenceÔa iÕmodulo 2 n if and only if we
have a 0a pÔmod 2 nÕ, a 1a p 1Ômod 2 nÕ, that is,
a 0a pF p¡1a 0 F p a 1F pÔa 1¡a F p 1a 0Ômod 2 nÕ,
a 1a p 1F p a 0 F p 1a 1Ômod 2 nÕ.
(1)
Now, If p is a period ofÔF iÕthen we have F pF 00Ômod 2 nÕand F p 1F 11Ômod 2 nÕ, which
by (1) implies that p is a period ofÔa iÕas well.
0Õ
Conversely, suppose that p is a period ofÔa iÕ. Combining the relations of (1) we
0Õ 1Õ
get
0¨
0a 1¤a 0¡a 0¤a 1a 1 F pÔa 1¡a F p 1a 0¨¡a 0ÔF p a 0 F p 1a
F pÔa 2 1¡a 1 a 0¡a 2 0ÕÔmod 2 nÕ,
a 2 1¡a 1 a 0¡a 2 0Ôa 1¡a 0Õa 1¡a 0¤a 0Ôa 1¡a 0ÕÔF p a 0 F p 1a 1Õ¡a 0 F pÔa 1¡a F p 1a
F p 1Ôa 2 1¡a 1 a 0¡a 2 0ÕÔmod 2 nÕ.
Since at least one of the numbers a 0 , a 1 is odd, the number a 2 1¡a 1 a 0¡a2 0 is odd as well. Therefore the
previous relations are equivalent with F p0Ômod 2 nÕand F p 11Ômod 2 nÕ, which means exactly
that p is a period ofÔF 0 ,F 1 ,...Õmodulo 2 n .
So, the sets of periods ofÔa iÕandÔF iÕcoincide, and hence the minimal periods coincide as well.