Imo2012 sl

Shortlisted Problems with Solutions
53rd
International Mathematical Olympiad
Mar del Plata, Argentina 2012

Note of Confidentiality
The shortlisted problems should be kept
strictly confidential until IMO 2013
Contributing Countries
The Organizing Committee and the Problem Selection Committee of IMO 2012 thank the
following 40 countries for contributing 136 problem proposals:
Australia, Austria, Belarus, Belgium, Bulgaria, Canada, Cyprus,
Czech Republic, Denmark, Estonia, Finland, France, Germany,
Greece, Hong Kong, India, Iran, Ireland, Israel, Japan,
Kazakhstan, Luxembourg, Malaysia, Montenegro, Netherlands,
Norway, Pakistan, Romania, Russia, Serbia, Slovakia, Slovenia,
South Africa, South Korea, Sweden, Thailand, Ukraine,
United Kingdom, United States of America, Uzbekistan
Problem Selection Committee
Mart´ın Avendaño
Carlos di Fiore
Géza Kós
Svetoslav Savchev

4
Algebra
A1. Find all the functions f : Z → Z such that
f(a)2
+ f(b)2
+ f(c)2
= 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a)
for all integers a, b, c satisfying a + b + c = 0.
A2. Let Z and Q be the sets of integers and rationals respectively.
a) Does there exist a partition of Z into three non-empty subsets A, B, C such that the sets
A + B, B + C, C + A are disjoint?
b) Does there exist a partition of Q into three non-empty subsets A, B, C such that the sets
Here X + Y denotes the set {x + y | x ∈ X, y ∈ Y }, for X, Y ⊆ Z and X, Y ⊆ Q.
A3. Let a2, . . . , an be n − 1 positive real numbers, where n ≥ 3, such that a2a3 · · ·an = 1.
Prove that
(1 + a2)2
(1 + a3)3
· · · (1 + an)n
> nn
.
A4. Let f and g be two nonzero polynomials with integer coeﬃcients and deg f > deg g.
Suppose that for inﬁnitely many primes p the polynomial pf + g has a rational root. Prove
that f has a rational root.
A5. Find all functions f : R → R that satisfy the conditions
f(1 + xy) − f(x + y) = f(x)f(y) for all x, y ∈ R
and f(−1) = 0.
A6. Let f : N → N be a function, and let fm
be f applied m times. Suppose that for
every n ∈ N there exists a k ∈ N such that f2k
(n) = n + k, and let kn be the smallest such k.
Prove that the sequence k1, k2, . . . is unbounded.
A7. We say that a function f : Rk
→ R is a metapolynomial if, for some positive integers m
and n, it can be represented in the form
f(x1, . . . , xk) = max
i=1,...,m
min
j=1,...,n
Pi,j(x1, . . . , xk)
where Pi,j are multivariate polynomials. Prove that the product of two metapolynomials is also
a metapolynomial.

5
Combinatorics
C1. Several positive integers are written in a row. Iteratively, Alice chooses two adjacent
numbers x and y such that x > y and x is to the left of y, and replaces the pair (x, y) by either
(y + 1, x) or (x − 1, x). Prove that she can perform only finitely many such iterations.
C2. Let n ≥ 1 be an integer. What is the maximum number of disjoint pairs of elements of the
set {1, 2, . . ., n} such that the sums of the different pairs are different integers not exceeding n?
C3. In a 999 × 999 square table some cells are white and the remaining ones are red. Let T
be the number of triples (C1, C2, C3) of cells, the first two in the same row and the last two in
the same column, with C1 and C3 white and C2 red. Find the maximum value T can attain.
C4. Players A and B play a game with N ≥ 2012 coins and 2012 boxes arranged around a
circle. Initially A distributes the coins among the boxes so that there is at least 1 coin in each
box. Then the two of them make moves in the order B, A, B, A, . . . by the following rules:
• On every move of his B passes 1 coin from every box to an adjacent box.
• On every move of hers A chooses several coins that were not involved in B’s previous
move and are in different boxes. She passes every chosen coin to an adjacent box.
Player A’s goal is to ensure at least 1 coin in each box after every move of hers, regardless of
how B plays and how many moves are made. Find the least N that enables her to succeed.
C5. The columns and the rows of a 3n × 3n square board are numbered 1, 2, . . . , 3n. Every
square (x, y) with 1 ≤ x, y ≤ 3n is colored asparagus, byzantium or citrine according as the
modulo 3 remainder of x + y is 0, 1 or 2 respectively. One token colored asparagus, byzantium
or citrine is placed on each square, so that there are 3n2
tokens of each color.
Suppose that one can permute the tokens so that each token is moved to a distance of
at most d from its original position, each asparagus token replaces a byzantium token, each
byzantium token replaces a citrine token, and each citrine token replaces an asparagus token.
Prove that it is possible to permute the tokens so that each token is moved to a distance of at
most d + 2 from its original position, and each square contains a token with the same color as
the square.
C6. Let k and n be fixed positive integers. In the liar’s guessing game, Amy chooses integers
x and N with 1 ≤ x ≤ N. She tells Ben what N is, but not what x is. Ben may then repeatedly
ask Amy whether x ∈ S for arbitrary sets S of integers. Amy will always answer with yes or no,
but she might lie. The only restriction is that she can lie at most k times in a row. After he
has asked as many questions as he wants, Ben must specify a set of at most n positive integers.
If x is in this set he wins; otherwise, he loses. Prove that:
a) If n ≥ 2k
then Ben can always win.
b) For sufficiently large k there exist n ≥ 1.99k
such that Ben cannot guarantee a win.
C7. There are given 2500
points on a circle labeled 1, 2, . . . , 2500
in some order. Prove that
one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums
of the pairs of numbers at the endpoints of the chosen chords are equal.

6
Geometry
G1. In the triangle ABC the point J is the center of the excircle opposite to A. This excircle
is tangent to the side BC at M, and to the lines AB and AC at K and L respectively. The
lines LM and BJ meet at F, and the lines KM and CJ meet at G. Let S be the point of
intersection of the lines AF and BC, and let T be the point of intersection of the lines AG
and BC. Prove that M is the midpoint of ST.
G2. Let ABCD be a cyclic quadrilateral whose diagonals AC and BD meet at E. The
extensions of the sides AD and BC beyond A and B meet at F. Let G be the point such that
ECGD is a parallelogram, and let H be the image of E under reflection in AD. Prove that
D, H, F, G are concyclic.
G3. In an acute triangle ABC the points D, E and F are the feet of the altitudes through A,
B and C respectively. The incenters of the triangles AEF and BDF are I1 and I2 respectively;
the circumcenters of the triangles ACI1 and BCI2 are O1 and O2 respectively. Prove that I1I2
and O1O2 are parallel.
G4. Let ABC be a triangle with AB = AC and circumcenter O. The bisector of ∠BAC
intersects BC at D. Let E be the reflection of D with respect to the midpoint of BC. The lines
through D and E perpendicular to BC intersect the lines AO and AD at X and Y respectively.
Prove that the quadrilateral BXCY is cyclic.
G5. Let ABC be a triangle with ∠BCA = 90◦
, and let C0 be the foot of the altitude
from C. Choose a point X in the interior of the segment CC0, and let K, L be the points on
the segments AX, BX for which BK = BC and AL = AC respectively. Denote by M the
intersection of AL and BK. Show that MK = ML.
G6. Let ABC be a triangle with circumcenter O and incenter I. The points D, E and F on
the sides BC, CA and AB respectively are such that BD + BF = CA and CD + CE = AB.
The circumcircles of the triangles BFD and CDE intersect at P = D. Prove that OP = OI.
G7. Let ABCD be a convex quadrilateral with non-parallel sides BC and AD. Assume
that there is a point E on the side BC such that the quadrilaterals ABED and AECD are
circumscribed. Prove that there is a point F on the side AD such that the quadrilaterals
ABCF and BCDF are circumscribed if and only if AB is parallel to CD.
G8. Let ABC be a triangle with circumcircle ω and ℓ a line without common points with ω.
Denote by P the foot of the perpendicular from the center of ω to ℓ. The side-lines BC, CA, AB
intersect ℓ at the points X, Y, Z different from P. Prove that the circumcircles of the triangles
AXP, BY P and CZP have a common point different from P or are mutually tangent at P.

7
Number Theory
N1. Call admissible a set A of integers that has the following property:
If x, y ∈ A (possibly x = y) then x2
+ kxy + y2
∈ A for every integer k.
Determine all pairs m, n of nonzero integers such that the only admissible set containing both m
and n is the set of all integers.
N2. Find all triples (x, y, z) of positive integers such that x ≤ y ≤ z and
x3
(y3
+ z3
) = 2012(xyz + 2).
N3. Determine all integers m ≥ 2 such that every n with m
3
≤ n ≤ m
2
divides the binomial
coefficient n
m−2n
.
N4. An integer a is called friendly if the equation (m2
+ n)(n2
+ m) = a(m − n)3
has a
solution over the positive integers.
a) Prove that there are at least 500 friendly integers in the set {1, 2, . . ., 2012}.
b) Decide whether a = 2 is friendly.
N5. For a nonnegative integer n define rad(n) = 1 if n = 0 or n = 1, and rad(n) = p1p2 · · · pk
where p1 < p2 < · · · < pk are all prime factors of n. Find all polynomials f(x) with nonnegative
integer coefficients such that rad(f(n)) divides rad(f(nrad(n)
)) for every nonnegative integer n.
N6. Let x and y be positive integers. If x2n
− 1 is divisible by 2n
y + 1 for every positive
integer n, prove that x = 1.
N7. Find all n ∈ N for which there exist nonnegative integers a1, a2, . . . , an such that
1
2a1
+
1
2a2
+ · · · +
1
2an
=
1
3a1
+
2
3a2
+ · · · +
n
3an
= 1.
N8. Prove that for every prime p > 100 and every integer r there exist two integers a and b
such that p divides a2
+ b5
− r.

8
Algebra
A1. Find all the functions f : Z → Z such that
f(a)2
+ f(b)2
+ f(c)2
= 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a)
for all integers a, b, c satisfying a + b + c = 0.
Solution. The substitution a = b = c = 0 gives 3f(0)2
= 6f(0)2
, hence
f(0) = 0. (1)
The substitution b = −a and c = 0 gives ((f(a) − f(−a))2
= 0. Hence f is an even function:
f(a) = f(−a) for all a ∈ Z. (2)
Now set b = a and c = −2a to obtain 2f(a)2
+ f(2a)2
= 2f(a)2
+ 4f(a)f(2a). Hence
f(2a) = 0 or f(2a) = 4f(a) for all a ∈ Z. (3)
If f(r) = 0 for some r ≥ 1 then the substitution b = r and c = −a−r gives (f(a+r)−f(a))2
= 0.
So f is periodic with period r, i. e.
f(a + r) = f(a) for all a ∈ Z.
In particular, if f(1) = 0 then f is constant, thus f(a) = 0 for all a ∈ Z. This function clearly
satisfies the functional equation. For the rest of the analysis, we assume f(1) = k = 0.
By (3) we have f(2) = 0 or f(2) = 4k. If f(2) = 0 then f is periodic of period 2, thus
f(even) = 0 and f(odd) = k. This function is a solution for every k. We postpone the
verification; for the sequel assume f(2) = 4k = 0.
By (3) again, we have f(4) = 0 or f(4) = 16k. In the first case f is periodic of period 4, and
f(3) = f(−1) = f(1) = k, so we have f(4n) = 0, f(4n+1) = f(4n+3) = k, and f(4n+2) = 4k
for all n ∈ Z. This function is a solution too, which we justify later. For the rest of the analysis,
we assume f(4) = 16k = 0.
We show now that f(3) = 9k. In order to do so, we need two substitutions:
a = 1, b = 2, c = −3 =⇒ f(3)2
− 10kf(3) + 9k2
= 0 =⇒ f(3) ∈ {k, 9k},
a = 1, b = 3, c = −4 =⇒ f(3)2
− 34kf(3) + 225k2
= 0 =⇒ f(3) ∈ {9k, 25k}.
Therefore f(3) = 9k, as claimed. Now we prove inductively that the only remaining function is
f(x) = kx2
, x ∈ Z. We proved this for x = 0, 1, 2, 3, 4. Assume that n ≥ 4 and that f(x) = kx2
holds for all integers x ∈ [0, n]. Then the substitutions a = n, b = 1, c = −n−1 and a = n−1,
b = 2, c = −n − 1 lead respectively to
f(n + 1) ∈ {k(n + 1)2
, k(n − 1)2
} and f(n + 1) ∈ {k(n + 1)2
, k(n − 3)2
}.
Since k(n − 1)2
= k(n − 3)2
for n = 2, the only possibility is f(n + 1) = k(n + 1)2
. This
completes the induction, so f(x) = kx2
for all x ≥ 0. The same expression is valid for negative
values of x since f is even. To verify that f(x) = kx2
is actually a solution, we need to check
the identity a4
+ b4
+ (a + b)4
= 2a2
b2
+ 2a2
(a + b)2
+ 2b2
(a + b)2
, which follows directly by
expanding both sides.

9
Therefore the only possible solutions of the functional equation are the constant function
f1(x) = 0 and the following functions:
f2(x) = kx2
f3(x) =
0 x even
k x odd
f4(x) =



0 x ≡ 0 (mod 4)
k x ≡ 1 (mod 2)
4k x ≡ 2 (mod 4)
for any non-zero integer k. The verification that they are indeed solutions was done for the
first two. For f3 note that if a + b + c = 0 then either a, b, c are all even, in which case
f(a) = f(b) = f(c) = 0, or one of them is even and the other two are odd, so both sides of
the equation equal 2k2
. For f4 we use similar parity considerations and the symmetry of the
equation, which reduces the verification to the triples (0, k, k), (4k, k, k), (0, 0, 0), (0, 4k, 4k).
They all satisfy the equation.
Comment. We used several times the same fact: For any a, b ∈ Z the functional equation is a
quadratic equation in f(a + b) whose coefficients depend on f(a) and f(b):
f(a + b)2
− 2(f(a) + f(b))f(a + b) + (f(a) − f(b))2
= 0.
Its discriminant is 16f(a)f(b). Since this value has to be non-negative for any a, b ∈ Z, we conclude
that either f or −f is always non-negative. Also, if f is a solution of the functional equation, then
−f is also a solution. Therefore we can assume f(x) ≥ 0 for all x ∈ Z. Now, the two solutions of the
quadratic equation are
f(a + b) ∈ f(a) + f(b)
2
, f(a) − f(b)
2
for all a, b ∈ Z.
The computation of f(3) from f(1), f(2) and f(4) that we did above follows immediately by setting
(a, b) = (1, 2) and (a, b) = (1, −4). The inductive step, where f(n + 1) is derived from f(n), f(n − 1),
f(2) and f(1), follows immediately using (a, b) = (n, 1) and (a, b) = (n − 1, 2).

10
A2. Let Z and Q be the sets of integers and rationals respectively.
a) Does there exist a partition of Z into three non-empty subsets A, B, C such that the sets
b) Does there exist a partition of Q into three non-empty subsets A, B, C such that the sets
Here X + Y denotes the set {x + y | x ∈ X, y ∈ Y }, for X, Y ⊆ Z and X, Y ⊆ Q.
Solution 1. a) The residue classes modulo 3 yield such a partition:
A = {3k | k ∈ Z}, B = {3k + 1 | k ∈ Z}, C = {3k + 2 | k ∈ Z}.
b) The answer is no. Suppose that Q can be partitioned into non-empty subsets A, B, C as
stated. Note that for all a ∈ A, b ∈ B, c ∈ C one has
a + b − c ∈ C, b + c − a ∈ A, c + a − b ∈ B. (1)
Indeed a+b−c /∈ A as (A+B)∩(A+C) = ∅, and similarly a+b−c /∈ B, hence a+b−c ∈ C. The
other two relations follow by symmetry. Hence A+B ⊂ C +C, B +C ⊂ A+A, C +A ⊂ B +B.
The opposite inclusions also hold. Let a, a′
∈ A and b ∈ B, c ∈ C be arbitrary. By (1)
a′
+ c − b ∈ B, and since a ∈ A, c ∈ C, we use (1) again to obtain
a + a′
− b = a + (a′
+ c − b) − c ∈ C.
So A + A ⊂ B + C and likewise B + B ⊂ C + A, C + C ⊂ A + B. In summary
B + C = A + A, C + A = B + B, A + B = C + C.
Furthermore suppose that 0 ∈ A without loss of generality. Then B = {0} + B ⊂ A + B
and C = {0} + C ⊂ A+ C. So, since B + C is disjoint with A+ B and A+ C, it is also disjoint
with B and C. Hence B + C is contained in Z (B ∪ C) = A. Because B + C = A + A, we
obtain A + A ⊂ A. On the other hand A = {0} + A ⊂ A + A, implying A = A + A = B + C.
Therefore A + B + C = A + A + A = A, and now B + B = C + A and C + C = A + B yield
B + B + B = A+ B + C = A, C + C + C = A+ B + C = A. In particular if r ∈ Q = A∪B ∪C
is arbitrary then 3r ∈ A.
However such a conclusion is impossible. Take any b ∈ B (B = ∅) and let r = b/3 ∈ Q.
Then b = 3r ∈ A which is a contradiction.
Solution 2. We prove that the example for Z from the ﬁrst solution is unique, and then use
this fact to solve part b).
Let Z = A ∪ B ∪ C be a partition of Z with A, B, C = ∅ and A + B, B + C, C + A disjoint.
We need the relations (1) which clearly hold for Z. Fix two consecutive integers from diﬀerent
sets, say b ∈ B and c = b+1 ∈ C. For every a ∈ A we have, in view of (1), a−1 = a+b−c ∈ C
and a + 1 = a + c − b ∈ B. So every a ∈ A is preceded by a number from C and followed by a
number from B.
In particular there are pairs of the form c, c + 1 with c ∈ C, c + 1 ∈ A. For such a pair and
any b ∈ B analogous reasoning shows that each b ∈ B is preceded by a number from A and
followed by a number from C. There are also pairs b, b−1 with b ∈ B, b−1 ∈ A. We use them
in a similar way to prove that each c ∈ C is preceded by a number from B and followed by a
number from A.
By putting the observations together we infer that A, B, C are the three congruence classes
modulo 3. Observe that all multiples of 3 are in the set of the partition that contains 0.

11
Now we turn to part b). Suppose that there is a partition of Q with the given properties.
Choose three rationals ri = pi/qi from the three sets A, B, C, i = 1, 2, 3, and set N = 3q1q2q3.
Let S ⊂ Q be the set of fractions with denominators N (irreducible or not). It is obtained
through multiplication of every integer by the constant 1/N, hence closed under sums and
differences. Moreover, if we identify each k ∈ Z with k/N ∈ S then S is essentially the set Z
with respect to addition. The numbers ri belong to S because
r1 =
3p1q2q3
N
, r2 =
3p2q3q1
N
, r3 =
3p3q1q2
N
.
The partition Q = A ∪ B ∪ C of Q induces a partition S = A′
∪ B′
∪ C′
of S, with A′
= A ∩ S,
B′
= B ∩ S, C′
= C ∩ S. Clearly A′
+ B′
, B′
+ C′
, C′
+ A′
are disjoint, so this partition has
the properties we consider.
By the uniqueness of the example for Z the sets A′
, B′
, C′
are the congruence classes mod-
ulo 3, multiplied by 1/N. Also all multiples of 3/N are in the same set, A′
, B′
or C′
. This holds
for r1, r2, r3 in particular as they are all multiples of 3/N. However r1, r2, r3 are in different sets
A′
, B′
, C′
since they were chosen from different sets A, B, C. The contradiction ends the proof.
Comment. The uniqueness of the example for Z can also be deduced from the argument in the first
solution.

12
A3. Let a2, . . . , an be n − 1 positive real numbers, where n ≥ 3, such that a2a3 · · ·an = 1.
Prove that
(1 + a2)2
(1 + a3)3
· · · (1 + an)n
> nn
.
Solution. The substitution a2 =
x2
x1
, a3 =
x3
x2
, . . . , an =
x1
xn−1
transforms the original problem
into the inequality
(x1 + x2)2
(x2 + x3)3
· · · (xn−1 + x1)n
> nn
x2
1x3
2 · · · xn
n−1 (∗)
for all x1, . . . , xn−1 > 0. To prove this, we use the AM-GM inequality for each factor of the
left-hand side as follows:
(x1 + x2)2
≥ 22
x1x2
(x2 + x3)3
= 2 x2
2
+ x3
3
≥ 33 x2
2
2
x3
(x3 + x4)4
= 3 x3
3
+ x4
4
≥ 44 x3
3
3
x4
...
...
...
(xn−1 + x1)n
= (n − 1) xn−1
n−1
+ x1
n
≥ nn xn−1
n−1
n−1
x1.
Multiplying these inequalities together gives (*), with inequality sign ≥ instead of >. However
for the equality to occur it is necessary that x1 = x2, x2 = 2x3, . . . , xn−1 = (n − 1)x1, implying
x1 = (n − 1)!x1. This is impossible since x1 > 0 and n ≥ 3. Therefore the inequality is strict.
Comment. One can avoid the substitution ai = xi/xi−1. Apply the weighted AM-GM inequality to
each factor (1 + ak)k, with the same weights like above, to obtain
(1 + ak)k
= (k − 1)
1
k − 1
+ ak
k
≥
kk
(k − 1)k−1
ak.
Multiplying all these inequalities together gives
(1 + a2)2
(1 + a3)3
· · · (1 + an)n
≥ nn
a2a3 · · · an = nn
.
The same argument as in the proof above shows that the equality cannot be attained.

13
A4. Let f and g be two nonzero polynomials with integer coefficients and deg f > deg g.
Suppose that for infinitely many primes p the polynomial pf + g has a rational root. Prove
that f has a rational root.
Solution 1. Since deg f > deg g, we have |g(x)/f(x)| < 1 for sufficiently large x; more
precisely, there is a real number R such that |g(x)/f(x)| < 1 for all x with |x| > R. Then for
all such x and all primes p we have
pf(x) + g(x) ≥ f(x) p −
|g(x)|
|f(x)|
> 0.
Hence all real roots of the polynomials pf + g lie in the interval [−R, R].
Let f(x) = anxn
+ an−1xn−1
+ · · · + a0 and g(x) = bmxm
+ bm−1xm−1
+ · · · + b0 where
n > m, an = 0 and bm = 0. Upon replacing f(x) and g(x) by an−1
n f(x/an) and an−1
n g(x/an)
respectively, we reduce the problem to the case an = 1. In other words one can assume that f
is monic. Then the leading coefficient of pf + g is p, and if r = u/v is a rational root of pf + g
with (u, v) = 1 and v > 0, then either v = 1 or v = p.
First consider the case when v = 1 infinitely many times. If v = 1 then |u| ≤ R, so there
are only finitely many possibilities for the integer u. Therefore there exist distinct primes p
and q for which we have the same value of u. Then the polynomials pf + g and qf + g share
this root, implying f(u) = g(u) = 0. So in this case f and g have an integer root in common.
Now suppose that v = p infinitely many times. By comparing the exponent of p in the
denominators of pf(u/p) and g(u/p) we get m = n − 1 and pf(u/p) + g(u/p) = 0 reduces to
an equation of the form
un
+ an−1pun−1
+ . . . + a0pn
+ bn−1un−1
+ bn−2pun−2
+ . . . + b0pn−1
= 0.
The equation above implies that un
+ bn−1un−1
is divisible by p and hence, since (u, p) = 1,
we have u + bn−1 = pk with some integer k. On the other hand all roots of pf + g lie in the
interval [−R, R], so that
|pk − bn−1|
p
=
|u|
p
< R,
|k| < R +
|bn−1|
p
< R + |bn−1|.
Therefore the integer k can attain only finitely many values. Hence there exists an integer k
such that the number pk−bn−1
p
= k − bn−1
p
is a root of pf + g for infinitely many primes p. For
these primes we have
f k − bn−1
1
p
+
1
p
g k − bn−1
1
p
= 0.
So the equation
f (k − bn−1x) + xg (k − bn−1x) = 0 (1)
has infinitely many solutions of the form x = 1/p. Since the left-hand side is a polynomial, this
implies that (1) is a polynomial identity, so it holds for all real x. In particular, by substituting
x = 0 in (1) we get f(k) = 0. Thus the integer k is a root of f.
In summary the monic polynomial f obtained after the initial reduction always has an
integer root. Therefore the original polynomial f has a rational root.

14
Solution 2. Analogously to the first solution, there exists a real number R such that the
complex roots of all polynomials of the form pf + g lie in the disk |z| ≤ R.
For each prime p such that pf + g has a rational root, by Gauss’ lemma pf + g is the
product of two integer polynomials, one with degree 1 and the other with degree deg f − 1.
Since p is a prime, the leading coefficient of one of these factors divides the leading coefficient
of f. Denote that factor by hp.
By narrowing the set of the primes used we can assume that all polynomials hp have the
same degree and the same leading coefficient. Their complex roots lie in the disk |z| ≤ R, hence
Vieta’s formulae imply that all coefficients of all polynomials hp form a bounded set. Since
these coefficients are integers, there are only finitely many possible polynomials hp. Hence there
is a polynomial h such that hp = h for infinitely many primes p.
Finally, if p and q are distinct primes with hp = hq = h then h divides (p − q)f. Since
deg h = 1 or deg h = deg f − 1, in both cases f has a rational root.
Comment. Clearly the polynomial h is a common factor of f and g. If deg h = 1 then f and g share a
rational root. Otherwise deg h = deg f − 1 forces deg g = deg f − 1 and g divides f over the rationals.
Solution 3. Like in the first solution, there is a real number R such that the real roots of all
polynomials of the form pf + g lie in the interval [−R, R].
Let p1 < p2 < · · · be an infinite sequence of primes so that for every index k the polynomial
pkf + g has a rational root rk. The sequence r1, r2, . . . is bounded, so it has a convergent
subsequence rk1 , rk2 , . . .. Now replace the sequences (p1, p2, . . . ) and (r1, r2, . . . ) by (pk1 , pk2 , . . .)
and (rk1 , rk2 , . . .); after this we can assume that the sequence r1, r2, . . . is convergent. Let
α = lim
k→∞
rk. We show that α is a rational root of f.
Over the interval [−R, R], the polynomial g is bounded, |g(x)| ≤ M with some fixed M.
Therefore
|f(rk)| = f(rk) −
pkf(rk) + g(rk)
pk
=
|g(rk)|
pk
≤
M
pk
→ 0,
and
f(α) = f lim
k→∞
rk = lim
k→∞
f(rk) = 0.
So α is a root of f indeed.
Now let uk, vk be relative prime integers for which rk = uk
vk
. Let a be the leading coefficient
of f, let b = f(0) and c = g(0) be the constant terms of f and g, respectively. The leading
coefficient of the polynomial pkf + g is pka, its constant term is pkb + c. So vk divides pka and
uk divides pkb + c. Let pkb + c = ukek (if pkb + c = uk = 0 then let ek = 1).
We prove that α is rational by using the following fact. Let (pn) and (qn) be sequences of
integers such that the sequence (pn/qn) converges. If (pn) or (qn) is bounded then lim(pn/qn) is
rational.
Case 1: There is an infinite subsequence (kn) of indices such that vkn divides a. Then (vkn )
is bounded, so α = limn→∞(ukn /vkn ) is rational.
Case 2: There is an infinite subsequence (kn) of indices such that vkn does not divide a.
For such indices we have vkn = pkn dkn where dkn is a divisor of a. Then
α = lim
n→∞
ukn
vkn
= lim
n→∞
pkn b + c
pkn dkn ekn
= lim
n→∞
b
dkn ekn
+ lim
n→∞
c
pkn dkn ekn
= lim
n→∞
b
dkn ekn
.
Because the numerator b in the last limit is bounded, α is rational.

15
A5. Find all functions f : R → R that satisfy the conditions
f(1 + xy) − f(x + y) = f(x)f(y) for all x, y ∈ R
and f(−1) = 0.
Solution. The only solution is the function f(x) = x − 1, x ∈ R.
We set g(x) = f(x) + 1 and show that g(x) = x for all real x. The conditions take the form
g(1 + xy) − g(x + y) = g(x) − 1 g(y) − 1 for all x, y ∈ R and g(−1) = 1. (1)
Denote C = g(−1) − 1 = 0. Setting y = −1 in (1) gives
g(1 − x) − g(x − 1) = C(g(x) − 1). (2)
Set x = 1 in (2) to obtain C(g(1) − 1) = 0. Hence g(1) = 1 as C = 0. Now plugging in x = 0
and x = 2 yields g(0) = 0 and g(2) = 2 respectively.
We pass on to the key observations
g(x) + g(2 − x) = 2 for all x ∈ R, (3)
g(x + 2) − g(x) = 2 for all x ∈ R. (4)
Replace x by 1 − x in (2), then change x to −x in the resulting equation. We obtain the
relations g(x) − g(−x) = C(g(1 − x) − 1), g(−x) − g(x) = C(g(1 + x) − 1). Then adding them
up leads to C(g(1 − x) + g(1 + x) − 2) = 0. Thus C = 0 implies (3).
Let u, v be such that u + v = 1. Apply (1) to the pairs (u, v) and (2 − u, 2 − v):
g(1 + uv) − g(1) = g(u) − 1 g(v) − 1 , g(3 + uv) − g(3) = g(2 − u) − 1 g(2 − v) − 1 .
Observe that the last two equations have equal right-hand sides by (3). Hence u+v = 1 implies
g(uv + 3) − g(uv + 1) = g(3) − g(1).
Each x ≤ 5/4 is expressible in the form x = uv + 1 with u + v = 1 (the quadratic function
t2
−t+(x−1) has real roots for x ≤ 5/4). Hence g(x+2)−g(x) = g(3)−g(1) whenever x ≤ 5/4.
Because g(x) = x holds for x = 0, 1, 2, setting x = 0 yields g(3) = 3. This proves (4) for x ≤ 5/4.
If x > 5/4 then −x < 5/4 and so g(2 − x) − g(−x) = 2 by the above. On the other hand (3)
gives g(x) = 2−g(2−x), g(x+2) = 2−g(−x), so that g(x+2)−g(x) = g(2−x)−g(−x) = 2.
Thus (4) is true for all x ∈ R.
Now replace x by −x in (3) to obtain g(−x) + g(2 + x) = 2. In view of (4) this leads to
g(x) + g(−x) = 0, i. e. g(−x) = −g(x) for all x. Taking this into account, we apply (1) to the
pairs (−x, y) and (x, −y):
g(1 − xy) − g(−x + y) = g(x) + 1 1 − g(y) , g(1 − xy) − g(x − y) = 1 − g(x) g(y) + 1 .
Adding up yields g(1 − xy) = 1 − g(x)g(y). Then g(1 + xy) = 1 + g(x)g(y) by (3). Now the
original equation (1) takes the form g(x + y) = g(x) + g(y). Hence g is additive.
By additvity g(1 + xy) = g(1) + g(xy) = 1 + g(xy); since g(1 + xy) = 1 + g(x)g(y) was
shown above, we also have g(xy) = g(x)g(y) (g is multiplicative). In particular y = x gives
g(x2
) = g(x)2
≥ 0 for all x, meaning that g(x) ≥ 0 for x ≥ 0. Since g is additive and bounded
from below on [0, +∞), it is linear; more exactly g(x) = g(1)x = x for all x ∈ R.
In summary f(x) = x − 1, x ∈ R. It is straightforward that this function satisﬁes the
requirements.
Comment. There are functions that satisfy the given equation but vanish at −1, for instance the
constant function 0 and f(x) = x2 − 1, x ∈ R.

16
A6. Let f : N → N be a function, and let fm
be f applied m times. Suppose that for
every n ∈ N there exists a k ∈ N such that f2k
(n) = n + k, and let kn be the smallest such k.
Prove that the sequence k1, k2, . . . is unbounded.
Solution. We restrict attention to the set
S = {1, f(1), f2
(1), . . .}.
Observe that S is unbounded because for every number n in S there exists a k > 0 such
that f2k
(n) = n + k is in S. Clearly f maps S into itself; moreover f is injective on S. Indeed
if fi
(1) = fj
(1) with i = j then the values fm
(1) start repeating periodically from some point
on, and S would be finite.
Define g : S → S by g(n) = f2kn
(n) = n + kn. We prove that g is injective too. Suppose
that g(a) = g(b) with a < b. Then a + ka = f2ka
(a) = f2kb (b) = b + kb implies ka > kb. So,
since f is injective on S, we obtain
f2(ka−kb)
(a) = b = a + (ka − kb).
However this contradicts the minimality of ka as 0 < ka − kb < ka.
Let T be the set of elements of S that are not of the form g(n) with n ∈ S. Note that 1 ∈ T
by g(n) > n for n ∈ S, so T is non-empty. For each t ∈ T denote Ct = {t, g(t), g2
(t), . . .};
call Ct the chain starting at t. Observe that distinct chains are disjoint because g is injective.
Each n ∈ ST has the form n = g(n′
) with n′
< n, n′
∈ S. Repeated applications of the same
observation show that n ∈ Ct for some t ∈ T, i. e. S is the disjoint union of the chains Ct.
If fn
(1) is in the chain Ct starting at t = fnt
(1) then n = nt + 2a1 + · · · + 2aj with
fn
(1) = gj
(fnt
(1)) = f2aj
(f2aj−1
(· · · f2a1
(fnt
(1)))) = fnt
(1) + a1 + · · · + aj.
Hence
fn
(1) = fnt
(1) +
n − nt
2
= t +
n − nt
2
. (1)
Now we show that T is infinite. We argue by contradiction. Suppose that there are only
finitely many chains Ct1 , . . . , Ctr , starting at t1 < · · · < tr. Fix N. If fn
(1) with 1 ≤ n ≤ N
is in Ct then fn
(1) = t + n−nt
2
≤ tr + N
2
by (1). But then the N + 1 distinct natural numbers
1, f(1), . . . , fN
(1) are all less than tr + N
2
and hence N + 1 ≤ tr + N
2
. This is a contradiction if
N is sufficiently large, and hence T is infinite.
To complete the argument, choose any k in N and consider the k + 1 chains starting at the
first k + 1 numbers in T. Let t be the greatest one among these numbers. Then each of the
chains in question contains a number not exceeding t, and at least one of them does not contain
any number among t+1, . . . , t+k. So there is a number n in this chain such that g(n)−n > k,
i. e. kn > k. In conclusion k1, k2, . . . is unbounded.

17
A7. We say that a function f : Rk
→ R is a metapolynomial if, for some positive integers m
and n, it can be represented in the form
f(x1, . . . , xk) = max
i=1,...,m
min
j=1,...,n
Pi,j(x1, . . . , xk)
where Pi,j are multivariate polynomials. Prove that the product of two metapolynomials is also
a metapolynomial.
Solution. We use the notation f(x) = f(x1, . . . , xk) for x = (x1, . . . , xk) and [m] = {1, 2, . . . , m}.
Observe that if a metapolynomial f(x) admits a representation like the one in the statement
for certain positive integers m and n, then they can be replaced by any m′
≥ m and n′
≥ n. For
instance, if we want to replace m by m + 1 then it is enough to define Pm+1,j(x) = Pm,j(x) and
note that repeating elements of a set do not change its maximum nor its minimum. So one can
assume that any two metapolynomials are defined with the same m and n. We reserve letters
P and Q for polynomials, so every function called P, Pi,j, Q, Qi,j, . . . is a polynomial function.
We start with a lemma that is useful to change expressions of the form min max fi,j to ones
of the form max min gi,j.
Lemma. Let {ai,j} be real numbers, for all i ∈ [m] and j ∈ [n]. Then
min
i∈[m]
max
j∈[n]
ai,j = max
j1,...,jm∈[n]
min
i∈[m]
ai,ji
,
where the max in the right-hand side is over all vectors (j1, . . . , jm) with j1, . . . , jm ∈ [n].
Proof. We can assume for all i that ai,n = max{ai,1, . . . , ai,n} and am,n = min{a1,n, . . . , am,n}.
The left-hand side is = am,n and hence we need to prove the same for the right-hand side.
If (j1, j2, . . . , jm) = (n, n, . . . , n) then min{a1,j1 , . . . , am,jm } = min{a1,n, . . . , am,n} = am,n which
implies that the right-hand side is ≥ am,n. It remains to prove the opposite inequality and
this is equivalent to min{a1,j1 , . . . , am,jm } ≤ am,n for all possible (j1, j2, . . . , jm). This is true
because min{a1,j1 , . . . , am,jm } ≤ am,jm ≤ am,n.
We need to show that the family M of metapolynomials is closed under multiplication, but
it turns out easier to prove more: that it is also closed under addition, maxima and minima.
First we prove the assertions about the maxima and the minima. If f1, . . . , fr are metapoly-
nomials, assume them defined with the same m and n. Then
f = max{f1, . . . , fr} = max{max
i∈[m]
min
j∈[n]
P1
i,j, . . . , max
i∈[m]
min
j∈[n]
Pr
i,j} = max
s∈[r],i∈[m]
min
j∈[n]
Ps
i,j.
It follows that f = max{f1, . . . , fr} is a metapolynomial. The same argument works for the
minima, but first we have to replace min max by max min, and this is done via the lemma.
Another property we need is that if f = max min Pi,j is a metapolynomial then so is −f.
Indeed, −f = min(− min Pi,j) = min max Pi,j.
To prove M is closed under addition let f = max min Pi,j and g = max min Qi,j. Then
f(x) + g(x) = max
i∈[m]
min
j∈[n]
Pi,j(x) + max
i∈[m]
min
j∈[n]
Qi,j(x)
= max
i1,i2∈[m]
(min
j∈[n]
Pi1,j(x) + min
j∈[n]
Qi2,j(x)) = max
i1,i2∈[m]
min
j1,j2∈[n]
Pi1,j1 (x) + Qi2,j2(x) ,
and hence f(x) + g(x) is a metapolynomial.
We proved that M is closed under sums, maxima and minima, in particular any function
that can be expressed by sums, max, min, polynomials or even metapolynomials is in M.
We would like to proceed with multiplication along the same lines like with addition, but
there is an essential difference. In general the product of the maxima of two sets is not equal

18
to the maximum of the product of the sets. We need to deal with the fact that a < b and c < d
do not imply ac < bd. However this is true for a, b, c, d ≥ 0.
In view of this we decompose each function f(x) into its positive part f+
(x) = max{f(x), 0}
and its negative part f−
(x) = max{0, −f(x)}. Note that f = f+
− f−
and f+
, f−
∈ M if
f ∈ M. The whole problem reduces to the claim that if f and g are metapolynomials with
f, g ≥ 0 then fg it is also a metapolynomial.
Assuming this claim, consider arbitrary f, g ∈ M. We have
fg = (f+
− f−
)(g+
− g−
) = f+
g+
− f+
g−
− f−
g+
+ f−
g−
,
and hence fg ∈ M. Indeed, M is closed under addition, also f+
g+
, f+
g−
, f−
g+
, f−
g−
∈ M
because f+
, f−
, g+
, g−
≥ 0.
It remains to prove the claim. In this case f, g ≥ 0, and one can try to repeat the argument
for the sum. More precisely, let f = max min Pij ≥ 0 and g = max min Qij ≥ 0. Then
fg = max min Pi,j · max min Qi,j = max min P+
i,j · max min Q+
i,j = max min P+
i1,j1
· Q+
i2,j2
.
Hence it suffices to check that P+
Q+
∈ M for any pair of polynomials P and Q. This reduces
to the identity
u+
v+
= max{0, min{uv, u, v}, min{uv, uv2
, u2
v}, min{uv, u, u2
v}, min{uv, uv2
, v}},
with u replaced by P(x) and v replaced by Q(x). The formula is proved by a case-by-case
analysis. If u ≤ 0 or v ≤ 0 then both sides equal 0. In case u, v ≥ 0, the right-hand side is
clearly ≤ uv. To prove the opposite inequality we use that uv equals
min{uv, u, v} if 0 ≤ u, v ≤ 1,
min{uv, uv2
, u2
v} if 1 ≤ u, v,
min{uv, u, u2
v} if 0 ≤ v ≤ 1 ≤ u,
min{uv, uv2
, v} if 0 ≤ u ≤ 1 ≤ v.
Comment. The case k = 1 is simpler and can be solved by proving that a function f : R → R is a
metapolynomial if and only if it is a piecewise polinomial (and continuos) function.
It is enough to prove that all such functions are metapolynomials, and this easily reduces to the
following case. Given a polynomial P(x) with P(0) = 0, the function f defined by f(x) = P(x) for
x ≥ 0 and 0 otherwise is a metapolynomial. For this last claim, it suffices to prove that (x+)n is a
metapolynomial, and this follows from the formula (x+)n = max{0, min{xn−1, xn}, min{xn, xn+1}}.

19
Combinatorics
C1. Several positive integers are written in a row. Iteratively, Alice chooses two adjacent
numbers x and y such that x > y and x is to the left of y, and replaces the pair (x, y) by either
(y + 1, x) or (x − 1, x). Prove that she can perform only finitely many such iterations.
Solution 1. Note first that the allowed operation does not change the maximum M of the
initial sequence. Let a1, a2, . . . , an be the numbers obtained at some point of the process.
Consider the sum
S = a1 + 2a2 + · · · + nan.
We claim that S increases by a positive integer amount with every operation. Let the operation
replace the pair (ai, ai+1) by a pair (c, ai), where ai > ai+1 and c = ai+1+1 or c = ai−1. Then the
new and the old value of S differ by d = (ic+(i+1)ai)−(iai +(i+1)ai+1) = ai −ai+1 +i(c−ai+1).
The integer d is positive since ai − ai+1 ≥ 1 and c − ai+1 ≥ 0.
On the other hand S ≤ (1 + 2 + · · · + n)M as ai ≤ M for all i = 1, . . . , n. Since S increases
by at least 1 at each step and never exceeds the constant (1 + 2 + · · · + n)M, the process stops
after a finite number of iterations.
Solution 2. Like in the first solution note that the operations do not change the maximum M
of the initial sequence. Now consider the reverse lexicographical order for n-tuples of integers.
We say that (x1, . . . , xn) < (y1, . . . , yn) if xn < yn, or if xn = yn and xn−1 < yn−1, or if xn = yn,
xn−1 = yn−1 and xn−2 < yn−2, etc. Each iteration creates a sequence that is greater than
the previous one with respect to this order, and no sequence occurs twice during the process.
On the other hand there are finitely many possible sequences because their terms are always
positive integers not exceeding M. Hence the process cannot continue forever.
Solution 3. Let the current numbers be a1, a2, . . . , an. Define the score si of ai as the number
of aj’s that are less than ai. Call the sequence s1, s2, . . . , sn the score sequence of a1, a2, . . . , an.
Let us say that a sequence x1, . . . , xn dominates a sequence y1, . . . , yn if the first index i
with xi = yi is such that xi < yi. We show that after each operation the new score sequence
dominates the old one. Score sequences do not repeat, and there are finitely many possibilities
for them, no more than (n − 1)n
. Hence the process will terminate.
Consider an operation that replaces (x, y) by (a, x), with a = y + 1 or a = x − 1. Suppose
that x was originally at position i. For each j < i the score sj does not increase with the
change because y ≤ a and x ≤ x. If sj decreases for some j y
and y ≤ a ≤ x, we see that si decreases by at least 1. This concludes the proof.
Comment. All three proofs work if x and y are not necessarily adjacent, and if the pair (x, y) is
replaced by any pair (a, x), with a an integer satisfying y ≤ a ≤ x. There is nothing special about
the “weights” 1, 2, . . . , n in the definition of S = n
i=1 iai from the first solution. For any sequence
w1 < w2 < · · · < wn of positive integers, the sum n
i=1 wiai increases by at least 1 with each operation.
Consider the same problem, but letting Alice replace the pair (x, y) by (a, x), where a is any positive
integer less than x. The same conclusion holds in this version, i. e. the process stops eventually. The
solution using the reverse lexicographical order works without any change. The first solution would
require a special set of weights like wi = Mi for i = 1, . . . , n.
Comment. The first and the second solutions provide upper bounds for the number of possible
operations, respectively of order Mn2 and Mn where M is the maximum of the original sequence.
The upper bound (n − 1)n in the third solution does not depend on M.

20
C2. Let n ≥ 1 be an integer. What is the maximum number of disjoint pairs of elements of the
set {1, 2, . . ., n} such that the sums of the different pairs are different integers not exceeding n?
Solution. Consider x such pairs in {1, 2, . . . , n}. The sum S of the 2x numbers in them is at
least 1+2+· · ·+2x since the pairs are disjoint. On the other hand S ≤ n+(n−1)+· · ·+(n−x+1)
because the sums of the pairs are different and do not exceed n. This gives the inequality
2x(2x + 1)
2
≤ nx −
x(x − 1)
2
,
which leads to x ≤ 2n−1
5
. Hence there are at most 2n−1
5
pairs with the given properties.
We show a construction with exactly 2n−1
5
pairs. First consider the case n = 5k + 3 with
k ≥ 0, where 2n−1
5
= 2k + 1. The pairs are displayed in the following table.
Pairs
3k + 1 3k · · · 2k + 2 4k + 2 4k + 1 · · · 3k + 3 3k + 2
2 4 · · · 2k 1 3 · · · 2k − 1 2k + 1
Sums 3k + 3 3k + 4 · · · 4k + 2 4k + 3 4k + 4 · · · 5k + 2 5k + 3
The 2k+1 pairs involve all numbers from 1 to 4k+2; their sums are all numbers from 3k+3
to 5k + 3. The same construction works for n = 5k + 4 and n = 5k + 5 with k ≥ 0. In these
cases the required number 2n−1
5
of pairs equals 2k + 1 again, and the numbers in the table
do not exceed 5k + 3. In the case n = 5k + 2 with k ≥ 0 one needs only 2k pairs. They can
be obtained by ignoring the last column of the table (thus removing 5k + 3). Finally, 2k pairs
are also needed for the case n = 5k + 1 with k ≥ 0. Now it suffices to ignore the last column
of the table and then subtract 1 from each number in the first row.
Comment. The construction above is not unique. For instance, the following table shows another
set of 2k + 1 pairs for the cases n = 5k + 3, n = 5k + 4, and n = 5k + 5.
Pairs
1 2 · · · k k + 1 k + 2 · · · 2k + 1
4k + 1 4k − 1 · · · 2k + 3 4k + 2 4k · · · 2k + 2
Sums 4k + 2 4k + 1 · · · 3k + 3 5k + 3 5k + 2 · · · 4k + 3
The table for the case n = 5k + 2 would be the same, with the pair (k + 1, 4k + 2) removed. For the
case n = 5k + 1 remove the last column and subtract 2 from each number in the second row.

21
C3. In a 999 × 999 square table some cells are white and the remaining ones are red. Let T
be the number of triples (C1, C2, C3) of cells, the ﬁrst two in the same row and the last two in
the same column, with C1 and C3 white and C2 red. Find the maximum value T can attain.
Solution. We prove that in an n × n square table there are at most 4n4
27
such triples.
Let row i and column j contain ai and bj white cells respectively, and let R be the set of
red cells. For every red cell (i, j) there are aibj admissible triples (C1, C2, C3) with C2 = (i, j),
therefore
T =
(i,j)∈R
aibj.
We use the inequality 2ab ≤ a2
+ b2
to obtain
T ≤
1
2
(i,j)∈R
(a2
i + b2
j ) =
1
2
n
i=1
(n − ai)a2
i +
1
2
n
j=1
(n − bj)b2
j .
This is because there are n − ai red cells in row i and n − bj red cells in column j. Now we
maximize the right-hand side.
By the AM-GM inequality we have
(n − x)x2
=
1
2
(2n − 2x) · x · x ≤
1
2
2n
3
3
=
4n3
27
,
with equality if and only if x = 2n
3
. By putting everything together, we get
T ≤
n
2
4n3
27
+
n
2
4n3
27
=
4n4
27
.
If n = 999 then any coloring of the square table with x = 2n
3
= 666 white cells in each row
and column attains the maximum as all inequalities in the previous argument become equalities.
For example color a cell (i, j) white if i − j ≡ 1, 2, . . . , 666 (mod 999), and red otherwise.
Therefore the maximum value T can attain is T = 4·9994
27
.
Comment. One can obtain a better preliminary estimate with the Cauchy-Schwarz inequality:
T =
(i,j)∈R
aibj ≤


(i,j)∈R
a2
i


1
2
·


(i,j)∈R
b2
j


1
2
=
n
i=1
(n − ai)a2
i
1
2
·


n
j=1
(n − bj)b2
j


1
2
.
It can be used to reach the same conclusion.

22
C4. Players A and B play a game with N ≥ 2012 coins and 2012 boxes arranged around a
circle. Initially A distributes the coins among the boxes so that there is at least 1 coin in each
box. Then the two of them make moves in the order B, A, B, A, . . . by the following rules:
• On every move of his B passes 1 coin from every box to an adjacent box.
• On every move of hers A chooses several coins that were not involved in B’s previous
move and are in different boxes. She passes every chosen coin to an adjacent box.
Player A’s goal is to ensure at least 1 coin in each box after every move of hers, regardless of
how B plays and how many moves are made. Find the least N that enables her to succeed.
Solution. We argue for a general n ≥ 7 instead of 2012 and prove that the required minimum N
is 2n − 2. For n = 2012 this gives Nmin = 4022.
a) If N = 2n − 2 player A can achieve her goal. Let her start the game with a regular
distribution: n − 2 boxes with 2 coins and 2 boxes with 1 coin. Call the boxes of the two kinds
red and white respectively. We claim that on her first move A can achieve a regular distribution
again, regardless of B’s first move M. She acts according as the following situation S occurs
after M or not: The initial distribution contains a red box R with 2 white neighbors, and R
receives no coins from them on move M.
Suppose that S does not occur. Exactly one of the coins c1 and c2 in a given red box X
is involved in M, say c1. If M passes c1 to the right neighbor of X, let A pass c2 to its left
neighbor, and vice versa. By doing so with all red boxes A performs a legal move M′
. Thus
M and M′
combined move the 2 coins of every red box in opposite directions. Hence after M
and M′
are complete each neighbor of a red box X contains exactly 1 coin that was initially
in X. So each box with a red neighbor is non-empty after M′
. If initially there is a box X
with 2 white neighbors (X is red and unique) then X receives a coin from at least one of them
on move M since S does not occur. Such a coin is not involved in M′
, so X is also non-empty
after M′
. Furthermore each box Y has given away its initial content after M and M′
. A red
neighbor of Y adds 1 coin to it; a white neighbor adds at most 1 coin because it is not involved
in M′
. Hence each box contains 1 or 2 coins after M′
. Because N = 2n−2, such a distribution
is regular.
Now let S occur after move M. Then A leaves untouched the exceptional red box R. With
all remaining red boxes she proceeds like in the previous case, thus making a legal move M′′
.
Box R receives no coins from its neighbors on either move, so there is 1 coin in it after M′′
.
Like above M and M′′
combined pass exactly 1 coin from every red box different from R to
each of its neighbors. Every box except R has a red neighbor different from R, hence all boxes
are non-empty after M′′
. Next, each box Y except R loses its initial content after M and M′′
.
A red neighbor of Y adds at most 1 coin to it; a white neighbor also adds at most 1 coin as
it does not participate in M′′
. Thus each box has 1 or 2 coins after M′′
, and the obtained
distribution is regular.
Player A can apply the described strategy indefinitely, so N = 2n−2 enables her to succeed.
b) For N ≤ 2n − 3 player B can achieve an empty box after some move of A. Let α be a
set of ℓ consecutive boxes containing a total of N(α) coins. We call α an arc if ℓ ≤ n − 2 and
N(α) ≤ 2ℓ − 3. Note that ℓ ≥ 2 by the last condition. Moreover if both extremes of α are
non-empty boxes then N(α) ≥ 2, so that N(α) ≤ 2ℓ − 3 implies ℓ ≥ 3. Observe also that if
an extreme X of α has more than 1 coin then ignoring X yields a shorter arc. It follows that
every arc contains an arc whose extremes have at most 1 coin each.
Given a clockwise labeling 1, 2, . . . , n of the boxes, suppose that boxes 1, 2, . . . , ℓ form an
arc α, with ℓ ≤ n − 2 and N(α) ≤ 2ℓ − 3. Suppose also that all n ≥ 7 boxes are non-empty.
Then B can move so that an arc α′
with N(α′
) < N(α) will appear after any response of A.

23
One may assume exactly 1 coin in boxes 1 and ℓ by a previous remark. Let B pass 1 coin
in counterclockwise direction from box 1 and box n, and in clockwise direction from each
remaining box. This leaves N(α)−2 coins in the boxes of α. In addition, due to 3 ≤ ℓ ≤ n−2,
box ℓ has exactly 1 coin c, the one received from box ℓ − 1.
Let player A’s next move M pass k ≤ 2 coins to boxes 1, 2, . . . , ℓ from the remaining ones.
Only boxes 1 and ℓ can receive such coins, at most 1 each. If k < 2 then after move M boxes
1, 2, . . . , ℓ form an arc α′
with N(α′
) < N(α). If k = 2 then M adds a coin to box ℓ. Also
M does not move coin c from ℓ because c is involved in the previous move of B. In summary
boxes 1, 2, . . . , ℓ contain N(α) coins like before, so they form an arc. However there are 2 coins
now in the extreme ℓ of the arc. Ignore ℓ to obtain a shorter arc α′
with N(α′
) < N(α).
Consider any initial distribution without empty boxes. Since N ≤ 2n − 3, there are at
least 3 boxes in it with exactly 1 coin. It follows from n ≥ 7 that some 2 of them are the
extremes of an arc α. Hence B can make the move described above, which leads to an arc α′
with N(α′
) < N(α) after A’s response. If all boxes in the new distribution are non-empty he
can repeat the same, and so on. Because N(α) cannot decrease indeﬁnitely, an empty box will
occur after some move of A.

24
C5. The columns and the rows of a 3n × 3n square board are numbered 1, 2, . . . , 3n. Every
square (x, y) with 1 ≤ x, y ≤ 3n is colored asparagus, byzantium or citrine according as the
modulo 3 remainder of x + y is 0, 1 or 2 respectively. One token colored asparagus, byzantium
or citrine is placed on each square, so that there are 3n2
tokens of each color.
Suppose that one can permute the tokens so that each token is moved to a distance of
at most d from its original position, each asparagus token replaces a byzantium token, each
byzantium token replaces a citrine token, and each citrine token replaces an asparagus token.
Prove that it is possible to permute the tokens so that each token is moved to a distance of at
most d + 2 from its original position, and each square contains a token with the same color as
the square.
Solution. Without loss of generality it suffices to prove that the A-tokens can be moved to
distinct A-squares in such a way that each A-token is moved to a distance at most d + 2 from
its original place. This means we need a perfect matching between the 3n2
A-squares and the
3n2
A-tokens such that the distance in each pair of the matching is at most d + 2.
To find the matching, we construct a bipartite graph. The A-squares will be the vertices in
one class of the graph; the vertices in the other class will be the A-tokens.
Split the board into 3 × 1 horizontal triminos; then each trimino contains exactly one A-
square. Take a permutation π of the tokens which moves A-tokens to B-tokens, B-tokens to
C-tokens, and C-tokens to A-tokens, in each case to a distance at most d. For each A-square S,
and for each A-token T, connect S and T by an edge if T, π(T) or π−1
(T) is on the trimino
containing S. We allow multiple edges; it is even possible that the same square and the same
token are connected with three edges. Obviously the lengths of the edges in the graph do not
exceed d + 2. By length of an edge we mean the distance between the A-square and the A-token
it connects.
Each A-token T is connected with the three A-squares whose triminos contain T, π(T)
and π−1
(T). Therefore in the graph all tokens are of degree 3. We show that the same is true
for the A-squares. Let S be an arbitrary A-square, and let T1, T2, T3 be the three tokens on
the trimino containing S. For i = 1, 2, 3, if Ti is an A-token, then S is connected with Ti; if Ti
is a B-token then S is connected with π−1
(Ti); finally, if Ti is a C-token then S is connected
with π(Ti). Hence in the graph the A-squares also are of degree 3.
Since the A-squares are of degree 3, from every set S of A-squares exactly 3|S| edges start.
These edges end in at least |S| tokens because the A-tokens also are of degree 3. Hence every
set S of A-squares has at least |S| neighbors among the A-tokens.
Therefore, by Hall’s marriage theorem, the graph contains a perfect matching between
the two vertex classes. So there is a perfect matching between the A-squares and A-tokens
with edges no longer than d + 2. It follows that the tokens can be permuted as specified in the
problem statement.
Comment 1. In the original problem proposal the board was infinite and there were only two colors.
Having n colors for some positive integer n was an option; we chose n = 3. Moreover, we changed
the board to a finite one to avoid dealing with infinite graphs (although Hall’s theorem works in the
infinite case as well).
With only two colors Hall’s theorem is not needed. In this case we split the board into 2 × 1
dominos, and in the resulting graph all vertices are of degree 2. The graph consists of disjoint cycles
with even length and infinite paths, so the existence of the matching is trivial.
Having more than three colors would make the problem statement more complicated, because we
need a matching between every two color classes of tokens. However, this would not mean a significant
increase in difficulty.

25
Comment 2. According to Wikipedia, the color asparagus (hexadecimal code #87A96B) is a tone
of green that is named after the vegetable. Crayola created this color in 1993 as one of the 16 to
be named in the Name The Color Contest. Byzantium (#702963) is a dark tone of purple. Its ﬁrst
recorded use as a color name in English was in 1926. Citrine (#E4D00A) is variously described as
yellow, greenish-yellow, brownish-yellow or orange. The ﬁrst known use of citrine as a color name in
English was in the 14th century.

26
C6. Let k and n be fixed positive integers. In the liar’s guessing game, Amy chooses integers
x and N with 1 ≤ x ≤ N. She tells Ben what N is, but not what x is. Ben may then repeatedly
ask Amy whether x ∈ S for arbitrary sets S of integers. Amy will always answer with yes or no,
but she might lie. The only restriction is that she can lie at most k times in a row. After he
has asked as many questions as he wants, Ben must specify a set of at most n positive integers.
If x is in this set he wins; otherwise, he loses. Prove that:
a) If n ≥ 2k
then Ben can always win.
b) For sufficiently large k there exist n ≥ 1.99k
such that Ben cannot guarantee a win.
Solution. Consider an answer A ∈ {yes, no} to a question of the kind “Is x in the set S?”
We say that A is inconsistent with a number i if A = yes and i ∈ S, or if A = no and i ∈ S.
Observe that an answer inconsistent with the target number x is a lie.
a) Suppose that Ben has determined a set T of size m that contains x. This is true initially
with m = N and T = {1, 2, . . . , N}. For m > 2k
we show how Ben can find a number y ∈ T
that is different from x. By performing this step repeatedly he can reduce T to be of size 2k
≤ n
and thus win.
Since only the size m > 2k
of T is relevant, assume that T = {0, 1, . . . , 2k
, . . . , m−1}. Ben
begins by asking repeatedly whether x is 2k
. If Amy answers no k + 1 times in a row, one
of these answers is truthful, and so x = 2k
. Otherwise Ben stops asking about 2k
at the first
answer yes. He then asks, for each i = 1, . . . , k, if the binary representation of x has a 0 in
the ith digit. Regardless of what the k answers are, they are all inconsistent with a certain
number y ∈ {0, 1, . . . , 2k
− 1}. The preceding answer yes about 2k
is also inconsistent with y.
Hence y = x. Otherwise the last k + 1 answers are not truthful, which is impossible.
Either way, Ben finds a number in T that is different from x, and the claim is proven.
b) We prove that if 1 < λ < 2 and n = (2 − λ)λk+1
− 1 then Ben cannot guarantee a win.
To complete the proof, then it suffices to take λ such that 1.99 < λ < 2 and k large enough so
that
n = (2 − λ)λk+1
− 1 ≥ 1.99k
.
Consider the following strategy for Amy. First she chooses N = n+1 and x ∈ {1, 2, . . . , n+1}
arbitrarily. After every answer of hers Amy determines, for each i = 1, 2, . . . , n + 1, the
number mi of consecutive answers she has given by that point that are inconsistent with i. To
decide on her next answer, she then uses the quantity
φ =
n+1
i=1
λmi
.
No matter what Ben’s next question is, Amy chooses the answer which minimizes φ.
We claim that with this strategy φ will always stay less than λk+1
. Consequently no expo-
nent mi in φ will ever exceed k, hence Amy will never give more than k consecutive answers
inconsistent with some i. In particular this applies to the target number x, so she will never lie
more than k times in a row. Thus, given the claim, Amy’s strategy is legal. Since the strategy
does not depend on x in any way, Ben can make no deductions about x, and therefore he cannot
guarantee a win.
It remains to show that φ < λk+1
at all times. Initially each mi is 0, so this condition holds
in the beginning due to 1 < λ < 2 and n = (2 − λ)λk+1
− 1. Suppose that φ < λk+1
at some
point, and Ben has just asked if x ∈ S for some set S. According as Amy answers yes or no,
the new value of φ becomes
φ1 =
i∈S
1 +
i/∈S
λmi+1
or φ2 =
i∈S
λmi+1
+
i/∈S
1.

27
Since Amy chooses the option minimizing φ, the new φ will equal min(φ1, φ2). Now we have
min(φ1, φ2) ≤
1
2
(φ1 + φ2) =
1
2 i∈S
1 + λmi+1
+
i/∈S
λmi+1
+ 1 =
1
2
(λφ + n + 1).
Because φ < λk+1
, the assumptions λ < 2 and n = (2 − λ)λk+1
− 1 lead to
min(φ1, φ2) <
1
2
(λk+2
+ (2 − λ)λk+1
) = λk+1
.
The claim follows, which completes the solution.
Comment. Given a ﬁxed k, let f(k) denote the minimum value of n for which Ben can guarantee a
victory. The problem asks for a proof that for large k
1.99k
≤ f(k) ≤ 2k
.
A computer search shows that f(k) = 2, 3, 4, 7, 11, 17 for k = 1, 2, 3, 4, 5, 6.

28
C7. There are given 2500
points on a circle labeled 1, 2, . . . , 2500
in some order. Prove that
one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums
of the pairs of numbers at the endpoints of the chosen chords are equal.
Solution. The proof is based on the following general fact.
Lemma. In a graph G each vertex v has degree dv. Then G contains an independent set S of
vertices such that |S| ≥ f(G) where
f(G) =
v∈G
1
dv + 1
.
Proof. Induction on n = |G|. The base n = 1 is clear. For the inductive step choose a vertex v0
in G of minimum degree d. Delete v0 and all of its neighbors v1, . . . , vd and also all edges with
endpoints v0, v1, . . . , vd. This gives a new graph G′
. By the inductive assumption G′
contains
an independent set S′
of vertices such that |S′
| ≥ f(G′
). Since no vertex in S′
is a neighbor
of v0 in G, the set S = S′
∪ {v0} is independent in G.
Let d′
v be the degree of a vertex v in G′
. Clearly d′
v ≤ dv for every such vertex v, and also
dvi
≥ d for all i = 0, 1, . . . , d by the minimal choice of v0. Therefore
f(G′
) =
v∈G′
1
d′
v + 1
≥
v∈G′
1
dv + 1
= f(G) −
d
i=0
1
dvi
+ 1
≥ f(G) −
d + 1
d + 1
= f(G) − 1.
Hence |S| = |S′
| + 1 ≥ f(G′
) + 1 ≥ f(G), and the induction is complete.
We pass on to our problem. For clarity denote n = 2499
and draw all chords determined by
the given 2n points. Color each chord with one of the colors 3, 4, . . . , 4n − 1 according to the
sum of the numbers at its endpoints. Chords with a common endpoint have diﬀerent colors.
For each color c consider the following graph Gc. Its vertices are the chords of color c, and two
chords are neighbors in Gc if they intersect. Let f(Gc) have the same meaning as in the lemma
for all graphs Gc.
Every chord ℓ divides the circle into two arcs, and one of them contains m(ℓ) ≤ n − 1 given
points. (In particular m(ℓ) = 0 if ℓ joins two consecutive points.) For each i = 0, 1, . . . , n − 2
there are 2n chords ℓ with m(ℓ) = i. Such a chord has degree at most i in the respective graph.
Indeed let A1, . . . , Ai be all points on either arc determined by a chord ℓ with m(ℓ) = i and
color c. Every Aj is an endpoint of at most 1 chord colored c, j = 1, . . . , i. Hence at most
i chords of color c intersect ℓ.
It follows that for each i = 0, 1, . . . , n − 2 the 2n chords ℓ with m(ℓ) = i contribute at
least 2n
i+1
to the sum c f(Gc). Summation over i = 0, 1, . . . , n − 2 gives
c
f(Gc) ≥ 2n
n−1
i=1
1
i
.
Because there are 4n − 3 colors in all, averaging yields a color c such that
f(Gc) ≥
2n
4n − 3
n−1
i=1
1
i
>
1
2
n−1
i=1
1
i
.
By the lemma there are at least 1
2
n−1
i=1
1
i
pairwise disjoint chords of color c, i. e. with the same
sum c of the pairs of numbers at their endpoints. It remains to show that 1
2
n−1
i=1
1
i
≥ 100 for
n = 2499
. Indeed we have
n−1
i=1
1
i
>
2400
i=1
1
i
= 1 +
400
k=1
2k
i=2k−1+1
1
i
> 1 +
400
k=1
2k−1
2k
= 201 > 200.
This completes the solution.

29
Geometry
G1. In the triangle ABC the point J is the center of the excircle opposite to A. This excircle
is tangent to the side BC at M, and to the lines AB and AC at K and L respectively. The
lines LM and BJ meet at F, and the lines KM and CJ meet at G. Let S be the point of
intersection of the lines AF and BC, and let T be the point of intersection of the lines AG
and BC. Prove that M is the midpoint of ST.
Solution. Let α = ∠CAB, β = ∠ABC and γ = ∠BCA. The line AJ is the bisector of ∠CAB,
so ∠JAK = ∠JAL = α
2
. By ∠AKJ = ∠ALJ = 90◦
the points K and L lie on the circle ω
with diameter AJ.
The triangle KBM is isosceles as BK and BM are tangents to the excircle. Since BJ is the
bisector of ∠KBM, we have ∠MBJ = 90◦
− β
2
and ∠BMK = β
2
. Likewise ∠MCJ = 90◦
− γ
2
and ∠CML = γ
2
. Also ∠BMF = ∠CML, therefore
∠LFJ = ∠MBJ − ∠BMF = 90◦
−
β
2
−
γ
2
=
α
2
= ∠LAJ.
Hence F lies on the circle ω. (By the angle computation, F and A are on the same side of BC.)
Analogously, G also lies on ω. Since AJ is a diameter of ω, we obtain ∠AFJ = ∠AGJ = 90◦
.
A
B C
GF
S T
K
M
L
ω
J
β γ
α
2
α
2
α
2
α
2
The lines AB and BC are symmetric with respect to the external bisector BF. Because
AF ⊥ BF and KM ⊥ BF, the segments SM and AK are symmetric with respect to BF,
hence SM = AK. By symmetry TM = AL. Since AK and AL are equal as tangents to the
excircle, it follows that SM = TM, and the proof is complete.
Comment. After discovering the circle AFKJLG, there are many other ways to complete the solu-
tion. For instance, from the cyclic quadrilaterals JMFS and JMGT one can ﬁnd ∠TSJ = ∠STJ = α
2 .
Another possibility is to use the fact that the lines AS and GM are parallel (both are perpendicular
to the external angle bisector BJ), so MS
MT = AG
GT = 1.

30
G2. Let ABCD be a cyclic quadrilateral whose diagonals AC and BD meet at E. The
extensions of the sides AD and BC beyond A and B meet at F. Let G be the point such that
ECGD is a parallelogram, and let H be the image of E under reflection in AD. Prove that
D, H, F, G are concyclic.
Solution. We show first that the triangles FDG and FBE are similar. Since ABCD is cyclic,
the triangles EAB and EDC are similar, as well as FAB and FCD. The parallelogram ECGD
yields GD = EC and ∠CDG = ∠DCE; also ∠DCE = ∠DCA = ∠DBA by inscribed angles.
Therefore
∠FDG = ∠FDC + ∠CDG = ∠FBA + ∠ABD = ∠FBE,
GD
EB
=
CE
EB
=
CD
AB
=
FD
FB
.
It follows that FDG and FBE are similar, and so ∠FGD = ∠FEB.
A B
D
G
E
F
H
C
Since H is the reflection of E with respect to FD, we conclude that
∠FHD = ∠FED = 180◦
− ∠FEB = 180◦
− ∠FGD.
This proves that D, H, F, G are concyclic.
Comment. Points E and G are always in the half-plane determined by the line FD that contains
B and C, but H is always in the other half-plane. In particular, DHFG is cyclic if and only if
∠FHD + ∠FGD = 180◦.

31
G3. In an acute triangle ABC the points D, E and F are the feet of the altitudes through A,
B and C respectively. The incenters of the triangles AEF and BDF are I1 and I2 respectively;
the circumcenters of the triangles ACI1 and BCI2 are O1 and O2 respectively. Prove that I1I2
and O1O2 are parallel.
Solution. Let ∠CAB = α, ∠ABC = β, ∠BCA = γ. We start by showing that A, B, I1
and I2 are concyclic. Since AI1 and BI2 bisect ∠CAB and ∠ABC, their extensions beyond I1
and I2 meet at the incenter I of the triangle. The points E and F are on the circle with
diameter BC, so ∠AEF = ∠ABC and ∠AFE = ∠ACB. Hence the triangles AEF and ABC
are similar with ratio of similitude AE
AB
= cos α. Because I1 and I are their incenters, we obtain
I1A = IA cos α and II1 = IA − I1A = 2IA sin2 α
2
. By symmetry II2 = 2IB sin2 β
2
. The law of
sines in the triangle ABI gives IA sin α
2
= IB sin β
2
. Hence
II1 · IA = 2 IA sin α
2
2
= 2 IB sin β
2
2
= II2 · IB.
Therefore A, B, I1 and I2 are concyclic, as claimed.
O2
O1
C
A F B
I2Q
I
E
D
I3
I1
In addition II1 · IA = II2 · IB implies that I has the same power with respect to the
circles (ACI1), (BCI2) and (ABI1I2). Then CI is the radical axis of (ACI1) and (BCI2); in
particular CI is perpendicular to the line of centers O1O2.
Now it suﬃces to prove that CI ⊥ I1I2. Let CI meet I1I2 at Q, then it is enough to check
that ∠II1Q + ∠I1IQ = 90◦
. Since ∠I1IQ is external for the triangle ACI, we have
∠II1Q + ∠I1IQ = ∠II1Q + (∠ACI + ∠CAI) = ∠II1I2 + ∠ACI + ∠CAI.
It remains to note that ∠II1I2 = β
2
from the cyclic quadrilateral ABI1I2, and ∠ACI = γ
2
,
∠CAI = α
2
. Therefore ∠II1Q + ∠I1IQ = α
2
+ β
2
+ γ
2
= 90◦
, completing the proof.
Comment. It follows from the ﬁrst part of the solution that the common point I3 = C of the
circles (ACI1) and (BCI2) is the incenter of the triangle CDE.

32
G4. Let ABC be a triangle with AB = AC and circumcenter O. The bisector of ∠BAC
intersects BC at D. Let E be the reflection of D with respect to the midpoint of BC. The lines
through D and E perpendicular to BC intersect the lines AO and AD at X and Y respectively.
Prove that the quadrilateral BXCY is cyclic.
Solution. The bisector of ∠BAC and the perpendicular bisector of BC meet at P, the midpoint
of the minor arc BC (they are different lines as AB = AC). In particular OP is perpendicular
to BC and intersects it at M, the midpoint of BC.
Denote by Y ′
the reflexion of Y with respect to OP. Since ∠BY C = ∠BY ′
C, it suffices to
prove that BXCY ′
is cyclic.
A
D
Y ′Y
B C
X
M
O
E
P
We have
∠XAP = ∠OPA = ∠EY P.
The first equality holds because OA = OP, and the second one because EY and OP are both
perpendicular to BC and hence parallel. But {Y, Y ′
} and {E, D} are pairs of symmetric points
with respect to OP, it follows that ∠EY P = ∠DY ′
P and hence
∠XAP = ∠DY ′
P = ∠XY ′
P.
The last equation implies that XAY ′
P is cyclic. By the powers of D with respect to the
circles (XAY ′
P) and (ABPC) we obtain
XD · DY ′
= AD · DP = BD · DC.
It follows that BXCY ′
is cyclic, as desired.

33
G5. Let ABC be a triangle with ∠BCA = 90◦
, and let C0 be the foot of the altitude
from C. Choose a point X in the interior of the segment CC0, and let K, L be the points on
the segments AX, BX for which BK = BC and AL = AC respectively. Denote by M the
intersection of AL and BK. Show that MK = ML.
Solution. Let C′
be the reﬂection of C in the line AB, and let ω1 and ω2 be the circles
with centers A and B, passing through L and K respectively. Since AC′
= AC = AL and
BC′
= BC = BK, both ω1 and ω2 pass through C and C′
. By ∠BCA = 90◦
, AC is tangent
to ω2 at C, and BC is tangent to ω1 at C. Let K1 = K be the second intersection of AX and
ω2, and let L1 = L be the second intersection of BX and ω1.
A
K
M
L1
K1
ω3
C
L
BC0
C′
X
ω2
ω1
By the powers of X with respect to ω2 and ω1,
XK · XK1 = XC · XC′
= XL · XL1,
so the points K1, L, K, L1 lie on a circle ω3.
The power of A with respect to ω2 gives
AL2
= AC2
= AK · AK1,
indicating that AL is tangent to ω3 at L. Analogously, BK is tangent to ω3 at K. Hence MK
and ML are the two tangents from M to ω3 and therefore MK = ML.

34
G6. Let ABC be a triangle with circumcenter O and incenter I. The points D, E and F on
the sides BC, CA and AB respectively are such that BD + BF = CA and CD + CE = AB.
The circumcircles of the triangles BFD and CDE intersect at P = D. Prove that OP = OI.
Solution. By Miquel’s theorem the circles (AEF) = ωA, (BFD) = ωB and (CDE) = ωC
have a common point, for arbitrary points D, E and F on BC, CA and AB. So ωA passes
through the common point P = D of ωB and ωC.
Let ωA, ωB and ωC meet the bisectors AI, BI and CI at A = A′
, B = B′
and C = C′
respectively. The key observation is that A′
, B′
and C′
do not depend on the particular choice
of D, E and F, provided that BD + BF = CA, CD + CE = AB and AE + AF = BC hold
true (the last equality follows from the other two). For a proof we need the following fact.
Lemma. Given is an angle with vertex A and measure α. A circle ω through A intersects the
angle bisector at L and sides of the angle at X and Y . Then AX + AY = 2AL cos α
2
.
Proof. Note that L is the midpoint of arc XLY in ω and set XL = Y L = u, XY = v. By
Ptolemy’s theorem AX · Y L+ AY · XL = AL· XY , which rewrites as (AX + AY )u = AL· v.
Since ∠LXY = α
2
and ∠XLY = 180◦
− α, we have v = 2 cos α
2
u by the law of sines, and the
claim follows.
X
L
u
u
v
A
Y
Apply the lemma to ∠BAC = α and the circle ω = ωA, which intersects AI at A′
. This
gives 2AA′
cos α
2
= AE + AF = BC; by symmetry analogous relations hold for BB′
and CC′
.
It follows that A′
, B′
and C′
are independent of the choice of D, E and F, as stated.
We use the lemma two more times with ∠BAC = α. Let ω be the circle with diameter AI.
Then X and Y are the tangency points of the incircle of ABC with AB and AC, and hence
AX = AY = 1
2
(AB + AC − BC). So the lemma yields 2AI cos α
2
= AB + AC − BC. Next,
if ω is the circumcircle of ABC and AI intersects ω at M = A then {X, Y } = {B, C}, and so
2AM cos α
2
= AB + AC by the lemma. To summarize,
2AA′
cos α
2
= BC, 2AI cos α
2
= AB + AC − BC, 2AM cos α
2
= AB + AC. (*)
These equalities imply AA′
+ AI = AM, hence the segments AM and IA′
have a common
midpoint. It follows that I and A′
are equidistant from the circumcenter O. By symmetry
OI = OA′
= OB′
= OC′
, so I, A′
, B′
, C′
are on a circle centered at O.
To prove OP = OI, now it suﬃces to show that I, A′
, B′
, C′
and P are concyclic. Clearly
one can assume P = I, A′
, B′
, C′
.
We use oriented angles to avoid heavy case distinction. The oriented angle between the lines l
and m is denoted by ∠(l, m). We have ∠(l, m) = −∠(m, l) and ∠(l, m) + ∠(m, n) = ∠(l, n)
for arbitrary lines l, m and n. Four distinct non-collinear points U, V, X, Y are concyclic if and
only if ∠(UX, V X) = ∠(UY, V Y ).

35
M
C
B′
I
O
E
P
C′
A
B
A′
ωA
D
F
ωB
ωC
Suppose for the moment that A′
, B′
, P, I are distinct and noncollinear; then it is enough to
check the equality ∠(A′
P, B′
P) = ∠(A′
I, B′
I). Because A, F, P, A′
are on the circle ωA, we have
∠(A′
P, FP) = ∠(A′
A, FA) = ∠(A′
I, AB). Likewise ∠(B′
P, FP) = ∠(B′
I, AB). Therefore
∠(A′
P, B′
P) = ∠(A′
P, FP) + ∠(FP, B′
P) = ∠(A′
I, AB) − ∠(B′
I, AB) = ∠(A′
I, B′
I).
Here we assumed that P = F. If P = F then P = D, E and the conclusion follows similarly (use
∠(A′
F, B′
F) = ∠(A′
F, EF) + ∠(EF, DF) + ∠(DF, B′
F) and inscribed angles in ωA, ωB, ωC).
There is no loss of generality in assuming A′
, B′
, P, I distinct and noncollinear. If ABC
is an equilateral triangle then the equalities (*) imply that A′
, B′
, C′
, I, O and P coincide, so
OP = OI. Otherwise at most one of A′
, B′
, C′
coincides with I. If say C′
= I then OI ⊥ CI
by the previous reasoning. It follows that A′
, B′
= I and hence A′
= B′
. Finally A′
, B′
and I
are noncollinear because I, A′
, B′
, C′
are concyclic.
Comment. The proposer remarks that the locus γ of the points P is an arc of the circle (A′B′C′I).
The reﬂection I′ of I in O belongs to γ; it is obtained by choosing D, E and F to be the tangency
points of the three excircles with their respective sides. The rest of the circle (A′B′C′I), except I,
can be included in γ by letting D, E and F vary on the extensions of the sides and assuming signed
lengths. For instance if B is between C and D then the length BD must be taken with a negative
sign. The incenter I corresponds to the limit case where D tends to inﬁnity.

36
G7. Let ABCD be a convex quadrilateral with non-parallel sides BC and AD. Assume
that there is a point E on the side BC such that the quadrilaterals ABED and AECD are
circumscribed. Prove that there is a point F on the side AD such that the quadrilaterals
ABCF and BCDF are circumscribed if and only if AB is parallel to CD.
Solution. Let ω1 and ω2 be the incircles and O1 and O2 the incenters of the quadrilater-
als ABED and AECD respectively. A point F with the stated property exists only if ω1
and ω2 are also the incircles of the quadrilaterals ABCF and BCDF.
D
C
E
B
O1
O2
AF1F2O
Let the tangents from B to ω2 and from C to ω1 (other than BC) meet AD at F1 and F2
respectively. We need to prove that F1 = F2 if and only if AB CD.
Lemma. The circles ω1 and ω2 with centers O1 and O2 are inscribed in an angle with vertex O.
The points P, S on one side of the angle and Q, R on the other side are such that ω1 is the
incircle of the triangle PQO, and ω2 is the excircle of the triangle RSO opposite to O. Denote
p = OO1 · OO2. Then exactly one of the following relations holds:
OP · OR p > OQ · OS, OP · OR = p = OQ · OS.
Proof. Denote ∠OPO1 = u, ∠OQO1 = v, ∠OO2R = x, ∠OO2S = y, ∠POQ = 2ϕ. Because
PO1, QO1, RO2, SO2 are internal or external bisectors in the triangles PQO and RSO, we have
u + v = x + y (= 90◦
− ϕ). (1)
R
S
O1O2
x
y
Pu
v Q
O ϕ
ϕ
By the law of sines
OP
OO1
=
sin(u + ϕ)
sin u
and
OO2
OR
=
sin(x + ϕ)
sin x
.
Therefore, since x, u and ϕ are acute,
OP · OR ≥ p ⇔
OP
OO1
≥
OO2
OR
⇔ sin x sin(u + ϕ) ≥ sin u sin(x + ϕ) ⇔ sin(x − u) ≥ 0 ⇔ x ≥ u.
Thus OP · OR ≥ p is equivalent to x ≥ u, with OP · OR = p if and only if x = u.
Analogously, p ≥ OQ · OS is equivalent to v ≥ y, with p = OQ · OS if and only if v = y.
On the other hand x ≥ u and v ≥ y are equivalent by (1), with x = u if and only if v = y. The
conclusion of the lemma follows from here.

37
Going back to the problem, apply the lemma to the quadruples {B, E, D, F1}, {A, B, C, D}
and {A, E, C, F2}. Assuming OE · OF1 > p, we obtain
OE · OF1 > p ⇒ OB · OD p ⇒ OE · OF2 < p.
In other words, OE · OF1 > p implies
OB · OD p > OE · OF2.
Similarly, OE · OF1 p > OA · OC and OE · OF1 < p < OE · OF2.
In these cases F1 = F2 and OB · OD = OA · OC, so the lines AB and CD are not parallel.
There remains the case OE · OF1 = p. Here the lemma leads to OB · OD = p = OA · OC
and OE · OF1 = p = OE · OF2. Therefore F1 = F2 and AB CD.
Comment. The conclusion is also true if BC and AD are parallel. One can prove a limit case of
the lemma for the configuration shown in the figure below, where r1 and r2 are parallel rays starting
at O′ and O′′, with O′O′′ ⊥ r1, r2 and O the midpoint of O′O′′. Two circles with centers O1 and O2
are inscribed in the strip between r1 and r2. The lines PQ and RS are tangent to the circles, with
P, S on r1, and Q, R on r2, so that O, O1 are on the same side of PQ and O, O2 are on different sides
of RS. Denote s = OO1 + OO2. Then exactly one of the following relations holds:
O′
P + O′′
R < s < O′′
Q + O′
S, O′
P + O′′
R > s > O′′
Q + O′
S, O′
P + O′′
R = s = O′′
Q + O′
S.
O2
R
S
O1
Q
P
r1
r2
O
O′
O′′
Once this is established, the proof of the original statement for BC AD is analogous to the one
in the intersecting case. One replaces products by sums of relevant segments.

38
G8. Let ABC be a triangle with circumcircle ω and ℓ a line without common points with ω.
Denote by P the foot of the perpendicular from the center of ω to ℓ. The side-lines BC, CA, AB
intersect ℓ at the points X, Y, Z different from P. Prove that the circumcircles of the triangles
AXP, BY P and CZP have a common point different from P or are mutually tangent at P.
Solution 1. Let ωA, ωB, ωC and ω be the circumcircles of triangles AXP, BY P, CZP and ABC
respectively. The strategy of the proof is to construct a point Q with the same power with
respect to the four circles. Then each of P and Q has the same power with respect to ωA, ωB, ωC
and hence the three circles are coaxial. In other words they have another common point P′
or
the three of them are tangent at P.
We first give a description of the point Q. Let A′
= A be the second intersection of ω
and ωA; define B′
and C′
analogously. We claim that AA′
, BB′
and CC′
have a common point.
Once this claim is established, the point just constructed will be on the radical axes of the
three pairs of circles {ω, ωA}, {ω, ωB}, {ω, ωC}. Hence it will have the same power with respect
to ω, ωA, ωB, ωC.
ℓ
ωA
ωC
ω
ωB
X Y ZP
A
B′
Q
O
C′
B
A′C
Z′ Y ′
X′
P′
We proceed to prove that AA′
, BB′
and CC′
intersect at one point. Let r be the circumra-
dius of triangle ABC. Define the points X′
, Y ′
, Z′
as the intersections of AA′
, BB′
, CC′
with ℓ.
Observe that X′
, Y ′
, Z′
do exist. If AA′
is parallel to ℓ then ωA is tangent to ℓ; hence X = P
which is a contradiction. Similarly, BB′
and CC′
are not parallel to ℓ.
From the powers of the point X′
with respect to the circles ωA and ω we get
X′
P · (X′
P + PX) = X′
P · X′
X = X′
A′
· X′
A = X′
O2
− r2
,
hence
X′
P · PX = X′
O2
− r2
− X′
P2
= OP2
− r2
.
We argue analogously for the points Y ′
and Z′
, obtaining
X′
P · PX = Y ′
P · PY = Z′
P · PZ = OP2
− r2
= k2
. (1)
In these computations all segments are regarded as directed segments. We keep the same
convention for the sequel.
We prove that the lines AA′
, BB′
, CC′
intersect at one point by Ceva’s theorem. To avoid
distracting remarks we interpret everything projectively, i. e. whenever two lines are parallel
they meet at a point on the line at infinity.

39
Let U, V, W be the intersections of AA′
, BB′
, CC′
with BC, CA, AB respectively. The idea
is that although it is difficult to calculate the ratio BU
CU
, it is easier to deal with the cross-ratio
BU
CU
/BX
CX
because we can send it to the line ℓ. With this in mind we apply Menelaus’ theorem
to the triangle ABC and obtain BX
CX
· CY
AY
· AZ
BZ
= 1. Hence Ceva’s ratio can be expressed as
BU
CU
·
CV
AV
·
AW
BW
=
BU
CU
/
BX
CX
·
CV
AV
/
CY
AY
·
AW
BW
/
AZ
BZ
.
ℓ
ω
X Y P
A
V
Q
W
U
B
C
Z′ ZX′ Y ′
Project the line BC to ℓ from A. The cross-ratio between BC and UX equals the cross-ratio
between ZY and X′
X. Repeating the same argument with the lines CA and AB gives
BU
CU
·
CV
AV
·
AW
BW
=
ZX′
Y X′
/
ZX
Y X
·
XY ′
ZY ′
/
XY
ZY
·
Y Z′
XZ′
/
Y Z
XZ
and hence
BU
CU
·
CV
AV
·
AW
BW
= (−1) ·
ZX′
Y X′
·
XY ′
ZY ′
·
Y Z′
XZ′
.
The equations (1) reduce the problem to a straightforward computation on the line ℓ.
For instance, the transformation t → −k2
/t preserves cross-ratio and interchanges the points
X, Y, Z with the points X′
, Y ′
, Z′
. Then
BU
CU
·
CV
AV
·
AW
BW
= (−1) ·
ZX′
Y X′
/
ZZ′
Y Z′
·
XY ′
ZY ′
/
XZ′
ZZ′
= −1.
We proved that Ceva’s ratio equals −1, so AA′
, BB′
, CC′
intersect at one point Q.
Comment 1. There is a nice projective argument to prove that AX′, BY ′, CZ′ intersect at one point.
Suppose that ℓ and ω intersect at a pair of complex conjugate points D and E. Consider a projective
transformation that takes D and E to [i; 1, 0] and [−i, 1, 0]. Then ℓ is the line at infinity, and ω is
a conic through the special points [i; 1, 0] and [−i, 1, 0], hence it is a circle. So one can assume that
AX, BY, CZ are parallel to BC, CA, AB. The involution on ℓ taking X, Y, Z to X′, Y ′, Z′ and leaving
D, E fixed is the involution changing each direction to its perpendicular one. Hence AX, BY, CZ are
also perpendicular to AX′, BY ′, CZ′.
It follows from the above that AX′, BY ′, CZ′ intersect at the orthocenter of triangle ABC.
Comment 2. The restriction that the line ℓ does not intersect the circumcricle ω is unnecessary.
The proof above works in general. In case ℓ intersects ω at D and E point P is the midpoint of DE,
and some equations can be interpreted differently. For instance
X′
P · X′
X = X′
A′
· X′
A = X′
D · X′
E,
and hence the pairs X′X and DE are harmonic conjugates. This means that X′, Y ′, Z′ are the
harmonic conjugates of X, Y, Z with respect to the segment DE.

40
Solution 2. First we prove that there is an inversion in space that takes ℓ and ω to parallel
circles on a sphere. Let QR be the diameter of ω whose extension beyond Q passes through P.
Let Π be the plane carrying our objects. In space, choose a point O such that the line QO is
perpendicular to Π and ∠POR = 90◦
, and apply an inversion with pole O (the radius of the
inversion does not matter). For any object T denote by T ′
the image of T under this inversion.
The inversion takes the plane Π to a sphere Π′
. The lines in Π are taken to circles through O,
and the circles in Π also are taken to circles on Π′
.
O
ℓ
P RQ
Q′
R′
ω
ℓ′
Π
P′
Π′
ω′
Since the line ℓ and the circle ω are perpendicular to the plane OPQ, the circles ℓ′
and ω′
also are perpendicular to this plane. Hence, the planes of the circles ℓ′
and ω′
are parallel.
Now consider the circles A′
X′
P′
, B′
Y ′
P′
and C′
Z′
P′
. We want to prove that either they
have a common point (on Π′
), diﬀerent from P′
, or they are tangent to each other.
H
C′
O
B1
X′
A′
W
Y ′
P′
Z′
Π′
ℓ′
ω′
A1
C1
B′
The point X′
is the second intersection of the circles B′
C′
O and ℓ′
, other than O. Hence,
the lines OX′
and B′
C′
are coplanar. Moreover, they lie in the parallel planes of ℓ′
and ω′
.
Therefore, OX′
and B′
C′
are parallel. Analogously, OY ′
and OZ′
are parallel to A′
C′
and A′
B′
.
Let A1 be the second intersection of the circles A′
X′
P′
and ω′
, other than A′
. The segments
A′
A1 and P′
X′
are coplanar, and therefore parallel. Now we know that B′
C′
and A′
A1 are
parallel to OX′
and X′
P′
respectively, but these two segments are perpendicular because OP′
is a diameter in ℓ′
. We found that A′
A1 and B′
C′
are perpendicular, hence A′
A1 is the altitude
in the triangle A′
B′
C′
, starting from A.
Analogously, let B1 and C1 be the second intersections of ω′
with the circles B′
P′
Y ′
and C′
P′
Z′
, other than B′
and C′
respectively. Then B′
B1 and C′
C1 are the other two al-
titudes in the triangle A′
B′
C′
.

41
Let H be the orthocenter of the triangle A′
B′
C′
. Let W be the second intersection of the
line P′
H with the sphere Π′
, other than P′
. The point W lies on the sphere Π′
, in the plane
of the circle A′
P′
X′
, so W lies on the circle A′
P′
X′
. Similarly, W lies on the circles B′
P′
Y ′
and C′
P′
Z′
as well; indeed W is the second common point of the three circles.
If the line P′
H is tangent to the sphere then W coincides with P′
, and P′
H is the common
tangent of the three circles.

42
Number Theory
N1. Call admissible a set A of integers that has the following property:
If x, y ∈ A (possibly x = y) then x2
+ kxy + y2
∈ A for every integer k.
Determine all pairs m, n of nonzero integers such that the only admissible set containing both m
and n is the set of all integers.
Solution. A pair of integers m, n fulfills the condition if and only if gcd(m, n) = 1. Suppose
that gcd(m, n) = d > 1. The set
A = {. . . , −2d, −d, 0, d, 2d, . . .}
is admissible, because if d divides x and y then it divides x2
+ kxy + y2
for every integer k.
Also m, n ∈ A and A = Z.
Now let gcd(m, n) = 1, and let A be an admissible set containing m and n. We use the
following observations to prove that A = Z:
(i) kx2
∈ A for every x ∈ A and every integer k.
(ii) (x + y)2
∈ A for all x, y ∈ A.
To justify (i) let y = x in the definition of an admissible set; to justify (ii) let k = 2.
Since gcd(m, n) = 1, we also have gcd(m2
, n2
) = 1. Hence one can find integers a, b such
that am2
+ bn2
= 1. It follows from (i) that am2
∈ A and bn2
∈ A. Now we deduce from (ii)
that 1 = (am2
+ bn2
)2
∈ A. But if 1 ∈ A then (i) implies k ∈ A for every integer k.

43
N2. Find all triples (x, y, z) of positive integers such that x ≤ y ≤ z and
x3
(y3
+ z3
) = 2012(xyz + 2).
Solution. First note that x divides 2012·2 = 23
·503. If 503 | x then the right-hand side of the
equation is divisible by 5033
, and it follows that 5032
| xyz + 2. This is false as 503 | x. Hence
x = 2m
with m ∈ {0, 1, 2, 3}. If m ≥ 2 then 26
| 2012(xyz + 2). However the highest powers
of 2 dividing 2012 and xyz + 2 = 2m
yz + 2 are 22
and 21
respectively. So x = 1 or x = 2,
yielding the two equations
y3
+ z3
= 2012(yz + 2), and y3
+ z3
= 503(yz + 1).
In both cases the prime 503 = 3 · 167 + 2 divides y3
+ z3
. We claim that 503 | y + z. This
is clear if 503 | y, so let 503 ∤ y and 503 ∤ z. Then y502
≡ z502
(mod 503) by Fermat’s little
theorem. On the other hand y3
≡ −z3
(mod 503) implies y3·167
≡ −z3·167
(mod 503), i. e.
y501
≡ −z501
(mod 503). It follows that y ≡ −z (mod 503) as claimed.
Therefore y + z = 503k with k ≥ 1. In view of y3
+ z3
= (y + z) (y − z)2
+ yz the two
equations take the form
k(y − z)2
+ (k − 4)yz = 8, (1)
k(y − z)2
+ (k − 1)yz = 1. (2)
In (1) we have (k − 4)yz ≤ 8, which implies k ≤ 4. Indeed if k > 4 then 1 ≤ (k − 4)yz ≤ 8,
so that y ≤ 8 and z ≤ 8. This is impossible as y + z = 503k ≥ 503. Note next that y3
+ z3
is even in the ﬁrst equation. Hence y + z = 503k is even too, meaning that k is even. Thus
k = 2 or k = 4. Clearly (1) has no integer solutions for k = 4. If k = 2 then (1) takes the form
(y + z)2
− 5yz = 4. Since y + z = 503k = 503 · 2, this leads to 5yz = 5032
· 22
− 4. However
5032
· 22
− 4 is not a multiple of 5. Therefore (1) has no integer solutions.
Equation (2) implies 0 ≤ (k − 1)yz ≤ 1, so that k = 1 or k = 2. Also 0 ≤ k(y − z)2
≤ 1,
hence k = 2 only if y = z. However then y = z = 1, which is false in view of y + z ≥ 503.
Therefore k = 1 and (2) takes the form (y − z)2
= 1, yielding z − y = |y − z| = 1. Combined
with k = 1 and y + z = 503k, this leads to y = 251, z = 252.
In summary the triple (2, 251, 252) is the only solution.

44
N3. Determine all integers m ≥ 2 such that every n with m
3
≤ n ≤ m
2
divides the binomial
coefficient n
m−2n
.
Solution. The integers in question are all prime numbers.
First we check that all primes satisfy the condition, and even a stronger one. Namely, if p
is a prime then every n with 1 ≤ n ≤ p
2
divides n
p−2n
. This is true for p = 2 where n = 1 is
the only possibility. For an odd prime p take n ∈ [1, p
2
] and consider the following identity of
binomial coefficients:
(p − 2n) ·
n
p − 2n
= n ·
n − 1
p − 2n − 1
.
Since p ≥ 2n and p is odd, all factors are non-zero. If d = gcd(p − 2n, n) then d divides p,
but d ≤ n < p and hence d = 1. It follows that p − 2n and n are relatively prime, and so the
factor n in the right-hand side divides the binomial coefficient n
p−2n
.
Next we show that no composite number m has the stated property. Consider two cases.
• If m = 2k with k > 1, pick n = k. Then m
3
≤ n ≤ m
2
but n
m−2n
= k
0
= 1 is not divisible
by k > 1.
• If m is odd then there exist an odd prime p and an integer k ≥ 1 with m = p(2k + 1).
Pick n = pk, then m
3
≤ n ≤ m
2
by k ≥ 1. However
1
n
n
m − 2n
=
1
pk
pk
p
=
(pk − 1)(pk − 2) · · ·(pk − (p − 1))
p!
is not an integer, because p divides the denominator but not the numerator.

45
N4. An integer a is called friendly if the equation (m2
+ n)(n2
+ m) = a(m − n)3
has a
solution over the positive integers.
a) Prove that there are at least 500 friendly integers in the set {1, 2, . . ., 2012}.
b) Decide whether a = 2 is friendly.
Solution. a) Every a of the form a = 4k − 3 with k ≥ 2 is friendly. Indeed the numbers
m = 2k − 1 > 0 and n = k − 1 > 0 satisfy the given equation with a = 4k − 3:
(m2
+ n)(n2
+ m) = (2k − 1)2
+ (k − 1) (k − 1)2
+ (2k − 1) = (4k − 3)k3
= a(m − n)3
.
Hence 5, 9, . . . , 2009 are friendly and so {1, 2, . . . , 2012} contains at least 502 friendly numbers.
b) We show that a = 2 is not friendly. Consider the equation with a = 2 and rewrite its
left-hand side as a diﬀerence of squares:
1
4
(m2
+ n + n2
+ m)2
− (m2
+ n − n2
− m)2
= 2(m − n)3
.
Since m2
+ n − n2
− m = (m − n)(m + n − 1), we can further reformulate the equation as
(m2
+ n + n2
+ m)2
= (m − n)2
8(m − n) + (m + n − 1)2
.
It follows that 8(m − n) + (m + n − 1)2
is a perfect square. Clearly m > n, hence there is an
integer s ≥ 1 such that
(m + n − 1 + 2s)2
= 8(m − n) + (m + n − 1)2
.
Subtracting the squares gives s(m + n − 1 + s) = 2(m − n). Since m + n − 1 + s > m − n,
we conclude that s < 2. Therefore the only possibility is s = 1 and m = 3n. However then
the left-hand side of the given equation (with a = 2) is greater than m3
= 27n3
, whereas its
right-hand side equals 16n3
. The contradiction proves that a = 2 is not friendly.
Comment. A computer search shows that there are 561 friendly numbers in {1, 2, . . . , 2012}.

46
N5. For a nonnegative integer n define rad(n) = 1 if n = 0 or n = 1, and rad(n) = p1p2 · · · pk
where p1 < p2 < · · · < pk are all prime factors of n. Find all polynomials f(x) with nonnegative
integer coefficients such that rad(f(n)) divides rad(f(nrad(n)
)) for every nonnegative integer n.
Solution 1. We are going to prove that f(x) = axm
for some nonnegative integers a and
m. If f(x) is the zero polynomial we are done, so assume that f(x) has at least one positive
coefficient. In particular f(1) > 0.
Let p be a prime number. The condition is that f(n) ≡ 0 (mod p) implies
f(nrad(n)
) ≡ 0 (mod p). (1)
Since rad(nrad(n)k
) = rad(n) for all k, repeated applications of the preceding implication show
that if p divides f(n) then
f(nrad(n)k
) ≡ 0 (mod p) for all k.
The idea is to construct a prime p and a positive integer n such that p − 1 divides n and p
divides f(n). In this case, for k large enough p − 1 divides rad(n)k
. Hence if (p, n) = 1 then
nrad(n)k
≡ 1 (mod p) by Fermat’s little theorem, so that
f(1) ≡ f(nrad(n)k
) ≡ 0 (mod p). (2)
Suppose that f(x) = g(x)xm
with g(0) = 0. Let t be a positive integer, p any prime factor
of g(−t) and n = (p−1)t. So p−1 divides n and f(n) = f((p − 1)t) ≡ f(−t) ≡ 0 (mod p), hence
either (p, n) > 1 or (2) holds. If (p, (p−1)t) > 1 then p divides t and g(0) ≡ g(−t) ≡ 0 (mod p),
meaning that p divides g(0).
In conclusion we proved that each prime factor of g(−t) divides g(0)f(1) = 0, and thus the
set of prime factors of g(−t) when t ranges through the positive integers is finite. This is known
to imply that g(x) is a constant polynomial, and so f(x) = axm
.
Solution 2. Let f(x) be a polynomial with integer coefficients (not necessarily nonnegative)
such that rad(f(n)) divides rad(f(nrad(n)
)) for any nonnegative integer n. We give a complete
description of all polynomials with this property. More precisely, we claim that if f(x) is such
a polynomial and ξ is a root of f(x) then so is ξd
for every positive integer d.
Therefore each root of f(x) is zero or a root of unity. In particular, if a root of unity ξ is
a root of f(x) then 1 = ξd
is a root too (for some positive integer d). In the original problem
f(x) has nonnegative coefficients. Then either f(x) is the zero polynomial or f(1) > 0 and
ξ = 0 is the only possible root. In either case f(x) = axm
with a and m nonnegative integers.
To prove the claim let ξ be a root of f(x), and let g(x) be an irreducible factor of f(x) such
that g(ξ) = 0. If 0 or 1 are roots of g(x) then either ξ = 0 or ξ = 1 (because g(x) is irreducible)
and we are done. So assume that g(0), g(1) = 0. By decomposing d as a product of prime
numbers, it is enough to consider the case d = p prime. We argue for p = 2. Since rad(2k
) = 2
for every k, we have
rad(f(2k
)) | rad(f(22k
)).
Now we prove that g(x) divides f(x2
). Suppose that this is not the case. Then, since g(x)
is irreducible, there are integer-coefficient polynomials a(x), b(x) and an integer N such that
a(x)g(x) + b(x)f(x2
) = N. (3)
Each prime factor p of g(2k
) divides f(2k
), so by rad(f(2k
))|rad(f(22k
)) it also divides f(22k
).
From the equation above with x = 2k
it follows that p divides N.

47
In summary, each prime divisor of g(2k
) divides N, for all k ≥ 0. Let p1, . . . , pn be the odd
primes dividing N, and suppose that
g(1) = 2α
pα1
1 · · · pαn
n .
If k is divisible by ϕ(pα1+1
1 · · · pαn+1
n ) then
2k
≡ 1 (mod pα1+1
1 · · · pαn+1
n ),
yielding
g(2k
) ≡ g(1) (mod pα1+1
1 · · · pαn+1
n ).
It follows that for each i the maximal power of pi dividing g(2k
) and g(1) is the same, namely pαi
i .
On the other hand, for large enough k, the maximal power of 2 dividing g(2k
) and g(0) = 0
is the same. From the above, for k divisible by ϕ(pα1+1
1 · · · pαn+1
n ) and large enough, we obtain
that g(2k
) divides g(0) · g(1). This is impossible because g(0), g(1) = 0 are ﬁxed and g(2k
) is
arbitrarily large.
In conclusion, g(x) divides f(x2
). Recall that ξ is a root of f(x) such that g(ξ) = 0; then
f(ξ2
) = 0, i. e. ξ2
is a root of f(x).
Likewise if ξ is a root of f(x) and p an arbitrary prime then ξp
is a root too. The argument
is completely analogous, in the proof above just replace 2 by p and “odd prime” by “prime
diﬀerent from p.”
Comment. The claim in the second solution can be proved by varying n (mod p) in (1). For instance,
we obtain
f(nrad(n+pk)
) ≡ 0 (mod p)
for every positive integer k. One can prove that if (n, p) = 1 then rad(n+pk) runs through all residue
classes r (mod p − 1) with (r, p − 1) squarefree. Hence if f(n) ≡ 0 (mod p) then f(nr) ≡ 0 (mod p)
for all integers r. This implies the claim by an argument leading to the identity (3).

48
N6. Let x and y be positive integers. If x2n
− 1 is divisible by 2n
y + 1 for every positive
integer n, prove that x = 1.
Solution. First we prove the following fact: For every positive integer y there exist infinitely
many primes p ≡ 3 (mod 4) such that p divides some number of the form 2n
y + 1.
Clearly it is enough to consider the case y odd. Let
2y + 1 = pe1
1 · · · per
r
be the prime factorization of 2y + 1. Suppose on the contrary that there are finitely many
primes pr+1, . . . , pr+s ≡ 3 (mod 4) that divide some number of the form 2n
y + 1 but do not
divide 2y + 1.
We want to find an n such that pei
i ||2n
y+1 for 1 ≤ i ≤ r and pi ∤ 2n
y+1 for r+1 ≤ i ≤ r+s.
For this it suffices to take
n = 1 + ϕ(pe1+1
1 · · · per+1
r p1
r+1 · · · p1
r+s),
because then
2n
y + 1 ≡ 2y + 1 (mod pe1+1
1 · · · per+1
r p1
r+1 · · · p1
r+s).
The last congruence means that pe1
1 , . . . , per
r divide exactly 2n
y + 1 and no prime pr+1, . . . , pr+s
divides 2n
y + 1. It follows that the prime factorization of 2n
y + 1 consists of the prime powers
pe1
1 , . . . , per
r and powers of primes ≡ 1 (mod 4). Because y is odd, we obtain
2n
y + 1 ≡ pe1
1 · · · per
r ≡ 2y + 1 ≡ 3 (mod 4).
This is a contradiction since n > 1, and so 2n
y + 1 ≡ 1 (mod 4).
Now we proceed to the problem. If p is a prime divisor of 2n
y + 1 the problem statement
implies that xd
≡ 1 (mod p) for d = 2n
. By Fermat’s little theorem the same congruence
holds for d = p − 1, so it must also hold for d = (2n
, p − 1). For p ≡ 3 (mod 4) we have
(2n
, p − 1) = 2, therefore in this case x2
≡ 1 (mod p).
In summary, we proved that every prime p ≡ 3 (mod 4) that divides some number of the
form 2n
y + 1 also divides x2
− 1. This is possible only if x = 1, otherwise by the above x2
− 1
would be a positive integer with infinitely many prime factors.
Comment. For each x and each odd prime p the maximal power of p dividing x2n
− 1 for some n is
bounded and hence the same must be true for the numbers 2ny + 1. We infer that p2 divides 2p−1 − 1
for each prime divisor p of 2ny +1. However trying to reach a contradiction with this conclusion alone
seems hopeless, since it is not even known if there are infinitely many primes p without this property.

49
N7. Find all n ∈ N for which there exist nonnegative integers a1, a2, . . . , an such that
1
2a1
+
1
2a2
+ · · · +
1
2an
=
1
3a1
+
2
3a2
+ · · · +
n
3an
= 1.
Solution. Such numbers a1, a2, . . . , an exist if and only if n ≡ 1 (mod 4) or n ≡ 2 (mod 4).
Let n
k=1
k
3ak
= 1 with a1, a2, . . . , an nonnegative integers. Then 1·x1+2·x2+· · ·+n·xn = 3a
with x1, . . . , xn powers of 3 and a ≥ 0. The right-hand side is odd, and the left-hand side has
the same parity as 1+2+· · ·+n. Hence the latter sum is odd, which implies n ≡ 1, 2 (mod 4).
Now we prove the converse.
Call feasible a sequence b1, b2, . . . , bn if there are nonnegative integers a1, a2, . . . , an such that
1
2a1
+
1
2a2
+ · · · +
1
2an
=
b1
3a1
+
b2
3a2
+ · · · +
bn
3an
= 1.
Let bk be a term of a feasible sequence b1, b2, . . . , bn with exponents a1, a2, . . . , an like above,
and let u, v be nonnegative integers with sum 3bk. Observe that
1
2ak+1
+
1
2ak+1
=
1
2ak
and
u
3ak+1
+
v
3ak+1
=
bk
3ak
.
It follows that the sequence b1, . . . , bk−1, u, v, bk+1, . . . , bn is feasible. The exponents ai are the
same for the unchanged terms bi, i = k; the new terms u, v have exponents ak + 1.
We state the conclusion in reverse. If two terms u, v of a sequence are replaced by one
term u+v
3
and the obtained sequence is feasible, then the original sequence is feasible too.
Denote by αn the sequence 1, 2, . . . , n. To show that αn is feasible for n ≡ 1, 2 (mod 4), we
transform it by n − 1 replacements {u, v} → u+v
3
to the one-term sequence α1. The latter is
feasible, with a1 = 0. Note that if m and 2m are terms of a sequence then {m, 2m} → m, so
2m can be ignored if necessary.
Let n ≥ 16. We prove that αn can be reduced to αn−12 by 12 operations. Write n = 12k + r
where k ≥ 1 and 0 ≤ r ≤ 11. If 0 ≤ r ≤ 5 then the last 12 terms of αn can be partitioned into
2 singletons {12k − 6}, {12k} and the following 5 pairs:
{12k − 6 − i, 12k − 6 + i}, i = 1, . . . , 5 − r; {12k − j, 12k + j}, j = 1, . . . , r.
(There is only one kind of pairs if r ∈ {0, 5}.) One can ignore 12k − 6 and 12k since αn
contains 6k − 3 and 6k. Furthermore the 5 operations {12k − 6 − i, 12k − 6 + i} → 8k − 4 and
{12k − j, 12k + j} → 8k remove the 10 terms in the pairs and bring in 5 new terms equal
to 8k − 4 or 8k. All of these can be ignored too as 4k − 2 and 4k are still present in the
sequence. Indeed 4k ≤ n − 12 is equivalent to 8k ≥ 12 − r, which is true for r ∈ {4, 5}. And if
r ∈ {0, 1, 2, 3} then n ≥ 16 implies k ≥ 2, so 8k ≥ 12 − r also holds. Thus αn reduces to αn−12.
The case 6 ≤ r ≤ 11 is analogous. Consider the singletons {12k}, {12k + 6} and the 5 pairs
{12k − i, 12k + i}, i = 1, . . . , 11 − r; {12k + 6 − j, 12k + 6 + j}, j = 1, . . . , r − 6.
Ignore the singletons like before, then remove the pairs via operations {12k − i, 12k + i} → 8k
and {12k + 6 − j, 12k + 6 + j} → 8k + 4. The 5 newly-appeared terms 8k and 8k + 4 can be
ignored too since 4k + 2 ≤ n − 12 (this follows from k ≥ 1 and r ≥ 6). We obtain αn−12 again.
The problem reduces to 2 ≤ n ≤ 15. In fact n ∈ {2, 5, 6, 9, 10, 13, 14} by n ≡ 1, 2 (mod 4).
The cases n = 2, 6, 10, 14 reduce to n = 1, 5, 9, 13 respectively because the last even term of αn
can be ignored. For n = 5 apply {4, 5} → 3, then {3, 3} → 2, then ignore the 2 occurrences
of 2. For n = 9 ignore 6 ﬁrst, then apply {5, 7} → 4, {4, 8} → 4, {3, 9} → 4. Now ignore
the 3 occurrences of 4, then ignore 2. Finally n = 13 reduces to n = 10 by {11, 13} → 8 and
ignoring 8 and 12. The proof is complete.

50
N8. Prove that for every prime p > 100 and every integer r there exist two integers a and b
such that p divides a2
+ b5
− r.
Solution 1. Throughout the solution, all congruence relations are meant modulo p.
Fix p, and let P = {0, 1, . . . , p − 1} be the set of residue classes modulo p. For every r ∈ P,
let Sr = (a, b) ∈ P × P : a2
+ b5
≡ r , and let sr = |Sr|. Our aim is to prove sr > 0 for
all r ∈ P.
We will use the well-known fact that for every residue class r ∈ P and every positive
integer k, there are at most k values x ∈ P such that xk
≡ r.
Lemma. Let N be the number of quadruples (a, b, c, d) ∈ P4
for which a2
+ b5
≡ c2
+ d5
. Then
N =
r∈P
s2
r (a)
and
N ≤ p(p2
+ 4p − 4). (b)
Proof. (a) For each residue class r there exist exactly sr pairs (a, b) with a2
+ b5
≡ r and sr
pairs (c, d) with c2
+ d5
≡ r. So there are s2
r quadruples with a2
+ b5
≡ c2
+ d5
≡ r. Taking the
sum over all r ∈ P, the statement follows.
(b) Choose an arbitrary pair (b, d) ∈ P and look for the possible values of a, c.
1. Suppose that b5
≡ d5
, and let k be the number of such pairs (b, d). The value b can be
chosen in p diﬀerent ways. For b ≡ 0 only d = 0 has this property; for the nonzero values of b
there are at most 5 possible values for d. So we have k ≤ 1 + 5(p − 1) = 5p − 4.
The values a and c must satisfy a2
≡ c2
, so a ≡ ±c, and there are exactly 2p − 1 such
pairs (a, c).
2. Now suppose b5
≡ d5
. In this case a and c must be distinct. By (a − c)(a + c) = d5
− b5
,
the value of a − c uniquely determines a + c and thus a and c as well. Hence, there are p − 1
suitable pairs (a, c).
Thus, for each of the k pairs (b, d) with b5
≡ d5
there are 2p − 1 pairs (a, c), and for each of
the other p2
− k pairs (b, d) there are p − 1 pairs (a, c). Hence,
N = k(2p − 1) + (p2
− k)(p − 1) = p2
(p − 1) + kp ≤ p2
(p − 1) + (5p − 4)p = p(p2
+ 4p − 4).
To prove the statement of the problem, suppose that Sr = ∅ for some r ∈ P; obviously
r ≡ 0. Let T = x10
: x ∈ P {0} be the set of nonzero 10th powers modulo p. Since each
residue class is the 10th power of at most 10 elements in P, we have |T| ≥ p−1
10
≥ 4 by p > 100.
For every t ∈ T, we have Str = ∅. Indeed, if (x, y) ∈ Str and t ≡ z10
then
(z−5
x)2
+ (z−2
y)5
≡ t−1
(x2
+ y5
) ≡ r,
so (z−5
x, z−2
y) ∈ Sr. So, there are at least p−1
10
≥ 4 empty sets among S1, . . . , Sp−1, and there
are at most p − 4 nonzero values among s0, s2, . . . , sp−1. Then by the AM-QM inequality we
obtain
N =
r∈PrT
s2
r ≥
1
p − 4


r∈PrT
sr


2
=
|P × P|2
p − 4
=
p4
p − 4
> p(p2
+ 4p − 4),
which is impossible by the lemma.

51
Solution 2. If 5 ∤ p − 1, then all modulo p residue classes are complete ﬁfth powers and the
statement is trivial. So assume that p = 10k + 1 where k ≥ 10. Let g be a primitive root
modulo p.
We will use the following facts:
(F1) If some residue class x is not quadratic then x(p−1)/2
≡ −1 (mod p).
(F2) For every integer d, as a simple corollary of the summation formula for geometric pro-
gressions,
2k−1
i=0
g5di
≡
2k if 2k d
0 if 2k | d
(mod p).
Suppose that, contrary to the statement, some modulo p residue class r cannot be expressed
as a2
+b5
. Of course r ≡ 0 (mod p). By (F1) we have (r −b5
)(p−1)/2
= (r −b5
)5k
≡ −1 (mod p)
for all residue classes b.
For t = 1, 2 . . . , k − 1 consider the sums
S(t) =
2k−1
i=0
r − g5i 5k
g5ti
.
By the indirect assumption and (F2),
S(t) =
2k−1
i=0
r − (gi
)5 5k
g5ti
≡
2k−1
i=0
(−1)g5ti
≡ −
2k−1
i=0
g5ti
≡ 0 (mod p)
because 2k cannot divide t.
On the other hand, by the binomial theorem,
S(t) =
2k−1
i=0
5k
j=0
5k
j
r5k−j
− g5i j
g5ti
=
5k
j=0
(−1)j 5k
j
r5k−j
2k−1
i=0
g5(j+t)i
≡
≡
5k
j=0
(−1)j 5k
j
r5k−j 2k if 2k j + t
0 if 2k | j + t
(mod p).
Since 1 ≤ j + t < 6k, the number 2k divides j + t only for j = 2k − t and j = 4k − t. Hence,
0 ≡ S(t) ≡ (−1)t 5k
2k − t
r3k+t
+
5k
4k − t
rk+t
· 2k (mod p),
5k
2k − t
r2k
+
5k
4k − t
≡ 0 (mod p).
Taking this for t = 1, 2 and eliminating r, we get
0 ≡
5k
2k − 2
5k
2k − 1
r2k
+
5k
4k − 1
−
5k
2k − 1
5k
2k − 2
r2k
+
5k
4k − 2
=
5k
2k − 2
5k
4k − 1
−
5k
2k − 1
5k
4k − 2
=
(5k)!2
(2k − 1)!(3k + 2)!(4k − 1)!(k + 2)!
(2k − 1)(k + 2) − (3k + 2)(4k − 1)
=
−(5k)!2
· 2k(5k + 1)
(2k − 1)!(3k + 2)!(4k − 1)!(k + 2)!
(mod p).
But in the last expression none of the numbers is divisible by p = 10k + 1, a contradiction.

52
Comment 1. The argument in the second solution is valid whenever k ≥ 3, that is for all primes
p = 10k + 1 except p = 11. This is an exceptional case when the statement is not true; r = 7 cannot
be expressed as desired.
Comment 2. The statement is true in a more general setting: for every positive integer n, for all
suﬃciently large p, each residue class modulo p can be expressed as a2 + bn. Choosing t = 3 would
allow using the Cauchy-Davenport theorem (together with some analysis on the case of equality).
In the literature more general results are known. For instance, the statement easily follows from
the Hasse-Weil bound.

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