Language: English

Wednesday, May 10, 2023

Problem 1. Find all functions f : R → R such that for all x, y ∈ R,

xf (x + f (y)) = (y − x)f (f (x)).

Solution. Answer: For any real c, f (x) = c − x for all x ∈ R and f (x) = 0 for all
x ∈ R.
Let P (x, y) denote the proposition that x and y satisfy the given equation. P (0, 1)
gives us f (f (0)) = 0.
From P (x, x) we get that xf (x+f (x)) = 0 for all x ∈ R, which together with f (f (0)) =
0 gives us f (x + f (x)) = 0 for all x.
Now let t be any real number such that f (t) = 0. If y is any number, we have from
P (t − f (y), y) the equality

(y + f (y) − t)f (f (t − f (y))) = 0

for all y and all t such that f (t) = 0. So, by taking y = f (0) we obtain

(f (0) − t)f (f (t)) = 0 and hence (f (0) − t)f (0) = 0. (A1-1)

Recall that as t with f (t) = 0 we can take x + f (x) for any real number x. If for all
reals x we have x + f (x) = f (0), then f (x) must be of the form f (x) = c − x for some real
c. It is straightforward that all functions of this form are indeed solutions.
Otherwise we can find some a 6= 0 so that a + f (a) 6= f (0). If t = a + f (a) in (A1-1),
then f (0) must be equal to 0. Now P (x, 0) gives us f (f (x)) = −f (x) for all real x. From
here P (x, x + f (x)) gives us xf (x) = −f (x)2 for all real x, which means for every x either
f (x) = 0 or f (x) = −x.
Let us assume that in this case there is some b 6= 0 so that f (b) = −b. For any y we
get from P (b, y) and f (f (b)) = −f (b) = b the equality bf (b + f (y)) = (y − b)b, which gives
us f (b + f (y)) = y − b for all y. If y 6= b, then the right hand side in the previous equality
is not zero, so we must have f (b + f (y)) = −b − f (y), which means that −b − f (y) = y − b,
or that f (y) = −y for all real y. But we already covered this solution (take c = 0 above).
If there is no such b, then f (x) = 0 for all x, which gives us the final solution.
Thus, all such functions are of the form c − x for real c or the zero function.
Problem 2. In triangle ABC, the incircle touches sides BC, CA, AB at D, E, F respec-
tively. Assume there exists a point X on the line EF such that
∠XBC = ∠XCB = 45◦ .
Let M be the midpoint of the arc BC on the circumcircle of ABC not containing A. Prove
that the line M D passes through E or F .
Solution 1. We first state a well-known lemma.
Lemma: In triangle ABC, let D, E, F be the points of tangency of the incircle to the
sides BC, CA, AB and let I be the incenter. Then the intersection of EF and BI lies on
the circle of diameter BC.
Proof: Let S be the intersection of EF and BI. ∠BIC = ∠EF C, hence S lies on the
circumcircle of F IC in which IC is a diameter. Thus ∠ISC = 90◦ , hence ∠BSC = 90◦ .
Returning to the problem, let I be the incenter. The lemma implies that the two
intersection points of EF with the circle of diameter BC are precisely the intersection
points of EF with BI and CI. We have ∠BXC = 90◦ , therefore either BX or CX is an
internal angle bisector, which means either ∠B = 90◦ or ∠C = 90◦ .
Assume, without loss of generality, that ∠B = 90◦ . Then we have ∠AM C = 90◦ , so
M is the second intersection point of AI with the circle of diameter AC, thus the lemma
implies that M lies on DF .
Solution 2.
Let I be the incenter of 4ABC and let K be the foot of the perpendicular from D to
EF . We begin by proving that BKXC is cyclic, which can be done in two ways:
∠C ∠B
First Way. Note that ∠KF D = 90◦ − and ∠KED = 90◦ − , so by using
2 2
FK tan ∠C 2 ∠B ∠C
KD ⊥ EF , we have = ∠B
. Similarly, since ∠IBD = and ∠ICD = ,
ED tan 2 2 2
BF BD tan ∠C ∠A
by using ID ⊥ BC, we have = = 2
∠B
. Then, since ∠BF K = 90◦ + =
EC DC tan 2 2
∠KEC, we conclude that 4BF K and 4CEK are similar, so ∠F KB = ∠CKE which
shows line EF is the external-angle bisector of ∠BKC. Therefore, X lies on both the
perpendicular bisector of the segment BC and the external angle bisector of ∠BKC (and
these lines are distinct) thus it lies on the circumcirle of 4BKC (in particular the midpoint
of arc BKC).
Second Way. Let T be the intersection of EF and BC, and N be the midpoint of
the segment BC. It is well-known that (T, D; B, C) is harmonic and T B · T C = T D · T N .
On the other hand, since XB = XC, we have ∠XN D = 90◦ = ∠XKD, so XKDN is
cyclic and T D · T N = T K · T X. Therefore, we have T K · T X = T B · T C, which implies
BKXC is cyclic.
Now we will show that either ∠B = 90◦ or ∠C = 90◦ . Note that ∠BKC = ∠BXC =
◦
90 = ∠F KD = ∠EKD and ∠F KB = ∠EKC. Then we have
∠F KB = ∠BKD = ∠DKC = ∠CKE = 45◦ .
Hence BK bisects ∠F KD, but B also lies on the perpendicular bisector of DF . Therefore,
either F KDB is cyclic or KF = KD while the former implies that ∠B = 180◦ −∠F KD =
∠C
90◦ . In the latter case, we have KB ⊥ F D, which gives 90◦ − = ∠KF D = 90◦ −
2
∠F KB = 45◦ and so ∠C = 90◦ as desired.
We consider, without loss of generality, the case where ∠B = 90◦ . Observing that
A, I, M are collinear we get:
∠CDI = 90◦ = ∠CBA = ∠CM A = ∠CM I
Hence M DIC is cyclic so:
∠B
 
∠M DC = ∠M IC = 180◦ − ∠CIA = 180◦ − 90◦ + = 45◦
2
We also have ∠F DB = 90◦ − ∠B
2 = 45◦ so ∠F DB = ∠M DC and thus M, D, F are
collinear as required.
Problem 3. For each positive integer n, denote by ω(n) the number of distinct prime
divisors of n (for example, ω(1) = 0 and ω(12) = 2). Find all polynomials P (x) with
integer coefficients, such that whenever n is a positive integer satisfying ω(n) > 20232023 ,
then P (n) is also a positive integer with

ω(n) ≥ ω(P (n)).

Solution. Answer: All polynomials of the form f (x) = xm for some m ∈ Z+ and
f (x) = c for some c ∈ Z+ with ω(c) ≤ 20232023 + 1.
First of all we prove the following (well-known) Lemma. Lemma: Let f (x) be a non-
constant polynomial with integer coefficients. Then, the number of primes p such that
p|f (n) for some n is infinite.
Proof: If f (0) = 0, then the Lemma is obvious. Otherwise, define the polynomial

f (xf (0))
g(x) = ,
f (0)

which has integer coefficients. Observe that g(0) = 1 and if g satisfies the property of the
Lemma, then so does f . So, we need to prove that there are infinitely many primes p such
that p|g(n) for some n. Suppose, for the sake of contradiction that the number of such
primes is finite, and let those be p1 , ..., pk . Then, set n = N p1 · · · pk for some large N ,
such that |g(n)| > 1. It is evident that g(n) has a prime divisor, but it is none of the pi ’s.
This is a contradiction and therefore the result follows.
Let M = 20232023 + 1. Observe that constant polynomials f (x) = c with c ∈ N such
that ω(c) ≤ M satisfy the conditions of the problem. On the other hand, if f (x) = c
with ω(c) > M , we can choose some n such that ω(n) = M to see that the condition of
the problem is not satisfied. Next, we look for non-constant polynomials that satisfy the
conditions of the problem. Let f (x) = xm g(x), where m ≥ 0 and g(x) is a polynomial
with g(0) 6= 0. We claim that g is a constant polynomial. Indeed, if it is not, then
(due to the Lemma) there exist pairwise distinct primes q1 , ..., qM +1 and non-zero integers
n1 , ..., nM +1 such that qi > |g(0)| and qi |g(ni ) for i = 1, 2, ..., M + 1. Set n = p1 p2 · · · pM ,
where p1 , ..., pM are distinct primes such that

p 1 ≡ ni (mod qi ), ∀i = 1, 2, ..., M + 1

and
pj ≡ 1 (mod qi ), ∀i = 1, 2, ..., M + 1, ∀j = 2, 3, ..., M.
Observe that since qi > |g(0)|, it is impossible to have qi |ni , so the existence of such primes
is guaranteed by the Chinese Remainder Theorem and the Dirichlet’s Theorem. Now, for
every i = 1, 2, ..., M + 1 we can see that n = p1 · · · pM ≡ ni (mod qi ), which means that

g(n) ≡ g(ni ) ≡ 0 (mod qi ) ∀i = 1, 2, ..., M + 1.

Thus, ω(f (n)) ≥ ω(g(n)) ≥ M + 1 > M = ω(n), which gives the desired contradiction.
Therefore, f (x) = cxm , for some m ≥ 1 (since f was non-constant). If c < 0, take some
n with ω(n) = M to see that f (n) is negative and so, does not satisfy the conditions of
the problem. If c > 1, choose some n with ω(n) = M and gcd(n, c) = 1 to observe that
f cannot satisfy the conditions of the problem. This means that f (x) = xm (which is
for sure a solution to the problem) for some m ≥ 1 and f (x) = c for some c ∈ Z+ with
ω(c) ≤ M are the only polynomials that satisfy the conditions of the problem.
Problem 4. Find the greatest integer k ≤ 2023 for which the following holds: whenever
Alice colours exactly k numbers of the set {1, 2, . . . , 2023} in red, Bob can colour some of
the remaining uncoloured numbers in blue, such that the sum of the red numbers is the
same as the sum of the blue numbers.
Solution. Answer: 592.
For k ≥ 593, Alice can color the greatest 593 numbers 1431, 1432, . . . , 2023 and any
other (k − 593) numbers so that their sum s would satisfy
2023 · 2024 1430 · 1431 1 2023 · 2024
 
s≥ − > · ,
2 2 2 2
thus anyhow Bob chooses his numbers, the sum of his numbers will be less than Alice’s
numbers’ sum.
We now show that k = 592 satisfies the condition. Let s be the sum of Alice’s 592
numbers; note that s < 12 · 2023·2024
2 . Below is a strategy for Bob to find some of the
remaining 1431 numbers so that their sum is
2023.2024 1 2023 · 2024
   
s0 = min s, − 2s 6 · ,
2 3 2
(Clearly, if Bob finds some numbers whose sum is 2023.2024
2 − 2s, then the sum of remaining
numbers will be s).
Case 1. s0 ≥ 2024. Let s0 = 2024a + b, where 0 6 b 6 2023. Bob finds two of the
remaining numbers with sum b or 2024 + b, then he finds a (or a −1) pairs among the
remaining numbers with sum 2024. Note that a ≤ 337 since s0 ≤ 13 · 2023·20242 .
j k
b−1
The 2 pairs

b−1 b−1
   
(1, b − 1), (2, b − 2), . . . , ,b − ,
2 2
j k
2023+b
have sum of their components equal to b and the 2 − b pairs

2023 + b 2023 + b
    
(2023, b + 1), (2022, b + 2), . . . , 2024 + b − ,
2 2
have sum of their components equal to 2024 + b. The total number of these pairs is
2023 + b b−1 2022 + b b − 2 2020
   
−b+ ≥ + −b= = 1010 > 592,
2 2 2 2 2
hence some of these pairs have no red-colored components, so Bob can choose one of these
pairs and color those two numbers in blue. Thus 594 numbers are colored so far.
Further, the 1011 pairs
(1, 2023), (2, 2022), . . . , (1011, 1013)
have sum of the components equal to 2024. Among these, at least 1011 − 594 = 417 >
337 ≥ a pairs have no components colored, so Bob can choose a (or a − 1) uncolored pairs
and color them all blue to achieve a collection of blue numbers with their sum equal to s0 .
Case 2. s0 ≤ 2023. Note that s ≥ 1 + 2 + . . . + 592 > 2023, thus we have s0 =
2023·2024
2 − 2s, i.e. s = 2023·2024
4 − s20 .
If s0 > 2 · 593, at least one of the 593 pairs
(1, s0 − 1), (2, s0 − 2), . . . , (593, s0 − 593)
have no red-colored components, so Bob can choose these two numbers and immediately
achieve the sum of s0 . And if s0 ≤ 2 · 593, then
2023 · 2024 s0
s= − ≥ (1432 + 1433 + . . . + 2023) − 593 = 839 + (1434 + 1435 + . . . + 2023),
4 2
hence Alice cannot have colored any of the numbers 1, 2, . . . , 838. Then Bob can easily
choose one or two of these numbers having the sum of s0 .