2025 TEST 1 seniors— Solution to Problems 2, 3 and 4
Problem 2. Let ABC be a scalene acute triangle with incentre I and circumcentre O.
Let AI cross BC at D. On circle ABC, let X and Y be the mid-arc points of ABC and BCA,
respectively. Let DX cross CI at E and let DY cross BI at F . Prove that the lines F X, EY
and IO are concurrent on the external bisector of ∠BAC.
David-Andrei Anghel
Solution. The argument hinges on the claim below:
Claim. The lines AE and BI are perpendicular; similarly, AF and CI are perpendicular
Proof. Let α = ∠BAC, β = ∠CAB and γ = ∠ACB. Let DX cross the circle ADC again
at D′ . Note that ∠ECX = ∠ACX − ∠ACE = 90◦ − β/2 − γ/2 = α/2 = ∠DAC = ∠DD′ C =
∠XD′ C. As ∠CXE = ∠CXD′ , triangles XCD′ and XEC are similar, so XD′ · XE = XC 2 .
As XA = XC, it follows that XD′ · XE = XA2 , so triangles XD′ A and XAE are similar,
so ∠XAE = ∠AD′ X = ∠AD′ D = ∠ACD = γ.
Finally, note that ∠XAD = ∠XAC − ∠DAC = 90◦ − β/2 − α/2 = γ/2, so ∠IAE =
∠DAE = ∠XAE − ∠XAD = γ/2. As ∠AIB = 90◦ + γ/2, the claim follows.
Let W be the mid-arc point of CAB and let I ′ be the reflection of I across O. As W , X, Y
are the mid-arc points of CAB, ABC, BCA, respectively, their reflections across O are the
mid-arc points opposite. These latter form a triangle with orthocentre I, so I ′ is the ortho-
centre of triangle W XY .
Reflection across O maps lines XV , W Y and W X to the perpendicular bisectors of AI, BI
and CI, respectively, so XY ⊥ AI, W Y ⊥ BI and W X ⊥ CI. By the claim, AE ⊥ IF and
AF ⊥ IE, so I is the orthocentre of triangle AEF and hence EF ⊥ AI as well.
Triangles AEF and W XY have therefore corresponding parallel sides, so they are homoth-
etic from some point R. This homothety maps I to I ′ , as they are corresponding orthocentres.
Hence the lines AW , EY , F Y and II ′ are concurrent at R. As I, O and I ′ are collinear and AW
is the external bisector of ∠BAC, the conclusion follows.
Problem 3. Determine all polynomials P with integer coefficients, satisfying 0 ≤ P (n) ≤ n!
for all non-negative integers n.
Andrei Chiriţă
Solution. The required polynomials are P = 0, P = 1, P = (X −1)2 , P = X(X −1) · · · (X −k)
and P = X(X − 1) · · · (X − k)(X − k − 2)2 for some non-negative integer k. The verification
is routine and is hence omitted.
Let P be a polynomial satisfying the condition in the statement. Clearly, P (0) = 0 or
P (0) = 1.
We first deal with the case P (0) = 1. The polynomials P1 = 1 and P2 = (X − 1)2 both
satisfy the condition in the statement and P1 (0) = P2 (0) = 1.
We will prove that either P = P1 or P = P2 . Consider an index i such that P (1) = Pi (1)
and let P̃ = P − Pi .
Induct on n to show that P̃ (n) = 0 for all non-negative integers n. The base cases n = 0
and n = 1 are clear. For the inductive step, assume P̃ (m) = 0 for all non-negative integers
m < n. Then X(X − 1) · · · X − (n − 1) divides P̃ , so n! divides P̃ (n). As 0 < Pi (n) < n!, it
follows that |P̃ (n)| = |P (n) − Pi (n)| < n!, so P̃ (n) = 0.
Consequently, P̃ has infinitely many roots, so it vanishes identically; that is, P = Pi , as
desired.
Finally, we deal with the case P (0) = 0. Assume P is non-zero. Let P (X) = XQ(X − 1),
where Q has integer coefficients. Then 0 ≤ Q(n) ≤ n! for all non-negative integers n. If
Q(0) = 0, repeat the argument for Q and so on and so forth, all the way down to some
polynomial with a non-zero constant term — this is clearly the case, as P is non-zero and
degrees strictly decrease in the process. By the preceding, such a polynomial is either 1 or
(X − 1)2 . An obvious induction then shows that P has one of the last two forms mentioned
in the beginning.
Problem 4. Determine the sets S of positive integers satisfying the following two conditions:
(a) For any positive integers a, b, c, if ab + bc + ca is in S, then so are a + b + c and abc; and
(b) The set S contains an integer N ≥ 160 such that N − 2 is not divisible by 4.
Bogdan Blaga, United Kingdom
Solution. We will prove that S is the set of all positive integers. The argument hinges on
the three facts below:
(1) The set S contains an integer M ≥ 40 divisible by 4.
(2) If 4k belongs to S for some integer k ≥ 2, then so does 4m for all positive integers m < k.
(3) The set S contains 4k for all integers k ≥ 10.
Assume the three for the moment and argue as follows: By (1) and (2), S contains all
positive multiples of 4 at most 40, and by (3) it contains all multiples of 4 at least 40, so S
contains all positive multiples of 4.
Let a be any positive integer and let b = c = 2. By the preceding, S contains 4a + 4, so it
contains a + 4, by (a). Hence S also contains all positive integers congruent to a modulo 4.
Combine with the preceding paragraph to deduce that S contains all integers at least 4.
Let a = 3 and let b = c = 1. As 7 is in S, so is 3, by (a). Repeat the argument for
a = b = c = 1 to deduce that S contains 1.
Finally, let a = 2 and let again b = c = 1. As 5 lies in S, so does 2. Combining with the
previous paragraphs, if follows that S exhausts all positive integers, as stated.
Proof of (1). If N is divisible by 4, choose M = N . If N = 4k + 1, set a = 2k and b = c = 1
in (a) to deduce that 2k and 2k + 2 are both in S. As N ≥ 160, the numbers 2k and 2k + 2
are both at least 80 > 40. Note that exactly one of 2k and 2k + 2 is divisible by 4 and let M
be that number.
If N = 4k + 3, set a = 2k + 1 and b = c = 1 in (a) to deduce that 2k + 1 and 2k + 3 both
lie in S. Next, set a = k or k + 1 and b = c = 1 in (a) to deduce that k, k + 1, k + 2 and k + 3
are all in S. Exactly one of these numbers is divisible by 4. As k ≥ 40, choose M to be that
number.
Proof of (2). Let b = c = 2. By (a), if 4a + 4 is in S, then so is 4a. Beginning with M
provided by (1), statement (2) now follows by backward recursion.
Proof of (3). Let b = c = 4. By (a), if 8a + 16 is in S, then so is 16a. Note that
16a = 8(2a − 2) + 16 with 2a − 2 > a for a ≥ 3.
As 40 ∈ S, then we S contains a strictly increasing subsequence of multiples of 8 (and thus
of multiples of 4.
Reference to (2) concludes the proof and completes the solution.
