olymmath

BG-MO 2013

Bulgarian Mathematical Olympiad 2013.

Other problems

Problem 2

Find all f:R→Rf : \mathbb{R}\to \mathbb{R} , bounded in (0,1)(0,1) and satisfying: x2f(x)−y2f(y)=(x2−y2)f(x+y)−xyf(x−y)x^2 f(x) - y^2 f(y) = (x^2-y^2) f(x+y) -xy f(x-y) for all x,y∈Rx,y \in \mathbb{R} Proposed by Nikolay Nikolov

Problem 3

The integer lattice in the plane is colored with 3 colors. Find the least positive real SS with the property: for any such coloring it is possible to find a monochromatic lattice points A,B,CA,B,C with S△ABC=SS_{\triangle ABC}=S .

Proposed by Nikolay Beluhov

EDIT: It was the problem 3 (not 2), corrected the source title.

Problem 4

Suppose α,β,γ∈[0.π/2)\alpha,\beta,\gamma \in [0.\pi/2) and tan⁡α+tan⁡β+tan⁡γ≤3\tan \alpha + \tan\beta + \tan \gamma \leq 3 . Prove that: cos⁡2α+cos⁡2β+cos⁡2γ≥0\cos 2\alpha + \cos 2\beta + \cos 2\gamma \ge 0

Proposed by Nikolay Nikolov

Problem 5

Consider acute △ABC\triangle ABC with altitudes AA1,BB1AA_1, BB_1 and CC1CC_1 ( A1∈BC,B1∈AC,C1∈ABA_1 \in BC,B_1 \in AC,C_1 \in AB ). A point C′C' on the extension of B1A1B_1A_1 beyond A1A_1 is such that A1C′=B1C1A_1C' = B_1C_1 . Analogously, a point B′B' on the extension of A 1C1_1C_1 beyond C1C_1 is such that C1B′=A1B1C_1B' = A_1B_1 and a point A′A' on the extension of C1B1C_1B_1 beyond B1B_1 is such that B1A′=C1A1B_1A' = C_1A_1 . Denote by A′′,B′′,C′′A'', B'', C'' the symmetric points of A′,B′,C′A' , B' , C' with respect to BC,CABC, CA and ABAB respectively. Prove that if R,R′R, R' and R'' are circumradiii of △ABC,△A′B′C′\triangle ABC, \triangle A'B'C' and △A′′B′′C′′\triangle A''B''C'' , then R,R′R, R' and R′′R'' are sidelengths of a triangle with area equals one half of the area of △ABC\triangle ABC .

Problem 6

Given m∈Nm\in\mathbb{N} and a prime number pp , p>mp>m , let M={n∈N∣m2+n2+p2−2mn−2mp−2np  is a perfect square}M=\{n\in\mathbb{N}\mid m^2+n^2+p^2-2mn-2mp-2np \,\, \text{is a perfect square} \} Prove that ∣M∣|M| does not depend on pp .

Proposed by Aleksandar Ivanov