MathDB

2017 CentroAmerican

Part of CentroAmerican

Subcontests

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Centroamerican Math Olympiad - 2017 P2

We call a pair (a,b)(a,b) of positive integers, a<391a<391, pupusa if
lcm(a,b)>lcm(a,391)\textup{lcm}(a,b)>\textup{lcm}(a,391)
Find the minimum value of bb across all pupusa pairs.
Fun Fact: OMCC 2017 was held in El Salvador. Pupusa is their national dish. It is a corn tortilla filled with cheese, meat, etc.

Centroamerican Math Olympiad 2017 - P5

Susana and Brenda play a game writing polynomials on the board. Susana starts and they play taking turns.
1) On the preparatory turn (turn 0), Susana choose a positive integer n0n_0 and writes the polynomial P0(x)=n0P_0(x)=n_0.
2) On turn 1, Brenda choose a positive integer n1n_1, different from n0n_0, and either writes the polynomial
P1(x)=n1x+P0(x) or P1(x)=n1x−P0(x)P_1(x)=n_1x+P_0(x) \textup{ or } P_1(x)=n_1x-P_0(x)
3) In general, on turn kk, the respective player chooses an integer nkn_k, different from n0,n1,…,nk−1n_0, n_1, \ldots, n_{k-1}, and either writes the polynomial
Pk(x)=nkxk+Pk−1(x) or Pk(x)=nkxk−Pk−1(x)P_k(x)=n_kx^k+P_{k-1}(x) \textup{ or } P_k(x)=n_kx^k-P_{k-1}(x)
The first player to write a polynomial with at least one whole whole number root wins. Find and describe a winning strategy.
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Centroamerican Math Olympiad - 2017 P1

The figure below shows a hexagonal net formed by many congruent equilateral triangles. Taking turns, Gabriel and Arnaldo play a game as follows. On his turn, the player colors in a segment, including the endpoints, following these three rules:
i) The endpoints must coincide with vertices of the marked equilateral triangles.
ii) The segment must be made up of one or more of the sides of the triangles.
iii) The segment cannot contain any point (endpoints included) of a previously colored segment.
Gabriel plays first, and the player that cannot make a legal move loses. Find a winning strategy and describe it.

Centroamerican Math Olympiad 2017 - P4

ABCABC is a right-angled triangle, with ∠ABC=90∘\angle ABC = 90^{\circ}. B′B' is the reflection of BB over ACAC. MM is the midpoint of ACAC. We choose DD on BM→\overrightarrow{BM}, such that BD=ACBD = AC. Prove that B′CB'C is the angle bisector of ∠MB′D\angle MB'D.
NOTE: An important condition not mentioned in the original problem is AB<BCAB<BC. Otherwise, ∠MB′D\angle MB'D is not defined or B′CB'C is the external bisector.