Problem 1. Let Γ be the circumcircle of acute-angled triangle ABC. Points D and E lie on
segments AB and AC, respectively, such that AD = AE. The perpendicular bisectors of BD and
CE intersect the minor arcs AB and AC of Γ at points F and G, respectively. Prove that the lines DE and FG are parallel (or are the same line).
Problem 2. Find all integers n ≥ 3 for which there exist real numbers a1, a2, …, an+2, such that an+1 = a1 and an+2 = a2, and
aiai+1 + 1 = ai+2
for i = 1, 2, …, n.
Problem 3. An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following array is an anti-Pascal triangle with four rows which contains every integer from 1 to 10.
5 7 1
8 3 1 9
Does there exist an anti-Pascal triangle with 2018 rows which contains every integer from 1 to 1 + 2 + · · · + 2018?
Problem 4. A site is any point (x, y) in the plane such that x and y are both positive integers less than or equal to 20.
Initially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance between any two sites occupied by red stones is not equal to p5. On his turn, Ben places a new blue stone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from any other occupied site.) They stop as soon as a player cannot place a stone.
Find the greatest K such that Amy can ensure that she places at least K red stones, no matter how Ben places his blue stones.
Problem 5. Let a1, a2, … … be an infinite sequence of positive integers. Suppose that there is an integer N > 1 such that, for each n ≥ N, the number is an integer.
Prove that there is a positive integer M such that am = am+1 for all m ≥ M.
Problem 6. A convex quadrilateral ABCD satisfies AB · CD = BC · DA. Point X lies inside ABCD so that ∠XAB = ∠XCD and ∠XBC = ∠XDA:
Prove that ∠BXA + ∠DXC = 1800.