## IMO 2018 Problem 1

Let Γ be the circumcircle of acute triangle ABC. Point D and E are on segments AB and AC respectively such that AD = AE. The perpendicular bisectors of BD and CE intersect minor arcs AB and AC of Γ at points F and G respectively. Prove that lines DE and FG are either parallel or they are the same line.

## IMO 2018 Problem 2

Find all integers n ≥ 3 for which there exists real numbers a1, a2, a3, …,an+2, satisfying an+1 = a1, an+2 = a2 and aiai+1, = ai+2, for i = 1, 2, 3, …,n.

## IMO 2018 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 absolute value of the difference of the two numbers immediately below it. For example, the following is an anti Pascal triangle with four rows which contains every integer from 1 to 10.

Does there exist an anti Pascal triangle with 2018 rows which contains every integer from 1 to 1 + 2 + 3 + … + 2018?

## IMO 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 an occupied. 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 root 5. 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 and ensure that she places at least K red stones, no matter how Ben places his blue stones.

## IMO 2018 Problem 5

Let a1, a2, a3, … be an infinite sequence of positive integers. Suppose that there is an integer N > 1 such that, for each n ≥ N the number (a1/a2) + (a2/a3) + … + (an-1/an) + (an/a1) is an integer. Prove that there is a positive integer M such that am = am+1 for all m ≥ M.

## IMO 2018 Problem 6

A convex quadrilateral ABCD satisfies AB⋅CD = BC⋅DA. Point X lies inside ABCD show that angle(XAB) = angle (XCD) and angle(XBC) = angle (XDA). Prove that angle(BXA) + angle (DXC) = 180°.