CMO 2019

Amy has drawn three points in a plane, $A, B$, and $C$, such that $A B=B C=C A=6$. Amy is allowed to draw a new point if it is the circumcenter of a triangle whose vertices she has already drawn. For example, she can draw the circumcenter $O$ of triangle $A B C$, and then afterwards she can draw the circumcenter of triangle $A B O$.

(a) Prove that Amy can eventually draw a point whose distance from a previously drawn point is greater than 7 .

(b) Prove that Amy can eventually draw a point whose distance from a previously drawn point is greater than 2019.

(Recall that the circumcenter of a triangle is the center of the circle that passes through its three vertices.)

Let $a$ and $b$ be positive integers such that $a+b^{3}$ is divisible by $a^{2}+3 a b+3 b^{2}-1$. Prove that $a^{2}+3 a b+3 b^{2}-1$ is divisible by the cube of an integer greater than 1 .

Let $m$ and $n$ be positive integers. A $2 m \times 2 n$ grid of squares is coloured in the usual chessboard fashion. Find the number of ways of placing $m n$ counters on the white squares, at most one counter per square, so that no two counters are on white squares that are diagonally adjacent. An example of a way to place the counters when $m=2$ and $n=3$ is shown below.

Let $n$ be an integer greater than 1 , and let $a_{0}, a_{1}, \ldots, a_{n}$ be real numbers with $a_{1}=a_{n-1}=0$. Prove that for any real number $k$,

$$\left|a_{0}\right|-\left|a_{n}\right| \leq \sum_{i=0}^{n-2}\left|a_{i}-k a_{i+1}-a_{i+2}\right|$$

David and Jacob are playing a game of connecting $n \geq 3$ points drawn in a plane. No three of the points are collinear. On each player's turn, he chooses two points to connect by a new line segment. The first player to complete a cycle consisting of an odd number of line segments loses the game. (Both endpoints of each line segment in the cycle must be among the $n$ given points, not points which arise later as intersections of segments.) Assuming David goes first, determine all $n$ for which he has a winning strategy.