Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively, such that $AD = AE$. The perpendicular bisectors of $\overline{BD}$ and $\overline{CE}$ intersect the minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$, respectively. Prove that the lines $DE$ and $FG$ are parallel.
Find all integers $n \geq 3$ for which there exist real numbers $a_1, a_2, \dots, a_n$ satisfying $$
a_i a_{i+1} +1 = a_{i+2}
$$ for $i=1,2, \dots, n$, where indices are taken modulo $n$.
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 $10$.
Does there exist an anti-Pascal triangle with $2018$ rows which contains every integer from $1$ to $1+2+\dots +2018$?
A *site* is any point $(x,y)$ in the plane for which $x,y \in \{1, \dots, 20\}$. Initially, each of the $400$ sites is unoccupied. Amy and Ben take turns placing stones on unoccupied sites, with Amy going first; Amy has the additional restriction that no two of her stones may be at a distance equal to $\sqrt5$. They stop once either player cannot move. Find the greatest $K$ such that Amy can ensure that she places at least $K$ stones. Amy places red stones and Ben places blue stones; Ben has no restriction on distances between occupied sites. Amy must guarantee the stated number regardless of Ben's moves.
Let $a_1$, $a_2$, … be an infinite sequence of positive integers, and $N\gt1$ an integer. Suppose that for all integers $n \ge N$, the expression $$
\frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots
+ \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}
$$ is an integer. Prove that $(a_n)$ is eventually constant.
A convex quadrilateral $ABCD$ satisfies $AB \cdot CD = BC \cdot DA$. Point $X$ lies inside $ABCD$ so that $$
\angle XAB=\angle XCD \quad \text{ and } \quad \angle XBC=\angle XDA.
$$ Prove that $\angle BXA + \angle DXC=180^\circ$.
