Let $ABC$ be a triangle and $J$ the center of the $A$-excircle. This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G$. Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC$. Prove that $M$ is the midpoint of $\overline{ST}$.
The liar's guessing game is a game played between two players $A$ and $B$. The rules of the game depend on two fixed positive integers $k$ and $n$ which are known to both players.
At the start of the game $A$ chooses integers $x$ and $N$ with $1 \le x \le N$. Player $A$ keeps $x$ secret, and truthfully tells $N$ to player $B$. Player $B$ now tries to obtain information about $x$ by asking player $A$ questions as follows: each question consists of $B$ specifying an arbitrary set $S$ of positive integers (possibly one specified in some previous question), and asking $A$ whether $x$ belongs to $S$. Player $B$ may ask as many questions as he wishes. After each question, player $A$ must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any $k+1$ consecutive answers, at least one answer must be truthful.
After $B$ has asked as many questions as he wants, he must specify a set $X$ of at most $n$ positive integers. If $x$ belongs to $X$, then $B$ wins; otherwise, he loses. Prove that:
(a) If $n \ge 2^k$, then $B$ can guarantee a win.
(b) For all sufficiently large $k$, there exists an integer $n \ge (1.99)^k$ such that $B$ cannot guarantee a win.
Find all functions $f \colon \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a$, $b$, $c$ that satisfy $a+b+c=0$, the following equality holds: $$
f(a)^2+f(b)^2+f(c)^2 = 2f(a)f(b)+2f(b)f(c)+2f(c)f(a).
$$
Let $ABC$ be a triangle with $\angle BCA = 90^\circ$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. Let $M = \overline{AL} \cap \overline{BK}$. Prove that $MK = ML$.