HMMT-Feb 2024 guts, Problem 16

Let $A B C$ be an acute isosceles triangle with orthocenter $H$. Let $M$ and $N$ be the midpoints of sides $\overline{A B}$ and $\overline{A C}$, respectively. The circumcircle of triangle $M H N$ intersects line $B C$ at two points $X$ and $Y$. Given $X Y=A B=A C=2$, compute $B C^{2}$.

Solution: \section*{Solution:} Let $D$ be the foot from $A$ to $B C$, also the midpoint of $B C$. Note that $D X=D Y=M A=M B=$ $M D=N A=N C=N D=1$. Thus, $M N X Y$ is cyclic with circumcenter $D$ and circumradius $1 . H$ lies on this circle too, hence $D H=1$. If we let $D B=D C=x$, then since $\triangle H B D \sim \triangle B D A$, \[ B D^{2}=H D \cdot A D \Longrightarrow x^{2}=\sqrt{4-x^{2}} \Longrightarrow x^{4}=4-x^{2} \Longrightarrow x^{2}=\frac{\sqrt{17}-1}{2} \] Our answer is $B C^{2}=(2 x)^{2}=4 x^{2}=2(\sqrt{17}-1)$ $\fbox{2(\sqrt{17}-1)}$.

Answer: 2(\sqrt{17}-1)