Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Find the answer

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 $B C^{2}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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 $B C^{2}=(2x)^{2}=4x^{2}=2(\sqrt{17}-1)$ .

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