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, 2024

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

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

Solution

Solution:

Let $D$ be the foot from $A$ to $BC$ , also the midpoint of $BC$ . Note that $DX = DY = MA = MB = MD = NA = NC = ND = 1$ . Thus, $MNXY$ is cyclic with circumcenter $D$ and circumradius $1$ . $H$ lies on this circle too, hence $DH = 1$ .

If we let $DB = DC = x$ , then since $\triangle HBD \sim \triangle BDA$ ,
$BD^2 = HD \cdot AD \Longrightarrow x^2 = \sqrt{4 - x^2} \Longrightarrow x^4 = 4 - x^2 \Longrightarrow x^2 = \frac{\sqrt{17} - 1}{2}$
Our answer is $BC^2 = (2x)^2 = 4x^2 = 2(\sqrt{17} - 1)$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.