Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Ireland

Let $ABCD$ be a square. The line segment $AB$ is divided internally at $H$ so that $|AB| \cdot |BH| = |AH|^2$ . Let $E$ be the midpoint of $AD$ and $X$ be the midpoint of $AH$ . Let $Y$ be the point on $EB$ such that $XY$ is perpendicular to $BE$ . Prove that $|XY| = |XH|$ .

Solution

Let $ABCD$ have side length $2a$ and write $x = |AH|$ . Then, by assumption
$x^2 = 2a(2a - x).$

Because $|AB| = 2|EA|$ , Pythagoras gives $|BE|^2 = |EA|^2 + |AB|^2 = 5|EA|^2$ .

Observe that $\triangle BXY$ and $\triangle BEA$ are similar. Hence, $\frac{|BE|}{|EA|} = \frac{|BX|}{|XY|}$ and so
$\begin{aligned} 5|XY|^2 &= |BX|^2 = \left(2a - \frac{x}{2}\right)^2 = 4a^2 - 2ax + \frac{x^2}{4} \\ &= 2a(2a - x) + \frac{x^2}{4} = x^2 + \frac{x^2}{4} = 5\left(\frac{x}{2}\right)^2. \end{aligned}$
This implies $|XY| = \frac{x}{2} = |XH|$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.