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Harvard-MIT November Tournament

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with $AB=13$ , $BC=14$ , and $CA=15$ . Let $D$ be the foot of the altitude from $A$ to $BC$ . The inscribed circles of triangles $ABD$ and $ACD$ are tangent to $AD$ at $P$ and $Q$ , respectively, and are tangent to $BC$ at $X$ and $Y$ , respectively. Let $PX$ and $QY$ meet at $Z$ . Determine the area of triangle $XYZ$ .

Solution

Solution:

Answer: $\frac{25}{4}$

First, note that $AD=12$ , $BD=5$ , $CD=9$ .

By equal tangents, we get that $PD=DX$ , so $PDX$ is isosceles. Because $D$ is a right angle, we get that $\angle PXD=45^{\circ}$ . Similarly, $\angle XYZ=45^{\circ}$ , so $XYZ$ is an isosceles right triangle with hypotenuse $XY$ .

However, by tangents to the incircle, we get that $XD=\frac{1}{2}(12+5-13)=2$ and $YD=\frac{1}{2}(12+9-15)=3$ .

Hence, the area of $XYZ$ is $\frac{1}{4}(XY)^{2}=\frac{1}{4}(2+3)^{2}=\frac{25}{4}$ .

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