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Problem 1065

AMC 12 late, AIME early

Geometry Difficulty 5.0 Prove it HMMT November · United States · 2014

Let $ABC$ be a triangle with $AB = AC = 5$ and $BC = 6$ . Denote by $\omega$ the circumcircle of $ABC$ . We draw a circle $\Omega$ which is externally tangent to $\omega$ as well as to the lines $AB$ and $AC$ (such a circle is called an $A$ -mixtilinear excircle). Find the radius of $\Omega$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Let $M$ be the midpoint of $BC$ . Let $D$ be the point diametrically opposite $A$ on the circumcircle, and let the $A$ -mixtilinear excircle be tangent to lines $AB$ and $AC$ at $X$ and $Y$ . Let $O$ be the center of the $A$ -mixtilinear excircle.

Notice that $\triangle AOX \sim \triangle ABM$ . If we let $x$ be the desired radius, we have
$\frac{x + AD}{x} = \frac{5}{3}.$
We can compute $\frac{AD}{5} = \frac{5}{4}$ since $\triangle ADB \sim \triangle ABM$ , we derive $AD = \frac{25}{4}$ . From here it follows that $x = \frac{75}{8}$ .

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