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Brazilian Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Brazil

The angles of the triangle $ABC$ satisfy $\frac{\angle A}{\angle C} = \frac{\angle B}{\angle A} = 2$ . The incenter is $O$ . $K$ , $L$ are the excenters of the excircles opposite $B$ and $A$ respectively. Show that triangles $ABC$ and $OKL$ are similar.

Solution

$\angle AOC = 180^\circ - \angle A/2 - \angle C/2$ . But $\angle KAO = \angle KCO = 90^\circ$ , so $\angle AKC = \angle A/2 + \angle C/2$ . So considering triangle $AKL$ , $\angle ALK = 90^\circ - \angle A/2 - \angle C/2 = \angle B/2$ . Similarly, $\angle BLC = \angle B/2 + \angle C/2$ , so $\angle BKL = \angle A/2$ .

Now we use the given facts that $\angle C = \angle A/2$ and $\angle A = \angle B/2$ . So $\angle OKL = \angle BKL = \angle C$ , and $\angle OLK = \angle ALK = \angle A$ . Hence triangles $OKL$ and $BCA$ are similar.

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