Saudi Booklet

Geometry Difficulty 5.2 AIME, harder Prove it Saudi Arabia

Given an acute triangle $ABC$ inscribed in $(O)$ with $\angle A = 60^\circ$ and symmedian point $L$ . The tangents at $B, C$ of $(O)$ intersect $CA, AB$ at $E, F$ . Prove that the line $OL$ , two circumcircles of triangles $AEF$ and $BOC$ pass through the same point.

Solution

Let $D$ be the midpoint of the minor arc $BC$ of $(O)$ and $P$ be the intersection point of the tangents at $B, C$ of $(O)$ . We will show that $D$ and $P$ lie on $(AEF)$ .

Indeed, we have
$\angle AEP = \angle AEB = \angle BAC - \angle ABE = \angle ODB - \angle ADB = \angle ADO$
so $(ADP)$ passes through $E$ . Similarly, $(ADP)$ passes through $F$ . Now, let $J$ be the center of $(AEF)$ , and we will show that $JD \parallel OL$ .

From here, since $D$ is the center of circle $(BOC)$ then $JD \perp PX$ and $OX \perp PX$ , it follows that $O, X, L$ lie on the same line parallel to $JD$ . Since $(ABC)$ intersects $(AEF)$ at $D$ other than $A$ , then $\Delta DBE \sim \Delta DCF$ but $DB = DC$ so $DE = DF$ . On the other hand, $JE = JF$ since $J$ is the center of $(AEF)$ so $JD \perp EF$ . According to the familiar result, $EF \perp OL$ , so $JD \parallel OL$ which finishes the proof. $\square$

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