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Team Selection Tests for IMO, 2018

Geometry Difficulty 8.3 Shortlist Prove it Saudi Arabia

Let $ABC$ be an acute-angled triangle inscribed in circle $(O)$ . Let $G$ be a point on the small arc $AC$ of $(O)$ and $(K)$ be a circle passing through $A$ and $G$ . Bisector of $\angle BAC$ cuts $(K)$ again at $P$ . The point $E$ is chosen on $(K)$ such that $AE$ is parallel to $BC$ . The line $PK$ meets the perpendicular bisector of $BC$ at $F$ . Prove that $\angle EGF = 90^{\circ}$ .

Solution

Let $D$ be the second intersection of $AP$ and $(O)$ and $H$ be the intersection of $OD$ and $AE$ . Note that $D$ is the midpoint of the minor arc $BC$ of $(O)$ , then $OD$ is the perpendicular bisector of $BC$ . Since $AE \parallel BC$ , we have $\angle DHE = 90^{\circ}$ .

On the other hand,
$\angle DOG = 2\angle DAG = \angle PKG,$
which implies that two isosceles triangles $KPG$ and $ODG$ are similar. We get $\angle ODG = \angle KPG$ therefore $PFGD$ is cyclic.

From this, $\angle AEG = \angle DPG = \angle DFG$ . This means $HEGF$ is cyclic, we deduce $\angle EGF = 180^{\circ} - \angle EHF = 90^{\circ}$ .

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