Autumn Tournament

Geometry Difficulty 5.2 AIME, harder Prove it Bulgaria

Given an obtuse isosceles triangle $ABC$ with $CA = CB$ and circumcenter $O$ . The point $P$ on $AB$ is such that $AP < \frac{AB}{2}$ and $Q$ on $AB$ is such that $BQ = AP$ . The circle with diameter $CQ$ meets $(ABC)$ at $E$ and the lines $CE$ , $AB$ meet at $F$ . If $N$ is the midpoint of $CP$ and $ON$ , $AB$ meet at $D$ , show that $ODCF$ is cyclic.

Solution

Let $T$ be the midpoint of $CQ$ (it is anyway the center of the important circle with diameter $CQ$ ). Then $OC$ is the perpendicular bisector of $AB$ (as $AC = BC$ ), triangle $ONT$ is isosceles by symmetry (as $P$ and $Q$ are symmetric with respect to the midpoint $M$ of $AB$ and hence with respect to $CO$ , by the problem condition) and $OT$ is the perpendicular bisector of $CE$ , so $\angle OCF = 90^\circ - \angle OCT = 90^\circ - \angle OCD = \angle ODM = \angle ODF$ , done! $\square$

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