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Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let $ABC$ be an acute triangle. The altitudes $BE$ and $CF$ intersect at the orthocenter $H$ , and point $O$ denotes the circumcenter. Point $P$ is chosen so that $\angle APH = \angle OPE = 90^{\circ}$ , and point $Q$ is chosen so that $\angle AQH = \angle OQF = 90^{\circ}$ . Lines $EP$ and $FQ$ meet at point $T$ . Prove that points $A, T, O$ are collinear.

Solution

Solution:

Observe that $T$ is the radical center of the circles with diameter $OE$ , $OF$ , $AH$ . So $T$ lies on the radical axis of $(OE), (OF)$ which is the altitude from $O$ to $EF$ , hence passing through $A$ .

So $A T O$ are collinear, done.

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