HMIC

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:
Let $ABCD$ be a parallelogram, and let $O$ be a point inside $ABCD$ . Suppose the circumcircles of triangles $OAB$ and $OCD$ intersect at $P \neq O$ , and the circumcircles of triangles $OBC$ and $OAD$ intersect at $Q \neq O$ . Prove $\angle POQ$ equals one of the angles of quadrilateral $ABCD$ .

Solution

Solution:
In what follows, all angles are directed.

Claim 1. The points $P$ and $Q$ are symmetric over the center of $ABCD$ .

Proof. Note that
$\angle APB = \angle AOB = \angle OAD + \angle CBO = \angle OQD + \angle CQO = \angle CQD.$
Similar equalities hold for each pair of opposite sides, so $P$ and $Q$ are symmetric across the parallelogram's center. $\square$

Consequently, $AP$ and $CQ$ are parallel, so
$\angle POQ = \angle POB + \angle BOQ = \angle PAB + \angle BCQ = \angle CBA,$
as desired. (Once we undirect the angles, $\angle POQ$ is either $\angle B$ or $\pi - \angle B = \angle A$ .)

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