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Final Round of the 73rd Czech and Slovak Mathematical Olympiad (March 17–20, ), 2024

Geometry Difficulty 5.2 AIME, harder Prove it Czech Republic

Suppose that a point $P$ lying in the interior of a convex quadrilateral $ABCD$ satisfies
$\angle PAD = \angle ADP = \angle CBP = \angle PCB = \angle CPD.$
Let $O$ be the circumcentre of the triangle $CPD$ . Prove that $OA = OB$ .

Solution

From the given equalities, one sees that $PC \parallel AD$ and $PD \parallel BC$ . Since the lines $PC$ and $PD$ are distinct (we know that $\angle CPD \neq 0$ ), the lines $AD$ and $BC$ are not parallel, so they intersect at a unique point $X$ such that $PCXD$ is a rhombus.

Now, note that the quadrilateral $AXCP$ is an isosceles trapezoid, since $AX \parallel CP$ and $\angle PAX = \angle CXA$ . Therefore, the perpendicular bisectors of its bases $AX$ and $CP$ coincide. Since $O$ is the circumcentre of $CPD$ , it lies on the bisector of $CP$ , hence also on the bisector of $AX$ , and so we have $OX = OA$ . By considering the isosceles trapezoid $BXDP$ , we can also obtain $OX = OB$ , which gives us the desired equality.

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