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Problem 1177

AMC 12 late, AIME early

Geometry Difficulty 4.9 Prove it Ukrainian National Mathematical Olympiad, 3rd Round · Ukraine

On the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$ consider points $X$ and $Y$ such that $ABXY$ is a parallelogram. Prove that the circumradii of $BXD$ and $CYA$ are equal.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Since $ABCD$ is cyclic, then $\angle ABD = \angle ACD$ . Moreover, $\angle ABD = \angle ABY = \angle ABY$ . Hence, $CXYD$ is cyclic because $\angle XCD = \angle XYB$ (fig. 12).

This implies that $\angle XCY = \angle XDY$ , from what it follows that $\sin \angle ACY = \sin \angle BDX$ . Moreover, $BX = AY$ , and applying sine Law for triangles $BXD$ and $CAY$ we get:
$R_{BDX} = \frac{BX}{2\sin \angle BDX} = \frac{AY}{2\sin \angle ACY} = R_{CYA}$

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