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Problem 1405

AIME late

Geometry Difficulty 5.7 Prove it Austrian Mathematical Olympiad · Austria

Let $ABC$ be a triangle, and $O$ its circumcenter. The circumcircle of triangle $AOC$ shall intersect the segment $BC$ in points $C$ and $D$ and the segment $AB$ in points $A$ and $E$ .
Prove that triangles $BDE$ and $AOC$ have equal circumradii.

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

In the circumcircle of triangle $ABC$ we have $\angle COA = 2\angle CBA$ . In the circumcircle of $ADC$ we therefore have $\angle CDA = \angle COA = 2\angle CBA$ . The angle $\angle CDA$ is an external angle in triangle $ABD$ , and we therefore obtain $\angle CBA + \angle BAD = \angle CDA = 2\angle CBA$ , and thus $\angle BAD = \angle CBA$ . In the circumcircle of $AOC$ we obtain $\angle BAD = \angle EAD$ on the chord $ED$ . The angles $\angle CBA = \angle DBE$ are equal in the circumcircle of triangle $BDE$ on the same chord $ED$ . Since the chords and subtended angles are equal in both circles, they must have the same radii, as claimed.

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