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Source: 2021 MEMO T-6

September 5, 2021
geometrypower of a pointcircumcircleMEMO 2021memo

Problem Statement

Let ABCABC be a triangle and let MM be the midpoint of the segment BCBC. Let XX be a point on the ray ABAB such that 2∠CXA=∠CMA2 \angle CXA=\angle CMA. Let YY be a point on the ray ACAC such that 2∠AYB=∠AMB2 \angle AYB=\angle AMB. The line BCBC intersects the circumcircle of the triangle AXYAXY at PP and QQ, such that the points P,B,CP, B, C, and QQ lie in this order on the line BCBC. Prove that PB=QCPB=QC.
Proposed by Dominik Burek, Poland
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