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Problem rejected from IMO, EGMO, USAMO. But I still like it!

Source: Kyiv City MO 2024 Round 2, Problem 11.4

February 4, 2024
geometry

Problem Statement

Let ABCABC be an acute triangle with circumcenter OO and orthocenter HH. Rays AOAO, COCO intersect sides BC,BABC, BA in points A1,C1A_1, C_1 respectively, KK is the projection of OO onto the segment A1C1A_1C_1, MM is the midpoint of ACAC. Prove that ∠HMA=∠BKC1\angle HMA = \angle BKC_1.
Proposed by Anton Trygub
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