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HMIC 2024 Problem 5

    Let ABCA B C be an acute, scalene triangle with circumcenter OO and symmedian point KK. Let XX be the point on the circumcircle of triangle BOCB O C such that ∠AXO=90∘\angle A X O=90^{\circ}. Assume that X≠KX \neq K. The hyperbola passing through B,C,O,KB, C, O, K, and XX intersects the circumcircle of triangle ABCA B C at points UU and VV, distinct from BB and CC. Prove that UVU V is the perpendicular bisector of AXA X. The symmedian point of triangle ABCA B C is the intersection of the reflections of BB-median and CC-median across the angle bisectors of ∠ABC\angle A B C and ∠ACB\angle A C B, respectively.

    Dataset text NuminaMath olympiads. Quoted with source.

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