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Bulgaria competition problems, 2022

Geometry Difficulty 6.0 National Olympiad Prove it Bulgaria

Let $ABC$ be an acute triangle and $M$ is the midpoint of $AB$ . A circle through the vertices $B$ and $C$ intersects the segments $CM$ and $BM$ in points $P$ and $Q$ respectively. Let $K$ be symmetrical point to $P$ regarding $M$ . Let circumcircles of $\triangle AKM$ and $\triangle CQM$ meet again in point $X$ , and circumcircles of $\triangle AMC$ and $\triangle KMQ$ meet again in point $Y$ . The segments $BP$ and $CQ$ meet in $T$ . Prove that $MT$ is tangent to the circumcircle of $\triangle MXY$ .

Solution

Obviously $AKBP$ is paralellogram and from quadrilateral $BQPC$ is inscribed follows $\triangle AKC = \triangle KPB = \triangle AQC = \varphi$ . Hence $AKQC$ is inscribed quadrilateral with center $O$ . Then $\triangle AYC = (180^\circ - \triangle AYM) + (180^\circ - \triangle CYM) = \triangle AKM + \triangle CQM = 2\varphi = \triangle AOC$ and so $AOYC$ is inscribed. Therefore $\triangle OYM = \triangle CYM - \triangle CYO = (360^\circ - \triangle AYC - \triangle AYM) - \triangle CAO = 180^\circ - \varphi - (90^\circ - \varphi) = 90^\circ$ and analogously $\triangle OXM = 90^\circ$ . If $TM \cap AK = T'$ then $TM = MT'$ from paralellogram $AKBP$ and the reverse butterfly theorem for $AKQC$ gives us $OM \perp TT'$ and $OM \perp MT$ . Hence $MT$ is perpendicular to the diameter $OM$ of circumcircle of $\triangle MXY$ .

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