Bariz - Problem

Source: 2020 Czech-Polish-Slovak Match p1

Tags: geometry, parallelogram, circumcircle, Computer problems

parmenides51

08.10.2020 05:14

Let $ABCD$ be a parallelogram whose diagonals meet at $P$. Denote by $M$ the midpoint of $AB$. Let $Q$ be a point such that $QA$ is tangent to the circumcircle of $MAD$ and $QB$ is tangent to the circumcircle of $MBC$. Prove that points $Q,M,P$ are collinear. (Patrik Bak, Slovakia)

This ``aesthetic" solution took way too long. We construct a point $X$ such that $MX = AD$; it follows that $X$ lies on both $(AMD)$ and $(BMC)$ as $BMCX$ and $AMXD$ are isosceles trapezoids. Also, let $N$ be the midpoint of $\overline{CD}$; it suffices to show that $\overline{QN} \parallel \overline{BC}$. Claim. $BQAX$ is cyclic. Proof. \[\measuredangle BQA = \measuredangle QBA + \measuredangle QAB = \measuredangle BXM + \measuredangle MXA = \measuredangle BXA. \ \ \blacksquare\]But $BANX$ is an isosceles trapezoid and thus cyclic, hence $B, Q, A, N, X$ are all concyclic. It follows that $\measuredangle BQN = \measuredangle BXN= \measuredangle QBC$ which implies the result.