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Problems from Ukrainian Authors

Geometry Difficulty 6.3 National Olympiad Prove it Ukraine

Let $O$ be the circumcenter of triangle $ABC$ with $\angle A = 120^\circ$ . Let $P$ and $Q$ denote the projections of $B$ onto $CO$ and $AO$ , respectively. Let $M$ be the midpoint of $AO$ . Prove that the circumcircle of $\triangle MPQ$ touches $AC$ .

Solution

Let $R$ be the projection of $B$ onto $AC$ (fig. 23). Then $P, Q, R$ are the projections of $B$ onto the sides of $\triangle AOC$ . Let $B_1$ be the isogonal conjugate of $B$ with respect to this triangle. Since $\angle BAC = \angle BOC = 120^\circ$ , we have $\angle B_1AO = \angle B_1OA$ . Then the projection of $B_1$ onto $AC$ is $M$ . Then the points $P, Q, R$ , and $M$ are concyclic. Thus, it suffices to prove $\angle QRA = \angle QPR$ . We will use the cyclicity of $ARBQ, OQBP,$ and $CPBR$ . Indeed,
$\angle RPQ = \angle RPC - \angle QPC = \angle RBC - \angle QBO = (90^\circ - \angle BCA) - (90^\circ - \angle BOA) = \angle BCA,$
$\angle QRA = \angle ABQ = 90^\circ - \angle BAO = \angle BCA,$
completing the proof.

Fig. 23

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