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Problem 1755

National Olympiad, first round

Geometry Difficulty 6.5 Prove it Mongolian Mathematical Olympiad · Mongolia

Given an acute triangle $ABC$ . Let $(\omega, I)$ be the inscribed circle of $ABC$ , $(\Omega, O)$ be the circumscribed circle of $ABC$ and $A_0$ be the midpoint of $AH$ altitude. $\omega$ touches $BC$ at point $D$ . $A_0D \cap \omega = P$ and the perpendicular from $I$ to $A_0D$ intersects $BC$ at the point $M$ . $MR$ and $MS$ lines touch $\Omega$ at $R$ and $S$ respectively. Prove that the points $R$ , $P$ , $D$ and $S$ are concyclic.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $\omega$ touch $AB$ and $AC$ at $E$ and $F$ respectively. $(AD) \cap \omega = Q$ , $(ID) \cap \omega = N$ , $(EF) \cap (BC) = T$ . Since $E$ , $F$ are points of tangency $(DEQF)$ is harmonic division. This implies $TQ$ is tangent to $\omega$ . Since $AA_0 = A_0H$ , $(AA_0H\infty)$ is harmonic division. From this $(QPDN)$ is harmonic too. Hence $NP$ , $EF$ , $BC$ are concurrent. Considering $ND$ is diameter, we have $NT \perp PD$ . Then $IM \parallel NT$ and $DM = MT$ . From $(DEQF)$ is harmonic division, implies that $(BDCT)$ is harmonic division. Hence $MD^2 = MC \cdot MB$ . (This is easy to prove) By tangent-secant theorem $MS^2 = MR^2 = MC \cdot MB$ . Considering $IM \perp DP$ , $MD = MP$ . Hence $S$ , $D$ , $P$ are $R$ are on a circle, whose center is $M$ .

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