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Russian Mathematical Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it Russia

Let $M$ be a midpoint of the side $AC$ of an acute-angled triangle $ABC$ with $AB > BC$ . Let $\Omega$ be the circumcircle of the triangle $ABC$ . The tangents to $\Omega$ at points $A$ and $C$ meet at $P$ . The segments $BP$ and $AC$ meet at $S$ . Let $AD$ be the altitude in the triangle $ABP$ . The circumcircle $\omega$ of the triangle $CSD$ intersects $\Omega$ at point $K \neq C$ . Prove that $\angle CKM = 90^\circ$ .

Solution

Поскольку $\angle AMP = \angle ADP = 90^\circ$ , точки $M$ и $D$ лежат на окружности $\gamma$ с диаметром $AP$ . Поскольку $PA$ — касательная к $\Omega$ , имеем $\angle KAP = \angle ACK$ . Так как точки $C, K, D$ и $S$ лежат на окружности $\omega$ , имеем $\angle ACK = \angle KDP$ . Значит, $\angle KAP = \angle ACK = \angle KDP$ ,

то есть точки $A, D, K$ и $P$ лежат на одной окружности. Итак, точка $K$ лежит на $\gamma$ , и $\angle AKP = 90^\circ$ (см. рис. 8).

Отсюда имеем $\angle MKP = 180^\circ - \angle MAP = 180^\circ - \angle ABC = \angle AKC$ . Значит, $\angle MKC = \angle AKC - \angle AKM = \angle MKP - \angle AKM = \angle AKP = 90^\circ$ , что и требовалось доказать.

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