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Mathematical Olympiad in Ukraine, Fourth Round (March 24, ), 2010

Geometry Difficulty 6.2 National Olympiad Prove it Ukraine

Points $K$ and $M$ are chosen on the sides $AB$ and $BC$ of triangle $ABC$ in such a way, that $AK = KM = MC$ . Let $N$ be the point of intersection of $AM$ and $CK$ , $P$ – the feet of perpendicular from point $N$ to the line $KM$ , and $Q$ is such point inside segment $KM$ , that $MQ = KP$ . Prove, that incircle of $KMB$ touches $KM$ at point $Q$ .

Solution

Let $I$ be incenter of $KMB$ . Then, $KI$ and $MI$ are bisectors of $\angle BKM$ and $\angle BMK$ respectively. $AKM$ and $CMK$ are isosceles triangles, thus, $\angle KAM = \angle KMA = \angle MKI = \angle BKI$ and $\angle MKC = \angle MCK = \angle KMI = \angle BMI$ , moreover, these angles are acute (Fig.17).

Fig.17

This implies that $P$ belongs to the segment $KM$ . $\angle IKM = \angle KMA$ and $\angle IMK = \angle MKC$ , thus $KI \parallel AM$ and $MI \parallel CK$ , hence $KIMN$ is parallelogram. So $KN = IM$ . $KP = MQ$ (by the condition of the problem), therefore, triangles $PKN$ and $QMI$ are equal (they have two equal corresponding sides and angle between them). This implies that $\angle IQM = \angle NPK = 90^\circ$ or $Q$ is a touching point, what we wanted to prove.

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