50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010)

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.