Balkan 2012 shortlist

Let $A_{1}S$, $B_{1}T$ and $C_{1}U$ be the altitudes of $A_1{B}_1{C}_1$. The circle $k$ with the diameter $A_1{H}_1$ contains the points $A_1$, $H_1$, $T$, $U$, $P$ and $Q$. Let $A{A}_1$ intersect the incircle of $ABC$ for the second time at $V$. Assume that $\angle B\ge \angle C$.

Observe that $\angle C_1{A}_{1}V=\angle T{A}_{1}P$, $\angle C_1{B}_1{A}_1=\angle C_1{A}_{1}B=\angle T{A}_{1}Q$, and $\angle V{A}_1{B}_1=\angle P{A}_{1}U$. Considering the chords subtending these angles in the circle $k$ and in the incircle of the triangle $ABC$ we conclude that the cyclic quadrilaterals $PTQU$ and $V{C}_1{A}_1{B}_1$ are similar.

Let $M$ and $N$ be the midpoints of the line segments $B_1{C}_1$ and $A_1{H}_1$. Notice that $$\angle MTN=180^{\circ }-\angle C_{1}TM-\angle NT{A}_1=180^{\circ }-\angle T{C}_{1}M-\angle N{A}_{1}T=90^{\circ }$$ as $M$ and $N$ are the midpoints of the hypotenuses of the right triangles $C_{1}T{B}_1$ and $T{H}_1{A}_1$, respectively.

Therefore $MT$ and, similarly, $MU$, are tangent to $k$. It follows that the pentagons $PMTQU$ and $VA{C}_1{A}_1{B}_1$ are similar. Since the points $A_1$, $V$, $A$ lie on a line, so do the corresponding points $Q$, $P$, $M$.