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Mongolian Mathematical Olympiad

Geometry Difficulty 7.1 National Olympiad, round 2 Prove it Mongolia

Draw a circle $\gamma$ through point $M$ which is the middle point of the arc $BC$ not containing vertex $A$ and the center $I$ of incircle of the triangle $ABC$ . If the circle $\gamma$ intersects side $BC$ in points $D$ , $E$ and the lines $MD$ , $ME$ intersect the circle $\omega$ in points $P$ , $Q$ which are different from $M$ then prove that all possible lines $PQ$ pass through a constant point not depending from circle $\gamma$ .

Solution

Let's prove that all possible lines $PQ$ pass through the point $I$ .

In order to prove this we need to prove that $\angle PIM + \angle QIM = 180^\circ$ . It is easy see $MI = MB = MC$ . It follows from $\angle IBM = \angle BIM = \frac{\alpha+\beta}{2}$ , $\angle ICM = \angle CIM = \frac{\alpha+\gamma}{2}$ . Since $\angle BPM = \angle BCM = \angle DBM$ , by AA criterion $\triangle BPM \sim \triangle DBM$ and we get $\frac{PM}{BM} = \frac{BM}{DM}$ . Considering that $BM = IM$ from where follows $\frac{PM}{IM} = \frac{IM}{DM}$ .

Triangles $\triangle PMI$ , $\triangle IMD$ have the common angle $\angle PMI = \angle IMD$ so we have $\triangle PMI \sim \triangle IMD$ . Therefore we conclude that $\angle PIM = \angle IMD$ .

Similarly, $\triangle CQM \sim \triangle ECM$ implies $\frac{QM}{CM} = \frac{CM}{EM}$ and $\triangle QMI \sim \triangle IME$ consequently we get $\angle QIM = \angle IEM$ . Since the points $I$ , $E$ , $M$ , $D$ lie on the circle $\gamma$ , $\angle IDM + \angle IEM = 180^\circ$ and thus we have proved that $\angle PIM + \angle QIM = 180^\circ$ .

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