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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Geometry Difficulty 6.2 National Olympiad Prove it Romania

Let $ABC$ a triangle, $I$ the incenter, $D$ the contact point of the incircle with the side $BC$ and $E$ the foot of the bisector of the angle $A$ . If $M$ is the midpoint of the arc $BC$ which contains the point $A$ of the circumcircle of the triangle $ABC$ and $\{F\} = DI \cap AM$ , prove that $MI$ passes through the midpoint of $[EF]$ .

Alexandru Gîrban

Solution

The bisector of the angle $\hat{A}$ passes through the midpoint of the arc $BC$ which does not contain the point $A$ . Denote this point by $S$ . $MS$ is the perpendicular bisector of $[BC]$ , so $MS \parallel ID$ (both lines are perpendicular on $BC$ ).

Also, $\angle ASC = \angle ABC$ and $\angle SAC = \angle BAE$ proves that the triangles $SAC$ and $BAE$ are similar, so $\frac{AB}{BE} = \frac{AS}{SC}$ .

Using the angle bisector theorem, we get $\frac{AB}{BE} = \frac{AI}{IE}$ .

It is known that $SI = SC$ (one might compute the angles of triangle $SCI$ ). We have $\frac{AS}{SC} = \frac{AS}{SI} = \frac{AM}{MF}$ .

We proved $\frac{AI}{IE} = \frac{AM}{MF}$ .

Using Menelaus's theorem in the triangle $AEF$ and the transversal $I-X-M$ (where $\{X\} = IM \cap EF$ ), we have $\frac{AI}{IE} \cdot \frac{EX}{XF} \cdot \frac{MF}{MA} = 1$ .

Using the equality we proved before, we find $\frac{EX}{XF} = 1$ , which means that $X$ is the midpoint of $[EF]$ .

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