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Team selection test for 50. IMO

Geometry Difficulty 8.0 National Olympiad, round 2 Prove it Bulgaria

The excircles of $\triangle ABC$ touch the sides $AB$ , $BC$ and $CA$ at points $M$ , $N$ and $P$ , respectively. Let $I$ and $O$ be the incenter and the circumcenter of $\triangle ABC$ . Prove that if $AMNP$ is a cyclic quadrilateral, then:

a) the points $M$ , $P$ and $I$ are collinear;

b) the points $I$ , $O$ and $N$ are collinear.

Solution

By Carnot's theorem, the perpendiculars from the points $M$ , $N$ and $P$ to the lines $AB$ , $BC$ and $CA$ , respectively, have a common point $X$ . Then the quadrilateral $AMXP$ is cyclic. Now it is easy to see that $X = N$ and hence $AN$ is a diameter of the circumcircle of $AMNP$ .

a) Denote by $I_B$ and $I_C$ the respective excenters of $\triangle ABC$ . Then Pappus' theorem for the triples of points $(I_C, A, I_B)$ and $(B, N, C)$ implies the desired result.

b) Let the incircle of $\triangle ABC$ touches the sides $AB$ and $AC$ at point $R$ and $Q$ , respectively. Then the bisectors of $AB$ and $AC$ coincide with the bisectors of $RM$ and $QP$ , respectively, and pass through the midpoint of $IN$ .

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