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Estonian Math Competitions

Geometry Difficulty 5.6 AIME, harder Prove it Estonia

The incircle of triangle $ABC$ touches the sides $AB$ and $AC$ at points $K$ and $L$ , respectively. The line $BL$ intersects the incircle of triangle $ABC$ at point $M$ ( $M \neq L$ ). A circle passing through point $M$ touches the lines $AB$ and $BC$ at points $P$ and $Q$ , respectively, and intersects the incircle of triangle $ABC$ at point $N$ ( $N \neq M$ ). Prove that if $KM \parallel AC$ then points $P$ , $N$ and $L$ are collinear.

Solution

Note that the circles of the problem can be obtained from each other by homothetic transformation with center $B$ since both are tangent to sides $BA$ and $BC$ . Let $X$ be the other intersection point of the line $BL$ with the circumcircle of the triangle $MPQ$ .

The aforementioned homothety takes point $P$ to point $K$ , point $M$ to point $L$ , and point $X$ to point $M$ . By the homothety, $\angle PXM = \angle KML$ . Hence $\angle KML = \angle KLA = \angle LKM$ . Finally,
$\angle PNM + \angle LNM = (180^\circ - \angle PXM) + \angle LKM \\ = (180^\circ - \angle KML) + \angle KML = 180^\circ.$
This proves the desired claim that points $P$ , $N$ and $L$ are collinear.

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