IRL ABooklet

Geometry Difficulty 5.1 AIME, harder Prove it Ireland

Let $A$ , $B$ , $C$ be three points on a circle $\Gamma$ and consider points $M$ , $N$ on the side $BC$ , point $P$ on the side $AC$ and $Q$ on the side $BC$ , such that $MP$ is parallel to $BC$ and $NQ$ is parallel to $AC$ . Let $T$ be a point on line $MP$ , so that $TC$ is tangent to $\Gamma$ . The circumcircle of $\triangle TMN$ intersects $\Gamma$ at $K$ and $L$ . Prove that $K$ , $L$ , $P$ , $Q$ are collinear.

Solution

The quadrilateral $TCMA$ is cyclic. Indeed, the Alternate Segment Theorem applied to the tangent $TC$ , and $MP \parallel BC$ implies
$\angle TCA = \angle CBA = \angle TMA.$
Let $S$ be the intersection point of the lines $TC$ and $NQ$ . It then follows in a similar way that $SCNB$ is cyclic.

Moreover, $TMNS$ is cyclic as well since
$\angle CTM = \angle CAM = \angle CAB = \angle SNB.$
Then $P$ is the radical centre of $(ABC)$ , $(TAC)$ and $(TMN)$ while $Q$ is the radical centre of $(ABC)$ , $(SBC)$ and $(TMN)$ . Thus they are both on $KL$ which is the radical axis of $(ABC)$ and $(TMN)$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.