MMO Round 4

Geometry Difficulty 5.1 AIME, harder Prove it Mongolia

Let $K$ and $L$ be points on side $BC$ , $M$ a point on side $AC$ , and $N$ a point on side $AB$ of triangle $ABC$ . These points are chosen such that $\triangle KMC \sim \triangle ABC$ and $\triangle LBN \sim \triangle ABC$ . Suppose that the segments $NL$ and $KM$ intersect at point $P$ inside triangle $ABC$ . The circumcircle of triangle $AMP$ intersects the line $CP$ again at point $X$ , and the circumcircle of triangle $ANP$ intersects the line $BP$ again at point $Y$ . Prove that the points $B, C, X$ , and $Y$ lie on the same circle.

Solution

Since $\triangle KMC \sim \triangle ABC$ , we have $\angle MKC = \angle BAC$ , so quadrilateral $AMKB$ is cyclic. Hence we have, $BC \cdot KC = AC \cdot MC$ , and since $AMPX$ is cyclic, we have $AC \cdot MC = XC \cdot PC$ .

Combining the two, we get $XC \cdot PC = BC \cdot KC$ , so quadrilateral $BXPK$ is cyclic. Hence,
$\angle BXC = \angle MKC = \angle BAC.$
Similarly, we can show that $\angle BYC = \angle NLC = \angle BAC$ , which implies that $B, C, X$ , and $Y$ lie on a common circle.

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