HMMT February, 2020

Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:
Let $ABC$ be a triangle inscribed in a circle $\omega$ and $\ell$ be the tangent to $\omega$ at $A$ . The line through $B$ parallel to $AC$ meets $\ell$ at $P$ , and the line through $C$ parallel to $AB$ meets $\ell$ at $Q$ . The circumcircles of $ABP$ and $ACQ$ meet at $S \neq A$ . Show that $AS$ bisects $BC$ .

Solutions — 2

Solution 1

Solution:
In directed angles, we have
$\measuredangle CBP = \measuredangle BCA = \measuredangle BAP$
so $BC$ is tangent to the circumcircle of $ABP$ . Likewise, $BC$ is tangent to the circumcircle of $ACQ$ . Let $M$ be the midpoint of $BC$ . Then $M$ has equal power $MB^{2} = MC^{2}$ with respect to the circumcircles of $ABP$ and $ACQ$ , so the radical axis $AS$ passes through $M$ .

Solution 2

Solution:
Since
$\measuredangle CBP = \measuredangle BCA = \measuredangle BAP = \measuredangle CQP$
quadrilateral $BCQP$ is cyclic. Then $AS$ , $BP$ , and $CQ$ concur at a point $A'$ . Since $A'B \parallel AC$ and $A'C \parallel AB$ , quadrilateral $ABA'C$ is a parallelogram so line $ASA'$ bisects $BC$ .

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