Vietnam TST 2001

[shehab ahmed]

Two circles intersect at A and B.One of their common tangents touches the circles at P and Q.Tangents to the circumcircle of APQ at P and Q intersect at S.Let the reflection of B wrt PQ is H.Prove that A,S,H is collinear.

[zadid xcalibured]

at first prove that H lies on the circumcircle of APQ.then angle chasing yeilds H lies on the A symmedian.but we know AS is the A symmedian.actually AB is median.if AB cuts PQ at M.then PM=QM from power of point and radical axis.angle chasing helps as we have alternate segmental angles.

[tanmoy]

$\text{My solution}$:

All the angles below are directed mod $\Pi$.
Let $BA \cap PQ=M$.Then by $\text{Power of A Point theorem}$,we get that $BM \times MA=MP^{2}=MQ^{2} \Leftrightarrow MP=MQ$.By Zhao's lemma,we get that $SA$ is a symmedian of $\Delta ABC$.Let $SA \cap PQ=N$.
Now,$\measuredangle QAM=\measuredangle BAQ=\measuredangle BTQ=\measuredangle BQP=\measuredangle PQH$.
Again,$\measuredangle PAM=\measuredangle BAP=\measuredangle BUP=\measuredangle BPQ=\measuredangle QPH$.
$\therefore$ $\measuredangle PAQ=\measuredangle PQH+\measuredangle QPH=\measuredangle PHQ$.
So,$P,A,Q,H$ are cyclic
As $\measuredangle QAM=\measuredangle BTQ,\measuredangle PAN=\measuredangle BTQ=\measuredangle BQP=\measuredangle PQH=\measuredangle PAH$.
So,$A,N,H$ are collinear.But $S,A,N$ are collinear.This implies that $S,A,N,H$ or $S,A,H$ are collinear.