Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.4 Shortlist Prove it Hong Kong

Let $ABC$ be an acute triangle. Suppose a circle $\Gamma_1$ , with centre $O_1$ , touches the sides $BC$ produced at $E$ , $AC$ produced at $G$ , and $AB$ at $C'$ . Suppose also that another circle $\Gamma_2$ , with centre $O_2$ , touches the sides $AB$ produced at $H$ , $BC$ produced at $F$ , and $AC$ at $B'$ . Let the extensions of $EG$ and $FH$ intersect at $P$ . Prove that $PA \perp BC$ .

Solution

Let $EC'$ meet $FB'$ at $D$ . Note that $EC'$ is parallel to the internal angle bisector of $\angle CBA$ , which is $BO_2$ . Therefore, $ED \perp PF$ . Similarly, $FD \perp PE$ . This implies $D$ is the orthocentre of $\triangle PEF$ , and hence $PD \perp EF$ . It suffices to show $P, A, D$ are collinear, since this would imply $PA \perp BC$ .

Consider $(PGB')$ and $(PC'H)$ . Since $AG \times AB' = AC' \times AH$ , the point $A$ lies on the radical axis of these circles. Thus, $PA$ is the radical axis. It remains to show that $D$ also lies on this radical axis. Indeed, we shall prove that $D$ lies on both circles. We have
$\angle GB'D = \angle CB'F = \angle B'FC = \angle GPD.$
The last equality holds since $D$ is the orthocentre of $\triangle PEF$ . This implies $D$ lies on $(PGB')$ . The other assertion can be proved similarly. So we are done.

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