IMO Hk TST

Geometry Difficulty 8.3 Shortlist Prove it Hong Kong

In $\triangle ABC$ , let $D$ be a point on side $BC$ . Suppose the incircle $\omega_1$ of $\triangle ABD$ touches sides $AB$ and $AD$ at $E, F$ respectively, and the incircle $\omega_2$ of $\triangle ACD$ touches sides $AD$ and $AC$ at $F, G$ respectively. Suppose the segment $EG$ intersects $\omega_1$ and $\omega_2$ again at $P$ and $Q$ respectively. Show that line $AD$ , tangent of $\omega_1$ at $P$ and tangent of $\omega_2$ at $Q$ are concurrent.

Solution

Let $O$ be the circumcentre of $\triangle FPQ$ . First of all,
$\angle OFP = 90^\circ - \angle PQF = \angle FQG - 90^\circ = \frac{\angle FAG}{2} = \angle GEF = \angle AFP$
where the second last equality follows from the fact that $AE = AF = AG$ . This shows that $O$ lies on $AD$ .

Next, Let $I_1$ be the centre of $\omega_1$ . Note that $\triangle OFI_1 \cong \triangle OPI_1$ as their corresponding sides are equal. Thus $\angle I_1PO = 90^\circ$ , which means $OP$ is tangent to $\omega_1$ . Similarly $OQ$ is tangent to $\omega_2$ .
Hence, the line AD, the tangent to $\omega_1$ at P and the tangent to $\omega_2$ at Q are concurrent at O, as desired.

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