Competition problem — Geometry — Olympiad Maths Prep

Geometry Difficulty 8.4 Shortlist Prove it Bulgaria

Given is a $\triangle ABC$ with incircle $k$ touching the sides $BC$ and $CA$ in points $P$ and $Q$ respectively. Let denote with $J$ the center of the excircle at side $AB$ in $\angle ABC$ and with $T$ – the second intersection point of the circumcircles of $\angle JBP$ and $\angle JAQ$ . Prove that the circumcircle of $\angle ABT$ touches $k$ .

Solution

Let $R$ be the tangent point of $k$ to $AB$ . We will use the standard notations about the angles of $\triangle ABC$ . We have
$\angle PTJ = 180^{\circ} - \angle PBJ = 90^{\circ} - \frac{\beta}{2} = \angle PRB$
and
$\angle QTJ = 180^{\circ} - \angle QAJ = 90^{\circ} - \frac{\alpha}{2} = \angle QRA.$
Then
$\angle PTQ = \angle PTJ + \angle QTJ = 180^{\circ} - \left(\frac{\alpha}{2} + \frac{\beta}{2}\right), \text{ A}$
i.e. $\angle PTQ = 180^{\circ} - \angle PRQ$ and hence $T \in k$ , $R \in TJ$ .

$\angle ATJ = \angle AQJ = \angle BPJ = \angle BTJ,$
i.e. $S$ is the midpoint of $\overline{AB}$ .
But the homothety $h(T, R \rightarrow S)$ sends $k$ to $\omega$ . Therefore these circles are tangent at point $T$ .

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