Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National Olympiad Prove it Ireland

The point $P$ is on the side $BC$ of triangle $ABC$ . The incircles of $\triangle ABC$ , $\triangle ABP$ and $\triangle ACP$ are denoted by $K$ , $K_1$ and $K_2$ , respectively.
Prove that the angle at which the radical axis of $K$ and $K_1$ meets the radical axis of $K$ and $K_2$ does not depend on the position of $P$ . Find the measure of this angle if $\angle BAC = 100^\circ$ .

Solution

Let $I$ be the incentre of $K$ . The incentres of $K$ and $K_1$ both lie on the bisector of $\angle ABC$ . The incentres of $K$ and $K_2$ both lie on the bisector of $\angle ACB$ . Note that $\angle BIC = 180^\circ - \frac{1}{2}(\angle ABC + \angle ACB) = 90^\circ + \frac{1}{2}\angle BAC$ .

Because the radical axis of two circles is perpendicular to the line connecting the centres of these circles, the radical axis of $K$ and $K_1$ is perpendicular to the bisector of $\angle ABC$ and the radical axis of $K$ and $K_2$ is perpendicular to the bisector of $\angle ACB$ . Hence, one of the angles between these two radical axes is equal to $\angle BIC$ ; and this angle does not depend on the position of $P$ . If $\angle BAC = 100^\circ$ , we obtain $\angle BIC = 90^\circ + 50^\circ = 140^\circ$ . Hence, the two angles between the two radical axes are equal to $140^\circ$ and $40^\circ$ .

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