CAPS Match 2024

Let $\omega$ be the circumcircle of triangle $AEF$ and let $I_A$ be the A-excenter of triangle $ABC$. First, we prove that $I_A$ is the midpoint of the arc $EF$ of $\omega$ that does not contain point $A$ (see the left figure). To that end, note that since $I_A$ lies on the external angle bisector of $\angle B$ and $BE=BD$, triangles $BE{I}_A$ and $BD{I}_A$ are congruent (SAS). Similarly, triangles $CF{I}_A$ and $CD{I}_A$ are congruent, so in particular $I_{A}E=I_{A}F$. Moreover, $\angle BE{I}_A+\angle CF{I}_A=\angle BD{I}_A+\angle CD{I}_A=180^{\circ }$, hence the points $A,E,I_A,F$ lie on a single circle in this order.



Next, we prove that $P$ is the second intersection of $I_{A}D$ and $\omega$ (see the middle figure). Let $I$ be the incenter of triangle $ABC$. Then $ED\parallel BI$ and $DF\parallel IC$. Setting $\angle EAF=\alpha$, we get $\angle EDF=\angle BIC=90^{\circ }+\frac{1}{2}\alpha$, thus $\angle EPF=\alpha =\angle EAF$, so $P$ lies on $\omega$. Since $I_A$ is the midpoint of arc, it lies on the angle bisector $PD$, so $P$ lies both on $I_{A}D$ and on $\omega$ as claimed. Finally, we show that $P$ lies on that arc of $\omega$ which lies inside $\Omega$ (see the right figure). Let $M\ne A$ be the second intersection of $\omega$ and $\Omega$ (if they are tangent, we set $M=A$). Then $M$ is the center of the spiral similarity that maps $BE$ to $CF$ (alternatively, we angle-chase that triangles $MBE$ and $MCF$ are similar by AA). Thus $MB/MC=BE/CF=BD/DC$, so $MD$ is the angle bisector of $BMC$, and so it passes through the midpoint $S$ of the arc $BC$ of $\Omega$ that does not contain $A$. Now forget about points $B,C,E,F$ and focus on circles $\Omega ,\omega$ and on the points $A,M,I_A,S,D,P$. Circles $\Omega$ and $\omega$ share points $A$ and $M$. Being the A-excenter of $ABC$, point $I_A$ belongs to that arc $AM$ of $\omega$ which lies outside of $\Omega$ (e.g. since $A{I}_A>AS$). Point $S$ lies on the segment $A{I}_A$ and point $D$ lies on the segment $SM$, so point $D$ lies inside the angle $A{I}_{A}M$. Thus, point $P=I_{A}D\cap \omega$ belongs to the other arc $AM$ of $\omega$ than $I_A$, namely to the one which lies inside $\Omega$.