66th Belarusian Mathematical Olympiad

We show that the circumcircle of the triangle $I_1{I}_{2}P$ touches the line $BC$ at the point $P$, then, in particular, it follows that this circle touches the circle ${\omega }_A$. We use the following well-known lemma.

Lemma. Let excircle touch the side $BC$ of the triangle $ABC$ at $A_1$ and touch the prolongations of the sides $AB$ and $AC$ at $C_1$ and $B_1$, respectively. Then $$B{A}_1=B{C}_1=\frac{AC+CB-BA}{2}\text{and}C{A}_1=C{B}_1=\frac{CB+BA-AC}{2}.$$ Let ${\omega }_1$ and ${\omega }_2$ denote excircles of the triangles $ABP$ and $APC$, respectively. Let $I_1$ and $I_2$ be the centers of ${\omega }_1$ and ${\omega }_2$, respectively. Let $F_1$

and $F_2$ denote the tangent points of ${\omega }_1,{\omega }_2$ and the line $AP$, respectively. We show that $F_1$ and $F_2$ coincide. Indeed, from the lemma it follows that $$\begin{array}{rl}2P{F}_1=PB+BA-AP& =\frac{AC+CB-BA}{2}+BA-AP=\\ & =\frac{AC+CB+BA}{2}-AP,\end{array}$$ $$\begin{array}{rl}2P{F}_2=PC+CA-AP& =\frac{CB+BA-AC}{2}+AC-AP=\\ & =\frac{AC+CB+BA}{2}-AP.\end{array}$$ Hence, $P{F}_1=P{F}_2$, i.e., $F_1\equiv F_2$. So we may use $F$ to denote $F_1,F_2$. We have $I_{1}F\perp AP$ and $I_{2}F\perp AP$, so $I_1,F$, and $I_2$ lie on the same line.

Let $\angle BPF=2x$, then $\angle CPF=180^{\circ }-2x$. Since the circle ${\omega }_1$ touches the lines $PA$ and $PB$, it follows that the line $P{I}_1$ is the bisector of the angle $\angle BPF$. Similarly, $P{I}_2$ is the bisector of the angle $\angle CPF$. Therefore, $\angle I_1{I}_{2}P=90^{\circ }$ and we have $$\angle I_1{I}_{2}P=90^{\circ }-\angle FP{I}_2=90^{\circ }-(90^{\circ }-x)=x=\angle BP{I}_1.$$