Iranian Mathematical Olympiad

Throughout the solution, let $\omega$ and ${\omega }_a$ be the circumcircle of triangle $ABC$ and the excircle of $ABC$ tangent to side $BC$, respectively. Furthermore, $O$ and $I_a$ are the centers of $\omega$ and ${\omega }_a$, respectively. Under an inversion with center $A$ and power $AB\times AC$, and then a reflection with respect to the bisector of $\angle BAC$, $B^{\prime }$ is sent to some point of line $AB$, and $C^{\prime }$ to some point of line $AC$. Denote by $B^{\prime \prime }$ and $C^{\prime \prime }$ these two points, respectively. Note that $$A{B}^{\prime }\cdot A{C}^{\prime \prime }=A{C}^{\prime }\cdot A{B}^{\prime \prime }=AB\cdot AC=AI\cdot A{I}_a.$$ Therefore, the quadrilaterals $I{I}_a{B}^{\prime \prime }{C}^{\prime }$ and $I{I}_a{C}^{\prime \prime }{B}^{\prime }$ are cyclic. Since $\angle I_{a}I{B}^{\prime }=90^{\circ }$, we get $I_a{C}^{\prime \prime }\perp AC$ and hence $C^{\prime \prime }$ is the tangency point of ${\omega }_a$ and $AC$. Similarly, $B^{\prime \prime }$ is the tangency point of ${\omega }_a$ and $AB$. Now, we are going to find the image of $B_1$ and $C_1$ under the transformation mentioned above. Denote by $M$ and $N$ the midpoints of arcs $ACB$ and $ABC$ of circle $\omega$, respectively. Then $\angle BA{C}_1=\angle CAM$. We know that under the transformation, line $BC$ transforms to circle $\omega$. Hence $C_1$ is sent to $M$. Analogously, $B_1$ is sent to $N$. So $T$ is sent to the intersection point of lines $N{C}^{\prime \prime }$ and $M{B}^{\prime \prime }$, say $T^{\prime }$. Note that the tangent to $\omega$ at $M$ is parallel to the tangent to ${\omega }_a$ at $C^{\prime \prime }$. Thus the direct homothetic center of $\omega$ and ${\omega }_a$, lies on line $N{C}^{\prime \prime }$. Denote this point by $T^{\prime \prime }$. By a similar argument, $T^{\prime \prime }$ lies on $M{B}^{\prime \prime }$ and hence $T^{\prime \prime }=T^{\prime }$ lies on line $O{I}_a$. This implies that $T^{\prime }{I}_a$ is perpendicular to $\omega$. Since the transformation above preserves angles, we deduce that the circumcircle of $ATI$ (the transformation of $T^{\prime }{I}_a$) is perpendicular to $BC$ (the transformation of $\omega$) and hence the center of this circle lies on the side $BC$.