Problem 2000

Assume without loss of generality that $M$ lies on the same side of line $BC$ as the vertex $A$. In this case, we denote by $(I)$ the common incircle of triangles $ABC$, and $MX_1X_2$ with radius $r$, and let $D, E, F, Y_1, Y_2$ be the tangency points of $(I)$ with the sidelines $BC, CA, AB, MX_1, MX_2$.

Consider the inversion $\Psi$ with center $I$ and power $r^2$, which takes the vertices $A, B, C, M, X_1, X_2$ to the midpoints $A', B', C', M', X_1', X_2'$ of the sides $EF, FD, DE, Y_2Y_1, Y_1D, DY_2$ of the intouch triangles $DEF$, and $DY_2Y_1$. The circumcircles $(O), (P)$ of triangles $ABC$, and $MX_1X_2$ become the circumcircles $(O'), (P')$ of the triangles $A'B'C', M'X_1'X_2'$, which, because they coincide with the nine-point circles of two triangles of the same circumcircle, are congruent and have the common radius $\frac{r}{2}$. Since the sidelines $BC, CA, AB, MX_1, MX_2$ are tangent to the inversion circle $(I)$, their images under $\Psi$ are the congruent circles $\Gamma_a, \Gamma_b, \Gamma_c, \Omega_1, \Omega_2$ with diameters $ID, IE, IF, IX_1, IX_2$, respectively. The congruent circles $(O'), \Gamma_b, \Gamma_c$

with radii $\frac{r}{2}$ meet at point $A'$. Thus, a circle $(A', r)$ with center $A'$ and radius $r$ is tangent to all three at points diametrically opposite to $A'$.

The internal angle-bisector $AI$ of the angle $\angle A$ passing through the inversion center $I$ is carried into itself. The mixtilinear incircle $(K_a)$ and the mixtilinear excircle $(L_a)$ of the triangle $ABC$ in the angle $A$ are the only two circles centered on $AI$ and simultaneously tangent to $CA, AB$, and $(O)$. Since the inversion center $I$ is the similarity center of a circle and its inversion images, only the images $(K'_a), (L'_a)$ of $(K_a), (L_a)$ are centered on $AI$ and tangent to $\Gamma_b, \Gamma_c, (O')$. The mixtilinear excircle $(L_a)$, lying outside of the circumcircle $(O)$ and outside of the inversion circle $(I)$, has both intersections with $AI$ on the ray $\overrightarrow{IL_a}$. Since the inversion in $(I)$ has positive power $r^2$, its images $(L'_a)$ also has both intersections with $AI$ on the ray $\overrightarrow{IL_a}$ and it is centered on the ray $\overrightarrow{IA}$. It cannot be identical with the circle $(A', r)$ centered on the opposite ray $\overrightarrow{IA}$. Therefore, the image of the mixtilinear incircle $(K_a)$ in angle $A$ under $\Psi$ is the circle $(A', r)$. Furthermore, the inverse image of the tangency point $Z$ of the circles $(K_a)$, and $(O)$ is the tangency point $Z'$ of $(A', r)$ and $(O)$, the antipode of $A'$ with respect to the circumcircle $(O')$.

Let now $A_0, B_0, C_0, O_1, O_2$ be the centers of the congruent circles $\Gamma_a, \Gamma_b, \Gamma_c, \Omega_1, \Omega_2$. Since $\Gamma_a$ is the reflection of $(O')$ in $B'C'$ and since $O'$ and $I$ are isogonal conjugated with respect to triangle $A'B'C'$, the quadrilateral $B'Z'C'I$ is a parallelogram, and therefore, its diagonals $B'C', IZ'$ cut each other at half at the midpoint of segment $B'C'$. It now follows that the quadrilateral $IA_0Z'O'$ is also a parallelogram, and thus, $IA_0 = O'Z' = \frac{r}{2}$. Now since $A_0I = A_0X'_1 = O_1I = O_1X'_1 = \frac{r}{2}$, the quadrilateral $A_0IO_1X'_1$ is a rhombus, and since the segments $O_1X'_1, IA_0, O'Z'$ are parallel and congruent, the quadrilateral $O_1X'_1Z'O'$ is a parallelogram. In conclusion, the triangles $PX'_1Z'$, and $M'O_1O'$ are congruent and $P'Z' = M'O' = \frac{r}{2}$. Hence, $Z'$ lies on the circle $(P')$, which means that the second intersection of the circumcircle $(P)$ of triangle $MX_1X_2$ with the circumcircle $(O)$ of $ABC$ (different from $M$) coincides with the tangency point $Z$ of the mixtilinear incircle in angle $A$ with the circumcircle $(O)$.