Problem 1213

Solution. Let the circumcircle of triangle $ABC$ be denoted by $\Omega$. Suppose the bisector $CI$ intersects $\Omega$ again at point $S$. Then, as is known, $SA=SB=SI$, which means point $S$ is the center of circle $\Gamma$. By symmetry, point $Z$ lies on the line $SC$.

Let the common tangents to circles $\omega$ and $\Gamma$ touch $\Gamma$ at points $M$ and $N$ (see Fig. 10). The line of centers $SI$ is the perpendicular bisector of segment $MN$, so $\angle IMN = \angle INM = \angle IMZ$ (the last equality holds because line $MZ$ is tangent to $\Gamma$). Therefore, $MI$ is the bisector of angle $ZMN$, meaning the distances from $I$ to $ZM$ and $MN$ are equal. Since $\omega$ is tangent to $ZM$, it is also tangent to line $MN$ at some point $Z'$; by symmetry, this point lies on $SI$.

Right triangles $SZ'M$ and $SMZ$ are similar, so $SZ \cdot SZ' = SM^2$. This means that under inversion with respect to circle $\Gamma$, point $Z'$ maps to point $Z$. Therefore, the circle $\omega$, containing points $X, Y$, and $Z'$, maps to the circumcircle of triangle $XYZ$. Furthermore, under this inversion, line $AB$ maps to circle $\Omega$. Since $\omega$ and $AB$ are tangent, their images will also be tangent, as required.

Remark 1. The fact that the circumcircle of $XYZ$ maps to $\omega$ under inversion with respect to $\Gamma$ can also be proven differently. Let $r$ and $\rho$ be the radii of circles $\omega$ and $\Gamma$; let $Z'$ be the intersection of segment $IS$ with $\omega$. Then $SZ' = \rho - r$. On the other hand, from the homothety with center at $Z$,
mapping $\omega$ to $\Gamma$, we have $\frac{r}{\rho} = \frac{ZI}{ZS} = 1 - \frac{\rho}{ZS}$, from which $SZ = \frac{\rho^2}{\rho - r}$. Therefore, $SZ \cdot SZ' = \rho^2$.

Remark 2. Another solution can be obtained by performing an inversion with respect to circle $\omega$. Under this inversion: points $A, B, C$ map to the midpoints $A'', B'', C''$ of sides $B'C', C'A'$, $A'B'$ of the triangle with vertices at the points of tangency of $\omega$ with the sides; circle $\Gamma$ maps to line $XY$, which contains the midline $A''B''$ of triangle $A'B'C'$; the circumcircle of triangle $XYZ$ maps to circle $\omega'$, symmetric to $\omega$ with respect to $XY$. Therefore, we need to prove that $\omega'$ is tangent to the circumcircle of triangle $A''B''C''$. This is true because under symmetry with respect to $XY$, the latter circle maps to the circumcircle of $A''B''C'$, which is tangent to $\omega$.