2016 European Girls' Mathematical Olympiad

Let the line $X_k{T}_k$ and $\omega$ meet again at $X_k^{\prime }$, $k=1,2$, and notice that the tangent $t_k$ to ${\omega }_k$ at $X_k$ and the tangent $t_k^{\prime }$ to $\omega$ at $X_k^{\prime }$ are parallel. Since the ${\omega }_k$ have equal radii, the $t_k$ are parallel, so the $t_k^{\prime }$ are parallel, and consequently the points $X_1^{\prime }$ and $X_2^{\prime }$ coincide (they are not antipodal, since they both lie on the same side of the line $T_1{T}_2$). The conclusion follows.





Alternative Solution:

The circle $\omega$ is the image of ${\omega }_k$ under a homothety $h_k$ centered at $T_k$, $k=1,2$. The tangent to $\omega$ at $X_k^{\prime }=h_k(X_k)$ is therefore parallel to the tangent $t_k$ to ${\omega }_k$ at $X_k$. Since the ${\omega }_k$ have equal radii, the $t_k$ are parallel, so $X_1^{\prime }=X_2^{\prime }$; and since the points $X_k,T_k$ and $X_k^{\prime }$ are collinear, the conclusion follows.





Alternative Solution:

Invert from $X_1$ and use an asterisk to denote images under this inversion. Notice that ${\omega }_k^{\ast }$ is the tangent from $X_2^{\ast }$ to ${\omega }^{\ast }$ at $T_k^{\ast }$, and the pole $X_1$ lies on the bisectrix of the angle formed by the ${\omega }_k^{\ast }$, not containing ${\omega }^{\ast }$. Letting $X_1{T}_1^{\ast }$ and ${\omega }^{\ast }$ meet again at $Y$, standard angle chase shows that $Y$ lies on the circle $X_1{X}_2^{\ast }{T}_2^{\ast }$, and the conclusion follows.