Incircle $\omega$ of triangle $ABC$ with center in point $I$ touches $AB, BC, CA$ in points $C_{1}, A_{1}, B_{1}$. Сircumcircle of triangle $AB_{1}C_{1}$ intersects second time circumcircle of $ABC$ in point $K$. Point $M$ is midpoint of $BC$, $L$ midpoint of $B_{1}C_{1}$. Сircumcircle of triangle $KA_{1}M$ intersects second time $\omega$ in point $T$.
Prove, that сircumcircles of triangle $KLT$ and triangle $LIM$ touch.
On
Without loss of generality you can choose a coorinate system in such a way that $\omega$ is the unit circle and $A_1=(1,0)$. Then you can use a rational parametrization of the circle to describe $B_1$ and $C_1$ in terms of two parameters $b,c\in\mathbb R$, ending up with these coordinates:
\begin{align*} I &= \begin{pmatrix} 0\\ 0 \end{pmatrix} & A_1 &= \begin{pmatrix} 1\\ 0 \end{pmatrix} \\ B_1 &= \frac1{b^{2} + 1}\begin{pmatrix} b^{2} - 1\\ 2 b \end{pmatrix} & C_1 &= \frac1{c^{2} + 1}\begin{pmatrix} c^{2} - 1\\ 2 c \end{pmatrix} \end{align*}
From there on, you can do a brute force coordinate computation, yielding these intermediate points:
\begin{align*} A &= \frac1{b c + 1}\begin{pmatrix} b c - 1\\ b + c \end{pmatrix} \\ B &= \frac1{c}\begin{pmatrix} c\\ 1 \end{pmatrix} \\ C &= \frac1{b}\begin{pmatrix} b\\ 1 \end{pmatrix} \\ M &= \frac1{2 b c}\begin{pmatrix} 2 b c\\ b + c \end{pmatrix} \\ L &= \frac1{b^{2} c^{2} + b^{2} + c^{2} + 1}\begin{pmatrix} b^{2} c^{2} - 1\\ b^{2} c + b c^{2} + b + c \end{pmatrix} \\ K &= \frac1{b^{2} c^{2} + b^{2} + 4 b c + c^{2} + 1}\begin{pmatrix} b^{2} c^{2} + b^{2} + 2 b c + c^{2} - 1\\ 2 b + 2 c \end{pmatrix} \\ T &= {\scriptsize\frac1{16 b^{4} c^{4} + 9 b^{4} c^{2} + 18 b^{3} c^{3} + 9 b^{2} c^{4} + b^{4} + 6 b^{3} c + 10 b^{2} c^{2} + 6 b c^{3} + c^{4} + b^{2} + 2 b c + c^{2}}}\cdot\\ &\phantom={\scriptsize\begin{pmatrix} 16 b^{4} c^{4} + 7 b^{4} c^{2} + 14 b^{3} c^{3} + 7 b^{2} c^{4} + b^{4} + 2 b^{3} c + 2 b^{2} c^{2} + 2 b c^{3} + c^{4} - b^{2} - 2 b c - c^{2}\\ 8 b^{4} c^{3} + 8 b^{3} c^{4} + 2 b^{4} c + 14 b^{3} c^{2} + 14 b^{2} c^{3} + 2 b c^{4} + 2 b^{3} + 6 b^{2} c + 6 b c^{2} + 2 c^{3} \end{pmatrix}} \end{align*}
I'd like to paste equations for the circles as well, but I still have to work out how to format them since they are pretty big.
The circles $KLT$ and $LIM$ then have a common tangent:
\begin{align*}{\scriptsize \bigl((b + c) \cdot (2 b^{4} c^{4} - 3 b^{4} c^{2} + 4 b^{3} c^{3} - 3 b^{2} c^{4} - b^{4} - 8 b^{2} c^{2} - c^{4} - b^{2} - 4 b c - c^{2})\bigr)x} \\ {\scriptsize+ \bigl(5 b^{5} c^{3} + 2 b^{4} c^{4} + 5 b^{3} c^{5} + b^{5} c + b^{4} c^{2} + 16 b^{3} c^{3} + b^{2} c^{4} + b c^{5} - b^{4} + 3 b^{3} c + 3 b c^{3} - c^{4} - b^{2} - 2 b c - c^{2}\bigr)y} \\ {\scriptsize= 2\cdot c \cdot b \cdot (b + c) \cdot (b c + 1)^{3}} \end{align*}
HINT for a less tour-de-force approach (impressive though MvG, or his software, may be)
The obvious idea is to invert (as the question hints). The first thing to try is clearly the incircle. Denote the inverse point of $X$ as $X'$. So the line $B_1C_1$ becomes the circle $AB_1C_1$ and hence $L'=A$. Like $A'=L$, $K'$ lies on the line $B_1C_1$ and the inverse of the circumcircle, which is the circumcircle of the three midpoints of the sides of $A_1B_1C_1$. $T$ lies on the incircle, so $T'=T$.
Thus the inverse of the circle $KLT$ is the circle $K'AT$ and the inverse of the circle $LIM$ is the line $AM'$. So we need to show that $AM'$ is tangent to the circle $K'AT$.