# Task "Inversion" (geometry with many circles)

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.

• Why did you include the word “inversion” in the title? It suggests circle inversions, but I don't see any of these.
– MvG
Oct 3, 2014 at 21:13
• Thats the name of the task. Im not the author. Oct 4, 2014 at 7:27
• Then it would be helpful if you could indicate where this task originated. That would provide some context which in turn might suggest appropriate tools to tackle this.
– MvG
Oct 4, 2014 at 7:29
• The task was originated in one of ukrainian school team math competitions (this task is one of the hardest to solve) Oct 4, 2014 at 7:38
• The figure is misleading $A,L,I$ are collinear. Oct 4, 2014 at 14:33

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 $K'AT$.

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*}