Beautiful triangle problem Circle, inscribed in $ABC$, touches $BC, CA, AB$ in points $A', B', C'$. $AA' BB', CC'$ intersect at $G$. Circumcircle of $GA'B'$ crosses the second time lines $AC$ and $BC$ at $C_A$ and $C_B$. Points $ A_B, A_C, B_C,B_A, C_A, C_B$ are concyclic.

Looks straightforward,  but I'm struggling to get something out of the given. I tried proving that two points and one line and two points and some other are concyclic and then combining three similar resul together, but not much yet.
 A: This is a coordinate-based approach, making heavy use of tools from projective geometry.
Without loss of generality, you can choose the coordinate system in such a way that the inscribed circle is the unit circle. On that you can use a rational parametrization, i.e. choose $a,b,c\in\mathbb R$ such that $A'=(1-a^2,2a)/(1+a^2)$ and likewise for $B'$ and $C'$. Or, even better, use homogeneous coordinates for these:
$$
A'=\begin{pmatrix}1-a^2\\2a\\1+a^2\end{pmatrix}\qquad
B'=\begin{pmatrix}1-b^2\\2b\\1+b^2\end{pmatrix}\qquad
C'=\begin{pmatrix}1-c^2\\2c\\1+c^2\end{pmatrix}
$$
From that, everything else follows. The tangent at a point on the circle is simply its polar line, which you get by multiplication with the matrix of the unit circle, namely the matrix
$$U=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}$$
You can join points using lines, and intersect lines to obtain points, simply by computing the cross product.
$$A=(U\cdot B')\times(U\cdot C')$$
and likewise for $B$ and $C$, the other two corners of the triangle. Then you get
$$G=(A\times A')\times(B\times B')$$
A circle through three points can be constructed as a conic through these points and the special points $I=(1,i,0)$ and $J=(1,-i,0)$ which have complex coordinates and lie on the line at infinity. Since they also lie on every circle, they are often called the ideal circle points. So we construct the matrix of the circle $\bigcirc GA'B'$ in several steps:
\begin{align*}
M_1 &= (A'\times I)\cdot(B'\times J)^T &
M_2 &= (A'\times J)\cdot(B'\times I)^T \\
M_3 &= M_1 + M_1^T & M_4 &= M_2 + M_2^T \\
M_5 &= (G\cdot M_3\cdot G)M_4 - (G\cdot M_4\cdot G)M_3 &
M_C &= iM_5
\end{align*}
$M_1$ and $M_2$ describe degenerate conics through $A',B',I,J$. $M_3$ and $M_4$ are the same except using symmetric matrices. $M_5$ is a linear combination which also passes through $G$, so that's the circle. $M_C$ is a real matrix describing the same circle, avoiding all the purely imaginary entries of $M_5$.
To intersect that circle with $AC$ (which is the same line as $B'C$) you compute
$$C_A=(C^T\cdot M_C\cdot C)B'-2(B'^T\cdot M_C\cdot C)C$$
This is still a homogeneous coordinate vector of a point, e.g. some $(x,y,z)$. Dehomogenize that to $(x/z, y/z)$ then take the norm of that:
\begin{align*}
\lVert C_A\rVert =& \frac{\sqrt s}{t} \\
s =& \phantom+ (a^4b^4 + a^4c^4 + b^4c^4) \\&
- 2\,abc\,(a^3b^2 + a^2b^3 + a^3c^2 + b^3c^2 + a^2c^3 + b^2c^3) \\&
+ 3\,a^2b^2c^2\,(a^2 + b^2 + c^2) \\&
+ 11\,(a^4b^2 + a^2b^4 + a^4c^2 + b^4c^2 + a^2c^4 + b^2c^4) \\&
+ 16\,abc\,(a^2b + ab^2 + a^2c + b^2c + ac^2 + bc^2) \\&
- 20\,(a^3b^3 + a^3c^3 + b^3c^3) \\&
- 20\,abc\,(a^3 + b^3 + c^3) \\&
- 42\,a^2b^2c^2 \\&
- 2\,(a^3b + ab^3 + a^3c + b^3c + ac^3 + bc^3) \\&
+ 3\,(a^2b^2 + a^2c^2 + b^2c^2) \\&
+ (a^4 + b^4 + c^4) \\
t =&\phantom+ (a^2 + b^2 + c^2) \\&
- (ab + ac + bc) \\&
+ (a^2b^2 + a^2c^2 + b^2c^2) \\&
- abc\,(a + b + c)
\end{align*}
This is the radius for one of the six points of your claimed circle. The other five can be obtained using the same computation, starting from a permutation of the three initial points. So the result will be the same except for a permutation of the parameters $a,b,c$. But the formula stated above is invariant under such a permutation, therefore all six points lie on a circle as claimed. Its center is the center of the coordinate system, i.e. the incenter of the triangle. Its radius will be the fraction described by the lengthy expressions stated above.
A: Consider $A$'s relationship with various circles.
Since $B^\prime$ and $C^\prime$ are points of tangency of lines through $A$ with the incircle, we have
$$|AB^\prime| = |AC^\prime| \qquad (\star)$$
Further, by the "Power of a Point" Theorem,
$$|AA_B||AC^\prime| = \operatorname{pow}(A\;,\;\bigcirc GB^\prime C^\prime) = |AA_C||AB^\prime|$$
Also,
$$|AB_A||AC^\prime| = \operatorname{pow}(A\;,\;\bigcirc A^\prime G C^\prime) = |AG||AA^\prime| = \operatorname{pow}(A\;,\;\bigcirc A^\prime B^\prime G) = |AC_A||AB^\prime|$$
Together with $(\star)$, these imply
$$|AA_B| = |AA_C| \qquad\text{and}\qquad |AB_A| = |AC_A|$$
This makes $\square A_B A_C B_A C_A$ an isosceles trapezoid, whose bases, $A_BA_C$ and $B_AC_A$, have the bisector of $\angle A$ as their common perpendicular bisector. The incenter, $I$, of $\triangle ABC$ lies on that bisector, so we can write
$$|IA_B| = |IA_C| \qquad\text{and}\qquad |IB_A| = |IC_A|$$
Identical arguments considering points $B$ and $C$ lead to corresponding relations
$$|IB_C| = |IB_A| \qquad |IC_B| = |IA_B| \qquad\text{and}\qquad|IC_A|=|IC_B|\qquad|IA_C|=|IB_C|$$
Necessarily, $I$ is equidistant from all six points $A_B$, $A_C$, $B_C$, $B_A$, $C_A$, $C_B$. $\square$
