60-A Problem of Seven circle

Problem:

Solution: by Telv Cohl 

Denote $O$ as the center of black circle, $O_{12}, O_{23}, O_{31}$ the center of three green circles $O_1, O_2, O_3$ the center of three red circles (in Dao's picture), $(O_1)$ tangent $(O_{12}) (O_{31})$ at $X, Y$ respectively. 

Let $H_1$ be the external center of similitude of $(O_{12})$ and $(O_{31})$ similarity define $H_2$ and $H_3$ inverse with center $H_1$ and power $H_1T_2.H_1T_3$. Since $H_1$ is the external center of similitude of $(O_{12})$ and $(O_{31})$ so this inversion map $(O_{12})$ and $(O_{31})$ to each other ... (1)

Easy to see that this inversion map $(O)$ to itself ... (2). Since $X$ is the internal center of similitude of $(O_{12})$ and $(O_1)$ and Y the internal center of similitude of $(O_1)$ and $(O_{31})$. 

By D'Alembert theorem we get $X, Y, H_1$ are collinear, and $H_1X.H_1Y=H_1T_2.H_1T_3$ hence this inversion also map $(O_1)$ to itself ... (3)

From (1) (2) (3) we deduce that this inversion map $(O_2)$ and $(O_3)$ to each other so $H_1, O_2, O_3$ are collinear  ie. $H_1$ is the intersection of $O_2O_3$ and $O_{12}O_{31}$ ... (4) similarity $H_2$ is the intersection of $O_3O_1$ and $O_{12}O_{23}$ ... (5) $H_3$ is the intersection of $O_1O_2$ and $O_{23}O_{31}$ ... (6)

By D'Alembert theorem we get $H_1, H_2, H_3$ are collinear ... (7) From (4) (5) (6) (7) we get $T_{12}T_{23}T_{31}$ and $T_3T_1T_2$ are perspective.
By Desargue theorem we get $T_1T_{23}, T_2T_{31}, T_3T_{12}$ are concurrent. Q.E.D