The symmetric difference of two homothetic triangles $T_1$ and $T_2$ consists of six triangles $t_1, \ldots, t_6$ with circumcircles $\omega_1, \omega_2, \ldots, \omega_6$ (counterclockwise, no two intersect). Circle $\Omega_1$ with center $O_1$ is externally tangent to $\omega_1, \omega_3,$ and $\omega_5$; circle $\Omega_2$ with center $O_2$ is externally tangent to $\omega_2, \omega_4,$ and $\omega_6$; circle $\Omega_3$ with center $O_3$ is internally tangent to $\omega_1, \omega_3,$ and $\omega_5$; circle $\Omega_4$ with center $O_4$ is internally tangent to $\omega_2, \omega_4,$ and $\omega_6$. Prove that $O_1O_3 = O_2O_4$. Proposed by Ilya Zamotorin