Team selection tests for GMO 2018

First, note that $OM\perp AB$ then $OM\parallel XE$. But $OMS$ is isosceles triangle implies that triangle $XES$ is also isosceles, or $XE=XS$. Similarly, $YE=YT$. Denote $\left({\omega }_1\right),\left({\omega }_2\right)$ as the circle of center $X$, radius $XS$ and center $Y$, radius $YT$. Since $CS\perp XS$, we have $CS$ is tangent to $\left({\omega }_1\right)$ so ${\mathcal{P}}_{C/\left({\omega }_1\right)}=C{S}^2$. On the other hand, ${\mathcal{P}}_{C/\left({\omega }_2\right)}=C{T}^2$ and $CS=CT$ imply that $C$ belongs to radical axis of two circles $\left({\omega }_1\right),\left({\omega }_2\right)$.

By similar triangles, we get $M{A}^2=MS\cdot ME$ and $M{B}^2=MF\cdot MT$, but $MA=MB$ then $MS\cdot ME=MF\cdot MT$, thus ${\mathcal{P}}_{M/\left({\omega }_1\right)}={\mathcal{P}}_{M/\left({\omega }_2\right)}$. By combining these results, we obtain $CM$ is the radical axis of two circles $\left({\omega }_1\right),\left({\omega }_2\right)$; hence, $CM\perp XY$.