First Round of the 73rd Czech and Slovak Mathematical Olympiad (December 12th, 2023)

Since $B$ and $Y$ lie on the same circle centered at $O$, we have $OY=OB$. Moreover, by Thales' Theorem, we see that $\angle AYB=90^{\circ }$. This means that $M$ is the midpoint of the hypotenuse of a right triangle $XYB$, so by Thales' Theorem, it is the circumcentre of $XYB$, hence we have $MY=MB$. This means that both the points $M$ and $O$ lie on the perpendicular bisector of $BY$ and since they are clearly distinct, we immediately get the desired conclusion.