Question 39
Let $ABC$ be an acute, scalene triangle with circumcenter $O$ and symmedian point $K$. Let $X$ be the point on the circumcircle of triangle $BOC$ such that $\angle AXO = 90^{\circ}$. Assume that $X \neq K$. The hyperbola passing through $B, C, O, K$, and $X$ intersects the circumcircle of triangle $ABC$ at points $U$ and $V$, distinct from $B$ and $C$. Prove that $UV$ is the perpendicular bisector of $AX$.
The symmedian point of triangle $ABC$ is the intersection of the reflections of $B$-median and $C$-median across the angle bisectors of $\angle ABC$ and $\angle ACB$, respectively.
The symmedian point of triangle $ABC$ is the intersection of the reflections of $B$-median and $C$-median across the angle bisectors of $\angle ABC$ and $\angle ACB$, respectively.