Question 26
Let $ABC$ be an acute triangle with circumcircle $\omega$. Prove that there is a point $J$ with the following property: if $X$ is an interior point of $ABC$, the rays $AX,BX,CX$ meet $\omega$ again at $A_1,B_1,C_1$, and $A_2,B_2,C_2$ are respectively the reflections of $A_1,B_1,C_1$ in the midpoints of $\overline{BC},\overline{CA},\overline{AB}$, then the four points $A_2,B_2,C_2,J$ lie on a common circle.