Balti Tee 2018 · Ülesanne 15

We prove that the orthocenter $H$ of $\triangle ABC$ belongs to the radical axis of the two fixed circles.

Denote the circles by $s_{1}$ and $s_{2}$. Let the line $A_{1}A_{2}$ intersect $s_{1}$ and $s_{2}$ for the second time at points $X_{1}$ and $X_{2}$, respectively, and let the line $B_{1}B_{2}$ intersect the circles for the second time at points $Y_{1}$ and $Y_{2}$.

The lines $A_{1}Y_{1}$ and $A_{2}Y_{2}$ are parallel, because both are perpendicular to $B_{1}B_{2}$. Analogously, $B_{1}X_{1}$ and $B_{2}X_{2}$ are parallel. Hence these four lines form a parallelogram $KLMN$ (see the figure). The perpendicular from $A$ to $BC$ and the perpendicular from $B$ to $AC$ lie on the midlines of this parallelogram. Therefore $H$ is the center of $KLMN$ and coincides with the midpoint of $KM$.

It is therefore enough to prove that both $K$ and $M$ lie on the radical axis of $s_{1}$ and $s_{2}$.

The points $X_{1}$ and $Y_{2}$ lie on the circle $s_{3}$ with diameter $B_{1}A_{2}$. The line $B_{1}X_{1}$ is the radical axis of $s_{1}$ and $s_{3}$, while $A_{2}Y_{2}$ is the radical axis of $s_{2}$ and $s_{3}$. Thus $K$ is the radical center of these three circles and hence lies on the radical axis of $s_{1}$ and $s_{2}$. Analogously, $M$ lies on the same radical axis.