Two circles in the plane do not intersect and do not lie inside each other. We choose diameters $A_1B_1$ and $A_2B_2$ of these circles such that the segments $A_1A_2$ and $B_1B_2'$ intersect. Let $A$ and $B$ be the midpoints of the segments $A_1A_2$ and $B_1B_2$, and $C$ be the intersection point of these segments. Prove that the orthocenter of the triangle $ABC$ belongs to a fixed line that does not depend on the choice of diameters.
Problem
Source: Baltic Way 2018, Problem 15
Tags: geometry, radical axis, Radical center
leon.tyumen
06.11.2018 17:40
Let $\omega_1, \omega_2$ are our circles. Also let $H$ is orthocenter of $\triangle ABC, B_1B_2 \cap \omega_1=Z \neq B_1, B_1B_2 \cap \omega_2 = T \neq B_2$, $A_1A_2 \cap \omega_1 =X \neq A_1, A_1A_2 \cap \omega_2 = Y \neq A_2$. $\angle A_1ZB_2=\angle A_1YB_2=90 \implies A_1, Z, Y, B_2$ lies om a circle $\implies A_1Z \cap B_2Y=K$ where $K$ lies on line $l$ - radical axe of circles $\omega_1, \omega_2$. If $B_1X \cap A_2T=L$, similary we can prove that $L \in l$. Also we know that $AH \parallel A_1K \parallel A_2L, BH \parallel B_2K \parallel B_1L$, and $A, B$ are midpoints of segments $A_1A_2, B_1B_2$ respectively, then, it is easy to prove that $H$ is midpoint of $KL \implies H \in l$, and $l$ is fixed. QED
Mogmog8
03.06.2022 05:33
We claim the orthocenter lies on the radical axis of the two circles, $\ell.$ Notice each point of $\triangle ABC$ is the midpoint of the corresponding points of $\triangle A_1CB_2$ and $\triangle A_2CB_1,$ so the orthocenter of $\triangle ABC,$ $H$ lies on the line connecting the orthocenters of $\triangle A_1CB_2$ and $\triangle A_2CB_1,$ $H_1$ and $H_2$ respectively. We see $\ell$ is the Steiner line of $A_1B_1B_2A_2.$ Hence, $\ell=\overline{H_1H_2}$ contains $H.$ $\square$