, 2010

Let $AH_a$ be an altitude in the triangle $ABC$ and $P_a$ be its midpoint. Define analogously $H_b$, $P_b$, etc.

It is easy to see that the point $A'$ divides the segment $P_bP_c$ in the same ratio that $A_1$ divides $BC$. By Menelaus' theorem for the triangle $P_aP_bP_c$, claim (a) follows.

Consider an affine transformation mapping of the triangle $ABC$ onto the triangle $P_aP_bP_c$. When $l$ contains a fixed point $X$, $l'$ contains the fixed point $Y$ whose affine coordinates with respect to triangle $P_aP_bP_c$ equal the affine coordinates of $X$ with respect to $ABC$. We are now left to show that $X \equiv O \Leftrightarrow Y \equiv O_9$.

This is easiest to do by considering two special cases, say, when $l$ contains some vertex of the triangle $ABC$.

Another approach is this: Let $Z$ be the point whose affine coordinates with respect to triangle $H_aH_bH_c$ equal the affine coordinates of $X$ with respect to triangle $ABC$. Clearly, $Y$ is the midpoint of $XZ$. Let $O'$ be the circumcenter of triangle $H_aH_bH_c$. It is clear that $AO'HH_a$ is a straight line and by similar figures $AH_bH_cO' \sim ABCO$ we see that $AO$ divides $BC$ and $H_aH$ divides $H_bH_c$ in equal ratios. It follows that $X=O \Leftrightarrow Z=H$. $\square$