Macedonian Mathematical Olympiad

Let $O$ be the center of the circumscribed circle around $△ABC$. We will show that each of the lines $p_A,p_B,p_C$ passes through $O$.

Because of symmetry, it is enough to show that $OC\perp H_A{H}_B$.

Let $D$ be the point of intersection of these two lines. We restrict ourselves to the case where $△ABC$ is acute (since in the case of $△ABC$ being obtuse the argument is analogous). It is enough to use the fact that $\angle H_{A}CD=\angle BCO=90^{\circ }-\alpha$ and $\angle D{H}_{A}C=\angle H_B{H}_{A}C=\alpha$ (the last equality follows from the fact that $AB{H}_A{H}_B$ is inscribed).

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Alternative solution.

According to Carnot's theorem, it is sufficient (and necessary) to show that $$|A{H}_B{|}^2-|A{H}_C{|}^2+|B{H}_C{|}^2-|B{H}_A{|}^2+|C{H}_A{|}^2-|C{H}_B{|}^2=0.$$ For that purpose, it is sufficient to sum the obvious equalities: $|B{H}_C{|}^2-|A{H}_C{|}^2=|BC{|}^2-|AC{|}^2$, $|C{H}_A{|}^2-|B{H}_A{|}^2=|AC{|}^2-|AB{|}^2$, and $|A{H}_B{|}^2-|C{H}_B{|}^2=|AB{|}^2-|BC{|}^2$.