Problem

Source: 2015 XVIII All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p23

Tags: geometry, Locus, Ukrainian TYM



An acute-angled triangle $ABC$ is given, through the vertices $B$ and $C$ of which a circle $\Omega$, $A \notin \Omega$, is drawn. We consider all points $P \in \Omega$, that do not lie on none of the lines $AB$ and $AC$ and for which the common tangents of the circumscribed circles of triangles $APB$ and $APC$ are not parallel. Let $X_P$ be the point of intersection of such two common tangents. a) Prove that the locus of points $X_P$ lies to some two lines. b) Prove that if the circle $\Omega$ passes through the orthocenter of the triangle $ABC$, then one of these lines is the line $BC$.