Problem (2022EGMOp1): Let $ABC$ be an acute-angled triangle in which $BC<AB$ and $BC<CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and $BQ = BC = CP$. Let $T$ be the circumcenter of triangle $APQ$, $H$ the orthocenter of triangle $ABC$, and $S$ the point of intersection of the lines $BQ$ and $CP$. Prove that $T$, $H$, and $S$ are collinear.
AlphaGeometry formalization: a b c = triangle; q = on_line a c, on_circle b c; p = on_line a b, on_circle c b; t = circumcenter a p q; h = orthocenter a b c; s = on_line b q, on_line c p ? coll t h s
HAGeo formalization: A B C = acute_triangle; AB = line A B; (C) = circle_center_point C B; P = intersection AB (C); AC = line A C; (B) = circle_center_point B C; Q = intersection AC (B); T = circumcenter A P Q; H = orthocenter A B C; BQ = line B Q; CP = line C P; S = intersection BQ CP; Prove: collinear T H S
Difficulty score: 2.0