Problem (2023PlanetCupp10): In obtuse $\triangle ABC$ ($\angle BAC>90^\circ$) with centroid $G$, let the nine-point circle and circumcircle intersect at $P,Q$. Lines $PG$ and $QG$ meet $BC$ at $X,Y$ respectively. Prove that the intersection point of $PY$, $QX$, and $AG$ lies on the circumcircle of $\triangle ABC$.
AlphaGeometry formalization: a b c = triangle; m1 m2 m3 g = centroid a b c; o = circumcenter a b c; ni = circumcenter m1 m2 m3; p = on_circle o a, on_circle ni m1; q = on_circle o a, on_circle ni m1; x = on_line p g, on_line b c; y = on_line q g, on_line b c; k = on_line p y, on_line a g ? cyclic k a b c
HAGeo formalization: A B C = obtuse_triangle; G M1 M2 M3 = centroid A B C; O = circumcenter A B C; Ni = circumcenter M1 M2 M3; (O) = circle_center_point O A; (Ni) = circle_center_point Ni M1; P Q = intersection (O) (Ni); PG = line P G; BC = line B C; X = intersection PG BC; QG = line Q G; Y = intersection QG BC; PY = line P Y; AG = line A G; K = intersection PY AG; Prove: concyclic K A B C
Difficulty score: 6.5