Problem (2023RMMSLG3): A point $P$ is chosen inside a triangle $ABC$ with circumcircle $\Omega$. Let $\Gamma$ be the circle passing through the circumcenters of the triangles $APB$, $BPC$, and $CPA$. Let $\Omega$ and $\Gamma$ intersect at points $X$ and $Y$. Let $Q$ be the reflection of $P$ in the line $XY$ . Prove that $\angle BAP = \angle CAQ$.
AlphaGeometry formalization: a b c = triangle; p = free; o = circumcenter a b c; o1 = circumcenter a p b; o2 = circumcenter b p c; o3 = circumcenter c p a; o_g = circumcenter o1 o2 o3; x = on_circle o a, on_circle o_g o1; y = on_circle o a, on_circle o_g o1; q = reflect p x y ? eqangle b a a p q a a c
HAGeo formalization: A B C = triangle; O = circumcenter A B C; (O) = circumcircle A B C; P = point; O1 = circumcenter A P B; O2 = circumcenter B P C; O3 = circumcenter C P A; O4 = circumcenter O1 O2 O3; (O1O2O3) = circle_center_point O4 O1; X Y = intersection (O) (O1O2O3); XY = line X Y; Q = reflect_point_wrt_line P XY; Prove: equal_angle B A P Q A C
Difficulty score: 5.0