Problem (2023IranTSTp6): $ABC$ is an acute triangle with orthocenter $H$. Point $P$ is in triangle $BHC$ that $\angle HPC = 3 \angle HBC $ and $\angle HPB =3 \angle HCB $. Reflection of point $P$ through $BH,CH$ is $X,Y$. if $S$ is the center of circumcircle of $AXY$ , Prove that: $$\angle BAS = \angle CAP$$
AlphaGeometry formalization: a b c = triangle; h = orthocenter a b c; h1 = reflect h b c; c1 = reflect c h b; b1 = reflect b h c; p = eqangle3 p h b c b1 h1, eqangle3 p h c b c1 h1; x = reflect p b h; y = reflect p c h; s = circumcenter a x y ? eqangle b a a s p a a c
HAGeo formalization: A B C = triangle; H = orthocenter A B C; BC = line B C; H1 = reflect_point_wrt_line H BC; BH = line B H; C1 = reflect_point_wrt_line C BH; CH = line C H; B1 = reflect_point_wrt_line B CH; ω1 = angle_equal2 H B B1 C H1; ω2 = angle_equal2 H C C1 B H1; P = intersection ω1 ω2; X = reflect_point_wrt_line P BH; Y = reflect_point_wrt_line P CH; S = circumcenter A X Y; Prove: equal_angle B A S P A C
Difficulty score: 5.0