Problem (2013USAMOp1): In triangle $ABC$, points $P$, $Q$, $R$ lie on sides $BC$, $CA$, $AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X$, $Y$, $Z$, respectively, prove that $YX/XZ=BP/PC$.
AlphaGeometry formalization: a b c = triangle; p = on_line b c; q = on_line a c; r = on_line a b; x = on_line a p, on_circum a q r; y = on_line a p, on_circum b r p; z = on_line a p, on_circum c p q ? eqratio y x x z b p p c
HAGeo formalization: A B C = triangle; BC = line B C; P = on_line BC; AC = line A C; Q = on_line AC; AB = line A B; R = on_line AB; ωa = circumcircle A Q R; ωb = circumcircle B R P; ωc = circumcircle C P Q; AP = line A P; X = intersection AP ωa; Y = intersection AP ωb; Z = intersection AP ωc; Prove: equal_ratio X Y X Z B P C P
Difficulty score: 1.8