Problem (JCGV2p2(2023ShuZhiMiMOM9p2)): Given $\triangle ABC$ and $\triangle XYZ$ sharing a circumcircle and isogonal conjugates $P,Q$. Let $YZ\cap AB=W$, $YZ\cap AC=V$, and define $U,T,S,R$ similarly. The circumcircles of $\triangle ZRW$, $\triangle XST$, $\triangle YUV$ meet $\odot ABC$ again at $M,N,O$. Prove $AN$, $BO$, $CM$ concur at a point depending only on $P,Q$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; p = free; x = on_circle o a, on_line a p; y = on_circle o a, on_line b p; z = on_circle o a, on_line c p; w = on_line y z, on_line a b; v = on_line y z, on_line a c; s = on_line x z, on_line b c; r = on_line x z, on_line b a; u = on_line x y, on_line c a; t = on_line x y, on_line c b; o1 = circumcenter z r w; m = on_circle o1 z, on_circle o a; o2 = circumcenter x s t; n = on_circle o2 x, on_circle o a; o3 = circumcenter y u v; q = on_circle o3 y, on_circle o a; k = on_line a n, on_line b q ? coll k c m
HAGeo formalization: A B C = triangle; O = circumcenter A B C; P = point; (O) = circle_center_point O A; AP = line A P; X = intersection (O) AP; BP = line B P; Y = intersection (O) BP; CP = line C P; Z = intersection (O) CP; YZ = line Y Z; AB = line A B; W = intersection YZ AB; AC = line A C; V = intersection YZ AC; XZ = line X Z; BC = line B C; S = intersection XZ BC; R = intersection XZ AB; XY = line X Y; U = intersection XY AC; T = intersection XY BC; O1 = circumcenter Z R W; (O1) = circle_center_point O1 Z; M = intersection (O1) (O); O2 = circumcenter X S T; (O2) = circle_center_point O2 X; N = intersection (O2) (O); O3 = circumcenter Y U V; (O3) = circle_center_point O3 Y; Q = intersection (O3) (O); AN = line A N; BQ = line B Q; K = intersection AN BQ; Prove: collinear K C M
Difficulty score: 5.8