Problem (2023ARMOg11p4): Let $\omega$ be the circumcircle of triangle $ABC$ with $AB<AC$. Let $I$ be its incenter and let $M$ be the midpoint of $BC$. The foot of the perpendicular from $M$ to $AI$ is $H$. The lines $MH, BI, AB$ form a triangle $T_b$ and the lines $MH, CI, AC$ form a triangle $T_c$. The circumcircle of $T_b$ meets $\omega$ at $B'$ and the circumcircle of $T_c$ meets $\omega$ at $C'$. Prove that $B', H, C'$ are collinear.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; i = incenter a b c; m = midpoint b c; h = foot m a i; v1 = on_line m h, on_line b i; v2 = on_line m h, on_line a b; o2 = circumcenter v1 v2 b; b1 = on_circle o2 b, on_circle o a; w1 = on_line m h, on_line c i; w2 = on_line m h, on_line a c; o3 = circumcenter w1 w2 c; c1 = on_circle o3 c, on_circle o a ? coll b1 h c1
HAGeo formalization: A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; I = incenter A B C; M = midpoint B C; AI = line A I; H = foot M AI; HM = line H M; BI = line B I; AB = line A B; V1 = intersection HM BI; V2 = intersection HM AB; O2 = circumcenter V1 V2 B; (O2) = circle_center_point O2 B; B1 = intersection (O2) (O); CI = line C I; AC = line A C; W1 = intersection HM CI; W2 = intersection HM AC; O3 = circumcenter W1 W2 C; (O3) = circle_center_point O3 C; C1 = intersection (O) (O3); Prove: collinear B1 H C1
Difficulty score: 3.6