Problem (2023RMMSLG1): Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. The incircle of the triangle $ABC$ touches the sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. The circumcircle of triangle $ADI$ crosses $\omega$ again at $P$, and the lines $PE$ and $PF$ cross $\omega$ again at $X$and $Y$, respectively. Prove that the lines $AI$, $BX$ and $CY$ are concurrent.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; o = circumcenter a b c; d = foot i b c; e = foot i c a; f = foot i a b; o1 = circumcenter a d i; p = on_circle o a, on_circle o1 a; x = on_line p e, on_circle o a; y = on_line p f, on_circle o a; z = on_line a i, on_line b x ? coll z c y
HAGeo formalization: A B C = triangle; I = incenter A B C; O = circumcenter A B C; BC = line B C; D = foot I BC; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; O1 = circumcenter A D I; (O1) = circle_center_point O1 A; (O) = circle_center_point O A; P = intersection (O1) (O); EP = line E P; X = intersection EP (O); FP = line F P; Y = intersection FP (O); AI = line A I; BX = line B X; Z = intersection AI BX; Prove: collinear Z C Y
Difficulty score: 2.7