Problem (2020POGCHAMPp2): Triangle $ABC$ has circumcircle $\Omega$ and incircle $\omega$. Let $M$ be the midpoint of $\widehat{BAC}$ in $\Omega$ and let $\omega$ and $BC$ meet at $D$. Let $P$ be the unique point on $\omega$ such that $\angle MPD = 90^\circ$. The circumcircle of $\triangle MAP$ intersects $\omega$ again at $X$ and $BC$ at distinct points $Y$ and $Z$. Prove that either $AX$ and $MY$ intersect on $\Omega$ or $AX$ and $MZ$ intersect on $\Omega$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; i = incenter a b c; d = foot i b c; m = on_bline b c, on_circle o a; n = midpoint m d; p = on_circle i d, on_circle n m; o1 = circumcenter m a p; x = on_circle o1 a, on_circle i d; y = on_line b c, on_circle o1 a; t = on_line a x, on_line m y ? cyclic t a b c
HAGeo formalization: A B C = triangle; O = circumcenter A B C; I = incenter A B C; (O) = circle_center_point O A; l = perpendicular_bisector B C; M = intersection (O) l; BC = line B C; D = foot I BC; (I) = circle_center_point I D; N = midpoint D M; (N) = circle_center_point N M; P = intersection (I) (N); O1 = circumcenter A M P; (O1) = circle_center_point O1 A; X = intersection (O1) (I); Y = intersection (O1) BC; AX = line A X; MY = line M Y; T = intersection AX MY; Prove: concyclic T A B C
Difficulty score: 3.0