Problem (2023SpFMOp5): In $\triangle ABC$, let $N,S$ be the north/south poles, $M$ the midpoint of $BC$, and $D\in NS$ with $ND=MS$. A circle centered at $D$ with radius $AD$ meets the midline parallel to $BC$ at $X,Y$. Prove $X$, the Feuerbach point $F_e$, $Y$, and the $A$-excircle's touchpoint $P$ with $BC$ are concyclic.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; s = on_bline b c, on_circle o a; n = on_bline b c, on_circle o a; m1 = midpoint b c; d = mirror m1 o; m3 = midpoint a b; m2 = midpoint a c; x = on_circle d a, on_line m2 m3; y = on_circle d a, on_line m2 m3; ni = circumcenter m1 m2 m3; i = incenter a b c; d1 = foot i b c; f_e = on_circle i d1, on_circle ni m1; i_a = excenter a b c; p = foot i_a b c ? cyclic x f_e y p
HAGeo formalization: A B C = triangle; O = circumcenter A B C; l1 = perpendicular_bisector B C; (O) = circle_center_point O A; S = intersection l1 (O); N = intersection l1 (O); M1 = midpoint B C; D = reflect_point_wrt_point M1 O; M3 = midpoint A B; M2 = midpoint A C; (D,A) = circle_center_point D A; M2M3 = line M2 M3; X = intersection (D,A) M2M3; Y = intersection (D,A) M2M3; Ni = circumcenter M1 M2 M3; I = incenter A B C; BC = line B C; D1 = foot I BC; (I) = circle_center_point I D1; (Ni) = circle_center_point Ni M1; Fe = intersection (I) (Ni); Ia = excenter A B C; P = foot Ia BC; Prove: concyclic X Fe Y P
Difficulty score: 5.4