Problem (2021ARMOg10p8): Given is a cyclic pentagon $ABCDE$, inscribed in a circle $k$. The line $CD$ intersects $AB$ and $AE$ in $X$ and $Y$ respectively. Segments $EX$ and $BY$ intersect again at $P$, and they intersect $k$ in $Q$ and $R$, respectively. Point $A'$ is reflection of $A$ across $CD$. The circles $(PQR)$ and $(A'XY)$ intersect at $M$ and $N$. Prove that $CM$ and $DN$ intersect on $(PQR)$.
AlphaGeometry formalization: a b c = triangle; d = on_circum a b c; e = on_circum a b c; o = circumcenter a b c; x = on_line c d, on_line a b; y = on_line c d, on_line a e; p = on_line e x, on_line b y; q = on_line e x, on_circle o a; r = on_line b y, on_circle o a; a1 = reflect a c d; o1 = circumcenter p q r; o2 = circumcenter a1 x y; m = on_circle o1 p, on_circle o2 x; n = on_circle o1 p, on_circle o2 x; z = on_line c m, on_line d n ? cyclic z p q r
HAGeo formalization: A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; D = on_circle (O); E = on_circle (O); CD = line C D; AB = line A B; X = intersection CD AB; AE = line A E; Y = intersection CD AE; EX = line E X; BY = line B Y; P = intersection EX BY; Q = intersection EX (O); R = intersection BY (O); A1 = reflect_point_wrt_line A CD; O1 = circumcenter P Q R; O2 = circumcenter A1 X Y; (PQR) = circle_center_point O1 P; (A1XY) = circle_center_point O2 X; M N = intersection (PQR) (A1XY); CM = line C M; DN = line D N; T = intersection CM DN; Prove: concyclic T P Q R
Difficulty score: 4.0