Problem (XinXingV28p2): For $\triangle ABC$ inscribed in $\odot O$ with unequal sides, let $J,K,L$ be midpoints of arcs $BAC,ABC,ACB$, and incircle $I$ touch sides at $D,E,F$. Extend $JD,KE,LF$ to meet $\odot O$ at $X,Y,Z$. Prove the radical center of $\odot XEF$, $\odot YFD$, $\odot ZDE$ lies on line $OI$.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; i = incenter a b c; j1 = on_line a i, on_circle o a; j = mirror j1 o; k1 = on_line b i, on_circle o a; k = mirror k1 o; l1 = on_line c i, on_circle o a; l = mirror l1 o; d = foot i b c; e = foot i a c; f = foot i a b; x = on_line j d, on_circle o a; y = on_line k e, on_circle o a; z = on_line l f, on_circle o a; o1 = circumcenter x e f; o2 = circumcenter y f d; o3 = circumcenter z d e; u = on_circle o1 f, on_circle o2 f; v = on_circle o1 e, on_circle o3 e; t = on_line u f, on_line v e ? coll t o i
HAGeo formalization: A B C = triangle; O = circumcenter A B C; I = incenter A B C; BC = line B C; AC = line A C; AB = line A B; AI = line A I; (O) = circle_center_point O A; J1 = intersection AI (O); J = reflect_point_wrt_point J1 O; BI = line B I; K1 = intersection BI (O); K = reflect_point_wrt_point K1 O; CI = line C I; L1 = intersection CI (O); L = reflect_point_wrt_point L1 O; D = foot I BC; E = foot I AC; F = foot I AB; JD = line J D; X = intersection JD (O); KE = line K E; Y = intersection KE (O); LF = line L F; Z = intersection LF (O); O1 = circumcenter X E F; O2 = circumcenter Y F D; O3 = circumcenter Z D E; (O1) = circle_center_point O1 F; (O2) = circle_center_point O2 F; U = intersection (O1) (O2); (O3) = circle_center_point O3 E; V = intersection (O1) (O3); UF = line U F; VE = line V E; T = intersection UF VE; Prove: collinear T O I
Difficulty score: 5.1