Problem (2024ShuZhiMiMOM3p5): In $\triangle ABC$, let $M$ be the midpoint of $BC$. The line $AM$ meets the circumcircle of $\triangle ABC$ at $E$. Denote by $N$ the midpoint of the arc $BAC$ (not containing $BC$). The line $NE$ intersects $BC$ at $L$. The line $AB$ meets $CN$ at $S$ and $MN$ at $T$. Let $\Gamma_1$ be the circle with diameter $BN$, and $\Gamma_2$ be the circumcircle of $\triangle BLT$; they intersect at point $U$. The lines $MS$ and $LT$ meet at $V$. Prove that the circumcircles of $\triangle CLV$ and $\triangle NTU$ are tangent to each other.
AlphaGeometry formalization: a b c = triangle; o = circumcenter a b c; m = midpoint b c; e = on_line a m, on_circle o a; n = on_bline b c, on_circle o a; l = on_line n e, on_line b c; s = on_line a b, on_line c n; t = on_line a b, on_line m n; o2 = circumcenter b l t; u = on_dia b n, on_circle o2 b; v = on_line m s, on_line l t; o3 = circumcenter c l v; o4 = circumcenter n t u; k = on_circle o3 c, on_circle o4 n ? coll k o3 o4
HAGeo formalization: A B C = triangle; O = circumcenter A B C; M = midpoint B C; AM = line A M; (O) = circle_center_point O A; E = intersection AM (O); l1 = perpendicular_bisector B C; N = intersection l1 (O); NE = line N E; BC = line B C; L = intersection NE BC; AB = line A B; CN = line C N; S = intersection AB CN; MN = line M N; T = intersection AB MN; O2 = circumcenter B L T; (BN) = circle_diameter B N; (O2) = circle_center_point O2 B; U = intersection (BN) (O2); MS = line M S; LT = line L T; V = intersection MS LT; O3 = circumcenter C L V; O4 = circumcenter N T U; (O3) = circle_center_point O3 C; (O4) = circle_center_point O4 N; K = intersection (O3) (O4); Prove: collinear K O3 O4
Difficulty score: 5.4