Problem (2017EuropeanMCupSp3): Let $ABC$ be a scalene triangle and let its incircle touch sides $BC$, $CA$ and $AB$ at points $D$, $E$ and $F$ respectively. Let line $AD$ intersect this incircle at point $X$. Point $M$ is chosen on the line $FX$ so that the quadrilateral $AFEM$ is cyclic. Let lines $AM$ and $DE$ intersect at point $L$ and let $Q$ be the midpoint of segment $AE$. Point $T$ is given on the line $LQ$ such that the quadrilateral $ALDT$ is cyclic. Let $S$ be a point such that the quadrilateral $TFSA$ is a parallelogram, and let $N$ be the second point of intersection of the circumcircle of triangle $ASX$ and the line $TS$. Prove that the circumcircles of triangles $TAN$ and $LSA$ are tangent to each other.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; x = on_line a d, on_circle i d; o1 = circumcenter a f e; m = on_line f x, on_circle o1 a; l = on_line a m, on_line d e; q = midpoint a e; o2 = circumcenter a l d; t = on_line l q, on_circle o2 a; s = parallelogram a t f s; o3 = circumcenter a s x; n = on_line t l, on_circle o3 a; u = circumcenter t a n; v = circumcenter l s a ? coll a u v
HAGeo formalization: A B C = triangle; I = incenter A B C; BC = line B C; D = foot I BC; (I) = circle_center_point I D; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; AD = line A D; X = intersection AD (I); FX = line F X; O1 = circumcenter A F E; (AEF) = circle_center_point O1 A; M = intersection FX (AEF); AM = line A M; DE = line D E; L = intersection AM DE; Q = midpoint A E; LQ = line L Q; O2 = circumcenter A L D; (ADL) = circle_center_point O2 A; T = intersection LQ (ADL); S = parallelogram A T F; O3 = circumcenter A S X; (ASX) = circle_center_point O3 A; ST = line S T; N = intersection (ASX) ST; U = circumcenter A N T; V = circumcenter A L S; Prove: collinear A U V
Difficulty score: 3.5