Problem (2016ARMOg10p8): In acute triangle $ABC$,$AC<BC$,$M$ is midpoint of $AB$ and $\Omega$ is it's circumcircle.Let $C^\prime$ be antipode of $C$ in $\Omega$. $AC^\prime$ and $BC^\prime$ intersect with $CM$ at $K,L$,respectively.The perpendicular drawn from $K$ to $AC^\prime$ and perpendicular drawn from $L$ to $BC^\prime$ intersect with $AB$ and each other and form a triangle $\Delta$.Prove that circumcircles of $\Delta$ and $\Omega$ are tangent.
AlphaGeometry formalization: a b c = triangle; m = midpoint a b; o = circumcenter a b c; c1 = mirror c o; k = on_line a c1, on_line c m; l = on_line b c1, on_line c m; p = on_line a b, on_tline k a c1; q = on_line a b, on_tline l b c1; r = on_tline k a c1, on_tline l b c1; o1 = circumcenter p q r; t = on_circle o a, on_circle o1 p ? coll o o1 t
HAGeo formalization: A B C = acute_triangle; M = midpoint A B; O = circumcenter A B C; (ABC) = circle_center_point O A; C1 = reflect_point_wrt_point C O; AC1 = line A C1; CM = line C M; K = intersection AC1 CM; BC1 = line B C1; L = intersection BC1 CM; p1 = perpendicular_line K AC1; p2 = perpendicular_line L BC1; AB = line A B; X = intersection p1 AB; Y = intersection p2 AB; Z = intersection p1 p2; O1 = circumcenter X Y Z; (XYZ) = circle_center_point O1 X; T = intersection (ABC) (XYZ); Prove: collinear O O1 T
Difficulty score: 4.0