Problem (2019IOMetrop3): In a non-equilateral triangle $ABC$ point $I$ is the incenter and point $O$ is the circumcenter. A line $s$ through $I$ is perpendicular to $IO$. Line $\ell$ symmetric to like $BC$ with respect to $s$ meets the segments $AB$ and $AC$ at points $K$ and $L$, respectively ($K$ and $L$ are different from $A$). Prove that the circumcenter of triangle $AKL$ lies on the line $IO$.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; o = circumcenter a b c; x = on_tline i i o; b1 = reflect b i x; c1 = reflect c i x; k = on_line b1 c1, on_line a b; l = on_line b1 c1, on_line a c; o1 = circumcenter a k l ? coll o1 i o
HAGeo formalization: A B C = triangle; I = incenter A B C; O = circumcenter A B C; IO = line I O; s = perpendicular_line I IO; BC = line B C; B1 = reflect_point_wrt_line B s; C1 = reflect_point_wrt_line C s; l = line B1 C1; AB = line A B; K = intersection AB l; AC = line A C; L = intersection AC l; M = circumcenter A K L; Prove: collinear M I O
Difficulty score: 4.2