Problem (2024KoMaLA857): Let $ABC$ be a given acute triangle, in which $BC$ is the longest side. Let $H$ be the orthocenter of the triangle, and let $D$ and $E$ be the feet of the altitudes from $B$ and $C$, respectively. Let $F$ and $G$ be the midpoints of sides $AB$ and $AC$, respectively. $X$ is the point of intersection of lines $DF$ and $EG$. Let $O_1$ and $O_2$ be the circumcenters of triangles $EFX$ and $DGX$, respectively. Finally, $M$ is the midpoint of line segment $O_1O_2$. Prove that points $X, H$ and $M$ are collinear.
AlphaGeometry formalization: a b c = triangle; h = orthocenter a b c; d = foot b a c; e = foot c a b; f = midpoint a b; g = midpoint a c; x = on_line d f, on_line e g; o1 = circumcenter e f x; o2 = circumcenter d g x; m = midpoint o1 o2 ? coll x h m
HAGeo formalization: A B C = acute_triangle; H = orthocenter A B C; AC = line A C; D = foot B AC; AB = line A B; E = foot C AB; F = midpoint A B; G = midpoint A C; DF = line D F; EG = line E G; X = intersection DF EG; O1 = circumcenter E F X; O2 = circumcenter D G X; M = midpoint O1 O2; Prove: collinear X H M
Difficulty score: 3.1