Problem (2018G4): A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$.
AlphaGeometry formalization: a b c = triangle; t = free; a1 = reflect t b c; b1 = reflect t a c; c1 = reflect t a b; o = circumcenter a1 b1 c1; a2 = on_line a1 t, on_circle o a1; b2 = on_line b1 t, on_circle o b1; c2 = on_line c1 t, on_circle o c1; x = on_line a a2, on_circle o a1 ? coll x b b2
HAGeo formalization: A B C = triangle; T = point; BC = line B C; A1 = reflect_point_wrt_line T BC; AC = line A C; B1 = reflect_point_wrt_line T AC; AB = line A B; C1 = reflect_point_wrt_line T AB; O = circumcenter A1 B1 C1; (O) = circle_center_point O A1; A1T = line A1 T; A2 = intersection A1T (O); B1T = line B1 T; B2 = intersection B1T (O); C1T = line C1 T; C2 = intersection C1T (O); AA2 = line A A2; X = intersection AA2 (O); Prove: collinear X B B2
Difficulty score: 3.3