Correct answer: Bc41.Rc4? (2.Be2#[A]/Bc2#[Б]) 1...R:c4[a]/Ba4 2.Be2#[A] 1...B:c4/[б]//Rge8 2.Bc2#[Б] 1...Rce8 2.Rc1#/Bc2# but 1...Rg2! 1.
