Correct answer: Be11.Bc4[A]? (2.Bf1#) 1...Bb5/[б]/ 2.Be1#[Б] but 1...Bc6[a]! 1.Bd5[C]? (2.Bg2#) 1...Bc6[a] 2.Be1#[Б] but 1...Bc2! 1.Be1[Б]! (2.
