Correct answer: Qc51...Kf4[a]/Bb2[c]/Ba3 2.Qh4#[A] 1...Bf4/[б]/ 2.Q:b7#[Б] 1...e:f5 2.Q:e5# 1.Q:e6? (2.Q:e5#) 1...Bf4/[б]/ 2.Qd5#[C] 1...Bb2[c] 2.<
