Correct answer: Qf51...Kc4[a] 2.Qd4#/Qc3#[A] 1...Kc6/[б]//Nc4/[б]/ 2.Qc7#[Б] 1...Nc6[c] 2.Qc3#[A] 1...Nb3/Nb7 2.Qc3#[A]/Qc7#[Б] 1.Qf4? zz 1...<
