Correct answer: Ra41...Ke3[a]/Kc3/[б]/ 2.Nd5#[A] 1...Ke5 2.Nd3#[Б] 1.Ra6? zz 1...Ke3[a] 2.Nd5#[A] 1...Ke5/Kc5[c] 2.Nd3#[Б] but 1...Kc3/[б]/! 1
