Victor Aberman (USA)  
Выигрыш 3+3

b) Qh5 → g5, c) pd6 → c7

<p>b) Qh5 → g5, c) pd6 → c7</p> <p>a) <br /> 1.Be4-d5 Ka2-a1 2.Qb7-b3 Qh5-e2 3.Bd5-c4! <br /> I. 3… Qe2-b2 4.Qb3-d1+ Qb2-b1 5.Qd1-a4+ Ka1-b2 6.Qa4-b3+ Kb2-c1 <br /> 7.Qb3-e3+ Kc1-b2 8.Ka5-b4 Kb2-a1+ 9.Bc4-b3 +- (Centurini final) <br /> II. 3… Qe2-e5 4.Ka5-a6! Qe5-b2 5.Qb3-d1+ Qb2-b1 6.Qd1-d2 d6-d5 <br /> 7.Qd2-c3+ Qb1-b2 8.Qc3-a5+ Ka1-b1 9.Bc4-d3+ Kb1-c1 10.Qa5-e1# <br /> (Horwitz and Kling final) <br /> </p> <p>b) <br /> 1.Ka5-a4! Qg5-c1 2.Qb7-d5+! Ka2-b2 3.Qd5-b5+ Kb2-a1 [3…Kb2-a2 4.Qb5-e2+ <br /> 4… Qc1-b2 5.Qe2-c4+ Ka2-a1 6.Qc4-f1+ Ka1-a2 7.Be4-d5+] 4.Qb5-a6! Qc1-e3 <br /> 5.Qa6-f1+ Ka1-b2 6.Qf1-b1+ Kb2-c3 7.Qb4# <br /> </p> <p>c) <br /> 1.Ka5-b4! c7-c5+ 2.Kb4-c3 Qh5-e5+ 3.Kc3-c2 Qe5-h2+ 4.Be4-g2+-  (Author)</p>