E e

a) 8...b4 9 &a4 ±b7 10 Ag5 is the same as 8...&b7 9 &g5 b4 10 t&a4.

10 &xe6 0-0 11 ±b3 ±; 9...b4 10 £ibd7 11 0-0-0 Wa5 12 £xf6 <&xf6 13 f3 <*> Radulov-Garcia Martinez, Havana 1969) 10 0-0-0 (10 &xe6?! fxe6

bl) Not 10...£k:5? 11 e5. b2) 10...b4?! 11 £kl5! exd5 12 exd5+ and now 12...i.e7 13 £if5 &f8

b3) 10.. .®b6 is best met by 11 f4!. b4) 10...2c8 and then: b41) 11 2hel 2xc3! (Il...h6 12 i.h4 g5 13 ±g3 2xc3 14 bxc3 ®a5

15 f3 d5?!, Rossetto-Panno, Buenos

Aires 1968, 16 &xe6!) 12 bxc3 Wa5 with good compensation, Hendriks-Mirumian, Groningen 1997.

b42) 11 <£)d5 exd5 and rather than 12 exd5+ We7!? +, White could try 12

Was this article helpful?

0 0

Post a comment