11th December

Is the a-pawn just queening?

7 thoughts on “11th December”

  1. Thomas Schou-Moldt

    Ok, first candidate was 1…Nc5, since 2.Rxc5, Rxc5 and black will stop the pawn from c8. But after 2.a7, Nb6 3.Rb6, Na8 4.Rb8, it seems like black will have to give up both the bishop and the knight for the pawn.
    Second try was the more tricky 1…Rg4+, with the idea to sac the rook on g3/h2, and then take on c7, stopping the pawn with the discovered check. If 2.Kh3, Rg3+ and if 3.Kh4, g5+ and the pawn will be stopped from h8 of by the discovered check trick. If 2.Kf3, Rg3+ and if the white king goes to f2 in either of the variants, Rh3 looks good, h2 and h8 are the key squares here of course. The endgame with bishop and three pawns vs rook must win for black.
    But then I was wondering whether white would be mated after 1…Bxc7 2.a7, Nc4 3.a8Q, Ne3+ It looks pretty nasty…and I think the mate is coming 4.Kg1/h1 is mate in 1, and if 4.Kh3, Rf3+ 5.Kh4, Nf5+ 6.Kg4, Rg3#

  2. The winning variation is 1. …, Bxc7;2.a7, Nc4 but, in my opinion, not the only one.
    Black could also play the interesting 2. …, Rg4+;3.Kh3 (3.Kf1, Rg1+;4.Kxg1, Bb6+;5.Rf2, Nf3+;6.Kg2, Bxa7;7.Rxf3+, Kg7 and the endgame should be winning for Black), Rg3+!;4.Kxg3, Nc6+;5.Kg4, Nxa7;6.Rxa7, Bd6;7.Rd7, Bb4 and also in this case Black should be winning.
    Obviously the mate variation is more attractive, but I like also the endgame variation.

  3. Laznicka – Bulski, Czech League 3rd Nov 2013. However, it seems that Black lost on time after 59…Bxc7 60.a7 – or the result was incorrectly reported. If anyone knows I would be very interested in knowing for sure.

  4. @Jacob According to the official website the result is 1-0 (Laznicka won)

    You can translate it to english and select 2013-14 in “Season” and Select Round 2 in “ChessBoard”
    Without translation select 2013-14 in Sezóny and 2.kolo- 3.11.2013 in Šachovnice

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