4th door of the calendar

How did Black fail to win this position? No, this is not the question: the question is how he would win it.

6 thoughts on “4th door of the calendar”

  1. My suggestion is 1…Rg3 2.Ke4 (2.Nc2 Rg2 3.Ne3 c2 -+) 2…Ke6! (2…Rxg6? 3.Kd4=) 3.Kd4 (3.Kd3 Kf6 and I don’t see how white can win the black pawn due to the pin of his knight. E.g. 4.Ke2 Kxg6 5.Nd5 c2! 6.Nf4+ (6.Kd2 Rd3+ -+) 6…Kf5 7.Nd3 Rxd3 -+) 3…Rh3! (3…Kf6? 4.Nf5 Kxg7 5.Nxc3 =) 4.g7 Kf7 -+. But I’m probably overseeing something as so often happens :-).

  2. Ray, I did more or less the same analysis, only in the variation with 3.Kd3 the direct 3. …c2 is probably a more direct win…

  3. after 1.Rg3 Nc2 2.Rg2 Ne3 3.c2.. how about Nxc2 Rxc2 g7…. i can’t see how to stop the pawn. So maybe 3. Rf2+ kg5 and only then 4.c2

  4. 1…Rg3 2.Ke4 Kc5! is simple. The basic idea is 3.Kd3 c2 -+.

    So White has to play 2.Nc2 Rg2 3.Ne3, where 3…c2?? 4.Nxc2 is a draw.

    Instead Black plays 3…Rf2+! 4.Kg5 Re2 or 4.Ke4 c2!, winning.

    But in the game Black played:

    Morozevich – Van Kampen, Warsaw 2013

    54… Rg3 55. Nc2 Rg2 56. Ne3 Rf2+ 57. Ke4 Ke6?? (c2-+) 58. g7 Kf7 59. Kd3 Rf3 60. Kd4
    Surprisingly this is a fortress!
    61. Nc2 Kf6 62. Kc4 Ke5 63. Nb4 Ke4 64. Nc2 Rh3 65. Nb4 Rg3 66. Nc2 Rf3
    67. Nb4 Kf5 68. Nc2 Kg4 69. Nd4 Rg3 70. Kb3 Re3 71. Nb5 Kf4 72. Nxc3 Ke5 73.
    Kc4 Rh3 74. Nb5 Rh8 75. Nc3 Rc8+ 76. Kd3 Rd8+ 77. Kc4 Rd4+ 78. Kc5 Rd3 79. Kc4
    Rxc3+ 80. Kxc3 1/2-1/2

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