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      Home  >  Chess Improvement • Chess Puzzles  >  An amazing coincidence – Part 2

      An amazing coincidence – Part 2

      endgame, Puzzle Solving


      This position took place also on move 36 between Sasikiran and Shirov shortly after the Ivanchuk – Aronian game. Sasikiran said he remembered the analysis of the Ivanchuk – Aronian game where 36.Ba6!! would be the winning move. So he played 36.Ba6, and lost. In this position, what would be the winning move for White?

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      14 Comments

      1. Anonymous Reply
        August 25, 2007 at 2:10 pm

        Zsuzsanna, can you please post less difficult positions? This is world class level, something 99.9% of us will never understand. Thank you

      2. anon Reply
        August 25, 2007 at 2:29 pm

        my grade is 145ECF which is about 1650 Elo, any chance of some easier puzzles alongside these ?
        I sometimes get these right…. but they are a bit hard Susan sorry!

        how about move the rook down to c5 allowing the rook e5 move?

      3. Anonymous Reply
        August 25, 2007 at 3:26 pm

        The adjacency of the two positions is a really good find: how do you do that!
        Still no idea why Ba6 was best before: presumably it’s not best now – I guess for Bxc8, but why it works now and not before is another mystery.

      4. jbennett Reply
        August 25, 2007 at 3:27 pm

        In this case, I think you can take
        the bishop right away. After

        1. Bxc8 Rxc8
        2. Kc3 Kf8
        3. Kxb3 Ke7
        4. Kc4 Kd7
        5. Kd5

        If black pushes f6, to keep the
        white king out, then white can
        play Rxf6, and force a trade of
        rooks if black takes c7.

        If black takes c7 right away, then
        white can trade rooks and then
        march his king to f6, which wins.

        P.S. To first anon, Susan does
        post easier puzzles too. Although
        sometimes people complain (not me)
        that they are too easy.

        I’m a class B player, and some of
        the puzzles are too hard for me,
        but the comments are always
        interesting.

        Jim

      5. mayadi Reply
        August 25, 2007 at 3:53 pm

        The clue is that after 1.Kc3 1…Re6 doesn’t work, because white can play 2.Rc4. In Part1 this wasn’t possible because of 2…b2 but now the black rook on e6 is hanging after 3.Bxc8 b1Q 4.Bxe6 and the pawn will promote after 4…Qc2+ 5.Kd4 Qxc6 6.c8Q+
        Now in part2 the rook on c6 prevents the black king to approach via f6, hence 1…Kf8 2.Kxb3 Ke7 3.Bxc8 Rxc8 4.Kc4 Kd7 5.Kd5 and white has a winning position.

      6. Anonymous Reply
        August 25, 2007 at 4:02 pm

        It’s hard to solve this puzle for me, but I can see the difference between two.
        In the second one the c6 is blocked by own rook for white and the movement of bishop a6-b5 is not targeting e8 rook, which is possible in the Ivanchuk-Aronian game. So if Ba6 then rook a8 will target bishop and go to help b3 pawn.

        Ashot

      7. Jochen Reply
        August 25, 2007 at 4:03 pm

        1. Ba6? b2!
        2. Kc2 Re6! wins

        The rook on e6 is defended and attacks the bishop on a6 through the white rook (x-ray attack).
        That’s the reason why Ba6 loses here instead of bringing the win as in the other example….

        I think Jim had the correct winning idea which didn’t work in the other position because black just could play Kd6 there instead of the weaker Kd7 here.

        Very interesting positions to compare. Thank you very much, Susan!

        Jochen

      8. Anonymous Reply
        August 25, 2007 at 4:07 pm

        I have to add that in the second one white rook can’t go to “a” if a Ba6 is made.

        Ashot

      9. Anonymous Reply
        August 25, 2007 at 4:22 pm

        Thanks Jochen
        The Re6 is really killer!

        Ashot

      10. egaion Reply
        August 25, 2007 at 5:00 pm

        Mayadi I did not get the following. I think that it is not clear about which example of the two you are talking. If you don’t mind write it again for my sake

        mayadi said…

        The clue is that after 1.Kc3 1…Re6 doesn’t work, because white can play 2.Rc4. In Part1 this wasn’t possible because of 2…b2 but now the black rook on e6 is hanging after 3.Bxc8 b1Q 4.Bxe6 and the pawn will promote after 4…Qc2+ 5.Kd4 Qxc6 6.c8Q+

      11. Pyada Reply
        August 25, 2007 at 7:40 pm

        BxC8 appears to be the move. First white can not lose as he is not down anything. But its also winning. Idea is to keep the rook on C8 file while bring the white’s king to kingside black pawn. Black king is far enough so as to give ample time to white king to capture pass b pawn and come back to d5 square.
        I think only BxC8 wins. I am surprised what analysis Sasi did to get Ba6.

      12. Anonymous Reply
        August 25, 2007 at 10:11 pm

        I think in this position white can just take the rook. The resulting endgame will be won as white’s king can make it to d5 in time and with the white rook on c6 black can not play f6 as in the previous position.

        36Ba6? just loses to 36…b2! 37 Kc2 Re6 -+

        -Justin Daniel

      13. Anonymous Reply
        August 25, 2007 at 10:50 pm

        I’m 1623 USCF, and this seems rather straightforward to me(I haven’t seen the other example yet). I would definately play Kc3 here, because both of black’s rooks are completely dead, can’t move at all, so white can just take the b pawn and then march on into black’s camp to help. Black’s c-file rook can’t move without being taken and then white queening, and black’s other rook can’t move anywhere except f8(a pointless square for the rook), without white taking the other rook for free. Therefore, white can just free his rook by taking the rogue pawn, all black can do is move his king and pawns.

      14. mayadi Reply
        August 25, 2007 at 10:54 pm

        @egaion:
        In this one (part2, with the white rook on c6) 1.Kc3 is possible, because if black plays 1…Re6 then 2.Rc4.
        While it wasn’t possible in the other one (part 1, with the white rook on c5), because after 1.Kc3 there would follow 1…Re5 and 2.Rc4 would lose because of 2…b2!

        In this one (part2), the same idea with 1…Re6 fails after 2.Rc4 b2 3.Bxc8 b1Q 4.Bxe6 Qc2+ 5.Kd4 Qxc6 6.c8Q+

        It seems that both moves, 1.Kc3 and 1.Bxc8 are possible. In a different order the same moves will follow.

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