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Chess problem 2007-10-24

chess problem 2007-10-24
White to mate in three

Add a comment to this post with your solution. I'll buy a beer for the first person to correctly solve the problem next Tuesday, during our regular session at Matchless.

Note—To win the beer, please show all possible variations.

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This entry was posted on Wednesday, October 24th, 2007 at 7:46 am and is filed under chess, problems. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

6 Responses to “Chess problem 2007-10-24”

  1. Ben Says:
    October 30th, 2007 at 2:22 pm

    1. d1 - e1
    2. b4 x c4
    3. e4 - f4 check
    4. g7 - e5
    5. e1 x e4 mate

  2. Jeffrey Says:
    November 5th, 2007 at 7:59 pm

    Nice try, Ben, but this is not the solution. Black is not forced to interpose the Bishop (2…Be5 3.Rxe5#), instead 2…Kxd5 3.Bf7+ e6 4.Bxe6+ Kxd4 5.Rd1#. It's still mate, but not mate in three.

  3. bicyclethief Says:
    November 13th, 2007 at 9:34 am

    1.Qe2
    1…Bxd5+ 2.Kf4+
    2…Be4 3.d5#
    2…Be5 3.Qxe5#
    1…Rb2 2.Kd3+ Rxd2 3.Nf4#

  4. bicyclethief Says:
    November 13th, 2007 at 10:03 am

    One more critical variation in there:

    1.Qe2 Rxd4+ 2.Kf3+ Re4 3.Nf4#

  5. bicyclethief Says:
    November 13th, 2007 at 1:26 pm

    1.Qe2 Rxd4 2.Kf3+ Kxd5 3.Qa2#

    Ugh. I’ve spent waaay too much time on this puzzle.

  6. checkersisfortramps Says:
    November 13th, 2007 at 1:43 pm

    I’m impressed! Nice work bicyclethief…If you’ve spent alot of time I would say it was time well spent.

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