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Today I was solving a chess problem as following picture :

[fen "r1bqr1k1/1p3nbp/2pp2p1/2n2P2/p1PN1P2/2N1B1PP/PPQ3B1/R3R1K1 b - - 0 1"]

The solution given was 1...Rxe3, but I did not get the actual benefit of this solution. Can anyone explain this?

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The point is that after 1...Rxe3 Black will be winning material, as the sole defender of the white knight has been removed, and a pin tactic is coming. Specifically, after 2.Rxe3 (anything other than this recapture means Black has just won a piece for nothing) and 2...Bxd4, White has no way to stop the follow-up 3...Bxe3+ since the rook is pinned to the king, and so Black has netted an extra piece anyway.

[FEN "r1bqr1k1/1p3nbp/2pp2p1/2n2P2/p1PN1P2/2N1B1PP/PPQ3B1/R3R1K1 b - - 0 1"]

1...Rxe3 2.Rxe3 Bxd4 *
  • We might need a few more moves, Ed. Black drops the exchange by Rxe3, and then wins the exchange after Bxd4. I don't see the white Rook dropping off the board for free. After Bxd4, White has 4 pieces that can defend e3 with one move. One little thing I see is that with the removal of the Knight on e4, Black can play Bxf5 with tempo and gain of a pawn. The solution you presented prevents the disruptive (but maybe bad...) 1. Nxc6 bxc6 2. Bxc6 Rook fork. Perhaps preventing that is the point of the puzzle. – Tony Ennis Feb 9 '15 at 11:56
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    @TonyEnnis In the position after Bxd4, White has two pieces for a rook, and will win an exchange on e3, thus he will be a piece up even if the Rook does not go for free. Am I missing something? – JiK Feb 9 '15 at 12:05
  • @JiK Yep, you're right - I miscounted it about 5 times. I neglected to tally the initial Bishop capture. – Tony Ennis Feb 9 '15 at 12:10

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