(2.Rxb7!? is also sufficient since after 2…Rxb7 [2…Re8 may be better] 3.Qc8+ Kg7 4.Qxb7 Qe2!? 5.Qb1! Bd4!?[5…f3? Qd1!]6.Re1! Qxc4 7.Qb3! forcing exchange of queens; the resulting endgame should be decisive in white’s favour. But White should avoid 2.Rxf2?? which allows 2…Bc8! with back rank threats.)
2…Qxd3 3.cxd3 Bd4 4.Rb3 followed by Rfb1 Should ultimately win.
Bxe4 Qxe4 Rxf2 wins a piece directly and threatens the Rxb7 continuation as a next move, forcing Black to defend. The idea is the Rxb7 attack will still be there after so why not take the free piece on f2 first?
Another possibility of searching for advantage seems to be the simple 1. Qf3. Best for black is 1. -, Nd2, I do not see something else against white’s threat 2. Bxe4 (if 1.-, f5 2. Bxe4 3. Qxf3). But after 2. Qxf2 black can take one of the rooks.
2.-, Nxf1 3. Rxf1 and white is getting c5 (after black’s Qxa4) and has two passed pawns or he is even getting a second pawn if black defends c5, then f4 falls and black cannot capture on a4 anymore (f7). 2. Nxb1 3. Rxb1 gives black the defense Qe3 but I don’t think he is going to be happy with the b-line (white exchanges queens and tries to get the bishop to that long diagonal and play d6 to win an exchange – I think that should be quick enough before black can get his kind over and free his rook).
So I think 1. Qf3 leads to a much better game, in black’s best case there’s only one extra pawn for white but there are these two pawns c4, d5 that are free and running…..
there some small material gain for white but black plays f3 which can be a problem for white as you can see. had to add in a move in had to correct my first solution. i have so many problems with e blogger. it gives me fits. keeps telling me my java cookies arnt enabled.
you’re right – Re8 seems to be a way to free the rook/bishop. After Re8 I’d play Be2, perhaps white is only slightly better but I think white can try to attack and win c5 or a6 with his rook on the open b-rank – I rethink that later.
Wolverine, do not propose 2. Rxf2 once more, please – read the comments above that is a losing move because of Bc8! and the weakness of the first line….
After Bxe4 Qe4; Rxf2 Bc8!! Rxb8 Qe1 ++ or Rf8 Bxh3; Rxb8+Kg7; white can’t capture B at h3.
Also Rxb7 Rxb7; Bxd4 Qxd4; Rxf2? loses to Rb1++ Rxf2 Nxf2+; Kg1 NxQh3 loses Bxe4 Qxe4; Rxb7 Rxb7; Qxc8+ Kg7; if Qxb7?, Qe1++
Try: Qf3 Nd2 (not Bd4; Bxe4) Qxf2 Nxb1 Rxb1 Qe3 (Qxa4; Qxc5 followed by Qc7) Rf1 Qxf2 Rxf2 g5?! g3 followed by gxf4 gxf4; Rxf4 if … f3; Rxf3 white is winning
A winning line for white with deflection:
Bxe4 Qxe4
Rxb7 Rxb7
Qc8+ Kg7
Qxb7
To continue, Qe2 black is OK?
1. Bxe4 Qxe4
2. Qd3!
(2.Rxb7!? is also sufficient since after 2…Rxb7 [2…Re8 may be better] 3.Qc8+ Kg7 4.Qxb7 Qe2!? 5.Qb1! Bd4!?[5…f3? Qd1!]6.Re1! Qxc4 7.Qb3! forcing exchange of queens; the resulting endgame should be decisive in white’s favour. But White should avoid 2.Rxf2?? which allows 2…Bc8! with back rank threats.)
2…Qxd3
3.cxd3 Bd4
4.Rb3 followed by Rfb1 Should ultimately win.
Regards
King.
1.Bxe4 Qxe4
2.Rxb7 Rxb7
3.Qc8+ Kg7
4.Qxb7 …
will lose for White because of the answer
4…f3!!
greetz
Bxe4 Qxe4
Rxf7 wins a piece directly and threatens the Rxb7 continuation
and it is hard after
1.Bxe4 Qxe4
2.Qd3 Qe3!
white is better – but is it enough for a win?
quote by adrenaline : “Bxe4 Qxe4
Rxf7 wins a piece directly and threatens the Rxb7 continuation” (it is Rxb7, of course)
and what about
1.Bxe4 Qxe4
2.Rxb7 Bc8!!
white will lose this!
correction:
1.Bxe4 Qxe4
2.Rxf2?? Bc8!!
sorry typo.
it should read Rxf2 not Rxf7
Bxe4 Qxe4
Rxf2 wins a piece directly and threatens the Rxb7 continuation as a next move, forcing Black to defend. The idea is the Rxb7 attack will still be there after so why not take the free piece on f2 first?
you are wron, adrenaline …
The Bc8-variation is really nice….
Another possibility of searching for advantage seems to be the simple 1. Qf3.
Best for black is 1. -, Nd2, I do not see something else against white’s threat 2. Bxe4 (if 1.-, f5 2. Bxe4 3. Qxf3).
But after 2. Qxf2 black can take one of the rooks.
2.-, Nxf1 3. Rxf1 and white is getting c5 (after black’s Qxa4) and has two passed pawns or he is even getting a second pawn if black defends c5, then f4 falls and black cannot capture on a4 anymore (f7).
2. Nxb1 3. Rxb1 gives black the defense Qe3 but I don’t think he is going to be happy with the b-line (white exchanges queens and tries to get the bishop to that long diagonal and play d6 to win an exchange – I think that should be quick enough before black can get his kind over and free his rook).
So I think 1. Qf3 leads to a much better game, in black’s best case there’s only one extra pawn for white but there are these two pawns c4, d5 that are free and running…..
Just another possibility to discuss.
Jochen
1. Bxe4 Qxe4
2. Rxb7 Rxb7
3. Qc8+ Kg7
4. Qxb7 Qxc2
5. Qb1 QxQ
6. RxQ
White rook on b1 will win the queen rook pawn. And with a passed pawn on the a file and another one on the d file, white will dominate black.
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Bxe4 Qxe4
Rxb7 Rxb7
Qc8+ Kg7
Qxb7 f3
g3 Bd4
h4 Qe2
Rxf3 Qxf3
Kh2 Qf2+
Kh3 Qf1+
Kh2 Bg1+
there some small material gain for white but black plays f3 which can be a problem for white as you can see. had to add in a move in had to correct my first solution. i have so many problems with e blogger. it gives me fits. keeps telling me my java cookies arnt enabled.
Bxe4 Qxe4
Rxf2
i think just getting the extra piece is enough for white to win the game.
If Qf3 Nd2; Qxf2 Nxb1; Rxb1 ..Qe3; Qxe3 fxe3; Kg1 Re8; Re1
Hello anonymous(5:27),
you’re right – Re8 seems to be a way to free the rook/bishop.
After Re8 I’d play Be2, perhaps white is only slightly better but I think white can try to attack and win c5 or a6 with his rook on the open b-rank – I rethink that later.
Wolverine, do not propose 2. Rxf2 once more, please – read the comments above that is a losing move because of Bc8! and the weakness of the first line….
Jochen
After Bxe4 Qe4; Rxf2 Bc8!!
Rxb8 Qe1 ++ or
Rf8 Bxh3; Rxb8+Kg7; white can’t capture B at h3.
Also
Rxb7 Rxb7; Bxd4 Qxd4; Rxf2? loses to Rb1++
Rxf2 Nxf2+; Kg1 NxQh3 loses
Bxe4 Qxe4; Rxb7 Rxb7; Qxc8+ Kg7; if Qxb7?, Qe1++
Try:
Qf3 Nd2 (not Bd4; Bxe4)
Qxf2 Nxb1
Rxb1 Qe3 (Qxa4; Qxc5 followed by Qc7)
Rf1 Qxf2
Rxf2 g5?!
g3 followed by gxf4 gxf4; Rxf4
if … f3; Rxf3 white is winning
Sorry g3 is an error.
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