Troitsky 1895, presented by Andreas
This is an absolutely brilliant puzzle. It is White to move an win. How can White do this? See if you can figure out without using computer assistance.
5B2/3p2p1/6Pk/3PRK2/6Pp/1P6/2P2p2/8 w – – 0 1
Chess Daily News from Susan Polgar
Typical elegant Troitzky humor. Not impossible to solve, either.
I think it’s Re1.
And Bd6 of course
I first found White’s draw with
1. Bxg7+ Kxg7
2. Re7+ Kg8 (or … Kh8, but not … Kf8 because of Rf7+ followed by Kg5, and certainly not … Kh6)
3. Re8+ with a perpetual.
Then it was White’s win with
1. Re1 fxe1(Q)
2. Bd6 Qf2+
3. Bf4+ Qxf4+
4. Kxf4 Kxg6 and White wins the ending.
This position made me miss supper.
Claudia
womancandidatemaster.blogspot.com
1. Te1! fxe1
2. Bd6!! and win!!
by stulzer
I found only the ideas but did not solve it the elegant way.
Brilliant puzzle!
I think Bxg7 also wins although I haven’t verified this with a computer…
if (1)
1.Bxg7+ Kxg7
2.Re7+ Kg8
3.g7 f1=Q+
4.Kg6 what can Black do?
if (2)
1.Bxg7+ Kxg7
2.Re7+ Kg8
3.g7 Kh7
4.g8=Q+ Kxg8
5.Kg6 Kf8
6.Rf7+ and pick up the Black pawn
which is about to Queen
if
1.Bxg7+ Kxg7
2.Re7+ Kh8
3.g7+ and it leads to positions in (1) or (2)
I think position (2) as highlighted by me above loses for White as after White’s Kg6, there is no Re8 mate, because after Kg6, Black can f1=Q and if White gives Re8+, then Black can play Qf8 and defend the Black King.
After Rxf8 Kxf8, White cannot prevent the pawn on the h-file from queening.
I think Bxg7 also wins although I haven’t verified this with a computer…
if (1)
1.Bxg7+ Kxg7
2.Re7+ Kg8
3.g7 f1=Q+
4.Kg6 what can Black do?
4…Qa6+.
if (2)
1.Bxg7+ Kxg7
2.Re7+ Kg8
3.g7 Kh7
4.g8=Q+ Kxg8
5.Kg6 Kf8
This is a blunder. 5…f1=Q wins.
DG – Sorry for the redundancy in (2). I see now that you caught your own mistake, usually not an easy thing to do.