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Qxc6 works as well
Stopped by 1...Bf5.
then 2. Rxf5 if pawn takes the rook then Qg6 mate, if instead of pawn takes, and the rook takes the Queen then Rf8+ and then rook on f1 to f7 will follow and is also checkmate. So I think that Qxc6 is still winning unless i missed something.
Oh yes, 1. Qxc6 Bf5 is still winning for white. 1...Bf5 stops white from reaching the puzzle goal of mate in three though, as it will be a mate in four.
good one indeed.
Laugh my f'n assss off @ the top comment
None of the above
Someone did post before the puzzle... I wonder how you do that, and why do people care so much about being on first page?
Forced mate in 3 !
wow!!! nice sacrifice!!!! GG
nice mate. what about QxQ? id RxQ the you Rf8 hes forced into a mate. If in stead of RxQ the Bxf1 to sav mate but u end up going down a queen for a bishop. Bxf1 Qxc8 Kxf7 QxRh8. Its a thought to consider. Am i correct in my analysis?
Qxc6 also a mate in 3
1.Qxc6 Rxc6 2.Rf8+ Kg7/kh7 3.R1f7#
1.Qxc6 Rd8 2.Rf8+ Rxf8 3.Qg6#
Puzzle very interesting.
If 1. Qxc6 Bxf1 then white has 2. Qxg6#.
Umm, correct me for being a noob, but Qxc6 is perfectly fine.
ex. 1. Qxc6 Rxc6 2. Rf8+ Kg7 3. Rf7#
ex. 1. Qxc6 and if Black decides to ignore this move Bf5 2. Qxc8+ Bxc8 3. Rf8+ Kg7 4. Rf7#
yay i got it!
Taking the queen also makes mate in 3, maybe in 2 if bishop takes rok