A practised solver should soon spot the idea of an interference on the crucial square of d3. 1.R6d3, for example, threatens 2.Nc4 mate and allows Bxd3 to be met by Nxd3 mate. 1...Qg8 or 1...Qc5 (or Qb4) are met by 2.Qd6 or 2.Qg7, while 1...Ne3 runs into 2.Qd4. But after 1...Nxg5! there's no mate.
Okay, so let's try 1.Bd3 with the same threat of Nc4. Now 1...Nxg5 does not help, 1...Bxd3 2.Nxd3 is still mate, and Qg8 allows Qxf5. But after 1...Ne3! there's no mate.
So it has to be 1.d3, when Ne3 meets with the clever 2.d4! But it's not that either, because after 1...Qg8! the d-pawn cuts off the bishop from f5.
The whole idea of interfering on d3 is a red herring. The right answer is 1.Rcxc6! (threatening Rd5) 1...bxc6 2.Nxc6 or 1...Ne3 2.d4 or 1...Qg8 2.Qxf5.
This one is a help-mate in two, which means you have to find a sequence: Black moves, White moves, Black moves, White mates. And when you have solved it, you must do it again, but with a white pawn instead of a black one on b3. Jonathan Mestel solved both versions within a few minutes. You have until tomorrow.Reuse content