On Monday I gave you this rather remarkable position. Last week you saw an endgame study in which no queens beat two queens. Here, if Black had found the correct move, no queens would have drawn with two queens.

The answer to my question is that Black can play 74… Qdxf2+! 75. Qgxf2 (75. Qfxf2 is met in the same way) 75… Qxf3+! 76. Kxf3 Qc3!+, when 77. Qxc3 is stalemate and other moves result in a drawn KQP v KQP ending.

If you found the answer, you can now consider yourself an expert in endings with three queens each!

Here’s the complete game if you’re interested. Click on any move for a pop-up window.