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Well I guess QxN, RxQ R-d8 wins…if Qf8 then gxh7+ followed by Rxf8 makes a shiny new queen.
“Well I guess QxN, RxQ R-d8 wins…if Qf8 then gxh7+ followed by Rxf8 makes a shiny new queen.”
After gxh7+ black will just take the pawn Kxh7, so no queen
But White does get a “shiny new queen” after
1. Qxe6+ Rxe6
2. Rd8+ Qf8
3. Rxf8+ Kxf8
4. gxh7
although I would need to analyze further to determine whether Black might survive after
4. … b3
4. h8=Q+ Ke7
Those two pawns, combined with the threat of an intermezzo rook check in some variations to protect a pawn to promote after White captures one of them, are troubling.
Is there something simpler that I’ve missed? Probably. 🙂
Nothing missed. 1. Qxe6+ is correct. This really is a type of back rank problem. As the knight on h5 and the pawn on g6 control all the escape squares of the black king. This forces the exchange of queens and the black pawn on f6 blocks the black rook from catching the white pawn after 4. gxh7. If black’s fourth move were to be b3 or even c2 for that matter, there are too many checking squares for the queen to use to eventually fork the king and those black pawns trying to promote.
4…c2 fails to the obvious Qh7+ followed by Qxc2.
4…b3 fails too
1.Qxe6 Rxe6 2.Rd8 Qf8 3.Rxf8+ Kxf8 4.h7, and White will queen in time to stop Black’s pawns.