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Rb2 would be played by cute and sexy players
1.Rxb2 Rxb2
2.c7 Rb3
3.Kc2 Rb4
4.c8=Q Rxe4
***Black resign
1.Rb2 Rb2
2.c7 Rb3
3.Kc2
OR
2….Rh8
3.c7 Rc8
4.Rc2
At first I thought:
1. c7 Rh8
2. Rb8 but Black can play Rc8, so…..
1. Rxb2 Rxb2
2. c7 Rb3
3. Kc2 Re3
4. c8=Q Rxe4 and it is one of these dreadful Q vs. R endings.
or
1. Rxb2 Rh3
2. Kc4 Ke7
3. Kb5 Kd8
4. Kb6 Kc8
5. Rb5 should win.
Rxb2
White can get a queen vs rook by sacrificing at b2. Now, black doesn’t have to accept the offer, but it won’t help:
1. Rb2 Rb2?
2. c7
And the pawn is unstoppable. This is, in fact, what Hammer played. After Hammer checked from b3 and played Rb4, the game reduced to a Q vs R+P endgame:
2. …..Rb3
3. Kc2
Here, Kc4 is met by Rb1 threatening a skewer if white queens the pawn, and black will likely win since he can sacrifice for the pawn, pick up e4 and win with the e-pawn. Continuing:
3. …..Rb4
4. c8(Q)Re4
And Inarkiev went on to win 28 moves later.
I checked the Nalimov tablebase to be certain, and the position after Re4 is a win in 38 moves.
Now, as I wrote, black isn’t forced to take at b2 on move 1:
1. Rb2 Rh8
2. Kc4 Ke7
3. Kc5 Rh1 (Rh4 4.Rb8 Re4 5.c7+-)
4. Kb6 and we have seen enough of these R+P vs R endgames to know this is clearly won for white, even with the two extra pawns on the board.