If the black R were not on a3, then Rh3 would be mate.
If the other black R were not on b1 then 1. b8Q would mate in 3 after 1… Kh7 2. Qg8+, Kh6. 3. Qg6#
So with 1. Nb3! white interferes with the action of both rooks and threatens mate in two ways. The black K can’t flee to h7 because then Rh3#. So the knight must be captured, but either capture allows one of the two threats to be carried out.
So, 1. … Rbxb3 2. Rh3+! Rxh3 3. b8Q+ and mates as shown above.
Or, 1. … Raxb3 2. b8Q+! and black can take hit choice which mate to suffer.
An interesting study, very hard to solve by conventional “tree” thinking (at least for me) and only solved by understanding first the functions of all the pieces in the starting position. Also quite attractive with the echo mates, and material sacrifices.
For some reason I noticed 1.Nb3 straight away threatening both to promote the b-pawn and mate on h3. Taking the knight seems forced after which white plays 2. b8Q+ winning (2…Rxb8 3.Rh3 mate)
Nice one. Nb3. one of the rooks must take N. Next play b8=Q. after RxQ, Rh3 mate. Do you understand this Breeown Laffertiii? Or should I explain it more slowly?
The answer is definitively 1. Nb3! provoking a Würzburg-Plachutta on the critical square. If 1. … Rbxb3 2. Rh3+ If 1. … Raxb3 2. b8=Q(R)+ and white wins
Very interesting position. This late at night, I am only able to take the Rook on a3 and hope for a trick after 1…Rxb7+. Phil
If the black R were not on a3, then Rh3 would be mate.
If the other black R were not on b1 then 1. b8Q would mate in 3 after 1… Kh7
2. Qg8+, Kh6. 3. Qg6#
So with 1. Nb3! white interferes with the action of both rooks and threatens mate in two ways. The black K can’t flee to h7 because then Rh3#. So the knight must be captured, but either capture allows one of the two threats to be carried out.
So, 1. … Rbxb3
2. Rh3+! Rxh3 3. b8Q+ and mates as shown above.
Or, 1. … Raxb3
2. b8Q+! and black can take hit choice which mate to suffer.
An interesting study, very hard to solve by conventional “tree” thinking (at least for me) and only solved by understanding first the functions of all the pieces in the starting position. Also quite attractive with the echo mates, and material sacrifices.
Nice one.
Cool!
nb3 followed by b8q and rh3
Nb3 good position
I quickly had a hunch that Nb3 was the decisive move, but it took me a while to work all the kinks out. Here, I think, is the solution:
1- Nb3; Rbxb3
(or … Raxb3 2-b8=Q+; Rxb8 3- Rh3#)
2- Rh3+; Rxh3
3- b8=Q+; kh7
4- Qg8+; kh6
5- Qg6#
From Spain…
1)Nb3!R3b3
2)b8=Q+,Kh7
3)Qh2++
1)Nb3,R3b3
2)b8=Q+,Rb8
3)Rh3++
1)Nb3,R1b3
2)Rh3+,Rh3
3)b8=Q++
Greeetings from Spain
For some reason I noticed 1.Nb3 straight away threatening both to promote the b-pawn and mate on h3. Taking the knight seems forced after which white plays 2. b8Q+ winning (2…Rxb8 3.Rh3 mate)
btw: if black takes the knight with his b-rook then white just switches move order (first Rh3+ and mating with b8Q)
Nb3! Rxb3
b8Q+! Rxb8
Rh3#
Nb3 🙂
black must take or white plays either Rh3 or b8Q
if he takes with the a-rook: b8Q+ Rxb8 Rh3#
if he takes with the b-rook: Rh3+ Rxh3 b8Q+ Kh7 Qg8+ Kh6 Qg6#
Looks like a double interception motive.
1.Nb3! (cutting one rook to protect h3, and the other one to protect b8)
if 1…Raxb3 2.b8=Q followed by Rh3
if 1…Rbxb3 2.Rh3 followed by b8=Q with mate to follow (Qg8 and Qg6)
1 Nb3 intersects the two rooks
Nb3
1.Nb3!
If 1.- Raxb3 then 2.b8=Q+ Rxb8 3.Rh3#
And if 1.- Rbxb3 then 2.Rh3+ Rxh3 3.b8=Q+ and black will be mated.
Nb3! and the black king get mated by the queen on h6
looks like Nb3! with b8Q and Rh3# or Rh3+ with b8Q and Qg8 with Qg6# wins
1.Nb3 mates
Nb3
Nice one. Nb3. one of the rooks must take N. Next play b8=Q. after RxQ, Rh3 mate. Do you understand this Breeown Laffertiii? Or should I explain it more slowly?
Very interesting position, but 1. Nb3!!
1. … Raxb3 2. b8Q+ Kh7 (2. … Rxb8 3. Rh3 mate) 3. Qg8+ Kh6 4. Qg6 mate
1. … Rbxb3 2. Rh3+ Rxh3 3. b8Q+ Kh7 4. Qg8+ Kh6 5. Qg6 mate.
Very interesting position.
I think 1. Nb3!!
1. … Raxb3 2. b8Q+ Kh7 (2. … Rxb8 3. Rh3#) 3. Qg8+ Kh6 4. Qg6#
1. … Rbxb3 2. Rh3+ Rxh3 3. b8Q+ Kh7 3. Qg8+ Kh6 4. Qg6#
Very interesting position.
I think 1. Nb3!!
1. … Raxb3 2. b8Q+ Kh7 (2. … Rxb8 3. Rh3#) 3. Qg8+ Kh6 4. Qg6#
1. … Rbxb3 2. Rh3+ Rxh3 3. b8Q+ Kh7 3. Qg8+ Kh6 4. Qg6#
The answer is definitively
1. Nb3! provoking a Würzburg-Plachutta on the critical square.
If 1. … Rbxb3 2. Rh3+
If 1. … Raxb3 2. b8=Q(R)+
and white wins