Mate in 2 with a paradoxical idea

Not so bad I think

1. Qe5 lets go of the pin on the pawn.  After 1. Qe5 Ra6+ 2. Rxa6+ is not mate due to  bxa6.  You need to protect both rooks while maintaining the pin and Qh1 is the only move that does both.

Then Ra6+

Edit: omnipaul Yeah, that's the idea

nice puzzle

Great one! is it from you?

I approve of this puzzle - very interesting Yosriv! 

I like it

Thanks guys

smurf12 No I found it in a book "Problèmes d'échecs en deux coups" (mates in 2) by J.P Boyer, and I wanted to share it with the members. All the mates in this book are from the romantic era.

But it's composed by Herlin!

sweet, took me less than a min,but many attempts (close to 10 moves try's,maybe slightly more)

i was bout to give up, haha quit! throw the towel!

Absolutely right, my friend. Th. Herlin, Schachzeitung, 1852.

@FishN4Fish It took me some time too

If I got it right, any of Qa4, Qd4, Qe5 moves is a mate in 1. Of course, besides Qh1.

What about Qe8 ? Then force mate in 1?

Nope.  Ra6+ and after something takes the rook, bxa6 keeps it from being a mate in 2.  Similarly, if the king moves out of check, it isn't a mate in 2.  You have to maintain the pin of the b pawn, which means the queen has to stay on the a8-h1 diagonal.

Very clever!

 I disagree.  Putting the Queen on either the a file or the 8th rank (i.e., Qa4 or Qe8) works too, notwithstanding releasing the pin on the pawn.  Point is, RxQ is met in either case by RxQ#, while if not RxQ (for example, a pawn move to either b6 or b5), then QxR#.  And if Qa4 is met by Re8+, then Rxe8#.  So it is a mate in two, but with more than one solution.

Edit:  nevermind.  On further analysis, any move other than Qh1 can be met by Ra6+, delaying mate.  Thus, Qh1 is the only way to mate in two.