tag:blogger.com,1999:blog-38193052.post8528778143817084423..comments2024-03-20T08:57:17.447-03:00Comments on Jornalheiros: Xadrez - Mate em 4! (Pierre Auguste d'Orville)PC Filhohttp://www.blogger.com/profile/16547063456626761789noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-38193052.post-21339344246485816332018-05-05T23:21:57.896-03:002018-05-05T23:21:57.896-03:00His problems are indeed very beautiful, with multi...His problems are indeed very beautiful, with multiple sacrifices, pure mates in the end...PC Filhohttps://www.blogger.com/profile/16547063456626761789noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-49388083462731650872018-05-05T22:49:51.563-03:002018-05-05T22:49:51.563-03:00What a brilliant problem!!
This man Pierre August...What a brilliant problem!!<br /><br />This man Pierre Auguste had a deal with Caissa...Unknownhttps://www.blogger.com/profile/06377619908080953145noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-32288813549709657922018-05-05T21:20:37.724-03:002018-05-05T21:20:37.724-03:00Well done, Jake!! What a beautiful way to see the ...Well done, Jake!! What a beautiful way to see the solution!!PC Filhohttps://www.blogger.com/profile/16547063456626761789noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-11902963284913731642018-05-05T20:50:02.807-03:002018-05-05T20:50:02.807-03:00If the b2-pawn had nothing standing in front of it...If the b2-pawn had nothing standing in front of it White would have mate in two with 1 Sc5+ Ka5 2 b4. However, a random rook move allows 1 ... S~! which creates a flight square for the black king. White solves this with <b>1 Rb4+! Ka5 2 Ra4+!</b>. After <b>2 ... Kxa4</b> the diagrammed position is restored, except with no white rook. So now <b>3 Sc5+ Ka5 4 b4</b> works.jrh150482https://www.blogger.com/profile/10502831081969372299noreply@blogger.com