tag:blogger.com,1999:blog-4454504925561903590.post7027836537733957828..comments2009-04-29T10:31:30.913-04:00chess_planethttp://www.blogger.com/profile/13978268675587340295noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4454504925561903590.post-66198397575797886112009-04-29T10:31:00.000-04:002009-04-29T10:31:00.000-04:00Well, I take it back 2 h3 runs into the same variation 2..Be4 3 Ne4 Qe4! 5 Qe4 Ng3+ winning a piece.<br /><br />So the problem has two acceptable solutions 1..Be4 and 1..Kg7 , even though the second relies on Bxe4 if White defends by 2h3.<br /><br />Well done everyone !chess_planethttp://www.blogger.com/profile/13978268675587340295noreply@blogger.comtag:blogger.com,1999:blog-4454504925561903590.post-86483139328673701672009-04-29T10:27:00.000-04:002009-04-29T10:27:00.000-04:001..Be4! is the right move. the first anonymous reader found it.<br /><br />1..Kg7 is an interesting try threatening 2..Rh2!+ <br />However I believe it is not right because of 2 h3! and now White has created an escape square in h2 for his king, and is back in the game.chess_planethttp://www.blogger.com/profile/13978268675587340295noreply@blogger.comtag:blogger.com,1999:blog-4454504925561903590.post-716724489690169342009-04-29T02:48:00.000-04:002009-04-29T02:48:00.000-04:00there is another variation: 1. Kg7 with the threat Rxh2+! and Rh8; if say g3 or g4, then follows: 2.Bxe4+<br />Nxe4; 3. Qxe4+ and 4. Nxg3+ with the same conclusionAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4454504925561903590.post-30529006182935851692009-04-29T02:19:00.000-04:002009-04-29T02:19:00.000-04:00well, that's easy: 1. Bxe4 and now if Qxe4, 2.Ng3 mates, if Nxe4 then 2.Qxe4 and black has an extra light piece, enough to won the gameAnonymousnoreply@blogger.com