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Oct 19, 2012 at 22:48 vote accept Cosmin Pohoata
May 29, 2012 at 21:42 answer added Jan Kyncl timeline score: 7
Nov 22, 2011 at 19:49 history edited François G. Dorais
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Nov 17, 2011 at 17:56 comment added Joseph O'Rourke I am still having difficulty understanding the phrase, "the convex polygon bounded by all the diagonals of $P$." In general, there is no convex polygon bounded by all the diagonals, if by "bounded" you mean, "forming the boundary of." There are many convex polygons, each bounded by a subset of the diagonals...
Nov 17, 2011 at 16:16 comment added Gerhard Paseman Also, the smallest convex lattice pentagon I can think of has only one interior lattice point, and that lies on the boundary of the interior pentagon. Does your result include such a case? Gerhard "Ask Me About System Design" Paseman, 2011.11.17
Nov 17, 2011 at 16:10 comment added Gerhard Paseman I was thinking of two diagonals of a 3x1 rectangle, cut off center by a diagonal of a perpendicular 4x1 rectangle, as some diagonals for my cex. Perhaps you can tell me how this fits in with your result? Gerhard "Ask Me About System Design" Paseman, 2011.11.17
Nov 17, 2011 at 15:23 comment added Cosmin Pohoata @Fedor: Sorry, it should have read "the convex polygon bounded by all the diagonals of $\mathcal{P}$.
Nov 17, 2011 at 15:23 history edited Cosmin Pohoata CC BY-SA 3.0
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Nov 17, 2011 at 13:10 comment added Igor Rivin Yes he does, in the first line.
Nov 17, 2011 at 11:41 comment added Fedor Petrov What do you mean by "the convex pentagon bounded by (all) the diagonals of P"? usually the do not bound a pentagon.
Nov 17, 2011 at 10:07 comment added Gerhard Paseman I suspect there are simple counterexamples for n=6, so I may be misunderstanding something. Can you say more about what interior region is supposed to have a lattice point? Gerhard "Ask Me About System Design" Paseman, 2011.11.17
Nov 17, 2011 at 9:26 history edited Cosmin Pohoata CC BY-SA 3.0
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Nov 17, 2011 at 7:17 history asked Cosmin Pohoata CC BY-SA 3.0