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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Apr 13, 2017 at 12:19 history edited CommunityBot
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Dec 8, 2016 at 4:48 answer added Gerhard Paseman timeline score: 3
Dec 8, 2016 at 3:05 history edited cats CC BY-SA 3.0
added 110 characters in body
Dec 8, 2016 at 3:04 comment added cats ah, you're right. But no, I do not mean to impose that condition in general. Thanks for the catch
Dec 8, 2016 at 3:00 comment added Noam D. Elkies "The classical result in the hexagonal case is that this is true iff $ace=bdf$" $-$ that cannot be right in general. Counterexample: $(0,1)$, $(1,1)$, $(2,0)$, $(0,-1)$, $(-1,-1)$, $(-1,0)$, with the main diagonals meeting at the origin. If I remember right the missing hypothesis is that the hexagon be inscribed in a circle. Do you mean to impose this condition in general?
Dec 8, 2016 at 1:58 history asked cats CC BY-SA 3.0