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By main diagonals I mean the diagonals $A_iA_{n+i},$ of which there are $n.$ One classical result in the hexagonal case is that this is true for cyclic hexagons with $ace = bdf.$ I'm wondering when this is true in general (although this might be rather difficult).

In particular, I asked in this question whether it is enough to have that the main diagonals are area bisectors. This holds for hexagons as per the argument given in the link, but I don't believe the answer there is true as it essentially says that all area bisectors must go through the center of mass. This seems like it should be false by the answers here, as if that were true, then the $2n$-gon would have to be centrally symmetric.

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  • $\begingroup$ "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? $\endgroup$
    – Noam D. Elkies
    Commented Dec 8, 2016 at 3:00
  • $\begingroup$ ah, you're right. But no, I do not mean to impose that condition in general. Thanks for the catch $\endgroup$
    – cats
    Commented Dec 8, 2016 at 3:04

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Draw n concurrent lines. Pick a point on one line a way from the concurrent point and, while maintaining convexity, draw lines to points around the concurrent point until you get the convex 2n-gon with the given lines. I do not see any nice characterization resulting from this. Indeed given n points that form half of the 2n-gon, you still have a lot of freedom in picking the concurrent point and the other n points.

Gerhard "Not Sure Of The Goal" Paseman, 2016.12.07.

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  • $\begingroup$ thanks for this; I guess I didn't expect a nice characterization for the more general question. $\endgroup$
    – cats
    Commented Dec 8, 2016 at 4:54
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    $\begingroup$ You can play around with this anyway. For example, draw the n lines, and place a circle somewhere around the concurrent point. The intersections determine a cyclic polygon, and there may be some relationship you can develop with opposing sides or pairs of sides as in cyclic quadrilaterals. Again, I don't see where you are going with this. Gerhard "But Have Some Fun Anyway" Paseman, 2016.12.07. $\endgroup$
    – Gerhard Paseman
    Commented Dec 8, 2016 at 5:09
  • $\begingroup$ Just for fun I guess :). The main question for me was whether or not main diagonals being area bisectors implies anything. $\endgroup$
    – cats
    Commented Dec 8, 2016 at 5:25
  • $\begingroup$ Even there you can play around. Start with a 2n-gon with area bisecting diagonals, color the vertices red and blue alternating, and then pull reds out and push blues in while maintain the bisection property. You might be able to prove something about the edge lengths then, but I don't know what. Gerhard "No Political Innuendo Intended. Really." Paseman, 2016.12.07. $\endgroup$
    – Gerhard Paseman
    Commented Dec 8, 2016 at 5:45
  • $\begingroup$ I don't really understand this last comment (in particular how you maintain the bisection property and get rid of the concurrency), but will chew on it. $\endgroup$
    – cats
    Commented Dec 9, 2016 at 15:14

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