Are you willing to allow degenerate polygons in which consecutive sides are parallel? Then all even N are possible; e.g., a 1-by-2 rectangle with an extra vertex at the midpoint of each of its length-2 sides can be construed as an equilateral hexagon. Jim Propp On Thursday, August 10, 2017, Keith F. Lynch <kfl@keithlynch.net> wrote:
James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons.
http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
Clever. But I did have to really work at it to makes sense of it since Safari rendered all the text like this:
The same argument works with larger regular polygons. The main point to realize is that for all regular [Math Processing Error] n-gons, where [Math Processing Error] n>4, when you construct the perpendicular on one of the sides, the resulting point is strictly inside the original polygon, and this is why the resulting regular [Math Processing Error] n-gon is strictly smaller than the original. This completes the proof for all [Math Processing Error] n-gons for [Math Processing Error] n>4.
Any idea why?
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
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