Unless there's something very wrong with my code, here's what I have so far. I'm assuming you want the "area" as defined by the winding rule, which means that some geometric areas may be counted multiple times. For instance, in a five-pointed star, the central pentagon, with winding number 2, counts double. Is this what you meant? Odd n=9 has the first solutions. Odd n=15 has many; here are some: 0 1 2 3 4 6 5 11 7 14 13 10 12 9 8 0 1 2 3 4 6 5 12 8 7 14 11 13 10 9 0 1 2 3 4 6 5 12 11 8 10 7 14 13 9 0 1 2 3 4 6 5 12 14 11 7 13 10 9 8 0 1 2 3 4 6 5 12 14 11 10 7 13 9 8 Odd n=21 has many; here are some: 0 1 2 3 4 5 6 7 8 16 10 15 14 13 12 20 19 18 17 11 9 0 1 2 3 4 5 6 7 8 16 14 13 12 20 19 18 17 11 10 15 9 0 1 2 3 4 5 6 7 8 16 15 13 12 20 14 19 18 17 11 10 9 0 1 2 3 4 5 6 7 8 16 15 14 12 20 19 13 18 17 11 10 9 0 1 2 3 4 5 6 7 8 16 15 14 13 18 12 20 19 17 11 10 9 So far I have not found any for any other odd values through n=31 (but I'll let them run overnight). Searching for symmetric solutions may well prove to find them more quickly than what I'm currently doing (just searching lexicographically). -tom On Fri, Jul 29, 2016 at 11:58 PM, Joerg Arndt <arndt@jjj.de> wrote:
The following might be helpful to determine which permutations of the roots of unity give a polygon with zero area.
Let A be the area, then 2 * A = = sum(i=0, n-1, x[i]*y[i+1] - x[i+1]*y[i] ) = sum(i=0, n-1, (x[i] + x[i+1]) * (y[i+1] - y[i]) ) = sum(i=1, n, x[i] * (y[i+1] - y[i-1]) ) where indices are taken modulo n This is from http://geomalgorithms.com/a01-_area.html which gives further references. I found this URL at
http://stackoverflow.com/questions/451426/how-do-i-calculate-the-area-of-a-2...
Best regards, jj
* James Propp <jamespropp@gmail.com> [Jul 30. 2016 08:11]:
For which n>2 does there exist a reentrant n-gon whose vertices are evenly spaced on a circle and whose signed area (aka algebraic area) is zero?
When n is even, there is an easy solution: draw a circuit from 1 to 2 to 3 to ... to n/2-1 to n/2 to n to n-1 to n-2 to ... to n/2+1 to 1 (here I'm numbering the n vertices cyclically).
For n=3, it's trivially impossible to find such an n-gon, and I've checked by brute force (using Mathematica) that there's no such n-gon when n=5 or n=7.
For n=9, there is such an n-gon (which I found through brute-force search): http://mathenchant.org/nonagon.pdf Mathematica assures me that 1/2 (-1/2 Sqrt[3] sin(\[Pi]/18)+1/2 sin(\[Pi]/9)-sin((2 \[Pi])/9)-2 sin(\[Pi]/18) sin((2 \[Pi])/9)-1/2 cos(\[Pi]/18)-1/2 Sqrt[3] cos(\[Pi]/9)+2 cos(\[Pi]/18) cos((2 \[Pi])/9)+2 sin(\[Pi]/9) cos(\[Pi]/9)) simplifies to 0, which implies the claim. Is there a nice (preferably trig-free) way to prove it? That is: is there a nice way to show that the total area of the three scalene black triangles equals the total area of the four nonscalene black triangles, maybe using dissection and area-preserving linear transformations, or maybe using algebraic tricks involving roots of unity?
I suspect that when n is prime, there's no such polygon. Can any of you find a proof?
Also, can any of you construct such a polygon with n=15? It'd be nice if the answer to my original question were "Precisely when n is composite".
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