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". Jim Propp