Tie your shoelace

By breaking down an $n$ sided regular polygon with side length $s$ into $n$ isosceles triangles, (in cases $n \neq 3$ each isosceles triangle with a vertex at the polygon's circumcenter), the area of such a polygon is

$\large \frac{ns^2}{4 \tan \frac{\pi}{n}}$ .

But by Pick's theorem , this area is an integer or half an integer. This means that (at very least) $\tan \frac{\pi}{n}$ is rational.

However, by Niven's theorem , $\tan \theta$ is only rational at $\theta = \frac{\pi}{4}$ and $\theta = 0$ (within the range we're working on here, $0 < \theta \leq \frac{\pi}{3}$ ).

So the only possible values of $n$ such that a regular polygon could be created with $n$ lattice points is $4$ , and we're given this works in the problem statement. Hence the answer is 'No'.

Extra: Well that's a (somewhat) high level solution (that generalises the $n=3$ case in the Pick's Theorem wiki article). Anyone want to find a nicer solution? :)