Regular polygon with vertices on grid Constructing regular polygons on a sheet of an orthogonal lined paper - a regular maths notebook - I pondered on what would it take for the polygon to have it's vertices at the intersection of the grid lines.
(We could consider the paper infinitely spanning in all directions)
 A: The only regular polygon that you can draw with vertices on the standard integer lattice is I'm afraid the square. I won't explain how to draw that!
We might as well consider polygons drawn in the Cartesian plane with vertices at rational points, and we might as well take the centre of
our polygon to be the origin. Let $P$ and $Q$ be adjacent vertices
with position vectors $v$ and $w$. Then $v\cdot v=w\cdot w=a$ say and $v\cdot w=a\cos(2\pi/n)=b$ say. We need $a$ and $b$ to be rational, so that $\cos(2\pi/ n)$ must be rational. But $2\cos(2\pi /n)$ is an algebraic
integer, so it must be an ordinary integer. We reduce to the cases
$n=3$, $4$ or $6$.
If we can draw a regular hexagon in $\mathbb{Q}^2$ we can draw an
equilateral triangle, so let's focus on that. Embed our plane in three
dimensional space, and consider the vector product $v\wedge w$.
Then $|v\wedge w|^2=(v\cdot v)(w\cdot w)-(v\cdot w)^2=a^2(1-\cos^2\pi/3)
=\frac34 a^2$. But $v\wedge w$ is a rational multiple of a unit vector
perpendicular to our plane, so $|v\wedge w|^2$ is a square of a rational. Oops!
I suppose one could look at good approximations to regular polygons
drawn on the integer lattice, or regular polygons in integer lattices
in dimensions $3$ or more.
