Rational regular polygons

Is it possible to create a regular polygon where all coordinates of vertices are rational numbers?

Equilateral triangle

Assume it is possible to create an equilateral triangle
If we call the points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ the surface area of the triangle is $\frac{1}{2}(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)$
But because the triangle is equilateral, the area also is:
$\frac{1}{2} bh=\frac{1}{2}\sqrt{3}b^2=\frac{\sqrt{3}}{2} \left(\sqrt{(x_1-x_2)(y_1-y_2)}\right)^2$
Putting both systems equal to eachother gives:
$x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2=\sqrt{3} \left(\sqrt{(x_1-x_2)(y_1-y_2)}\right)^2$

$\frac{x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2}{ \left|(x_1-x_2)(y_1-y_2)\right|}=\sqrt{3}$

As the left side is rational and the right side is irrational, this lead to a contradiction, so at least one of the coordinate points has to be irrational.

All polygons

The technique in general is the same. The surface area of a regular polygon is $\frac{1}{4} s^2 n \cot{\pi/n}$. As according to Pick’s theorem the area of a rational polygon is rational, $\cot{\pi/n}$ has to be rational. The only solution to this is $n=4$, so the only polygon that can be constructed with 4 rational points are squares.

Squares

Now that we know that it is possible to create a square with 4 rational points, we can ask ourselves the question, what about a square with 3, 2, 1 or zero rational points. 1 and 0 are trivially possible. But what about the others?
Assume you have a square with at least two rational points. We can translate our squares, so that one of our squares is located at $(0,0)$, and the other one at $(p,q)$. If those two points were adjacent, the other two points would be $(-q,p)$ and $(p-q,p+q)$. Another symmetric possibility is $(q,-p)$ and $(p+q,-p+q)$. If the points are at the opposite side of the triangle, the other points are $\left(\frac{p+q}{2},\frac{-p+q}{2}\right)$ and $\left(\frac{p-q}{2},\frac{p+q}{2}\right)$. All these points are rational, so it is a square has at least two rational points, it certainly has 4.

Expanding this property to other polygons will be done in a future blogpost