I would like to find the algebraic translation of the following problem:
Let $r>0$ be a real number. For some $r$, the regular hexagon with area $\pi r^2$ is not constructible (with straightedge and compass).
How do I reduce this statement to a condition of some number being algebraic over the rationals?
Regular hexagon with side length of $a$ has area $\frac{3\sqrt{3}}{2} a^2$. Therefore we want to find segment $a = \sqrt{\frac{2\pi r^2}{3\sqrt{3}}}$, in particular, this has to be an algebraic number.
(This is the case iff $\frac{3\sqrt{3}}{2}a^2 = \pi r^2$ is algebraic, which is easier to handle.)