Is it possible to construct a regular heptagon with just compass and straightedge? Is it possible to construct a regular heptagon (a figure with seven sides) with just compass and straightedge? If so, could you please give me directions for how to do this?
 A: No, it's not possible; in fact, the regular heptagon is the regular polygon with the least number of sides that is impossible to construct with compass and straightedge alone. It is, however, possible to construct it using a neusis ruler. A related question has been asked (and answered) here.
A: No, it is not possible. Because $7=2*3+1$, and there is a factor of 3 in there, you will need to trisect and angle in order to get an exact construction, something not possible with compass and straightedge alone.
The minimal polynomial for any compass&straightedge constructible number must have a degree that is a power of two (quadratic, fourth order, eighth order, 16th order, et cetera). Whereas the minimal polynomial for $\cos\frac{2\pi}{7}$ is $8x^3+4x^2-4x-1$.
And here is $\cos\frac{2\pi}{7}$ in expanded form:
$\displaystyle\cos\frac{2\pi}{7}=\frac{\sqrt[3]{\frac{7-21i\sqrt{3}}{2}}+\sqrt[3]{\frac{7+21i\sqrt{3}}{2}}-1}{6}.$
Here, the complex numbers in the numerator, when added, cancel out each other's imaginary parts leaving behind a real expression. Note that you CANNOT get rid of the complex numbers in the numerator, as the minimal polynomial that I pointed out earlier is casus irreducibilis.
And it doesn't end there. In fact, when dividing arbitrary angles into anything that isn't a power of two, a lot of times you will end up having to take cube roots, 5th roots, 7th roots, et cetera of complex numbers, which you cannot remove from the expression without introducing trigonometric functions. For example, while a regular pentagon is fairly easy to construct with compass and straightedge, constructing a regular 25-gon requires you to use an Archimedean spiral or something else in order to divide a 72° angle into five 14.4° angles. And the resulting expression for $\cos 14.4^\circ$?
$\displaystyle\cos14.4^\circ=\frac{\sqrt[5]{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}+\sqrt[5]{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}}{2\sqrt[5]{4}}.$
