## Argue that the regular 5-gon is constructible

Use the fact that $\alpha = 2\mathsf{cos}(2\pi/5)$ satisfies the polynomial $p(x) = x^2+x-1$ (to be proved later) to argue that the regular 5-gon is constructible using a straightedge and compass.

Using the rational root test, we can see that $p(x)$ is irreducible over $\mathbb{Q}$. Thus the roots of $p(x)$ lie in a degree 2 extension of $\mathbb{Q}$, and we have seen that all such numbers are constructible by straightedge and compass. So $2\mathsf{cos}(2\pi/5)$, and hence $\beta = \mathsf{cos}(2\pi/5)$, is constructible.

Recall that $\mathsf{sin}^2\ \theta + \mathsf{cos}^2\ \theta = 1$, and so (since $2\pi/5$ is in the first quadrant) $\mathsf{sin}(2\pi/5) = \sqrt{1 - \mathsf{cos}^2(2\pi/5)}$. In particular, $\mathsf{sin}(2\pi/5)$ is also constructible.

I’ll try to describe the rest in words, because my geometric diagrams tend to look like garbage unless I spend a couple of hours on them.

Suppose now that we have a line segment $AB$, which we want to be an edge of a regular 5-gon. Extend the line $AB$ and construct the line perpendicular to $AB$ at $B$. On $AB$, construct the point $X$ such that $ABX$ and $BX$ has measure $\mathsf{cos}(2\pi/5)$. On the perpendicular, construct the point $Y$ such that $BY$ has measure $\mathsf{sin}(2\pi/5)$. Now construct the perpendicular to $BX$ at $X$ and to $BY$ at $Y$, and let $Z$ be the intersection of these two lines. Finally, construct the point $C$ on $BZ$ such that either $BCZ$ or $BZC$ and such that $BC$ and $AB$ have the same measure. By construction, $\angle CBX$ has measure $2\pi/5$, so that $\angle ABC$ has measure $3\pi/5$. Repeat this construction with $BC$ (and so on), being careful with the orientation, to construct a regular 5-gon.