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.

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