The algebra

Consider five equally spaced line segments meeting at a point B. The angels formed will be $360/5 = 72^\circ$ angles. Thus we will need to construct $\cos(72^\circ)$ using our straight-edge, sring and chalk. Recall that in radians $72^\circ = \frac{2\pi}{5}$. Let

$$\alpha = e^{\frac{2\pi}{5}i} = \cos(\frac{2\pi}{5}) + \sin(\frac{2\pi}{5}),$$

then $\alpha^5 = (e^{\frac{2\pi}{5}i})^5 = e^{2\pi i} = 1$. Hence $\alpha$ is a root of the polynomial X⁵-1 = (X-1)(1+X+X²+X³+X⁴). Thus, because $\alpha\neq 1$, we have

$$1+\alpha + \alpha^2 + \alpha^3+ \alpha^4 = 0$$

Also $\alpha^4 \alpha = \alpha^5 =1$, So

$$\alpha^4 = \alpha^{-1} = \cos(-\frac{2\pi}{5}) + \sin(-\frac{2\pi}{5}) = \cos(\frac{2\pi}{5}) - \sin(\frac{2\pi}{5}),$$

because $\cos(-x) = \cos(x)$ and $\sin(-x) = -\sin(x)$. Let $\beta = \alpha + \alpha^{4}$. Then

$$\beta= 2\cos(\frac{2\pi}{5}) = 2\cos(72^\circ).$$

We need to construct $\beta/2$. Now

$$\beta^2+\beta - 1 =(\alpha + \alpha^{4})^2+(\alpha + \alpha^{... ...^{4} -1\\ =\alpha^2 + 1+ \alpha^{3}+\alpha + \alpha^{4}\\ =0$$

Therefor $\beta/2$ and hence $\cos(72^\circ)$ is a root of

4X²+2X-1

a quadratic. Thus $\cos(72^\circ)$ can be drawn bye inscribing straight lines and circles. Note in particular applying the quadratic formula we have

$$\cos(72^\circ) = \frac{\sqrt{5} -1}{4}.$$

Thus we need to construct this length.