Exercise 10.1.12

Proof.

(a)

If $M=(x,y)$ is a point of the spiral of Archimedes, then $$r=\sqrt{x^2+y^2}=𝜃.$$

If $𝜃=\pi ∕2$, then $x=0,y>0$, so $y=\pi ∕2$. So $\pi ∕2$ and $\pi$ is constructible, if we use the spiral of Archimede, and also $\sqrt{\pi }$ by Section 10.1. So the spiral of Archimedes enables one to square the circle.

(b)

If $0<{𝜃}_0<\pi$, let $M=(a,b)$ the corresponding point on the spiral, so $$r=\sqrt{a^2+b^2}={𝜃}_0.$$

We can obtain by the geometric construction of Section 10.1 the real number $r^{\prime }=r∕3$. Let $C$ the circle with center 0 and radius $r^{\prime }=r∕3$. The intersection of $C$ with the spiral gives is the point $M^{\prime }=((r∕3)\cos <!--\; FUNCTION\; APPLICATION\; -->({𝜃}_0∕3),(r∕3)\sin <!--\; FUNCTION\; APPLICATION\; -->({𝜃}_0∕3))$, so the measure of the angle $\widehat{({\stackrel{\to }{e}}_1,\stackrel{\to }{O{M}^{\prime }})}$ is ${𝜃}_0∕3$. So the spiral enables one to trisect ${𝜃}_0$.