Exercise 10.1.10

Proof.

(a)

For all $x\in ]0,\pi ]$, $$x\cot <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\mathit{\pi x}}{2}\right)=\frac{2}{\pi }\frac{\frac{\mathit{\pi x}}{2}}{\sin <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\mathit{\pi x}}{2}\right)}\cos <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\mathit{\pi x}}{2}\right).$$

As $\underset{u\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}\frac{\sin <!--\; FUNCTION\; APPLICATION\; -->u}{u}=1$ and $\underset{u\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}\cos <!--\; FUNCTION\; APPLICATION\; -->(u)=1$, then

$$\underset{x\to 0}{\lim <!--\; FUNCTION\; APPLICATION\; -->}x\cot <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\mathit{\pi x}}{2}\right)=\frac{2}{\pi }.$$

(b)

Let $ℍ$ be the set of Hippias-constructible points. Then by Section 10.1, $ℍ$ is a subfield of $ℂ$, and $𝒞$ is a subfield of $ℍ$.

As $(0,2∕\pi )$ is on the quadratrix and the $y$-axis is constructible, $(2∕\pi )i\in ℍ$. Since $i$ is constructible and since $ℍ$ is a field, $2∕\pi \in ℍ$, so $\pi ∕2$ and $\pi$ are in $ℍ$. Then the proof of Theorem 10.1.4 shows that $\sqrt{\pi }\in ℍ$. If $r\in {ℝ}_{\text{+}}^{\text{∗}}$ is a constructible number and the radius of a circle $C$, then $\sqrt{\pi }r\in ℍ$, so we can square every constructible circle with the quadratrix.

(c)

As $M=(a,b)$ is on the quadratrix, $b=a\cot <!--\; FUNCTION\; APPLICATION\; -->(\mathit{\pi a}∕2)$. By definition, $𝜃$ is a measure of the angle $\widehat{(\stackrel{\to }{\mathit{OM}},{\stackrel{\to }{e}}_2)}$ (where ${\stackrel{\to }{e}}_2=(0,1)$), then $$\frac{b}{a}=\cot <!--\; FUNCTION\; APPLICATION\; -->𝜃=\cot <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\pi }{2}a\right).$$

If $𝜃=0$, then $a=0$. Since the restriction $\cot <!--\; FUNCTION\; APPLICATION\; -->:]0,\pi ∕2]\to [0,+\infty [$ is strictly decreasing, hence is injective, and $𝜃\in ]0,\pi ∕2],\frac{\pi }{2}a\in ]0,\pi ∕2]$, then

$$𝜃=\frac{\pi }{2}a,0\le a\le 1.$$

(d)

Let $l$ a line containing the origin $O$, and $(a,b)$ the intersection of $l$ with the quadratrix. As in part (c), $$\frac{b}{a}=\cot <!--\; FUNCTION\; APPLICATION\; -->𝜃=\cot <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\pi }{2}a\right).$$

Let $(c,d)$ be the point on the quadratrix corresponding with the angle $𝜃∕3$. Then

$$\frac{d}{c}=\cot <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{𝜃}{3}\right)=\cot <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{\pi }{2}c\right),$$

therefore, as in part (c), $\frac{𝜃}{3}=\frac{\pi }{2}c$, so

$$c=\frac{a}{3}.$$

Starting from the line $l$, the quadratrix and the points $0,1$, We can construct with straightedge-and-compass $M=(a,b)$, and $c=a∕3$ by the construction of Section 10.1, Figure 1, thus we can construct $M^{\prime }=(c,d)$ at the intersection of the quadratrix and the vertical line passing through $(c,0)$, and this gives the constructive line $l^{\prime }=O{M}^{\prime }$ corresponding to the angle $𝜃∕3$. So we can trisect every angle $\widehat{(\stackrel{\to }{\mathit{OM}},{\stackrel{\to }{e}}_2)}$, with measure $𝜃,0<𝜃<\pi ∕2$.

(e)

Using construction given in Exercise 2 (a), we can transport with straightedge-and-compass constructions any given angle determined by $\alpha ,\beta ,\gamma$ (with measure $𝜃,0<𝜃<\pi ∕2$) on a congruent angle $\widehat{(\stackrel{\to }{\mathit{OM}},{\stackrel{\to }{e}}_2)}$. The inverse transport gives the trisection of the given angle. If $\pi ∕2\le 𝜃\le \pi$ (obtuse angle), then we can trisect the constructive angle with measure $𝜃-\pi ∕2$, then add the constructive angle with measure $\pi ∕6$. So we can trisect arbitrary angle with the quadratrix.

(f)

In the reasoning of parts(d) and (e), we can replace $𝜃∕3$ by $𝜃∕n$, where $n$ is any positive integer. So we can divide any angle in $n$ parts with the quadratrix. Thanks to the quadratrix!