Exercise 10.3.13

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

According to [15] (G.E.Martin, Geometric Constructions), Pappus used a marked ruler to trisect angles as follows. Given an angle $0<	heta < \pi/2$, write it as $	heta = \angle POA$, where:
\be
\item[$\bullet$] The distance between $P$ and $O$ is $1/2$.
\item[$\bullet$] The line $l_1$ determined by $P$ and $A$ is perpendicular to the line determined by $O$ and $A$.
\ee
Any angle $0 < 	heta < \pi/2$ can be put in this form by a marked-ruler construction. Finally, let $l_2$ be the line through $P$ that is perpendicular to $l_1$. Then verging with $O$ and the lines $l_1$ and $l_2$ gives points $Q \in l_1$ and $R \in l_2$ such that $Q$ and $R$ are one unit apart.

Prove that $\angle QOA = 	heta/3$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{figure}[htbp]
\begin{center}
\includegraphics[width=10.8cm,height=11.4cm]{markedRulers5.pdf}
\end{center}
\end{figure}

\begin{proof}
As $OP = \frac{1}{2}$, $P = \left(\frac{1}{2} \cos 	heta, \frac{1}{2} \sin 	heta ight)$, and $A = \left(\frac{1}{2} \cos 	heta,0 ight)$. The vertical line is not a solution. The equation of every nonvertical line  $l$ with slope $m$ passing through $Q$ is $$l : y = mx,$$ where $m = 	an \alpha$ and $\alpha\in ] -\frac{\pi}{2}, \frac{\pi}{2}[$ is a measure of $\angle AOR$.
The equations of $l_1,l_2$ are 
\begin{align*}
l_1  &: x = \frac{1}{2} \cos 	heta,\
l_2 &: y = \frac{1}{2} \sin 	heta.
\end{align*}
The intersection point $Q = (x_1,y_1)$ of $l$ with $l_1$ is given by
\begin{align*}
x_1 &= \frac{1}{2} \cos 	heta,\
y_1 &= m \frac{1}{2} \cos 	heta,
\end{align*}
and the intersection point $R$ of $l$ with $l_2$ is given by
\begin{align*}
x_2 &= \frac{1}{2m} \sin 	heta,\
y_2 & = \frac{1}{2} \sin 	heta.
\end{align*}
Therefore
\begin{align*}
QR = 1 &\iff 1 = (x_2-x_1)^2 + (y_2-y_1)^2\
&\iff 1 = \left ( \frac{1}{2m} \sin 	heta -  \frac{1}{2} \cos 	heta ight)^2 + \left( \frac{1}{2} \sin 	heta - m \frac{1}{2} \cos 	hetaight)^2\
&\iff 1 = \frac{1}{4m^2} (\sin 	heta - m \cos 	heta)^2 + \frac{1}{4} (\sin 	heta - m \cos 	heta)^2\
&\iff 4 m^2  = (\sin 	heta - m \cos 	heta)^2(m^2+1)\
&\iff 0 = \left (\sin 	heta - m \cos 	heta - \frac{2m}{\sqrt{m^2+1}}ight )\left (\sin 	heta - m \cos 	heta + \frac{2m}{\sqrt{m^2+1}}ight).
\end{align*}
Note that, since $m = 	an \alpha$,
$$ \frac{2m}{\sqrt{m^2+1}} = \frac{2 \, \frac{\sin \alpha}{\cos \alpha}} {\sqrt{\frac{\sin^2 \alpha}{\cos^2 \alpha} + 1}}  = 2 \sin \alpha.$$
Moreover 
$$\sin 	heta - m \cos 	heta = \sin 	heta - \frac{\sin \alpha}{\cos \alpha} \cos 	heta = \frac{\sin(	heta - \alpha)}{\cos \alpha}.$$
Therefore
\begin{align*}
 QR = 1 &\iff 0 = \left (\frac{\sin(	heta - \alpha)}{\cos \alpha} - 2 \sin \alpha ight) \left (\frac{\sin(	heta - \alpha)}{\cos \alpha} + 2 \sin \alpha ight)\
 &\iff 0 = (\sin(	heta - \alpha) - \sin (2\alpha))(\sin(	heta - \alpha) + \sin (2\alpha))\
 &\iff \sin(	heta - \alpha) =  \pm \sin (2\alpha)\
 &\iff 	heta -\alpha = 2 \alpha - k \pi  \quad \mathrm{or} \quad 	heta - \alpha =  \pi - 2 \alpha + k \pi \qquad (k\in \Z)\
 &\iff \alpha = \frac{	heta}{3} + k \frac{\pi}{3} \quad \mathrm{or} \quad \alpha = \pi - 	heta + k \pi  \qquad (k \in \Z).
\end{align*}
Since $0 <	heta  < \frac{\pi}{2}$ and $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$, we obtain finally
$$QR = 1 \iff \alpha = \frac{	heta}{3} \quad \mathrm{or} \quad \alpha = \frac{	heta + \pi}{3}  \quad \mathrm{or} \quad \alpha = \frac{	heta - \pi}{3} \quad \mathrm{or} \quad  \alpha = -	heta .$$

The only solution $0 < \alpha < 	heta$ is $\alpha = 	heta/3$.

In the example of the angle $	heta = 60^{\circ} = \pi/3$, where $	heta/3 = 20^{\circ} = \pi/9$ is not constructible by straightedge and compass, we obtain $\alpha \in \left \{ \frac{\pi}{9}, \frac{4\pi}{9},-\frac{2\pi}{9}, - \frac{\pi}{3}ight \}$, all constructible with the marked ruler (see figure).
\end{proof}

{\bf Solution 2}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=11.7cm,height=8.1cm]{markedRulers6.pdf}
\end{center}
\end{figure}

\begin{proof}
We follow the indication of David A. Cox and give the more geometric proof of [15] (G.E.Martin, Geometric Constructions):

``Let $\angle AOR$ have measure $t$. Then $\angle PRO$ has measure $t$. Let $M$ be the midpoint of $Q$ and $R$. Since $\angle RPQ$ is right, then $P$ lies on the circle with diameter $\overline{RQ}$ by the converse of the Theorem of Thales. So $MQ = MP = MR = OP = 1/2$, and $	riangle MPR$ and $	riangle POM$ are isosceles triangles. Hence, $\angle MPR$ has measure $t$ by the Pons Asinorum, $\angle OMP$ has measure $2t$ by the Exterior Angle Theorem, and $\angle POM$ has measure $2t$ by the Pons Asinorum. Therefore, $\overrightarrow{OR}$ trisects $\angle POA$."
\end{proof}

Exercise 10.3.13

Answers

Comments

Exercise 10.3.13

Answers

Comments

Add answer