Exercise 10.3.10

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}}

In Example 10.3.10, prove that $l$ meets $l_1$ and $l_2$ at the points $Q_1$ and $Q_2$ given in (10.13) and (10.14). Also draw the four lines whose slopes are the roots of (10.15).

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=8cm,height=8cm] {marquedRulers.pdf}
\end{center}
\end{figure}

\begin{proof}
The equation of the line $l$ with slope $m$ through $P=(\frac{1}{2},0)$ is
$$ l: y = m\left (x-\frac{1}{2}ight).$$

The intersection point $(x_1,y_1)$ of $l$ with $l_1: y=x$ is given by the system
\begin{align*}
y_1 &= x_1,\
y_1 &= m\left (x_1-\frac{1}{2}ight),
\end{align*}
which gives $x_1 = m\left (x_1-\frac{1}{2}ight)$,  $2x_1 = 2mx_1-m$, so
$$Q_1 = (x_1,y_1) = \left ( \frac{m}{2m-2}, \frac{m}{2m-2}ight).$$
The intersection point $(x_2,y_2)$ of $l$ with $l_2: y=-\frac{1}{2}x$ is given by the system
\begin{align*}
y_2 &= -\frac{1}{2}x_2,\
y_2 &= m\left (x_2-\frac{1}{2}ight),
\end{align*}
which gives $-\frac{1}{2}x_2 = m\left (x_2-\frac{1}{2}ight)$, so
$$Q_2 = (x_2,y_2) = \left ( \frac{m}{2m+1}, -\frac{m}{2(2m+1)}ight).$$
Therefore
\begin{align*}
Q_1Q_2 = 1 &\iff 1 = (x_2-x_1)^2 + (y_2-y_1)^2 \
&\iff 1 = \left(\frac{m}{2m+1} -  \frac{m}{2m-2}ight)^2 + \left(-\frac{m}{2(2m+1)} - \frac{m}{2m-2}ight)^2 \
&\iff 1 = m^2 \left[\left(\frac{2}{4m+2} - \frac{1}{2m-2}ight)^2 + \left(\frac{1}{4m+2} + \frac{1}{2m-2}ight)^2 ight]\
& \iff 1 = \frac{m^2}{(4m+2)^2(2m-2)^2} \left\{ [(2(2m-2)-(4m+2)]^2 + [2m-2+4m+2]^2ight \}\
&\iff (4m+2)^2(2m-2)^2 = m^2(36+36m^2)\
&\iff 16 (2m+1)^2(m-1)^2 = 36 m^2(1+m^2)\
&\iff 4(2m+1)^2(m-1)^2 = 9 m^2(1+m^2)\
&\iff 16 m^4-16m^3-12m^2+8m+ 4 = 9 m^4 + 9m^2\
&\iff 7m^4-16m^3-21m^2+8m+4 = 0.
\end{align*}

We obtain the four slopes and the figure with the following SAGE instructions:
\begin{verbatim}
m = var('m')
p = 7*m^4-16*m^3-21*m^2+8*m+4
l = solve(p,m)
sols = [eq.right() for eq in l]
slopes = [sol.n() for sol in sols]; slopes

[-1.06517627861170, -0.312773186089791, 0.551041848035361, 3.11262190238042]

P = (1/2,0)
g = plot(x,x,xmin,xmax,color = 'red')
g += plot(-1/2*x,x,xmin,xmax, color = 'red')
for m in slopes:
    x1,x2 = m/(2*m-2),m/(2*m+1)
    Q1,Q2 = (x1,x1),(x2,-1/2*x2)
    g += line([Q1,P], color = 'yellow')
    g += line([Q1,Q2], color = 'black')
    
texte1 = text("$l_1$",(0.8,0.9), fontsize=15, rgbcolor=(0,0,0))
texte2 = text("$l_2$",(0.8,-0.5), fontsize=15, rgbcolor=(0,0,0))
g += texte1 + texte2
g.show(aspect_ratio=1)
g.save('marquedRulers.pdf',aspect_ratio=1,xmin=-1,xmax=1,ymin=-1,ymax=1)
\end{verbatim}
\end{proof}

Exercise 10.3.10

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Exercise 10.3.10

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