The objective of this exercise is to find a method to construct a regular pentagon with straightedge and compass. In the complex plane equipped with a direct orthonormal coordinate system ( $\mathrm { O } , \vec { u } , \vec { v }$ ), we consider the regular pentagon $A _ { 0 } A _ { 1 } A _ { 2 } A _ { 3 } A _ { 4 }$, with center $O$ such that $\overrightarrow { O A _ { 0 } } = \vec { u }$. We recall that in the regular pentagon $A _ { 0 } A _ { 1 } A _ { 2 } A _ { 3 } A _ { 4 }$:
the five sides have the same length;
the points $A _ { 0 } , A _ { 1 } , A _ { 2 } , A _ { 3 }$ and $A _ { 4 }$ belong to the unit circle;
for any integer $k$ belonging to $\{ 0 ; 1 ; 2 ; 3 \}$ we have $\left( \overrightarrow { O A _ { k } } ; \overrightarrow { O A _ { k + 1 } } \right) = \frac { 2 \pi } { 5 }$.
We consider the points $B$ with affix $-1$ and $J$ with affix $\frac { \mathrm { i } } { 2 }$. The circle $\mathscr { C }$ with center $J$ and radius $\frac { 1 } { 2 }$ intersects the segment $[ B J ]$ at a point $K$. Calculate $B J$, then deduce $B K$.
a. Give in exponential form the affix of point $A _ { 2 }$. Justify briefly. b. Prove that $B A _ { 2 } { } ^ { 2 } = 2 + 2 \cos \left( \frac { 4 \pi } { 5 } \right)$. c. A computer algebra system displays the results below, which may be used without justification:
\multicolumn{2}{|l|}{Formal calculation}
1
\begin{tabular}{ l } $\cos \left( 4 ^ { * } \mathrm { pi } / 5 \right)$
\hline 2 & $\operatorname { sqrt } ( ( 3 - \operatorname { sqrt } ( 5 ) ) / 2 )$ \hline & $\rightarrow \frac { 1 } { 2 } ( \sqrt { 5 } - 1 )$ \hline \end{tabular} ``sqrt'' means ``square root'' Deduce, using these results, that $B A _ { 2 } = B K$.
In the coordinate system ( $\mathrm { O} , \vec { u } , \vec { v }$ ) provided in the appendix, construct a regular pentagon with straightedge and compass. Do not use a protractor or the ruler's graduations and leave the construction lines visible.
The objective of this exercise is to find a method to construct a regular pentagon with straightedge and compass.\\
In the complex plane equipped with a direct orthonormal coordinate system ( $\mathrm { O } , \vec { u } , \vec { v }$ ), we consider the regular pentagon $A _ { 0 } A _ { 1 } A _ { 2 } A _ { 3 } A _ { 4 }$, with center $O$ such that $\overrightarrow { O A _ { 0 } } = \vec { u }$.\\
We recall that in the regular pentagon $A _ { 0 } A _ { 1 } A _ { 2 } A _ { 3 } A _ { 4 }$:
\begin{itemize}
\item the five sides have the same length;
\item the points $A _ { 0 } , A _ { 1 } , A _ { 2 } , A _ { 3 }$ and $A _ { 4 }$ belong to the unit circle;
\item for any integer $k$ belonging to $\{ 0 ; 1 ; 2 ; 3 \}$ we have $\left( \overrightarrow { O A _ { k } } ; \overrightarrow { O A _ { k + 1 } } \right) = \frac { 2 \pi } { 5 }$.
\end{itemize}
\begin{enumerate}
\item We consider the points $B$ with affix $-1$ and $J$ with affix $\frac { \mathrm { i } } { 2 }$.
The circle $\mathscr { C }$ with center $J$ and radius $\frac { 1 } { 2 }$ intersects the segment $[ B J ]$ at a point $K$.\\
Calculate $B J$, then deduce $B K$.
\item a. Give in exponential form the affix of point $A _ { 2 }$. Justify briefly.\\
b. Prove that $B A _ { 2 } { } ^ { 2 } = 2 + 2 \cos \left( \frac { 4 \pi } { 5 } \right)$.\\
c. A computer algebra system displays the results below, which may be used without justification:
\begin{center}
\begin{tabular}{ | l | l | }
\hline
\multicolumn{2}{|l|}{Formal calculation} \\
\hline
1 & \begin{tabular}{ l }
$\cos \left( 4 ^ { * } \mathrm { pi } / 5 \right)$ \\
$\rightarrow \frac { 1 } { 4 } ( - \sqrt { 5 } - 1 )$ \\
\end{tabular} \\
\hline
2 & $\operatorname { sqrt } ( ( 3 - \operatorname { sqrt } ( 5 ) ) / 2 )$ \\
\hline
& $\rightarrow \frac { 1 } { 2 } ( \sqrt { 5 } - 1 )$ \\
\hline
\end{tabular}
\end{center}
``sqrt'' means ``square root''\\
Deduce, using these results, that $B A _ { 2 } = B K$.
\item In the coordinate system ( $\mathrm { O} , \vec { u } , \vec { v }$ ) provided in the appendix, construct a regular pentagon with straightedge and compass. Do not use a protractor or the ruler's graduations and leave the construction lines visible.
\end{enumerate}