The plane is equipped with an orthonormal coordinate system ( $\mathrm { O } , \vec { u } , \vec { v }$ ). For all integer $n \geqslant 4$, we consider $P _ { n }$ a regular polygon with $n$ sides, with centre $O$ and whose area is equal to 1. We admit that such a polygon is made up of $n$ triangles superimposable to a given triangle $\mathrm { OA } _ { n } \mathrm {~B} _ { n }$, isosceles at O. We denote $r _ { n } = \mathrm { OA } _ { n }$ the distance between the centre O and the vertex $\mathrm { A } _ { n }$ of such a polygon.

\textbf{Part A: study of the particular case $n = 6$}

\begin{enumerate}
  \item Justify the fact that the triangle $\mathrm { OA } _ { 6 } \mathrm {~B} _ { 6 }$ is equilateral, and that its area is equal to $\frac { 1 } { 6 }$.
  \item Express as a function of $r _ { 6 }$ the height of the triangle $\mathrm { OA } _ { 6 } \mathrm {~B} _ { 6 }$ from the vertex $\mathrm { B } _ { 6 }$.
  \item Deduce that $r _ { 6 } = \sqrt { \frac { 2 } { 3 \sqrt { 3 } } }$.
\end{enumerate}

\textbf{Part B: general case with $n \geqslant 4$}

In the method considered, we take as initial matrix the matrix $I = \left( \begin{array} { l l } 1 & 0 \\ 0 & 1 \end{array} \right)$.

\begin{enumerate}
  \item Determine the two missing matrices $A$ and $B$, in the third row of the Stern-Brocot tree.
  \item We associate to a matrix $M = \left( \begin{array} { l l } a & c \\ b & d \end{array} \right)$ of the Stern-Brocot tree the fraction $\frac { a + c } { b + d }$. Show that, in this association, the path ``left-right-left'' starting from the initial matrix in the tree, leads to a matrix corresponding to the fraction $\frac { 3 } { 5 }$.
  \item Let $M = \left( \begin{array} { l l } a & c \\ b & d \end{array} \right)$ be a matrix of the tree. We recall that $a , b , c , d$ are integers. We denote $\Delta _ { M } = a d - b c$, the difference of the diagonal products of this matrix.\\
  a. Show that if $a d - b c = 1$, then $d ( a + c ) - c ( b + d ) = 1$.\\
  b. Deduce that if $M = \left( \begin{array} { l l } a & c \\ b & d \end{array} \right)$ is a matrix of the Stern-Brocot tree such that $\Delta _ { M } = a d - b c = 1$, then $\Delta _ { M \times G } = 1$, that is, the difference of the diagonal products of the matrix $M \times G$ is also equal to 1. We similarly admit that $\Delta _ { M \times D } = 1$, and that all other matrices $N$ of the Stern-Brocot tree satisfy the equality $\Delta _ { N } = 1$.
  \item Deduce from the previous question that every fraction associated with a matrix of the Stern-Brocot tree is in lowest terms.
  \item Let $m$ and $n$ be two non-zero natural integers that are coprime. Thus the fraction $\frac { m } { n }$ is in lowest terms. We consider the following algorithm:
\begin{verbatim}
VARIABLES : m and n are non-zero natural integers and coprime
PROCESSING : While m = do
        If m<n
            Display "Left"
            n takes the value n-m
    Else
        Display "Right"
            m takes the value m-n
\end{verbatim}
  a. Copy and complete the following table, indicate what the algorithm displays when run with the values $m = 4$ and $n = 7$.
\begin{center}
\begin{tabular}{ | c | l | l | l | l | l | }
\hline
Display &  & $\ldots$ & $\ldots$ & $\ldots$ & $\ldots$ \\
\hline
$m$ & 4 & $\ldots$ & $\ldots$ & $\ldots$ & $\ldots$ \\
\hline
$n$ & 7 & $\ldots$ & $\ldots$ & $\ldots$ & $\ldots$ \\
\hline
\end{tabular}
\end{center}
  b. Conjecture the role of this algorithm. Verify by a matrix calculation the result provided with the values $m = 4$ and $n = 7$.
\end{enumerate}