%
%\begin{enumerate}
%\item $\begin{aligned}[t]
%\prodscal{AC}{AM} & = AC \times AM \times \cos\angleor{AC}{AM}\\
%& = AB\sqrt2 \times AM \times \cos\dfrac{\pi}{8}\\
%& = \sqrt2\times\dfrac{\sqrt{2+\sqrt2}}{2}AM\\
%& = \dfrac{\sqrt{4+2\sqrt2}}{2}AM.
%\end{aligned}$
%
%\item $\begin{aligned}[t]
%\prodscal{AC}{AM} & = \big(\vv{AD}+\vv{DC}\big)\cdot\big(\vv{AD}+\vv{DM}\big)\\
%& = \prodscal{AD}{AD} + \underbrace{\prodscal{AD}{DM}}_{=\vv{0}} + \underbrace{\prodscal{DC}{AD}}_{=\vv{0}} + \prodscal{DC}{DM}\\
%& = AD^2 + DC\times DM\\
%& = 1 + DM.
%\end{aligned}$
%
%\item Des deux questions précédentes, on déduit que:
%\[
%\dfrac{\sqrt{4+2\sqrt2}}{2}AM = 1 + DM.
%\]
% Or,
% \[
%AM = \sqrt{AD^2+DM^2} = \sqrt{1+DM^2}
%\]
%donc l'égalité devient:
%\begin{align*}
%\dfrac{\sqrt{4+2\sqrt2}}{2}AM= 1 + DM & \iff \dfrac{\sqrt{4+2\sqrt2}}{2}\sqrt{1+DM^2} = 1 + DM\\
%& \iff \left(\dfrac{\sqrt{4+2\sqrt2}}{2}\sqrt{1+DM^2} \right)^2 = (1 + DM)^2\\
%  & \iff \dfrac{(4+2\sqrt2)(1+DM^2)}{4} = 1 + 2DM + DM^2\\
%  & \iff \dfrac{2+\sqrt2}{2} + \dfrac{2+\sqrt2}{2}DM^2-1-2DM-DM^2=0\\
%  & \iff \dfrac{\sqrt2}{2}DM^2 - 2DM + \dfrac{\sqrt2}{2} = 0.
%\end{align*}
% 
%\item Le discriminant de l'équation précédente, d'inconnue $DM$, est:
%\[
%\Delta = (-2)^2 - 4\times\dfrac{\sqrt2}{2}\times\dfrac{\sqrt2}{2} = 2.
%\]
%Les solutions sont alors:
%\[
%\dfrac{2-\sqrt2}{\sqrt2}=\sqrt2-1\qquad\text{et}\qquad\dfrac{2+\sqrt2}{\sqrt2}=\sqrt2+1>1.
%\]
%$0 < DM < 1$ donc $DM = \sqrt2-1$.
%\end{enumerate}
