
\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}
