Five, let the base of the pyramid $M-ABCD$ be a square, and $MA=MD$, $MA \perp AB$. If the area of $\triangle AMD$ is 1, find the radius of the largest sphere that can fit inside this pyramid.

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The base of the pyramid $M-ABCD$ is a square, and $MA=MD$, $MA \perp AB$. If the area of $\triangle AMD$ is 1, try to find the radius of the largest sphere that can fit inside this pyramid.
Five, as shown in the figure, since $A B \perp A D, A B \perp M A$, therefore, $A B$ is perpendicular to the plane $M A D$, from which we know that the plane $M A D$ is perpendicular to the plane $A C$.
Let $E$ be the midpoint of $A D$, and $F$ be the midpoint of $B C$, then $M E \perp A D$, so, $M E$ is perpendicular to the plane $A C$, and $M E \perp E F$.
Let the sphere $O$ be the sphere that is tangent to the planes $M A D, A C, M B C$. Without loss of generality, we can assume that $O$ is on the plane $M E F$. Thus, $O$ is the incenter of $\triangle M E F$. Let the radius of the sphere $O$ be $r$, then
$$
r=\frac{2 S_{\triangle M E F}}{E F+E M+M F}
$$

Let $A D=E F=a$, since $S_{\triangle A M D}=1$, so
$$
\begin{array}{c}
M E=\frac{2}{a}, M F=\sqrt{a^{2}+\left(\frac{2}{a}\right)^{2}}, \\
r=\frac{2}{a+\frac{2}{a}+\sqrt{a^{2}+\left(\frac{2}{a}\right)^{2}}} \leqslant \frac{2}{2 \sqrt{2}+2}=\sqrt{2}-1,
\end{array}
$$

and when $a=\frac{2}{a}$, i.e., $a=\sqrt{2}$, the above inequality holds with equality, so when $A D=$ $M E=\sqrt{2}$, the radius of the sphere that is tangent to the three planes $M A D, A C, M B C$ is maximized, and this maximum radius is $\sqrt{2}-1$.

Construct $O G \perp M E$ at $G$, it is easy to prove that $O G \parallel$ the plane $M A B$, and the distance from $G$ to the plane $M A B$ is the distance from $O$ to the plane $M A B$. Draw $M H \perp M A$ at $H$, then $G H$ is the distance from $G$ to the plane $M A B$.
$$
\begin{array}{l}
\because \quad \triangle M H G \backsim \triangle M E A \quad \therefore \quad \frac{G H}{A E}=\frac{M G}{M A} . \\
\text { Also } \quad M G=\sqrt{2}-(\sqrt{2}-1)=1, A E=\frac{\sqrt{2}}{2}, \\
M A=\sqrt{\left(\frac{\sqrt{2}}{2}\right)^{2}+(\sqrt{2})^{2}}=\frac{\sqrt{10}}{2} \\
\therefore \quad H G=\frac{A E \cdot M G}{M A}=\frac{\frac{\sqrt{2}}{2} \cdot 1}{\frac{\sqrt{10}}{2}}=\frac{\sqrt{5}}{5} \\
\because \quad \frac{\sqrt{5}}{5}>\sqrt{2}-1,
\end{array}
$$
$\therefore$ the distance from $O$ to the plane $M A B$ is greater than the radius $r$ of the sphere $O$, similarly, the distance from $O$ to the plane $M C D$ is also greater than the radius $r$ of the sphere $O$,
$\therefore$ the sphere $O$ is inside the pyramid $M-A B C D$, and cannot be any larger.
$\therefore$ the maximum radius of the sphere is $\sqrt{2}-1$.