1. A triangular paper piece with side lengths of $\frac{3}{2}, \frac{\sqrt{5}}{2}, \sqrt{2}$ is folded along a line perpendicular to the side of length $\frac{3}{2}$. What is the maximum area of the overlapping part?
Solution 1: Let's assume in $\triangle ABC$, $a=\frac{3}{2}, b=\sqrt{2}$, $c=\frac{\sqrt{5}}{2}$. As shown in the figure, let the midpoint of $BC$ be $D$, and $AE \perp BC$. When $MN$ is folded, the overlapping area reaches its maximum. It is easy to see that $M$ is between $D$ and $E$. Let the symmetric point of $C$ with respect to $M I N$ be $C'$, and $C'N \cap AB = C$. Let $DM = x$, then $B' = 2x$.
In $\triangle ABC$, by the cosine rule,
$$
\begin{array}{l} 
\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{\sqrt{2}}{2}, \\
\cos B \\
= \frac{1}{\sqrt{5}} .
\end{array}
$$

Thus, $\angle C=45^{\circ}$,
$$
\sin B=\frac{2}{\sqrt{5}} .
$$

In $\triangle BC'G$,
$$
\begin{array}{l}
\sin G=\sin \left(\angle ABC - \angle C'\right) \\
=\sin (\angle ABC - \angle ACB) = \sin (B - C) \\
=\sin B \cos C - \cos B \sin C \\
=\frac{1}{\sqrt{10}} .
\end{array}
$$

By the sine rule,
$$
C'G=\frac{BC' \cdot \sin B}{\sin G}=4 \sqrt{2} x .
$$

Thus, $S_{\triangle BCG}=\frac{1}{2} \cdot BC' \cdot C'G \cdot \sin C' = 4 x^{2}$.
And $S_{\triangle CMN} = \frac{1}{2} \left(\frac{a}{2} + x\right)^{2}$,
Therefore, the overlapping area $MBGN$ is
$$
\begin{aligned}
S_{MBGN} & = S_{\triangle CMN} - S_{\triangle BCG} \\
& = \frac{1}{2} \left(\frac{a}{2} + x\right)^{2} - 4 x^{2} \\
& = -\frac{7}{2} \left(x - \frac{a}{14}\right)^{2} + \frac{a^{2}}{4} \leqslant \frac{a^{2}}{7} .
\end{aligned}
$$

Thus, when $x = \frac{a}{14} = \frac{3}{28}$, $S_{MBGN}$ reaches $\frac{a^{2}}{7} = \frac{9}{28}$. We now prove that when $x = \frac{3}{28}$, $M$ is between $D$ and $E$. In fact,
$$
AE = AC \cdot \sin C = 1 .
$$

Thus, $OE = CE - CD = AE - CD = \frac{1}{4}$.
Since $\frac{3}{28} < \frac{1}{4}$, $M$ is between $D$ and $E$. Therefore, when $DM = \frac{3}{28}$, the overlapping area reaches its maximum value $\frac{9}{28}$.

Solution 2: According to the diagram, the fold line $l$ lies between the perpendicular bisector of $A$ and $BC$ for the area $S$ to possibly reach its maximum. As shown in the figure, establish a coordinate system, and easily determine the coordinates of points $A$, $B$, and $C$. Let the fold line $I$ intersect $OC$ at $E$ and $AC$ at $H$, and the overlapping part is the quadrilateral $EFGH$.