18.3.37 ** Find all triangles with integer side lengths such that their perimeter equals their area.
From the problem, we have $\sqrt{p(p-a)(p-b)(p-c)}=2 p$, where $a, b, c$ are natural numbers, and $p=\frac{a+b+c}{2}$. Let $p-a=x, p-b=y, p-c=z$, then $x, y, z$ are either all natural numbers or all reduced fractions with a denominator of 2.
$\sqrt{(x+y+z) x y z}=2(x+y+z)$, squaring both sides gives
$$
x y z-4(x+y+z) \text {. }
$$
The right side of (1) is a positive integer, thus $x y z$ is a positive integer, and $x, y, z$ must all be positive integers.

Assume $x \geqslant y \geqslant z$, if $z \geqslant 4$, then from (1) we get $x y \leqslant x+y+z \leqslant 3 x, y \leqslant 312$. Thus $y=3$, but this leads to $x-\frac{24}{5}$ not being an integer.
Therefore, the solutions that satisfy the problem are
\begin{tabular}{ccccccc}
$x$ & $y$ & $z$ & $x+y+z=p$ & $a$ & $b$ & $c$ \\
24 & 5 & 1 & 30 & 6 & 25 & 29 \\
14 & 6 & 1 & 21 & 7 & 15 & 20 \\
9 & 8 & 1 & 18 & 9 & 10 & 17 \\
10 & 3 & 2 & 15 & 5 & 12 & 13 \\
6 & 4 & 2 & 12 & 6 & 8 & 10
\end{tabular}