$12 \cdot 90$ Find all triangles with integer side lengths and an incircle radius of 1.
(29th International Mathematical Olympiad Shortlist, 1988)
[Solution] Let the three sides of the triangle satisfy
$$
a \geqslant b \geqslant c, \quad a, b, c \in \mathbb{N}.
$$

Let $p=\frac{1}{2}(a+b+c)$, and the inradius $r=1$.
By the area formula, we have

which means
$$
\begin{array}{c}
p r=\sqrt{p(p-a)(p-b)(p-c)}, \\
p=(p-a)(p-b)(p-c), \\
4(a+b+c)=(b+c-a)(c+a-b)(a+b-c).
\end{array}
$$

Since the left side of (1) is even, the right side of (1) must also be even.
Furthermore, $b+c-a, \quad c+a-b, a+b-c$ have the same parity, so they must all be even.
Let $x=\frac{b+c-a}{2}, y=\frac{c+a-b}{2}, \quad z=\frac{a+b-c}{2}$.
Then $x, y, z$ are all positive integers, and $x \leqslant y \leqslant z$.
From (1), we have
$$
x y z=x+y+z.
$$

Since
$$
x \leqslant y \leqslant z,
$$

we have
$$
\begin{array}{c}
x y z=x+y+z \leqslant 3 z, \\
x y \leqslant 3.
\end{array}
$$
(1) If $x y=3$, then $x y z=3 z$,

thus
$$
x=y=z.
$$

However, from $x y=3$ we get $x=1, y=3$, which leads to a contradiction.
(2) If $x y=2$, then $x=1, y=2$, substituting into (2) gives
$$
z=3.
$$

In this case,
$$
a=5, \quad b=4, \quad c=3.
$$
(3) If $x y=1$, then $x=1, y=1$, substituting into (2) gives
$$
2+z=z,
$$

which leads to a contradiction.
Therefore, the only solution is: $a=5, \quad b=4, c=3$.