8. If the altitudes of $\triangle A B C$ with a perimeter of 1 can serve as the side lengths of another triangle, then the range of $\min \{A B, B C, C A\}$ is $\qquad$
8. $\left(\frac{3-\sqrt{5}}{4}, \frac{1}{3}\right)$.

Let $a$, $b$, and $c$ represent the lengths of $BC$, $CA$, and $AB$ respectively, and assume $a \leqslant b \leqslant c$.
Next, we find the range of $a$.
When $\triangle ABC$ is an equilateral triangle, it meets the conditions, and at this time, $a$ reaches its maximum value of $\frac{1}{3}$.
Now, we find the minimum value of $a$.
Let $c = xa$ and $b = ya$. Then $x \geqslant y \geqslant 1$.
Note that, at this time, $\triangle ABC$ is not an equilateral triangle.
Therefore, $x > 1$.
By the triangle inequality, we have
$$
a + b \geqslant c \Rightarrow y > x - 1.
$$

Let the area of $\triangle ABC$ be $S$. Then, by the given conditions, $\frac{2S}{c} + \frac{2S}{b} > \frac{2S}{a} \Rightarrow \frac{1}{b} > \frac{1}{a} - \frac{1}{c}$
$\Rightarrow y > 1$.
Also, $a(1 + x + y) = 1$, so
$a > \frac{1}{3 + \sqrt{5}} = \frac{3 - \sqrt{5}}{4}$.
When $\frac{3 - \sqrt{5}}{4} < a \leqslant \sqrt{5} - 2$, we can set
$b = \frac{1 + \sqrt{5}}{2} a$, $c = 1 - a - b$;
When $\sqrt{5} - 2 < a \leqslant \frac{1}{3}$, we can set $b = c = \frac{1 - a}{2}$.
It is not difficult to verify that all the given conditions are satisfied at this time.
In summary, the range of $a$ is $\frac{3 - \sqrt{5}}{4} < a \leqslant \frac{1}{3}$.