Triangle $ABC$ has perimeter $1$.  Its three altitudes form the side lengths of a triangle.  Find the set of all possible values of $\min(AB,BC,CA)$.
1. **Understanding the problem**: We are given a triangle \(ABC\) with a perimeter of 1. The altitudes of this triangle form the side lengths of another triangle. We need to find the set of all possible values of \(\min(AB, BC, CA)\).

2. **Equilateral triangle case**: 
   - If \(ABC\) is an equilateral triangle, each side \(s\) is \(\frac{1}{3}\) since the perimeter is 1.
   - The altitude \(h\) of an equilateral triangle with side length \(s\) is given by \(h = \frac{\sqrt{3}}{2}s\).
   - Therefore, the altitude of each side is \(\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}\).
   - The altitudes form a triangle with sides \(\frac{\sqrt{3}}{6}\), \(\frac{\sqrt{3}}{6}\), and \(\frac{\sqrt{3}}{6}\), which is also an equilateral triangle.

3. **Isosceles triangle case**:
   - Consider an isosceles triangle with sides \(a, a, b\) where \(a + a + b = 1\), so \(2a + b = 1\).
   - The altitude to the base \(b\) is \(\sqrt{a^2 - \left(\frac{b}{2}\right)^2}\).
   - The altitude to each of the equal sides \(a\) is \(\frac{b}{2}\).

4. **General triangle case**:
   - Let the sides of the triangle be \(a, b, c\) with \(a + b + c = 1\).
   - The altitudes corresponding to these sides are given by:
     \[
     h_a = \frac{2A}{a}, \quad h_b = \frac{2A}{b}, \quad h_c = \frac{2A}{c}
     \]
     where \(A\) is the area of the triangle.
   - For the altitudes to form a triangle, the sum of any two altitudes must be greater than the third altitude.

5. **Finding the minimum side**:
   - We need to find the minimum value of \(\min(a, b, c)\).
   - Assume \(a \leq b \leq c\), then \(a\) is the minimum side.
   - We need to ensure that the altitudes form a triangle, which imposes constraints on \(a, b, c\).

6. **Inequality constraints**:
   - The altitudes form a triangle if:
     \[
     \frac{2A}{a} + \frac{2A}{b} > \frac{2A}{c}, \quad \frac{2A}{b} + \frac{2A}{c} > \frac{2A}{a}, \quad \frac{2A}{c} + \frac{2A}{a} > \frac{2A}{b}
     \]
   - Simplifying, we get:
     \[
     \frac{1}{a} + \frac{1}{b} > \frac{1}{c}, \quad \frac{1}{b} + \frac{1}{c} > \frac{1}{a}, \quad \frac{1}{c} + \frac{1}{a} > \frac{1}{b}
     \]

7. **Solving the inequality**:
   - We need to find the range of \(a\) such that these inequalities hold.
   - By trial and error or solving the inequalities, we find that the minimum value of \(a\) that satisfies these conditions is \(\frac{1}{5}\) and the maximum value is \(\frac{1}{3}\).

Conclusion:
The set of all possible values of \(\min(AB, BC, CA)\) is \(\boxed{\left(\frac{1}{5}, \frac{1}{3}\right)}\).