Let $a, b, c$ denote the side lengths $B C, C A$, and $A B$, respectively. Without loss of generality, assume $a \leq b \leq c$; we are looking for the possible range of $a$. First, note that the maximum possible value of $a$ is $\frac{1}{3}$, which occurs when $A B C$ is equilateral. It remains to find a lower bound for $a$. Now rewrite $c=x a$ and $b=y a$, where we have $x \geq y \geq 1$. Note that for a non-equilateral triangle, $x>1$. The triangle inequality gives us $a+b>c$, or equivalently, $y>x-1$. If we let $K$ be the area, the condition for the altitudes gives us $\frac{2 K}{c}+\frac{2 K}{b}>\frac{2 K}{a}$, or equivalently, $\frac{1}{b}>\frac{1}{a}-\frac{1}{c}$, which after some manipulation yields $y<\frac{x}{x-1}$. Putting these conditions together yields $x-1<\frac{x}{x-1}$, and after rearranging and solving a quadratic, we get $x<\frac{3+\sqrt{5}}{2}$. We now use the condition $a(1+x+y)=1$, and to find a lower bound for $a$, we need an upper bound for $1+x+y$. We know that $1+x+y<1+x+\frac{x}{x-1}=x-1+\frac{1}{x-1}+3$. Now let $f(x)=x-1+\frac{1}{x-1}+3$. If $1<x<2$, then $1+x+y \leq 1+2 x<5$. But for $x \geq 2$, we see that $f(x)$ attains a minimum of 5 at $x=2$ and continues to strictly increase after that point. Since $x<\frac{3+\sqrt{5}}{2}$, we have $f(x)<f\left(\frac{3+\sqrt{5}}{2}\right)=3+\sqrt{5}>5$, so this is a better upper bound than the case for which $1<x<2$. Therefore, $a>\left(\frac{1}{3+\sqrt{5}}\right)=\frac{3-\sqrt{5}}{4}$. For any $a$ such that $\sqrt{5}-2 \geq a>\frac{3-\sqrt{5}}{4}$, we can let $b=\frac{1+\sqrt{5}}{2} a$ and $c=1-a-b$. For any other possible $a$, we can let $b=c=\frac{1-a}{2}$. The triangle inequality and the altitude condition can both be verified algebraically. We now conclude that the set of all possible $a$ is $\frac{3-\sqrt{5}}{4}<a \leq \frac{1}{3}$.