, 2013

Solution:

Answer: $\left(\frac{3-\sqrt{5}}{4}, \frac{1}{3}\right]$

Let $a, b, c$ denote the side lengths $BC, CA$, and $AB$, 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 $ABC$ 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 $\frac{2K}{c} + \frac{2K}{b} > \frac{2K}{a}$, 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 $1 + x + y \leq 1 + 2x < 5$. 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 $\frac{3-\sqrt{5}}{4} < a \leq \frac{1}{3}$.