, 2009

We begin with the standard expression:
$$\frac{1}{r} = \frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c},$$
where $r$ is the inradius and $h_a, h_b, h_c$ are the altitudes of the given triangle, in standard notation.

The arithmetic mean - harmonic mean inequality gives
$$\frac{1}{r} \ge \frac{9}{h_a + h_b + h_c} \ge \frac{9}{20}.$$
Thus $r \le 20/9$. It follows that $r = 1$ or $2$.

Case 1. Suppose $r = 1$. We may assume that $h_a \le h_b \le h_c$. This implies that $c \le b \le a$. Using $a < b + c$, we must have
$$\frac{1}{h_a} < \frac{1}{h_b} + \frac{1}{h_c}.$$
Thus $3h_a \le h_a + h_b + h_c \le 20$ and we get $h_a \le 6$. We may easily rule out $h_a = 1$. If $h_a = 2$, then we see that $1/h_a = 1/h_b + 1/h_c$, which is impossible.
If $h_a = 3$, then we get
$$\frac{1}{h_b} + \frac{1}{h_c} = 1 - \frac{1}{3} = \frac{2}{3}.$$
This leads to
$$(2h_b - 3)(2h_c - 3) = 9.$$
This gives $(h_b, h_c) = (2, 6)$ or $(3, 3)$. The first pair is not possible, since $h_b \ge h_a = 3$. In the second case we have $(h_a, h_b, h_c) = (3, 3, 3)$.
A similar analysis shows that there are no possible triples in the cases $h_a = 4, 5, 6$.

Case 2. Suppose $r = 2$. We see that $3 \le h_a \le 6$, in this case. Again a case by case analysis shows that there are no possible triangles when $h_a = 3$ or $4$. If $h_a = 5$, we have one possibility: $(h_a, h_b, h_c) = (5, 5, 10)$. If $h_a = 6$, we have the triple $(6, 6, 6)$.

In the cases $(h_a, h_b, h_c) = (3, 3, 3)$ and $(6, 6, 6)$, we have equilateral triangles. It is easy to get area in terms of the altitude: $\Delta = h^2/\sqrt{3}$. Thus we get areas equal to $3\sqrt{3}$ and $12\sqrt{3}$. In the other case $(h_a, h_b, h_c) = (5, 5, 10)$, we see that $a = b = 2c$. Heron's formula can be used to compute $\Delta$. We get $\Delta = 100/\sqrt{15}$.

Thus the possible values for $\Delta$ are $3\sqrt{3}$, $12\sqrt{3}$ and $100/\sqrt{15}$.