, 2024

Let $x, y$, and $z$ be the lengths of $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$, respectively. Let $r$ be the inradius of $\triangle ABC$. Then
$$\text{Area}(\triangle ABC) = \frac{1}{2}xa = \frac{1}{2}yb = \frac{1}{2}zc = \frac{1}{2}(x + y + z)r.$$
Therefore $x = \frac{2\text{Area}(\triangle ABC)}{a}$, $y = \frac{2\text{Area}(\triangle ABC)}{b}$, and $z = \frac{2\text{Area}(\triangle ABC)}{c}$, so
$$\text{Area}(\triangle ABC) = \frac{1}{2} \left( \frac{2\text{Area}(\triangle ABC)}{a} + \frac{2\text{Area}(\triangle ABC)}{b} + \frac{2\text{Area}(\triangle ABC)}{c} \right) r.$$

Dividing by Area($\triangle ABC$) gives $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})r = 1$, so
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{r}.$$
Because $a, b, c \le 9$, it follows that
$$\frac{1}{r} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{1}{3},$$
implying $r \le 3$. It is then possible to find the solutions $(a, b, c)$ by examining cases based on the value of $r$.
* If $r = 1$, then, because $a \le b \le c$, either $a = 2$ or $a = 3$. If $a = 2$, then $\frac{1}{b} + \frac{1}{c} = \frac{1}{2}$. Then, because $b \le c$, either $b = 3$ or $b = 4$. If $b = 3$, then $c = 6$; and if $b = 4$, then $c = 4$. If $a = 3$, then it must be that $b = c = 3$. So the solutions in this case are $(2, 3, 6)$, $(2, 4, 4)$, and $(3, 3, 3)$.
* If $r = 2$, then, because $a \le b \le c$, it follows that $a = 3, a = 4, a = 5$, or $a = 6$. If $a = 3$, then $\frac{1}{b} + \frac{1}{c} = \frac{1}{6}$, which has no solutions because $\frac{1}{9} + \frac{1}{9} > \frac{1}{6}$. If $a = 4$, then $\frac{1}{b} + \frac{1}{c} = \frac{1}{4}$. In this case, $b = c = 8$ is the only solution. If $a = 5$, then $\frac{1}{b} + \frac{1}{c} = \frac{3}{10}$, which gives no solutions. If $a = 6$, then it follows that $b = c = 6$. So the solutions in this case are $(4, 8, 8)$ and $(6, 6, 6)$.
* If $r = 3$, then $(a, b, c) = (9, 9, 9)$ is the only solution because if $a < 9$, then $c > 9$.
Finally, the altitude lengths must be checked to ensure that these lengths give dimensions for a valid triangle. In the $(2, 3, 6)$ case, the side lengths of the triangles become $(3t, 2t, t)$ for some $t$, which does not form a triangle. In the $(2, 4, 4)$ and $(4, 8, 8)$ cases, the side lengths of the triangles become $(2t, t, t)$ for some $t$, which again does not form a triangle. The rest of the cases, namely $(3, 3, 3)$, $(6, 6, 6)$, and $(9, 9, 9)$, do produce valid triangles because if all of the altitudes have length $h$, then it is possible to form an equilateral triangle with side length $\frac{2\sqrt{3}}{3}h$. Thus there are 3 ordered triples satisfying the given conditions.