How many ordered triples $(a,b,c)$ of positive integers, with $a\le b\le c\le12$, are there such that there exists a (non-degenerate) triangle $\triangle ABC$ with an integer inradius for which $a$, $b$, and $c$ are the lengths of the altitudes from $A$ to $\overline{BC}$, $B$ to $\overline{AC}$, and $C$ to $\overline{AB}$, respectively?

1. With area $K$, the altitudes satisfy $a=\frac{2K}{BC}$, etc., so the sides are proportional to $\frac1a,\frac1b,\frac1c$ and the inradius $r$ satisfies $\frac1r=\frac{BC+CA+AB}{2K}=\frac1a+\frac1b+\frac1c$.
2. The triangle is non-degenerate exactly when $\frac1a<\frac1b+\frac1c$, equivalent to $a>2r$. Also $\frac1r\le\frac3a$, i.e. $a\le3r$. So $2r
3. $r=1$: $a=3$, $\frac1b+\frac1c=\frac23$: $(3,3,3)$.
4. $r=2$: $a=5$: $\frac1b+\frac1c=\frac3{10}$ gives only $(b,c)=(5,10)$; $a=6$: $\frac1b+\frac1c=\frac13$ gives $(6,6)$. So $(5,5,10)$, $(6,6,6)$.
5. $r=3$: $a\in\{7,8,9\}$ and $\frac1b+\frac1c=\frac13-\frac1a$. For $a=7$ (sum $\frac4{21}$) there is no solution with $7\le b\le c\le12$ (for $b=7,8,9$ we get $c=21,\frac{168}{11},\frac{63}5$, for $b=10$ we get $c=\frac{210}{19}$, and $b\ge11$ gives $c
6. $r=4$: $8
7. In total there are $1+2+2+1=\boxed{6}$ triples.