Use $a$, $b$, and $c$. Clearly, $c=\sqrt{a^{2}+b^{2}}$. Set up the equation $ab=3a+3b+3\sqrt{a^{2}+b^{2}}$. Solve for the radical and we get $3\sqrt{a^{2}+b^{2}}=ab-3a-3b$. Square both sides to get $a^{2}b^{2}-6a^{2}b-6ab^{2}+9a^{2}+18ab+9b^{2}=9a^{2}+9b^{2}$, or $a^{2}b^{2}-6a^{2}b-6ab^{2}+18ab=0$. Factor to get $ab(ab-6a-6b+18)=0$. Clearly, $a$ and $b$ must be positive, so $ab\neq0$. Divide by it to get $ab-6a-6b+18=0$. Add 18 to both sides and factor to get $(a-6)(b-6)=18$. Since $a,b\in\mathbb{Z}$, $a-6$ and $b-6$ are integers as well. We have the following possibilities for $a-6$ and $b-6$: $\pm1\text{ and }\pm18$ $\pm2\text{ and }\pm9$ $\pm3\text{ and }\pm6$ No negative solution works, because if we have -6, -9, or -18, either $a$ or $b$ will not be a positive integer. Trying out the three that remain, we get $\boxed{7,24,25}$, $\boxed{8,15,17}$, and $\boxed{9,12,15}$. And guess what: THEY ALL WORK!

in other words we have: $2\triangle = 6s$ $\triangle = 3s$ $rs = 3s$ $r = 3$. Now draw the diagram, since $r = 3$, and tangent lines from the same point are congruent, we deduce that $a+b-6 = c = \sqrt{a^{2}+b^{2}}$. Squaring and using Simon's favorite trick we get $(a-6)(b-6) = 18$ and get the sides lengths: $(9, 12, 15)$ $(7, 24, 25)$ $(8, 15, 17)$