Summation of triangle sides!

This comes from summing a large number of GP series. Firstly, the triangle could be equilateral. Then $S_1 \; =\; \sum_{a=1}^\infty \frac{2^a}{3^a5^a} \; = \; \tfrac{2}{13}$ Next consider the cases where the triangle is isosceles, we have $\begin{aligned} S_{2,1} & = \; \sum_{(a,b,b) \in T \;,\; a < b}\frac{2^a}{3^b5^b} \; = \; \sum_{a=1}^\infty \sum_{b=a+1}^\infty \frac{2^a}{15^b} \; = \; \tfrac{1}{91} \\ S_{2,2} & = \; \sum_{(b,a,b)\in T\;,\; a < b} \frac{2^b}{3^a5^b} \; = \; \tfrac{4}{39} \\ S_{2,3} & = \; \sum_{(b,b,a)\in T\;,\; a < b} \frac{2^b}{3^b5^a} \; = \; \tfrac{4}{13} \\ S_{2,4} & = \; \sum_{(a,b,b) \in T \;,\; a > b}\frac{2^a}{3^b5^b} \; = \; \sum_{b=2}^\infty \sum_{a=b+1}^{2b-1} \frac{2^a}{15^b} \; = \; \tfrac{8}{143} \\ S_{2,5} & = \; \sum_{(b,a,b)\in T\;,\; a > b} \frac{2^b}{3^a5^b} \; = \; \tfrac{4}{559} \\ S_{2,6} & = \; \sum_{(b,b,a)\in T\;,\; a > b} \frac{2^b}{3^b5^a} \; = \; \tfrac{4}{949} \end{aligned}$ There are also six cases to consider when the triangle is scalene, with $\begin{aligned} S_{3,1} & = \; \sum_{(a,b,c) \in T\;,\; a < b < c} \frac{2^a}{3^b5^c} \; = \; \sum_{a=2}^\infty \sum_{b=a+1}^\infty \sum_{c=b+1}^{a+b-1} \frac{2^a}{3^b5^c} \; = \; \tfrac{2}{6643} \\ S_{3,2} & = \; \sum_{(a,b,c) \in T\;,\; a < c < b} \frac{2^a}{3^b5^c} \; = \; \tfrac{2}{3913} \\ S_{3,3} & = \; \sum_{(a,b,c) \in T\;,\; b < a < c} \frac{2^a}{3^b5^c} \; = \; \tfrac{8}{2847} \\ S_{3,4} & = \; \sum_{(a,b,c) \in T\;,\; b < c < a} \frac{2^a}{3^b5^c} \; = \; \tfrac{16}{429} \\ S_{3,5} & = \; \sum_{(a,b,c) \in T\;,\; c < a < b} \frac{2^a}{3^b5^c} \; = \; \tfrac{8}{559} \\ S_{3,6} & = \; \sum_{(a,b,c) \in T\;,\; c < b < a} \frac{2^a}{3^b5^c} \; = \; \tfrac{16}{143} \end{aligned}$ Thus the desired sum is $S_1 + \sum_{j=1}^6 S_{2,j} + \sum_{j=1}^6 S_{3,j} \; = \; \tfrac{17}{21}$ making the answer $17+21 = \boxed{38}$ .