An Infinity of Circles

Make the side of the triangle $2\sqrt{3}$ , for convenience, so that $A_{\triangle}=3\sqrt{3}$ . The radius of the largest circle is $r_1=1$ , and the second circle is the incircle of an equilateral triangle with height 1, so $r_2=\frac{1}{3}$ . Likewise, $r_n=\frac{1}{3^n}$ . Now $\sum_{n=1}^{\infty} \pi r_n^2=\frac{\pi}{1-\frac{1}{9}}==\frac{9\pi}{8}$ and the ratio we seek is $\frac{\sqrt{3}\pi}{8}$ . The answer is $\boxed{11}$ .

The height of $\triangle{ABC}$ is $H_{1} = \dfrac{\sqrt{3}}{2}a$ and $\dfrac{a}{2} = \dfrac{\sqrt{3}}{2}m_{1} \implies m_{1} = \dfrac{a}{\sqrt{3}} \implies h_{1} = \dfrac{a}{2\sqrt{3}}$

$H_{2} = H_{1} - 2h_{1} = \dfrac{a}{2\sqrt{3}}$ and $\dfrac{H_{1}}{H_{2}} = 3 = \dfrac{a}{l_{2}} \implies l_{2} = \dfrac{a}{3} \implies \dfrac{a}{6} = \dfrac{\sqrt{3}}{2}m_{2} \implies m_{2} = \dfrac{a}{3\sqrt{3}} \implies h_{2} = \dfrac{a}{6\sqrt{3}} = \dfrac{a}{3(2\sqrt{3})}$

Note: $\dfrac{H_{2}}{H_{1}} = \dfrac{1}{3} = \dfrac{l_{2}}{l_{1}}$ , so where actually done here. I.E; $h_{n} = (\dfrac{a}{2\sqrt{3}})(\dfrac{1}{3^{n - 1}})$

$H_{3} = H_{2} - 2h_{2} = \dfrac{a}{6\sqrt{3}}$ and $\dfrac{H_{2}}{H_{3}} = 3 = \dfrac{a}{3l_{3}} \implies l_{3} = \dfrac{a}{9} \implies \dfrac{a}{18} = \dfrac{\sqrt{3}}{2}m_{3} \implies m_{3} = \dfrac{a}{9\sqrt{3}} \implies h_{3} = \dfrac{a}{18\sqrt{3}} = \dfrac{a}{3^2(2\sqrt{3})}$

Containing in this manner we have:

$h_{n} = (\dfrac{a}{2\sqrt{3}})(\dfrac{1}{3^{n - 1}}) \implies A_{n} = \dfrac{a^2\pi}{12}(\dfrac{1}{9^{n - 1}})$

$\implies T = \sum_{n = 1}^{\infty} A_{n} = \dfrac{a^2\pi}{12} \sum_{n = 1}^{\infty} \dfrac{1}{9^{n - 1}} =$ $\dfrac{a^2\pi}{12}(\dfrac{9}{8}) = \dfrac{3}{32}\pi a^2$

and $A_{\triangle{ABC}} = \dfrac{\sqrt{3}}{4}a^2$

$\implies \dfrac{T}{A_{\triangle{ABC}}} = \dfrac{\sqrt{3}}{8}\pi = \dfrac{a}{b}\pi \implies a + b = \boxed{11}$

This is, in fact, the same solution as Dr. Otto Bretscher's. It is just expressed differently: $\frac{\pi \sum _{n=1}^{\infty } \left(\frac{1}{3^{n-1}}\right)^2}{3 \sqrt{3}}= \frac{\sqrt{3} \pi }{8}$ .