Nested Equilateral Triangles and Circles

Let $O$ be the centroid of the first equilateral $\triangle ABC$ and $M$ be the midpoint of $AC$ . Then the medium $BM$ passes through $O$ and is perpendicular to $AC$ . Since $O$ is the centroid, $BO=2OM$ .Therefore the linear dimension of the second triangle is half that of the first, and the third's is half of the second's and so on. Hence the area of $n$ th triangle and incircle are $\frac 14$ that of the preceding $(n-1)$ th ones respectively.

That is $A_T(n) = \dfrac 14 A_T(n-1)$ $\implies A_T(n) = \dfrac {A_T(1)}{4^{n-1}}$ . Similarly, $A_c(n) = \dfrac {A_c(1)}{4^{n-1}}$ .

Then we have:

$\begin{aligned} \frac {\sum_{n=1}^\infty \left(A_T(n) + A_c(n)\right)}{A_T(1)} & = \frac 1{A_T(1)} \sum_{\color{#3D99F6}n=1}^\infty \left(\frac {A_T(1)}{4^{\color{#3D99F6}n-1}} + \frac {A_c(1)}{4^{\color{#3D99F6}n-1}} \right) \\ & = \frac {A_T(1)+A_c(1)}{A_T(1)} \sum_{\color{#D61F06}n=0}^\infty \left(\frac 14\right)^{\color{#D61F06}n} \\ & = \left(1+\color{#3D99F6} \frac \pi{3\sqrt 3}\right) \frac 1{1-\frac 14} & \small \color{#3D99F6} \text{Note that }\frac {A_c(1)}{A_T(1)} = \frac \pi{3\sqrt 3} \\ & = \frac {4(3\sqrt 3+\pi}{9\sqrt 3} \end{aligned}$

Therefore, $a+b = 4+3 = \boxed 7$ .

Let $a_{1}$ be a side of equilateral $\triangle{ABC}$ .

$A_{T_{1}} = \boxed{\dfrac{\sqrt{3}}{4}a_{1}^2}$

$\dfrac{a_{1}}{2} = \dfrac{\sqrt{3}}{2}m_{1} \implies m_{1} = \dfrac{a_{1}}{\sqrt{3}} \implies h_{1} = \dfrac{a_{1}}{2\sqrt{3}} \implies A_{c}(1) = \dfrac{\pi}{12}a_{1}^2 = \boxed{\dfrac{\pi}{4 * 3}}$

$\dfrac{a_{2}}{2} = \dfrac{\sqrt{3}}{2}h_{1} \implies a_{2} = \sqrt{3}h_{1} = \sqrt{3}(\dfrac{a_{1}}{2\sqrt{3}}) = \dfrac{a_{1}}{2}$ and $h_{2} = \dfrac{\sqrt{3}}{2}a_{2} = \dfrac{\sqrt{3}}{2}(\dfrac{a_{1}}{2}) = \dfrac{\sqrt{3}}{4}a_{1} \implies A_{T_{2}} = \dfrac{\sqrt{3}}{16}a_{1}^2 = \boxed{\dfrac{\sqrt{3}}{4^2}a_{1}^2}$

$\dfrac{a_{2}}{2} = \dfrac{\sqrt{3}}{2}m_{2} \implies m_{2} = \dfrac{a_{2}}{\sqrt{3}} = \dfrac{a_{1}}{2\sqrt{3}} \implies$ $h_{3} = \dfrac{1}{2}(\dfrac{a_{1}}{2\sqrt{3}}) = \dfrac{a_{1}}{4\sqrt{3}}$ $\implies A_{c_{2}} = \boxed{\dfrac{\pi}{4^2 * 3}a_{1}^2}$

Continuing in this manner we have:

$\boxed{A_{T}(n) = \dfrac{\sqrt{3}}{4^n}a_{1}^2}$ and $\boxed{A_{c}(n) = \dfrac{\pi}{3 * 4^n}a_{1}^2}$

$T = \sum_{n = 1}^{\infty} A_{T}(n) + A_{c}(n) = \dfrac{1}{4}(\dfrac{3\sqrt{3} + \pi}{3})a_{1}^2 \sum_{n = 1}^{\infty} \dfrac{1}{4^{n - 1}} =\dfrac{1}{4}(\dfrac{3\sqrt{3} + \pi}{3})a_{1}^2(\dfrac{4}{3})a_{1}^2 =$ $(\dfrac{3\sqrt{3} + \pi}{9})a_{2}^2$

and

$\dfrac{T}{A_{T_{1}}} = \dfrac{4(3\sqrt{3} + \pi)}{9\sqrt{3}} = \dfrac{a(b\sqrt{b} + \pi)}{b^2\sqrt{b}} \implies a + b = \boxed{7}$