Circles And Circles And Circles

Let us just focus on the right side of the symmetrical figure. The figure above shows part of the largest circle and the next two largest circles with centers at $A$ , $B$ and $C$ respectively. Since the triangle is equilateral, then all its interior angles are equal to $\dfrac{180^\circ}3 = 60^\circ$ , so $\angle AGD = \dfrac{60^\circ}2 = 30^\circ$ and so $\angle GAD = 180^\circ - 90^\circ - 30^\circ = 60^\circ$ .

Thus, $\cos \angle GAD = \dfrac{AG}{AD} \Rightarrow \cos 60^\circ = \dfrac{AG}1 \Rightarrow AG = 2$ .

Let $r_1 > r_2 > r_3 > \cdots$ be all the radii of the subsequent circles. So in the figure, we see that $r_1 = BE$ and $r_2 = CF$ .

Then $BG = AG - AB = 2 - 1 - r_1 =1 - r_1$ . Because $\triangle ADG \sim \triangle BEG$ , then $\dfrac{AG}{AB} = \dfrac{BG}{BE} \quad \Rightarrow \quad \dfrac21 = \dfrac{1 - r_1}{r_1} \quad \Rightarrow \quad r_1 = \dfrac13 \; .$

Likewise, $CG = AG - AB - BC = 2 - (1 + r_1) + (r_1 + r_2) = 1 - 2r_1 - r_2$ and because $\triangle ADG \sim \triangle CFG$ , then

$\dfrac{AG}{AB} = \dfrac{BG}{BE} \quad \Rightarrow \quad \dfrac21 = \dfrac{1 - 2r_1 - r_2}{r_2} \quad \Rightarrow \quad r_2 = \dfrac19 \; .$

If we continue on this pattern, we see that $r_3 = \dfrac1{27} , r_4 = \dfrac1{81} , r_5 = \dfrac1{243}$ . This suggests that $r_n = \dfrac1{3^n}$ .

By looking at the working above, it's not hard to extract out the recursive formula, $\displaystyle 3r_n =1 - 2 \sum_{k=1}^{n-1} r_k$ , where $k = 1,2,3,\ldots$ . A simple induction shows that $r_n = \dfrac1{3^n}$ is indeed true.

Hence, the area in the right figure is equal to the sum of area of the semicircles and the areas of the circles with radii $r_1, r_2, \ldots$ . That is,

$\begin{aligned}\text{Area in the right figure} &=& \dfrac12 \pi R^2 + \pi r_1^2 + \pi r_2 ^2 + \pi r_3^2 + \cdots \\ &=& \dfrac12 \pi + \pi \sum_{k=1}^\infty r_k ^2 \\ &=& \dfrac12 \pi + \pi \sum_{k=1}^\infty \dfrac1{9^k} \\ &=& \dfrac12 \pi + \pi \cdot \dfrac{1/9}{1-1/9} = \dfrac58 \pi, \end{aligned}$

With the series in the penultimate step follows a geometric progression sum .

Doubling this number gives the desired area $\dfrac54 \pi$ , and so $a+b = 5 + 4 = \boxed9$ .

Relevant wiki: Similar Triangles - Problem Solving - Medium