Finding the area of circles in triangle In the figure there are infinitely many circles approaching the
vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of
length $1$, how can i find the total area occupied by the circles? 

 A: Hint: Imagine scaling up the bottom right sequence of circles, so the second largest lies on top of the largest. How much have you scaled up by? How much has the area changed? How large is the big central circle?
These should guide you to a complete solution.
A: Alt. hint: $\;\;\;\;\;\;\;\;\;\;\;\;\;\;$

A: In a equilateral triangle the ratio $\frac{A_{\text{incircle}}}{A_{\text{triangle}}}$ equals $\frac{\pi}{3\sqrt{3}}$. Step 0: we draw the incircle of the original triangle.
Step 1: we draw three incircles for equilateral triangles with side length $\frac{1}{3}$. Step $n$: we draw three incircles for equilateral triangles with side length $\frac{1}{3^n}$. The area occupied by the circles is so
$$ \frac{\pi}{3\sqrt{3}}\cdot \frac{\sqrt{3}}{4}\cdot 1^2 + \sum_{n\geq 1}\frac{\pi}{3\sqrt{3}}\cdot \frac{\sqrt{3}}{4}\cdot 3\left(\frac{1}{3^n}\right)^2 $$
which simplifies to
$$ \frac{\pi}{12}+\frac{\pi}{12}\sum_{n\geq 1}\frac{3}{9^n}=\frac{\pi}{12}+\frac{\pi}{12}\cdot\frac{3}{8}=\frac{11\pi}{96}. $$
A: Denote the circumradius by $R$ and the radii of the circles by $r_1,r_2,r_3,...$. 

We will find $R$ and $r_1$:
$$r_1=\frac{2S}{a+b+c}=\frac{2\cdot\frac{\sqrt{3}}{4}}{3}=\frac{1}{2\sqrt{3}} \\
R=\frac{abc}{4S}=\frac{1}{4\cdot \frac{\sqrt{3}}{4}}=\frac{1}{\sqrt{3}}.$$
From the similarity of the triangles:
$$\begin{align}\frac{r_2}{r_1}=\frac{R-r_1-r_2}{R} \Rightarrow r_2=&\frac{1}{2\cdot 3\sqrt{3}}, \\
\frac{r_3}{r_1}=\frac{R-r_1-2r_2-r_3}{R} \Rightarrow r_3=&\frac{1}{2\cdot 3^2\sqrt{3}}, \\ 
\frac{r_4}{r_1}=\frac{R-r_1-2r_2-2r_3-r_4}{R} \Rightarrow r_4=&\frac{1}{2\cdot 3^3\sqrt{3}}, \\
\vdots \\
r_n=&\frac{1}{2\cdot 3^{n-1}\sqrt{3}}
\end{align}$$ 
Hence:
$$\begin{align}S=&\pi r_1^2+3\cdot \pi \cdot \left(r_2^2+r_3^2+r_4^2+\cdots\right)=\\
&\frac{\pi}{12}+3\cdot \pi\cdot \left(\frac{1}{4\cdot 3^3}+\frac{1}{4\cdot 3^5}+\frac{1}{4\cdot 3^7}+\cdots\right)=\\
&\frac{\pi}{12}+\frac{3\pi}{4}\cdot \frac{\frac{1}{3^3}}{1-\frac{1}{3^2}}=\\
&\frac{\pi}{12}+\frac{\pi}{32}=\\
&\frac{11\pi}{96}.\end{align}$$
