Inscribed kissing circles in an equilateral triangle 
Triangle is equilateral (AB=BC=CA), I need to find AB and R.
Any hints? 
I was trying to make another triangle by connecting centers of small circles but didn't found anything

Let $a$ be the side of the triangle. If $A$ denotes the area and $P$ denotes the perimeter, then the radius of the incircle is given by $R = \frac{2A}{P} = \frac{2\sqrt{3} a^2/4}{3a} = \frac{\sqrt{3} a}{6}$
Let $x$ be the distance of the center of the smaller circle to the nearest vertex.
The altitude is $x + 8 + 2R = \frac{\sqrt{3}a}{2}$.
From similar triangle, we also have $\frac{x+4}{4} = \frac{x+8+R}{R} \Rightarrow \frac{x}{4} = \frac{x+8}{R}$
You have three equations in $x$,$R$ and $a$ solve them to get your $R$ and $a$.
EDIT
$x+8 = \frac{\sqrt{3}a}{2} - 2R = \frac{\sqrt{3}a}{2} - \frac{\sqrt{3}a}{3} = \frac{\sqrt{3}a}{6} = R$. Hence, $\frac{x}{4} = 1 \Rightarrow x = 4$.
$R = x + 8 = 12 \Rightarrow a = 2 \sqrt{3} R = 24 \sqrt{3}$.
