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
 A: Two hints:


*

*Draw the line segments from $A$ and $B$ through the center of the large circle and use the 30°-60°-90° triangles and similar triangles to get a relationship between $AB$ and $R$.

*Draw the line parallel to $\overline{BC}$ through the point of tangency of the large circle and the small circle closest to $A$ to get a small equilateral triangle, and use what you learned about the relationship between $AB$ and $R$ on the small triangle.



edit (related to comments below):

Above is the small triangle formed at the top of your diagram.  $DF=DE=4$, since both are radii of the small circle.  You can use the 30°-60°-90° triangles to find $AD$ and $AE$.  In particular, what do you find is $\frac{AD}{AF}$?

edit 2: Since the homework problem is now done, here's how I would actually have done the problem myself, though it is not the solution I would expect from a geometry student:  In an equilateral triangle, the median, altitude, angle bisector, perpendicular bisector, etc. are all the same segment.  The point of concurrency of the angle bisectors is the center of the inscribed circle and the point of concurrency of the medians divides the medians in the ratio $2:1$, so the height of the smaller triangle shown above is $R=3\cdot 4=12$ and the height of the large triangle is $3R=36$.  That height is the $\sqrt{3}$ side in a $1:\sqrt{3}:2$ right triangle (30°-60°-90°), where $AB$ is the $2$ side, so $AB=\frac{2}{\sqrt{3}}\cdot 36=\frac{72}{\sqrt{3}}=24\sqrt{3}$.
A: 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, $\frac{x}{4} = 1 \Rightarrow x = 4$.
$R = x + 8 = 12 \Rightarrow a = 2 \sqrt{3} R = 24 \sqrt{3}$.
