Three circles within a larger circle Let's say that the radius of the bigger circle is R. Every circle inside touches the perimeter of the bigger circle and two other circles. How to find the radius of the smaller circles (all are identical).

To illustrate:



Any tips and ideas would be awesome.
The middle stanza of Soddy's Kiss Precise gives the formula:
Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.
Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.
Applied here it says $$\frac 3{r^2}+\frac 1{R^2}=\frac 12 \left(\frac 3r-\frac 1R\right)^2\\\frac 3{r^2}+\frac 1{R^2}=\frac {9}{2r^2}+\frac 1{2R^2}-\frac 3{rR}$$  As all the we can get is the ratio, let $r=1$ and we have $$3+\frac 1{R^2}=\frac 92+\frac 1{2R^2}-\frac 3R\\0=3R^2-6R-1\\R=\frac 16(6\pm\sqrt{48})=\frac 13(3\pm2\sqrt{3})$$ and we want the plus sign.
Let $R$ be radius of enclosing circle and $r$ be the radii of the three internal circles. Since three internal circles kiss each other their centers form a  equilateral triangle with sides of $2r$. The centroid of this equilateral triangle is also center of enclosing Circle of radius $R$.There is a smaller space filling Circle that kisses the three inner circles. This Circle has radius $z$ where 

$$R=2r+z$$

Centroid of triangle is $2/3$ length from vertex to midpoint of opposite side. For equilateral triangle this is $\sqrt 3 r/2$ This gives $z$ as function of $r$. Also, $R = 2r + z$.