<s>[INST] Find the length of... Find the length of $\overline{AD}$ knowing that it is divided into three equal parts by to tangent circles with radius respectively  $3\sqrt{3}$ and $\sqrt{3}$ . Here's the graph: so the segments $\overline{AB}$, $\overline{BC}$ and $\overline{BC}$ are equal. 

Thank you very much! 
 [/INST] 
Let $AB=BC=CD=x$.
$O1M$, $O2E$ $⊥AD$. (Perpendicular bisectors)
$O1MEO2$ - rectangular trapezoid.
$AM=MB$ and $CE=ED$. Hence $ME = 2x$.
From right triangles $O1AM$ and $O2ED$ we got $O1M = \sqrt{R^2 - \frac{x^2}{4}}$ and $O2E = \sqrt{r^2 - \frac{x^2}{4}}$
Use Pythagorean theorem for triangle $O1FO2$ ($O2F ⊥ O1M$):
$(\sqrt{R^2 - \frac{x^2}{4}} - \sqrt{r^2 - \frac{x^2}{4}})^{2} + (2x)^2 = (R+r)^2$ 
$\equiv$
$R^2  - \frac{x^2}{4} + r^2 - \frac{x^2}{4} - 2\sqrt{R^2 - \frac{x^2}{4}} \sqrt{r^2 - \frac{x^2}{4}} + 4x^2 = R^2 + 2Rr+r^2$
$\equiv$
$\frac{7}{2}x^2-2Rr=2\sqrt{R^2 - \frac{x^2}{4}} \sqrt{r^2 - \frac{x^2}{4}}$
$\equiv$
$\frac{49}{4}x^4-14x^2Rr+4R^2r^2=4R^2r^2+\frac{x^4}{4}-x^2(R^2+r^2)$
$\equiv$
$12x^4+x^2(R^2+r^2-14Rr)=0$
Where we get $x=\frac{1}{2\sqrt{3}} \sqrt{14Rr-R^2-r^2} = 2\sqrt{2}$
By condition $AD=3x$. So $AD=6\sqrt{2}$ 
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