Five congruent circles inscribed in a regular pentagon

Label the bottom part of the diagram as follows:

As half the interior angle of a regular pentagon, $\angle CAE = 54°$ , and since $CE = r$ , $AE = r \cot 54° = \sqrt{5 - 2\sqrt{5}}r$ .

Since $AE + EF + FB = AB$ , $\sqrt{5 - 2\sqrt{5}}r + 2r + \sqrt{5 - 2\sqrt{5}}r = 6$ , which solves to $r = \cfrac{3}{1 + \sqrt{5 - 2\sqrt{5}}} \approx 1.73758$ .

Therefore, $\lfloor 10^4 r \rfloor = \boxed{17375}$ .

Label the diagram as seen in the figure, where $M$ is the midpoint of $AB$ , $K$ is the center of $o$ and $L$ is the common point of circles $o$ and $p$ . Then, $AM=3$ , $KL=LM=r$ .

$OM$ is the apothem of the regular pentagon, thus

$OM=\frac{AB}{2\sqrt{5-\sqrt{20}}}=\frac{6}{2\sqrt{5-\sqrt{20}}}=\frac{3}{\sqrt{5-\sqrt{20}}}$ Triangles $\triangle OKL$ and $\triangle OAM$ are similar, hence

$\frac{KL}{AM}=\frac{OL}{OM}\Rightarrow \frac{r}{3}=\frac{OM-r}{OM}\Rightarrow r=\frac{3OM}{3+OM}$ Substituting the value of $OM$ , we get $r=\frac{3}{\sqrt{5-\sqrt{20}}+1}\approx 1.73758$ For the answer, $\left\lfloor {{10}^{4}}r \right\rfloor =\boxed{17375}$ .