5. Six equal circles $\mathrm{O}, \mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}, \mathrm{O}_{4}$, $0_{5}$, on the surface of a sphere with radius 1, $\odot O$ is tangent to the other five circles and $\odot \mathrm{O}_{1}$ is tangent to $\odot \mathrm{O}_{2}$, $\odot \mathrm{O}_{2}$ is tangent to $\odot \mathrm{O}_{3}$, $\odot O_{3}$ is tangent to $\odot O_{4}$, $\odot O_{4}$ is tangent to $\odot O_{5}$, $\odot O_{5}$ is tangent to $\odot O_{1}$, find the radius of these circles.
Let the center of the sphere be $K$, and $\odot O$ and $\odot O_{1}$ touch at $A$. The common tangent line through $A$ is $l$. Then, $K A, O A, O_{1} A$ are all in the plane perpendicular to $l$ through $A$. In this plane, $K A=1$, $O A=O_{1} A=r$. By plane geometry (see Fig. (1)), $K O=K O_{1}=\sqrt{1-r}$,
$$
\begin{array}{l}
x\left(=O B=\frac{1}{2} O O_{1}\right)=\frac{r \sqrt{1-r^{2}}}{1} \\
=r \sqrt{1-r^{2}}, \\
y(=A B)=\frac{r^{2}}{1}=r^{*} \text {. } \\
\frac{1}{2} O_{1} C \times O K=S_{\triangle O O_{1 K}}=S_{\text {quadrilateral } O_{1} A \cup K} \\
-S_{\triangle O 14 O} \\
=r \sqrt{1-r^{2}}-x y=r \sqrt{1 \cdots r^{2}} \\
-r^{3} \sqrt{1-r}=r \sqrt{1-r^{2}}\left(1-r^{2}\right) \text {, } \\
z\left(=O_{1} C\right)=\approx r\left(1-r^{2}\right) \\
\frac{1}{2} h_{r}=S_{\angle O 1 A O}=r_{\}_{j}}, \quad h=\frac{2 \times y}{r} \\
=2 r \sqrt{1-r^{2}} \text {. } \\
\end{array}
$$

Let the plane of $\odot O$ be $Q$, then plane 2 is perpendicular to $K O$, and the distances from $O, O, O, O, O_{5}$ to plane 2 are all $h=2 r \sqrt{1-r}$, so $O_{1}, O_{2}, O_{3}, O_{4}, O_{6}$ are on the same plane $R$, and $K O \perp$ plane $R$, with the foot of the perpendicular being $C$ (see Fig. (2)).

The intersection of plane $R$ with the sphere is a circle (with center $C$), and it is easy to see that the pentagon $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4} \mathrm{O}_{0}$ is an inscribed regular pentagon of this circle, and
$$
C O_{1}=z, \quad O . O_{-}=2 x, \angle O_{1} C O_{\angle}=\frac{2 \pi}{5} .
$$

Therefore,
$$
\begin{array}{l}
\frac{1}{\sin ^{2} \frac{\pi}{5}}=\left(\frac{z}{x}\right)^{2}=\left(\frac{2\left(1-r^{2}\right)}{\sqrt{1-r^{2}}}\right)^{2} \\
=4\left(1-r^{2}\right),
\end{array}
$$

which implies
$$
4 r^{2}=4-\frac{1}{\sin ^{2} \frac{\pi}{5}}=3-\operatorname{ctg} \frac{2 \pi}{5} .
$$

Thus,
$$
r=\frac{1}{2} \sqrt{3-\operatorname{ctg}^{2} \frac{\pi}{5}} .
$$