(4) In space, there are four spheres with radii of $2, 2, 3, 3$, and each sphere is externally tangent to the other three. There is another smaller sphere that is externally tangent to these four spheres. Find the radius of the smaller sphere.
Solve: The centers of four balls form a tetrahedron $ABCD$, where $AB=AC=3+2=5$, $AD=3+3=6$, $BD=CD=3+2=5$, $BC=2+2=4$.
Let the midpoints of $BC$ and $AD$ be $E$ and $F$, respectively, then it is easy to know (Figure 1)
$$
\begin{array}{c}
AE=DE=\sqrt{AC^{2}-CE^{2}}=\sqrt{5^{2}-2^{2}} \\
EF=\sqrt{AE^{2}-AF^{2}}=\sqrt{5^{2}-2^{2}-3^{2}}=\sqrt{12}
\end{array}
$$

By symmetry, the center of the small ball $O$ lies on $EF$, and
$$
\begin{array}{l}
OE=\sqrt{OC^{2}-CE^{2}}=\sqrt{(r+2)^{2}-2^{2}} \\
OF=\sqrt{OA^{2}-AF^{2}}=\sqrt{(r+3)^{2}-3^{2}}
\end{array}
$$
(where $r$ is the radius of the small ball), thus
$$
\sqrt{(r+2)^{2}-2^{2}}+\sqrt{(r+3)^{2}-3^{2}}=\sqrt{12}
$$

That is
$$
\sqrt{r(r+4)}+\sqrt{r(r+6)}=\sqrt{12}
$$

Dividing the above equation by $r(r+6)-r(r+4)=2r$ gives
$$
\sqrt{r(r+6)}-\sqrt{r(r+4)}=\frac{r}{\sqrt{3}}
$$
(1) + (2) gives

Therefore
$$
\begin{array}{l}
2 \sqrt{r(r+6)}=\sqrt{12}+\frac{r}{\sqrt{3}}=\frac{6+r}{\sqrt{3}} \\
2 \sqrt{r}=\frac{\sqrt{6+r}}{\sqrt{3}}, 4 r=\frac{6+r}{3}, r=\frac{6}{11}
\end{array}
$$

Thus, the radius of the small ball is $\frac{6}{11}$.