On a smooth table, there are three wooden balls with radii of $1, 2, 4$ respectively, all externally tangent to each other. Additionally, there is another small wooden ball on the table with a radius less than 1, which is tangent to the table and externally tangent to all three balls. Find the radius of this small wooden ball.
It is known that the distance between the two points of tangency of two spheres with radii $r_{1}, r_{2}$, which are tangent to the table and externally tangent to each other, is $2 \sqrt{r_{1} r_{2}}$. Let the radius of the small wooden ball be $x$, and the points of tangency of the wooden balls with radii $1, 2, 4, x$ on the table be $A, B, C, D$ respectively. Thus,
$$
\begin{array}{l}
A B=2 \sqrt{2}, A C=4, B C=4 \sqrt{2}, \\
A D=2 \sqrt{x}, C D=4 \sqrt{x}, B D=2 \sqrt{2 x} .
\end{array}
$$
Let $\angle A D B=\alpha, \angle B D C=\beta$, then $\angle C D A=2 \pi-(\alpha+\beta)$. In $\triangle A D B$, by the cosine rule, we have $A B^{2}=A D^{2}+B D^{2}$ $-2 A D \cdot B D \cdot \cos \alpha$. Therefore,
$$
\cos \alpha=\frac{3 x-2}{2 \sqrt{2} x} .
$$

Similarly, using the cosine rule in $\triangle B D C$ and $\triangle C D A$, we have
$$
\begin{array}{l}
\cos \beta=\frac{3 x-4}{2 \sqrt{2} x}, \\
\cos [2 \pi-(\alpha+\beta)]=\frac{5 x-4}{4 x} .
\end{array}
$$

It is easy to get
$$
\begin{array}{l}
\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{\sqrt{12 x-4-x^{2}}}{2 \sqrt{2} x}, \\
\sin \beta=\sqrt{1-\cos ^{2} \beta}=\frac{\sqrt{24 x-16-x^{2}}}{2 \sqrt{2} x} .
\end{array}
$$

Using (2), (3), (4), and (5), we have
$$
\begin{array}{l}
\frac{5 x-4}{4 x}=\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta \\
=\frac{1}{8 x^{2}}[(3 x-2)(3 x-4) \\
\left.-\sqrt{\left(12 x-4-x^{2}\right)\left(24 x-16-x^{2}\right)}\right] .
\end{array}
$$

From (6), we have
$$
\begin{array}{l}
\sqrt{\left(12 x-4-x^{2}\right)\left(24 x-16-x^{2}\right)} \\
=(3 x-2)(3 x-4)-2 x(5 x-4) \\
=8-10 x-x^{2} .
\end{array}
$$
Squaring both sides of (7), we get
$$
\begin{array}{l}
\left(12 x-4-x^{2}\right)\left(24 x-16-x^{2}\right) \\
=\left(8-10 x-x^{2}\right)^{2} .
\end{array}
$$
Expanding and simplifying (8), we have
$$
7 x^{2}-28 x+16=0 \text {. }
$$

Solving (9), since $0<x<1$, we get
$$
x=2-\frac{2}{7} \sqrt{21} .
$$