Through the center of a sphere of radius $R$, three mutually perpendicular planes are drawn. Find the radius of the sphere that touches all these planes and the given sphere.
Let the specified planes intersect along the lines $O A, O B$, and $O C$, where $O$ is the center of a given sphere of radius $R$, and $O A=O B=O C$. If $Q$ is the center of a sphere of radius $x$ that touches the given sphere at point $P$, and also touches the planes $A O B, A O C$, and $B O C$, and the plane $A O B$ at point $M$, then $Q M \perp O M$, and $O Q=O P+Q P$ (external tangency) or $O Q=O P-Q P$ (internal tangency).

Let $\angle Q O M=\alpha$. In the right triangle $Q O M$, we have the relation $Q M=O Q \sin \alpha$ or $x=(R \pm x) \sin \alpha$. Therefore, to solve the problem, it is sufficient to find $\sin \alpha$.

For this, consider a cube with vertex $O$. Its diagonal, drawn from vertex $O$, forms an angle with the planes of the three faces containing vertex $O$, which is also equal to $\alpha$. If the edge of the cube is $c$, then its diagonal is $c \sqrt{3}$. Thus,

$$
\sin \alpha=\begin{aligned}
& \frac{c}{c \sqrt{3}} \\
& =\frac{1}{\sqrt{3}}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-43.jpg?height=1229&width=1503&top_left_y=0&top_left_x=569)

$$
\text { From the equation } x=\begin{array}{r}
\frac{R \pm x}{\sqrt{3}}
\end{array} \text {, we find that } x=\begin{aligned}
& \frac{(\sqrt{3} \pm 1) R}{2}
\end{aligned} .
$$