$A$ and $B$ are two opposite vertices of a unit cube. Determine the radius of the sphere that touches the faces passing through $A$ and the edges passing through $B$.
Solution. Let the sought radius be $r$. The sphere touches the faces of the cube passing through $A$, so the center $O$ is inside the cube, equidistant from these faces. The points on the space diagonal $AB$ of the cube satisfy these conditions, so $O$ lies on the space diagonal. Let the other vertices of the cube be denoted as shown in the first figure. The lines tangent to the sphere are perpendicular to the radii at the points of tangency, so if the point of tangency on the edge $BE_{2}$ is $T_{1}$, then $OT_{1}=r$ and $OT_{1} \perp BE_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a20d75211cd0c27268f2g-1.jpg?height=602&width=607&top_left_y=366&top_left_x=694)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_a20d75211cd0c27268f2g-1.jpg?height=469&width=681&top_left_y=985&top_left_x=711)

If the point of tangency of the sphere on the face $AE_{1}E_{2}E_{3}$ is denoted by $T_{2}$, then $OT_{2}$ is perpendicular to the plane of the face, so $T_{2}$ lies on the orthogonal projection of the space diagonal $AB$ onto the face, which is the face diagonal $AE_{2}$.

Consider the plane section $AE_{2}BE_{4}$ (Figure 2). In the quadrilateral $OT_{2}E_{2}T_{1}$, the angles at $T_{1}$, $T_{2}$, and $E_{2}$ are right angles, and two adjacent sides are of equal length $\left(OT_{1}=r=OT_{2}\right)$, so the quadrilateral is a square, hence $T_{1}E_{2}=T_{2}E_{2}=r$. From this, using that $AE_{2}=\sqrt{2}$, we get that $AT_{2}=\sqrt{2}-r$ and $BT_{1}=1-r$. The triangles $AT_{2}O$ and $OT_{1}B$ are similar because their corresponding sides are parallel. Therefore, the ratio of these sides is equal, so

$$
\frac{AT_{2}}{OT_{2}}=\frac{OT_{1}}{BT_{1}}, \quad \text { that is } \quad \frac{\sqrt{2}-r}{r}=\frac{r}{1-r}
$$

From this, bringing to a common denominator and rearranging, we get $\sqrt{2}-r(1+\sqrt{2})=0$, or the sought radius is:

$$
r=\frac{\sqrt{2}}{1+\sqrt{2}}=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}
$$