We drew the face diagonals of a unit-edged cube and drew a sphere around the center of the cube. The sphere intersected the diagonals at the vertices of a convex body, all of whose faces are regular. What was the radius of the sphere?
The smallest sphere that has a common point with the face diagonals is the one that just touches the squares at their center points. The radius of this sphere is 1/2, and the resulting convex body is an octahedron, which is bounded by 8 equilateral triangular faces.

The largest sphere that still has a common point with the face diagonals is the sphere circumscribed around the cube. The intersection points are the endpoints of the diagonals, so the resulting convex body is the original cube. The radius of the sphere is half the space diagonal of the cube, \(\frac{\sqrt{3}}{2}\).

![](https://cdn.mathpix.com/cropped/2024_05_02_c013603a8601658bef9bg-1.jpg?height=686&width=1195&top_left_y=427&top_left_x=454)

Now let's consider the case where the sphere intersects the face diagonals at two points different from their endpoints. Regardless of the sphere's radius, its center (which is also the center of the cube) is \(\frac{1}{2}\) units away from the center of any face. From the symmetry of the cube and the sphere, it follows that four intersection points are created on each face. Let's examine the convex body defined by these intersection points.
![](https://cdn.mathpix.com/cropped/2024_05_02_c013603a8601658bef9bg-1.jpg?height=510&width=980&top_left_y=1374&top_left_x=560)

Let \(A, B, C, D\) denote the vertices of the cube's base in a positive rotational direction, and \(A', B', C', D'\) the vertices above \(A, B, C, D\) respectively. On the adjacent faces \(A B B' A'\) and \(A' B' C' D'\), the vertices of the squares cut by the sphere are \(A_1, A_2, A_3, A_4\) and \(B_1, B_2, B_3, B_4\) respectively, listed in a positive rotational direction, such that \(A_1 A_4 \parallel A' B'\) and \(B_1 B_2 \parallel A' B'\), with \(A_1\) and \(B_1\) being closer to \(A' B'\) than \(A_2\) and \(B_4\).

By symmetry, the distance of point \(B_2\) from the plane \(C' B' B\) is the same as that of point \(A_4\), i.e., \(B_2 A_4 \parallel C' B' B\) plane. Since \(B_1 B_2 \parallel A' B' \parallel A_1 A_4\) and \(A' B'\) is perpendicular to the plane \(C' B' B\), it is also perpendicular to the line \(A_4 B_2\), so the quadrilateral \(A_1 A_4 B_2 B_1\) is generally a rectangle. The convex body is thus bounded by 6 squares, 12 rectangles, and 8 triangles. Since a \(120^\circ\) rotation around the space diagonals of the cube maps the cube and the sphere onto themselves, the latter are regular.

In our problem, we need to find the radius of the sphere that, when intersecting the cube, makes the sides of the rectangles equal.

Let the radius of the sphere that satisfies this condition be \(r\), and express the side of the squares formed on the cube's faces as a function of \(r\). The half-diagonal of the square is \(\sqrt{r^2 - \left(\frac{1}{2}\right)^2}\), so the side of the square is

\[
a = \frac{2}{\sqrt{2}} \sqrt{r^2 - \left(\frac{1}{2}\right)^2} = \sqrt{2r^2 - \frac{1}{2}}
\]

This must also be equal to the length of segment \(A_4 B_2\). The length of \(A_4 B_2\) can also be determined using the Pythagorean theorem. Construct a plane perpendicular to \(A_4 B_2\) that intersects \(A' B'\) at point \(M\). In the isosceles right triangle \(A_4 M B_2\),

\[
M B_2 = M A_4 \quad \text{and} \quad A_4 B_2 = \sqrt{2} \cdot M B_2 = a
\]

so

\[
M B_2 = \frac{a}{\sqrt{2}}
\]

The plane \(B_2 M A_4\) intersects the top face of the cube along a line parallel to \(B' C'\), so

\[
a + 2 \cdot \frac{a}{\sqrt{2}} = 1 \quad \text{and} \quad a = \sqrt{2r^2 - \frac{1}{2}}
\]

Using these equations, we get

\[
(\sqrt{2} + 1) \sqrt{2r^2 - \frac{1}{2}} = 1
\]

From this, since \(r > 0\), we find that the radius of the sphere is \(r = \sqrt{\frac{7}{4} - \sqrt{2}} = 0.579\).