Consider a unit cube and a concentric sphere with radius $r$ $(1 / 2 < r < 1)$. Let us examine the convex body whose vertices are the points where the sphere intersects the three face diagonals of the cube and the vertices of the cube. For what value of the radius $r$ will the dihedral angles of the body be equal?
Let the center of the cube be $O$, two faces of the cube be the squares $ABCD=N$ and $AD B^{*} C^{*}$, their centers be $K_{1}$ and $K_{2}$, and the reflections of $O$ on these be $O_{1}$ and $O_{2}$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_ed8bf5eeb2494a7615a0g-1.jpg?height=911&width=810&top_left_y=268&top_left_x=641)

Figure 1

Thus, the lines $O K_{i}$ $(i=1,2)$ are the face diagonals of the cube. Since $A D=1, O K_{i}=1 / 2$, and $O O_{i}=2 \cdot O K_{i}=1$, according to the constraint on the radius of the sphere, $O K_{i}<r<O O_{i}$, and the 6 new vertices of the body $T$ are each an internal point $E_{i}$ of the segment $K_{i} O_{i}$. Therefore, due to the convexity of $T$, the triangles $A D E_{i}$ are faces of $T$, because every other vertex is on one side of the plane $A D E_{i}$ (Figure 2), since the closest point on the ray $K_{i} O_{i}$, other than $O_{i}$, is not allowed to be $E_{i}$, and $A, D, O_{1}, O_{2}$ are clearly in the same plane.
![](https://cdn.mathpix.com/cropped/2024_05_02_ed8bf5eeb2494a7615a0g-1.jpg?height=738&width=376&top_left_y=1534&top_left_x=858)

Figure 3

Thus, the faces of $T$ are the lateral faces of a regular pyramid, whose base is a face of the cube, and $T$ inherits all the symmetries of the cube, because any rotation or reflection that maps the cube onto itself also maps the set of 6 new vertices onto itself.

Accordingly, $T$ has only two different dihedral angles for any allowed $r$: at the edges originating from the cube, the dihedral angle is the same as the angle between the faces $A D E_{1}$ and $A D E_{2}$ at the edge $A D$, and at the pyramid edges, the dihedral angle is the same as the angle between the faces $A E_{1} D$ and $A E_{1} B$ at the edge $A E_{1}$.

The first dihedral angle is equal to the angle at $F$ in the isosceles triangle $E_{1} E_{2} F$—where $F$ is the midpoint of the edge $A D$—because $F E_{1}$ and $F E_{2}$ are both perpendicular to $A D$. For the second dihedral angle, we use the symmetries of $T$: the lateral faces, the triangles $A E_{1} D$ and $A E_{1} B$, are reflections of each other across the diagonal plane $A C C^{*}$ of the cube, so the feet of the perpendiculars from $D$ and $B$ to $A E_{1}$ coincide. Denoting this point by $G$, the second type of dihedral angle is measured by the angle at $G$ in the isosceles triangle $D B G$.

Thus, the two dihedral angles are equal if and only if the triangle $D B G$ is similar to the triangle $E_{1} E_{2} F$, that is, if

$$
\frac{E_{1} F}{D G}=\frac{E_{1} E_{2}}{D B}
$$

Here, the left side is the ratio of two altitudes of the triangle $A D E_{1}$, which is equal to the reciprocal of the ratio of the bases perpendicular to them, $\frac{A E_{1}}{A D}$—specifically, $A E_{1}$—and the right side is the ratio of the hypotenuses of the isosceles right triangles $E_{1} E_{2} O$ and $D B A$, which is equal to the ratio of their legs, $\frac{E_{1} O}{D A}=E_{1} O=r$. Substituting these into (1) and squaring, we get

$$
A E_{1}^{2}=A K_{1}^{2}+K_{1} E^{2}=\frac{1}{2}+\left(r-\frac{1}{2}\right)^{2}=r^{2}
$$

and from this, the condition for the equality of the dihedral angles is:

$$
r=\frac{3}{4}
$$

Gál Péter (Budapest, Fazekas M. Gyak. Gimn., IV. o. t.)

Remark. The dihedral angle can also be calculated. For any allowed $r$,

$$
E_{1} F E_{2} \varangle=\varphi=90^{\circ}+2 \cdot E_{1} F K_{1} \varangle,
$$

and for this,

$$
\operatorname{tg} E_{1} F K_{1} \varangle=\frac{E_{1} K_{1}}{K_{1} F}=2 r-1,
$$

so for the determined value,

$$
\frac{\varphi-90^{\circ}}{2}=\operatorname{arctg}(2 r-1)=\operatorname{arctg} \frac{1}{2}=26^{\circ} 34^{\prime}, \quad \varphi=143^{\circ} 08^{\prime}
$$