A parallelepiped has faces that are congruent rhombuses. Furthermore, we know that the solid has a vertex where any two edges form the same angle. Express the volume of the solid in terms of appropriate defining data of the rhombus faces.
Solution. I. Let $A^{*}$ be a vertex from which the edges emanating form equal angles with each other. Place the parallelepiped $P$ on a horizontal table $H$ with a face $A^{*} B^{*} C^{*} D^{*}$ containing this vertex. Let the vertex adjacent to $A^{*}$ on the top face be $A$, and let $\angle A A^{*} B^{*} = \angle A A^{*} D^{*} = \angle B^{*} A^{*} D^{*} = \alpha$. We exclude the case $\alpha = 90^{\circ}$, as it leads to the well-known cube.

The height of $P$ is given by the distance of $A$ from its projection $A^{\prime}$ on $H$. We will show that $A^{\prime}$ lies on the line of the diagonal $A^{*} C^{*}$. By rotating the plane $A^{*} B^{*} A$ around the edge $A^{*} B^{*}$, and due to the angle equalities and $A A^{*} = D^{*} A^{*}$, $A$ passes through $D^{*}$. The plane $S$ of the rotation is perpendicular to $H$ because $H$ contains the axis. Therefore, the projection of $A$ on $H$ lies on the line $g$ perpendicular to $A^{*} B^{*}$ through $D^{*}$, so $A^{\prime}$ is also on $g$. Similarly, from $A A^{*} = B^{*} A^{*}$, it follows that $A^{\prime}$ lies on the line perpendicular to $A^{*} D^{*}$ through $B^{*}$. Thus, $A^{\prime}$ is the orthocenter of the triangle $A^{*} B^{*} D^{*}$. Consequently, $A^{\prime}$ lies on the altitude from $A^{*}$, which is the diagonal line $A^{*} C^{*}$ of the rhombus base. This is what we wanted to prove.

![](https://cdn.mathpix.com/cropped/2024_05_02_98b38cbe3f13bba4770bg-1.jpg?height=545&width=1071&top_left_y=627&top_left_x=516)

Let $S$ intersect the axis $A^{*} B^{*}$ at $E$, so the desired height is:

$$
m = A A^{\prime} = \sqrt{A E^{2} - A^{\prime} E^{2}}
$$

The right triangles $A A^{*} E$ and $D^{*} A^{*} E$ have angles $\alpha$ and $\alpha > 90^{\circ}$ at $A^{*}$, respectively, so $A E = a \sin \alpha$, $A^{*} E = |a \cos \alpha|$, where $a$ is the side length of the rhombus faces. The angle at $A^{*}$ in the right triangle $A^{\prime} A^{*} E$ is $\alpha / 2$, either equal to $\angle B^{*} A^{*} C^{*}$ or its supplementary angle, so $A^{\prime} E = A^{*} E \cdot \tan \alpha / 2$,

$$
m = \sqrt{a^{2} \sin^{2} \alpha - a^{2} \cos^{2} \alpha \tan^{2} \alpha / 2}
$$

The area of the base is $a^{2} \sin \alpha$, so the volume is

$$
V = V(a, \alpha) = a^{3} \sin \alpha \sqrt{\sin^{2} \alpha - \cos^{2} \alpha \tan^{2} \alpha / 2}
$$

The expression under the square root, $K$, can be transformed in several ways using trigonometric identities. Let $\cos \alpha / 2 = c$, $\tan \alpha / 2 = t$. Using the identities $1 + \tan^{2} x = 1 / \cos^{2} x$ and $\cos 2x = 2 \cos^{2} x - 1$,

$$
\begin{gathered}
K = 1 - (1 + t^{2}) \cos^{2} \alpha = 1 - \frac{\cos^{2} \alpha}{c^{2}} = \\
= \frac{1}{c^{2}} (-4 c^{4} + 5 c^{2} - 1) = \frac{-4}{c^{2}} (c^{2} - 1)(c^{2} - 1 / 4) = \\
= \frac{(1 - c^{2})(2 c + 1)(2 c - 1)}{c^{2}}
\end{gathered}
$$

Alternatively, using $\sin \alpha = 2 t / (1 + t^{2})$, $\cos \alpha = (1 - t^{2}) / (1 + t^{2})$,

$$
\begin{gathered}
K = 1 - \cos^{2} \alpha \left(1 + \frac{1 - \cos \alpha}{1 + \cos \alpha}\right) = \frac{-2 (\cos \alpha - 1)(\cos \alpha + 1 / 2)}{1 + \cos \alpha} = \\
= \frac{1 - \cos \alpha}{1 + \cos \alpha} \cdot (1 + 2 \cos \alpha).
\end{gathered}
$$

According to the last expression,

$$
V = a^{3} \sin \alpha \tan \frac{\alpha}{2} \sqrt{1 + 2 \cos \alpha} = 2 a^{3} \sin^{2} \frac{\alpha}{2} \sqrt{1 + 2 \cos \alpha}
$$

For $K > 0$, the requirement is that the rhombus faces must have angles $\alpha$ such that $2 c - 1 > 0$, $3 - t^{2} > 0$, or $\cos \alpha / 2 > 1 / 2$, $\sqrt{3} > \tan \alpha / 2 (> 0)$, and $1 + 2 \cos \alpha > 0$, $\cos \alpha > -1 / 2$. Therefore, $\alpha < 120^{\circ}$. If $\alpha \geq 120^{\circ}$, the rhombus cannot form a vertex of type $A^{*}$ with its obtuse angle.

It is easy to see that there are no other conditions for the construction. Thus, for $\alpha < 120^{\circ}$, the 6 congruent rhombus faces can form a vertex of type $A^{*}$ with either the acute or obtuse angle. If the rhombus has an obtuse angle $\geq 120^{\circ}$, only the acute angle can form such a vertex.

II. We can express $V$ in terms of the diagonals $A^{*} C^{*} = e$ and $B^{*} D^{*} = f$ of the rhombus. With these,

$$
\begin{aligned}
a & = \frac{1}{2} \sqrt{e^{2} + f^{2}}, \quad \tan \frac{\alpha}{2} = \frac{f}{e} \\
\sin \alpha & = \frac{m_{1}}{a} = \frac{a m_{1}}{a^{2}} = \frac{e f}{2 a^{2}} = \frac{2 e f}{e^{2} + f^{2}}
\end{aligned}
$$

($m_{1}$ is the height of the rhombus face), so from (3) and (1),

$$
V = V(e, f) = \frac{f}{8 e} (e^{2} + f^{2}) \sqrt{3 e^{2} - f^{2}}
$$

(The independent derivation of this expression from part I is left to the reader.)

Tihanyi László (Makó, József A. g. III. o. t.) Rosta László (Csongrád, Batsányi J. g. III. o. t.)

Remark. According to some solutions, the second sentence of the problem statement is redundant. Indeed, if $P$ has two different angles at any vertex, then two of the three angles are equal. By moving along the edge connecting the equal angles, we find a neighboring vertex of $P$ where all three angles are equal. - The second sentence aimed to avoid unnecessary case distinctions.