The gúla (a type of pyramid) inherits the symmetries of the underlying rhombus: the symmetry planes $A O E$ and $B O E$, with 2 out of the 4 dihedral angles being equal, and it is sufficient to calculate half of each type.

Let $O$ be the projection of $E$ onto the shorter side $A$ at $P$, and onto the longer side $B$ at $R$. Since $O B$ is perpendicular to the plane $A O E$, it is also perpendicular to the $O P$ within it, so the dihedral angle at edge $E A$ is exactly the angle $O P B$, and similarly, the dihedral angle at edge $E B$ is the angle $O R A$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e772adec77399bd660bbg-1.jpg?height=924&width=702&top_left_y=381&top_left_x=706)

We used the double expression for the area of the triangle $O A E$. Similarly,

$$
\operatorname{tg} \angle O R A = \frac{O A}{O B} \cdot \frac{B E}{O E}
$$

For the calculation, let $\frac{1+\sqrt{5}}{2}=a$ (the golden ratio - "aurea sectio" ratio), and choose the length unit as $O B$. Thus,

$$
\begin{gathered}
O A = \frac{1}{a} = \frac{\sqrt{5}-1}{2}, \quad O E = a, \\
A E^{2} = \left(\frac{1}{a}\right)^{2} + a^{2} = \frac{(3-\sqrt{5}) + (3+\sqrt{5})}{2} = 3, \\
B E^{2} = 1 + a^{2} = \frac{5+\sqrt{5}}{2}, \\
\operatorname{tg} \angle O P B = \sqrt{3}, \\
\operatorname{tg} \angle O R A = \sqrt{1 + a^{2}} \cdot \frac{1}{a^{2}} = \sqrt{5 - 2 \sqrt{5}},
\end{gathered}
$$

The half-angles are $60^{\circ}$ and $36^{\circ}$, so the dihedral angles at the edges of the pyramid are $120^{\circ}$ and $72^{\circ}$.

Remark. According to our result, 3 such pyramids can be assembled around $E$ and the vertices of type $A$, filling the space along the edge $E A$ without gaps and in a single layer. Similarly, 5 such pyramids can fill the space around the edge $B E$.

By assembling 30 such pyramids, we filled the neighborhood of $E$, with only the 30 rhombic bases visible outward. The overall picture is shown in the right lower diagram on the back cover of our February 1980 issue; according to the introduction, by painting the 30 faces with 5 colors appropriately, the planes of the 6 faces of the same color form 1 cube.
The base of the pyramid is the rhombus $ABCD$, and its vertex $E$ is erected perpendicularly at the intersection point $O$ of the diagonals, furthermore

$$
O A: O B=O B: O E=1: \frac{1+\sqrt{5}}{2} \text {. }
$$

Calculate the dihedral angles between any two adjacent lateral faces of the pyramid!