The roof of our cube-shaped house consists of two isosceles triangles and two symmetrical trapezoids, whose edges are equal, and any two adjacent sides form the same angle with each other. How many times the length of the roof's edge is the length of the cube's edge?
Solution. Let the edge of the cube be of length $a$, and the edge of the roof be of length $b$. The endpoints of the roof edges, as shown in Figure 1, are $A, B, C, D, E$, and $F$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0fb7a821c6c69669589ag-1.jpg?height=577&width=521&top_left_y=243&top_left_x=737)

Figure 1

The side panels $E B A, E B C F$, and $E F D A$ are adjacent, so their planes form equal angles pairwise. The intersection lines of these three planes also form equal angles pairwise, and each edge starting from $E$ is of length $b$. Based on this, the triangle $A B F$ (and similarly, the triangle $E C D$) is equilateral. The symmetric trapezoids $A E F D$ and $B C F E$ have diagonals and longer bases of length $a$, and shorter bases and legs of length $b$.

Consider one of the two symmetric trapezoids forming the roof, with vertices $A, E, F$, and $D$. By drawing the diagonals of the trapezoid, we obtain isosceles triangles whose angles, based on the diagram, are

$$
\angle M A D = \angle M D A = \angle M E F = \angle M F E = \angle M D F = \angle F A E = \alpha.
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_0fb7a821c6c69669589ag-1.jpg?height=246&width=463&top_left_y=1319&top_left_x=820)

Using the fact that $A D = A F = a$, we see that $\angle D F M = \angle F D A = 2 \alpha$. Therefore, the triangles $A D F$ and $F D M$ are isosceles and similar, and the ratio of their sides is $F M : F D = F D : A F$. This precisely means that $(a - b) : b = b : a$. Multiplying and rearranging, we get

$$
a^2 - a b = b^2
$$

Now, setting this to zero and dividing by $a^2$, we obtain a quadratic equation for the ratio $\frac{b}{a}$. The positive root of this equation is the famous golden ratio:

$$
\frac{\sqrt{5} - 1}{2}
$$

Remark. We get such a roof when we select an edge of a regular dodecahedron and cut the dodecahedron with the plane defined by the four vertices adjacent to the endpoints of the edge, forming a square.

Based on the work of Tekla Tóth.