We know about a convex pentagon that each of its sides is parallel to one of its diagonals. What can be the ratio of the length of a side and the length of the diagonal parallel to it?
Solution. Let the vertices of the pentagon be denoted as $A, B, C, D, E$ in some cyclic order. Let $M$ and $N$ be the intersection points of the diagonal $EC$ with the diagonals $AD$ and $BD$, respectively. Furthermore, let $AB = a$, $BC = b$, $DM = x$, and $EM = y$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_d7e126ce2aae46c1710ag-1.jpg?height=507&width=572&top_left_y=278&top_left_x=771)

From the parallelism of the sides and diagonals, it follows that the quadrilaterals $ABCM$ and $ABNE$ are parallelograms. Therefore, $AM = b$, $EN = MC = a$, and thus $MN = t = a - y$. The triangles $AMC$ and $DME$ are similar because their corresponding sides are parallel. Therefore, the ratios of their corresponding sides are equal, so $x : b = y : a$. Similarly, from the similarity of the triangles $AME$ and $DMN$, we get $x : b = t : y$. From these two equalities, we have $y^2 = at$, or $y^2 = a(a - y)$. Solving this quadratic equation, we get $y = \frac{-a \pm \sqrt{a^2 + 4a^2}}{2}$. Since $y > 0$, the ratio of the side $AB$ to the parallel diagonal $EC$ is

$$
AB : EC = a : (y + a) = a : \frac{(1 + \sqrt{5})a}{2} = \frac{2}{1 + \sqrt{5}} = \frac{\sqrt{5} - 1}{2}
$$

Clearly, this ratio is the same for any side and the parallel diagonal. Pentagons that satisfy the conditions do exist, for example, a regular pentagon.

Remark. Many tried to prove that the regularity of the pentagon follows from our conditions. However, this is not true. The conditions of the problem are satisfied not only by the regular pentagon but also by any affine regular pentagon. For more details on these polygons, interested readers can refer to Chapter 12 of István Reiman's book "Geometry and Its Frontier Areas."