We have a rhombus with an acute angle of $72^{\circ}$. Construct a regular pentagon inscribed in the rhombus and calculate its side length. (By inscribed, we mean that each vertex of the pentagon lies on one of the sides of the rhombus.)
Given the rhombus $ABCD=R$, at least two vertices of the regular pentagon $EFGHK=P$ fall on one of its sides, and it is clear that at most two, and only two adjacent, pentagon vertices can fall on one side. Let's choose the labeling such that the side $EF$ of the pentagon lies on the side $AB$ of the rhombus, and there is a $72^{\circ}$ angle at $A$. Thus, $G$ cannot be on the side $CD$, because then $H$ would be outside the rhombus, so it must be on $BC$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f3e80c3531dd8ad67c2bg-1.jpg?height=479&width=599&top_left_y=346&top_left_x=752)

Since $\angle ABC = 108^{\circ} = \angle EFG$, this can only happen if $F$ falls on vertex $B$. Therefore, the rhombus can be obtained from the pentagon by extending the sides $EF$ and $FG$ and intersecting them with the lines parallel to $EF$ through $H$ and parallel to $FG$ through $K$. We need to show that the resulting parallelogram is indeed a rhombus. This follows from the fact that reflecting the figure over the angle bisector of the pentagon at $F$, the sides $EF$ and $FG$ will be the reflections of each other, and the images of parallel lines are also parallel. Thus, the parallelogram maps onto itself, meaning the sides on the lines $EF$ and $FG$ are equal, so the quadrilateral is a rhombus (and the axis of reflection is a diagonal).

If $R$ is given, $P$ can be easily constructed by noting that $\triangle BHC$ and $\triangle ABK$ are isosceles triangles. Indeed, $BH \parallel EK$, so $\angle BHC = \angle KEA = 180^{\circ} - 108^{\circ} = 72^{\circ} = \angle BCH$, thus $BC = BH = BK$. Therefore, $H$ and $K$ can be found by intersecting the circle centered at $B$ with radius $BC$ with the sides $CD$ and $DA$, respectively, and then $E$ and $G$ can be found by drawing lines parallel to $BC$ through $H$ and parallel to $AB$ through $K$ on sides $AB$ and $BC$, respectively.

Let the side of the rhombus be $c$ and the side of the pentagon be $a$. From the isosceles triangle $\triangle BHC$, the similar triangle $\triangle HCG$ is cut off because $\angle HGC = 72^{\circ} = \angle HCG$, so $HC = HG = a$, and

$$
\frac{HC}{BC} = \frac{GC}{HC}, \quad HC^2 = BC \cdot GC, \quad \text{or} \quad a^2 = c(c - a)
$$

We rearrange this equation and solve for $a$ (taking the positive root):

$$
a^2 + ac - c^2 = 0, \quad a = \frac{\sqrt{5} - 1}{2} c
$$

Chikán Attila (Eger, Gárdonyi G. g. IV. o. t.)

Note. Observe that it was not necessary to specify that the given parallelogram is a rhombus; this followed from the knowledge of the angles and the fact that a regular pentagon can be inscribed in it. [^0]

[^0]:    ${ }^{1}$ We performed the same calculation to find $\cos 36^{\circ}$ in problem 1157. See page 118 of this issue.