We divide the sides of a regular hexagon in the ratio $p: q$. The points of division determine a new hexagon, and we again divide its sides in the same ratio, imagining this process to continue indefinitely. What is the sum of the areas of the hexagons? What must $p: q$ be for the sum of the areas of the inscribed hexagons to be four times the area of the original hexagon?
Let the side of the original hexagon be denoted by $a$, and the intersections of $a$ by $x$ and $y$. Then,

$$
x + x = a
$$

and

$$
x : y = p : q,
$$

from which

$$
x = \frac{a p}{p+q}, \quad y = \frac{a q}{p+q}
$$

If the side of the regular hexagon formed by the first (or $k$-th) division is $a_{1} (a_{k})$, then

$$
a_{1}^{2} = x^{2} + y^{2} - 2 x y \cos 120^{\circ} = x^{2} + y^{2} + x y = a^{2} \frac{p^{2} + q^{2} + p q}{(p+q)^{2}}
$$

or

$$
a_{k}^{2} = a_{k-1}^{2} \frac{p^{2} + q^{2} + p q}{(p+q)^{2}}
$$

Thus, the sum of the areas of the hexagons is

$$
\begin{gathered}
T = \frac{3 a^{2}}{2} \sqrt{3} + \frac{3 a_{1}^{2}}{2} \sqrt{3} + \ldots + \frac{3 a_{k}^{2}}{2} \sqrt{3} + \ldots = \\
= \frac{3}{2} \sqrt{3} \left(a^{2} + a_{1}^{2} + \ldots + a_{k}^{2} + \ldots \right)
\end{gathered}
$$

The series in the parentheses is an infinite geometric progression with a common ratio

$$
q = \frac{p^{2} + q^{2} + p q}{(p+q)^{2}} = \frac{(p+q)^{2} - p q}{(p+q)^{2}} = 1 - \frac{p q}{(p+q)^{2}} < 1
$$

Therefore,

$$
T = \frac{3}{2} \sqrt{3} \frac{a^{2}}{1 - \left\{1 - \frac{p q}{(p+q)^{2}}\right\}} = \frac{3 \sqrt{3}}{2} a^{2} \frac{(p+q)^{2}}{p q}
$$

If $T$ is four times the area of the original hexagon, then

$$
\frac{3 \sqrt{3}}{2} a^{2} \frac{(p+q)^{2}}{p q} = 4 \frac{3 \sqrt{3}}{2} a^{2}
$$

or

$$
\begin{gathered}
(p+q)^{2} = 4 p q \\
(p-q)^{2} = 0
\end{gathered}
$$

from which

$$
p = q
$$

(Jánosy Gyula, Budapest.)

The problem was also solved by: Blum J., Csada I., Erdős V., Fekete M., Fodor H., Földes R., Haar A., Heimlich P., Kiss J., Krampera Gy., Morvai O., Patz S., Pichler S., Ruvald S., Schuster Gy., Schwarz O.