Let's divide the sides $AB$, $BC$, and $CA$ of an equilateral triangle in the same ratio, and connect the division points. We again obtain an equilateral triangle. In what ratio should we divide the sides of the given triangle so that the area of the triangle formed by the division points is minimal?
Let one side of the triangle be $a$, one segment $x$, so the other is $a-x$; if the area of the inscribed triangle is $y$, then

$$
\begin{gathered}
y=\frac{a^{2}}{4} \sqrt{3}-\frac{3 x(a-x) \sin 60^{\circ}}{2}=\frac{a^{2}}{4} \sqrt{3}-\frac{3}{4} x(a-x) \sqrt{3} \\
y=\frac{a^{2}}{4} \sqrt{3}+\frac{3 \sqrt{3}}{4}\left(x^{2}-a x\right)
\end{gathered}
$$

$y$ is minimized when $x=\frac{a}{2}$, that is, when the sides of the original triangle are bisected.

(Weisz Pál, Nyitra.)

Number of solutions: 39.