We have an equilateral triangle. Divide this triangle into two parts with a line parallel to the base so that the ratio of the areas of the circles circumscribed around the parts is $\frac{1}{3}$.
Let one side of the equilateral triangle be $a$, one side of the cut-off triangle be $x$, the base of the trapezoid be $a$, the side parallel to it be $x$, and one of the legs be $a-x$. The radius of the circumcircle of the small triangle is

$$
r=\frac{x}{2 \sin 60^{\circ}}=\frac{x}{\sqrt{3}}
$$

The area of this circle is

$$
t_{1}=\frac{x^{2} \pi}{3}
$$

If one diagonal of the trapezoid is $d$, then the radius of the circumcircle of the trapezoid is

$$
R=\frac{d}{2 \sin 60^{\circ}}=\frac{d}{\sqrt{3}}
$$

The area of this circle is

$$
t_{2}=\frac{d^{2} \pi}{3}
$$

where

$$
d^{2}=a^{2}+(a-x)^{2}-2 a(a-x) \cos 60^{\circ}=a^{2}-a x+x^{2}
$$

and thus

$$
t_{2}=\frac{\left(a^{2}-a x+x^{2}\right) \pi}{3}
$$

According to the problem, $t_{2}=3 t_{1}$, so

$$
3 x^{2}=a^{2}-a x+x^{2}
$$

or

$$
2 x^{2}+a x-a^{2}=0
$$

from which the positive value of $x$ is

$$
x=\frac{a}{2}
$$

Therefore, the line drawn parallel to the base divides the two sides of the given triangle in half.