Let's draw an isosceles triangle with area $t$ inside the equilateral triangle $ABC$, such that the vertex $C_{1}$ of the isosceles triangle falls on the midpoint of side $AB$. At what distance from side $AB$ does the base $A_{1} B_{1}$ of the isosceles triangle intersect sides $AC$ and $BC$?
Let $C C_{1}$ be the height of the side $A_{1} B_{1}$ at point $C_{2}$, and let:

$$
\begin{gathered}
C C_{1}=m, \quad C_{1} C_{2}=x \\
A B=B C=C A=c \quad \text { and } \quad A_{1} B_{1}=c_{1}
\end{gathered}
$$

According to the problem:

$$
2 t=c_{1} x
$$

$$
c_{1}: c=(m-x): m
$$

In these equations, only $c_{1}$ and $x$ are unknowns. From (2):

$$
c_{1} x: c x=(m-x): m
$$

Thus:

$$
\begin{aligned}
& 2 t: \frac{2 m}{\sqrt{3}} x=(m-x): m \\
& \quad t: x=(m-x): \sqrt{3}
\end{aligned}
$$

And thus:

$$
x=\frac{m \pm \sqrt{m^{2}-4 t \sqrt{3}}}{2}
$$

Therefore, the problem generally has two solutions. If

$$
m^{2}-4 t \sqrt{3}=0
$$

then

$$
t=\frac{m^{2} \sqrt{3}}{12}=\frac{1}{4} \cdot A B C
$$

there is only one solution, and since in this case

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

$A_{1} B_{1} C_{1}$ is also an equilateral triangle.

Finally, there is no solution if:

$$
m^{2}-4 r \sqrt{3} > \frac{1}{4} \cdot A B C
$$

(Neidenbach Emil, Arad.)

The problem was also solved by: Baranyó A., Bartók I., Biró A., Braun J., Dömény E., Dömény I., Enyedi B., Haar A., Harsányi Z., Heimlich P., Hirschfeld Gy., Jánosy Gy., Kertész G., Kiss J., Kürti I., Liebner A., Losonczy J., Messer P., Pám M., Pichler S., Pivnyik J., Rássy P. Riesz K., Riesz M., Rosenberg J., Sáry B., Schöffer I., Schuster Gy., Schwemmer I., Schwarz Gy., Sonnenfeld J., Söpkéz Gy., Szávay Z., Szúcs A., Tóth B., Weisz P.