Let's draw a line parallel to the base $AB$ of triangle $ABC$ that divides the triangle into two regions of equal area. What is the height of the resulting trapezoid if the height of the original triangle is $m$?
Let $DE$ be the drawn parallel, $x$ the questioned height, and $t_{1}$ the area of $\triangle ABC$ and $t_{2}$ the area of $\triangle CDE$. Since $\triangle CDE \sim \triangle CAB$, we have

$$
m^{2}:(m-x)^{2}=t_{1}: t_{2}=2: 1
$$

or

$$
2 x^{2}-4 m x+m^{2}=0
$$

from which

$$
x=m \sqrt{2}\left(\frac{\sqrt{2} \pm 1}{2}\right)
$$

Of the two signs, the negative one must be taken.

(Leó Spitzer, Budapest.)

The problem was also solved by: Bauer E., Breuer P., Czigler I., Dénes M., Engler J., Erdős V., Fried E., Grün E., Koffler B., Köhler I., Neumann L., Sichermann F., Szilárd V., Vilcsek A., Viola R.