In the triangle $A B C$, we draw the line $D E$ parallel to the base through the midpoint of the height $B B_{1}$. How does the area of the triangle $B D E$ compare to the area of the trapezoid $A D E C$?
Let the area of triangle $B D E$ be denoted by $t_{1}$, the area of trapezoid $A D E C$ by $t_{2}$, and the area of triangle $A B C$ by $t$. Then,

$$
t_{1}=\frac{B B_{1}}{4} D E=\frac{B B_{1}}{8} A C=\frac{t}{4}
$$

and

$$
t_{2}=t-t_{1}=\frac{3}{4} t
$$

thus

$$
t_{1}: t_{2}=1: 3
$$

(Andor Vilcsek, Eger.)

The problem was also solved by: Baján A., Bauer E., Dénes M., Dőry V., Erdős V., Fried E., Gotléb I., Grünwald Gy., Klein G., Lendvai D., Neumann F., Neumann L., Rosenberg E., Schlesinger R., Schnabel L., Sichermann F., Spitzer L., Steiner L., Stern S., Szántó L., Szilárd V., Viola R., Weinberger Gy., Wellis E.