Let the intersection point of the medians be $O$, the area of trapezoid $B D E C$ be $t_{1}$, and the area of triangle $D O E$ be $t_{2}$. Since $\triangle B O C = \frac{1}{3} \triangle A B C$, it follows that

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

However, from the similarity of triangles $\triangle B O C$ and $\triangle D O E$, it follows that

$$
t_{2}=\frac{t}{48}
$$

and thus

$$
t_{1}=\frac{5 t}{16}
$$

(József Kiss, Pápa.)

The problem was also solved by: L. Bánó, P. Füstös, Z. Harsányi, P. Heimlich, Gy. Jánosy, S. Pichler, Gy. Schilhanek, Gy. Schuster, S. Szabó, B. Tóth.
In triangle $ABC$ - which has a given area $t$ - we draw the medians $BB_{1}$ and $CC_{1}$. If the midpoint of $BB_{1}$ is $D$ and the midpoint of $CC_{1}$ is $E$, what is the area of the quadrilateral $BDEC$?