The base of a trapezoid is $a$, the parallel side is $b$, and the height is $m$. Divide the height into 3 equal parts and draw lines parallel to the base through the division points. What are the areas of the resulting trapezoids?
Let the parallel sides of the trapezoids closer to the base be $a_{1}$, the next one $a_{2}$, and the areas of the resulting trapezoids in the same order be $t_{1}, t_{2}$, and $t_{3}$. Since

$$
a_{1}=\frac{2 a+b}{3} \text { and } a_{2}=\frac{2 b+a}{3}
$$

therefore

$$
t_{1}=\frac{m}{18}(5 a+b), \quad t_{2}=\frac{m}{6}(a+b), \quad t_{3}=\frac{m}{18}(a+5 b)
$$

(Ferencz Kertész, Szeged.)

Number of solutions: 69.