1. The base $AB$ of trapezoid $ABCD$ is three times longer than the base $CD$. Let $M$ be the midpoint of side $AB$ and $P$ the intersection of segment $DM$ with diagonal $AC$. Calculate the ratio of the areas of triangle $CDP$ and quadrilateral $MBCP$.
1. The calculation is based on two known rules: (1) If two triangles are similar with a similarity coefficient $k$, the ratio of their areas is equal to $k^{2}$. (2) If three points $X, Y, Z$ lie on the same line and point $V$ is outside it, the ratio of the areas of triangles $X Y V$ and $Y Z V$ is equal to the ratio $|X Y|:|Y Z|$.

From the congruence of alternate interior angles between the parallel lines $A B$ and $C D$, it follows that triangles $A M P$ and $C D P$ are similar by the theorem $u$, with a similarity coefficient of $|A M|:|C D|=\frac{3}{2}$. If we denote the area of triangle $C D P$ as $S$, the area of triangle $A M P$ is $\left(\frac{3}{2}\right)^{2} S=\frac{9}{4} S$. From the equalities $|A P|:|C P|=|M P|:|D P|=\frac{3}{2}$, it follows that the area of each of the triangles $A P D$ and $M P C$ is $\frac{3}{2}$ times the area of triangle $C D P$, i.e., $\frac{3}{2} S$. The areas of triangles $A M C$ and $B M C$ are the same, and thus equal to $\frac{9}{4} S+\frac{3}{2} S=\frac{15}{4} S$ (Fig. 1). Therefore, the area of quadrilateral $M B C P$ is $\frac{3}{2} S+\frac{15}{4} S=\frac{21}{4} S$, and the desired ratio is therefore $4: 21$.

![](https://cdn.mathpix.com/cropped/2024_04_17_05fdfd338c92e7f3acbdg-2.jpg?height=360&width=740&top_left_y=971&top_left_x=661)

Fig. 1

For a complete solution, award 6 points, of which 3 points for determining the ratio of the areas of triangles $A M P$ and $C D P$. The calculations of the areas may naturally differ depending on the choice of the base area used to express the areas of the other triangles in the figure. (The rules mentioned at the beginning of our solution will be derived by the solvers for specific pairs of triangles by expressing their areas using bases and heights.)