3. The base of the pyramid $P A B C D$ is a trapezoid $A B C D$, the base $A D$ of which is twice the length of the base $B C$. The segment $M N$ is the midline of the triangle $A B P$, parallel to the side $A B$. Find the ratio of the volumes of the solids into which the plane $D M N$ divides this pyramid.
5.II.3. Let's complete the pyramid $P A B C D$ to the tetrahedron $P A D F$ (figure).

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![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-097.jpg?height=428&width=697&top_left_y=383&top_left_x=682)

The section of the tetrahedron by the plane $D M N$ intersects its edge at some point $O$. The quadrilateral $D M N O$ is a section of the pyramid $P A B C D$ by the plane $D M N$. Since $B C \| A D$ and $A D = 2 B C$, $B C$ is the midline of triangle $A D F$, meaning that point $C$ is the midpoint of segment $D F$. On the other hand, by Thales' theorem, point $K$ is the midpoint of segment $P F$, hence point $O$ is the intersection of the medians of triangle $D F P$, so $P O : P C = 2 : 3$.

Let $V$ be the volume of the given pyramid. The ratio of the volumes of pyramids $P A C D$ and $P A B C$ is equal to the ratio of the areas of their bases, which is two. Therefore, $V_{P A C D} = \frac{2}{3} V$ and $V_{P A B C} = \frac{1}{3} V$. Further,

$$
\frac{V_{P M N O}}{V_{P A B C}} = \frac{P M \cdot P N \cdot P O}{P A \cdot P B \cdot P C} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6}, \text{ so } V_{P M N O} = \frac{1}{18} V
$$

Similarly,

$$
\frac{V_{P M D O}}{V_{P A D C}} = \frac{P M \cdot P D \cdot P O}{P A \cdot P D \cdot P C} = \frac{1}{2} \cdot \frac{1}{1} \cdot \frac{2}{3} = \frac{1}{3}, \text{ so } V_{P M D O} = \frac{2}{9} V.
$$

Thus, $V_{P M N O D} = \frac{1}{18} V + \frac{2}{9} V = \frac{5}{18} V$.

Answer: $\frac{5}{13}$.