3. (15 points) In parallelogram $A B C D$, a point $P$ is taken on side $B C$ such that $3 P B=2 P C$, and a point $Q$ is taken on side $C D$ such that $4 C Q=5 Q D$. Find the ratio of the area of triangle $A P Q$ to the area of triangle $P Q C$.

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# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_34be784d6fd67db418cfg-6.jpg?height=451&width=991&top_left_y=311&top_left_x=584)

From the equality $3 P B=2 P C$, it follows that $P B=2 x, P C=3 x$, where $x$ is the proportionality coefficient. Similarly, from $4 C Q=5 Q D$, it follows that $C Q=5 y, D Q=4 y$. The area of the parallelogram is divided by the diagonal into two equal parts, so $S_{A B C}=S_{A D C}=S_{B C D}=\frac{S}{2}$, where $S$ is the area of the parallelogram. According to the theorem on the ratio of areas of triangles with equal heights, $S_{\text {ABP }}=\frac{2}{5} \cdot \frac{S}{2}=\frac{1}{5} S$. Similarly, $S_{A D Q}=\frac{4}{9} \cdot \frac{S}{2}=\frac{2}{9} S$. By the theorem on the ratio of areas of triangles with one equal angle, $\frac{S_{P Q C}}{S_{B C D}}=\frac{P C \cdot C Q}{B C \cdot C D}$. Then $S_{P Q C}=\frac{1}{2} S$ and $S_{A P Q}=S-S_{A B P}-S_{P C Q}-S_{A D Q}=\frac{37}{90} S$. $\frac{S_{A P Q}}{S_{P Q C}}=\frac{37}{15}$

Answer: $\frac{37}{15}$.

## Criteria:

| Points | Conditions for awarding |
| :---: | :--- |
| 15 points | Complete, justified solution |
| 12 points | The solution method is correct, but a computational error has been made. |
| 10 points | The areas of all corner triangles are correctly found in terms of the area of the parallelogram, but further actions are not performed or are performed incorrectly. |
| 5 points | The areas of two corner triangles are correctly found in terms of the area of the parallelogram, but further actions are not performed or are performed incorrectly. |
| 0 points | The solution does not meet the criteria listed above |