3. (15 points) In triangle $P Q R$, the median $P A$ and the angle bisector $Q B$ (where $A$ and $B$ are the points of intersection with the corresponding sides of the triangle) intersect at point $O$. It is known that $3 P Q = 5 Q R$. Find the ratio of the area of triangle $P Q R$ to the area of triangle $P Q O$.

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# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_1ae08a41583c767d264cg-07.jpg?height=546&width=665&top_left_y=915&top_left_x=747)

From the equality $3 P Q=5 Q R$ and the fact that $P A$ is a median, it follows that $P Q=10 x, \quad Q A=A R=3 x$, where $x$ is a proportionality coefficient. By the property of the angle bisector, $\frac{P O}{O A}=\frac{P Q}{Q A}=\frac{10}{3}$. According to the theorem on the ratio of areas of triangles with the same height, $\frac{S_{P Q O}}{S_{Q O A}}=\frac{P O}{O A}=\frac{10}{3}$. Considering that the median $P A$ divides the area of triangle $P Q R$ into two equal parts, we get $\frac{S_{P Q R}}{S_{P Q O}}=\frac{26}{10}=2.6$

Answer: 2.6.

## 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 ratio of the areas of triangles $P Q O$ and $Q O A$ is correctly found, but further actions are not performed or are performed incorrectly. |
| 5 points | The point $O$ is correctly found to divide the segment $P A$, but further actions are not performed or are performed incorrectly. |
| 0 points | The solution does not meet the criteria listed above |