Points $P$ and $Q$ are located on side $B C$ of triangle $A B C$, such that $B P: P Q: Q C=1: 2: 3$. Point $R$ divides side $A C$ of this triangle so that

$A R: R C=1: 2$. What is the ratio of the area of quadrilateral $P Q S T$ to the area of triangle $A B C$, if $S$ and $T$ are the points of intersection of line $B R$ with lines $A Q$ and $A P$ respectively?
Find the ratios $A S: A Q$ and $A T: A P$.

## Solution

Draw a line through vertex $A$ parallel to $B C$, and extend $B R$ to intersect this line at point $K$. From the similarity of triangles $A R K$ and $C R B$, we find that $A K=1 / 2 B C=B Q$. Therefore, triangles $A S K$ and $Q S B$ are equal. From the similarity of triangles $A T K$ and $P T B$, we get $A T: T P = A K: B P = 3: 1$.

Therefore, $S_{P Q S T} = S_{A P Q} - S_{A T S} = S_{A P Q} - \frac{A S}{A Q} \cdot \frac{A P}{A P} S_{A P Q} = S_{A P Q} \left(1 - \frac{3}{4} \cdot \frac{1}{2}\right) = \frac{5}{8} S_{A P Q} = \frac{5}{8} \cdot \frac{1}{3} \cdot S_{A B C} = \frac{5}{24} S_{A B C}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-49.jpg?height=418&width=946&top_left_y=2412&top_left_x=562)

## Answer

## Problem