[ Properties of medians. The centroid of a triangle. ] [ The ratio of the areas of triangles with a common angle ]

Given triangle $A B C$, the area of which is 2. On the medians $A K, B L$ and $C N$ of triangle $A B C$, points $P, Q$ and $R$ are taken respectively such that $A P: P K=1, B Q: Q L=1: 2, C R: R N=5: 4$. Find the area of triangle $P Q R$.
Find the ratios $\frac{O P}{O A}, \frac{O Q}{O B}$ and $\frac{O R}{O C}$, where $O$ is the point of intersection of the medians of triangle $A B C$.

## Solution

Let $O$ be the point of intersection of the medians of triangle $A B C$. Then

$$
\frac{A O}{O K}=\frac{B O}{O L}=\frac{C O}{O N}=2
$$

Notice that

$$
S_{\triangle \mathrm{AOB}}=S_{\triangle \mathrm{AOC}}=S_{\Delta \mathrm{BOC}}=\frac{1}{3} \cdot S_{\Delta \mathrm{ABC}}=\frac{2}{3}
$$

Let $A K=6 a, B L=6 b, C N=9 c$. Then

$$
O P=O A-A P=4 a-3 a=a, O Q=O B-B Q=4 b-2 b=2 b, O R=6 c-5 c=c
$$

Therefore,

$$
\begin{gathered}
S_{\triangle \mathrm{PQR}}=S_{\triangle \mathrm{POQ}}+S_{\Delta \mathrm{POR}}+S_{\Delta \mathrm{ROQ}}= \\
=\frac{O P}{O A} \cdot \frac{O Q}{O B} \cdot S_{\Delta \mathrm{AOB}}+\frac{O P}{O A} \cdot \frac{O R}{O C} \cdot S_{\Delta \mathrm{AOC}}+\frac{O R}{O C} \cdot \frac{O Q}{O B} \cdot S_{\Delta \mathrm{BOC}}= \\
=\frac{a}{4 a} \cdot \frac{2 b}{4 b} \cdot S_{\Delta \mathrm{AOB}}+\frac{2 b}{4 b} \cdot \frac{c}{6 c} \cdot S_{\Delta \mathrm{AOC}}+\frac{c}{6 c} \cdot \frac{2 b}{4 b} \cdot S_{\Delta \mathrm{BOC}}= \\
=\frac{1}{4} \cdot \frac{1}{2} \cdot S_{\Delta \mathrm{AOB}}+\frac{1}{4} \cdot \frac{1}{6} \cdot S_{\Delta \mathrm{AOC}}+\frac{1}{6} \cdot \frac{1}{2} \cdot S_{\Delta \mathrm{BOC}}= \\
=\frac{2}{3}\left(\frac{1}{4} \cdot \frac{1}{2}+\frac{1}{4} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{2}\right)=\frac{2}{3}\left(\frac{1}{8}+\frac{1}{24}+\frac{1}{12}\right)=\frac{2}{3} \cdot \frac{1}{4}=\frac{1}{6}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-38.jpg?height=646&width=780&top_left_y=0&top_left_x=639)

## Answer

$\frac{1}{6}$.