Given a triangle $A B C$ with an area of 1. On the medians $A K, B L$, and $C N$, points $P, Q$, and $R$ are taken such that $A P=P K, B Q: Q L=1$ : $2, C R: R N=5: 4$. Find the area of triangle $P Q R$.
Let $M$ be the point of intersection of the medians of triangle $ABC$. Then $MP = (2/3 - 1/2) AK = 1/6 AK = 1/4 AM$, $MQ = 1/3 BL = 1/2 BM$, $MR = (2/3 - 5/9) NC = 1/9 NC = 1/6 MC$.

Therefore, $S_{MPQ} = 1/4 \cdot 1/2 S_{AMB} = 1/24$, $S_{QMR} = 1/2 \cdot 1/6 S_{BMC} = 1/36$, $S_{PMR} = 1/4 \cdot 1/6 S_{AMC} = 1/72$, $S_{PQR} = 1/24 + 1/36 + 1/72 = 1/12$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9617f685748a8dfcd82ag-21.jpg?height=412&width=984&top_left_y=0&top_left_x=540)

## Answer

$1/12$

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