10.353. On each median of a triangle, a point is taken that divides the median in the ratio $1: 3$, counting from the vertex. By what factor is the area of the triangle with vertices at these points smaller than the area of the original triangle
## Solution.

Let $O$ be the point of intersection of the medians $A N, B K, C M$ of triangle $A B C$ (Fig. 10.134),

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0810.jpg?height=503&width=604&top_left_y=1141&top_left_x=659)

Fig. 10.134

$$
\frac{A D}{D N}=\frac{B E}{E K}=\frac{C F}{F M}=\frac{1}{3} \cdot \text { Let } A D=x . \text { Then } D N=3 x, A N=4 x
$$

$A O=\frac{2}{3} A N=\frac{8}{3} x, O D \cdot=A O-O D=\frac{5}{3} x, \frac{O D}{O A}=\frac{5}{8}$. Therefore, the medians of $\triangle D E F$ are $5 / 8$ of the medians of $\triangle A B C, \triangle D E F \sim \triangle A B C$ and $\frac{S_{\triangle A B C}}{S_{\triangle D E F}}=\left(\frac{O A}{O D}\right)^{2}=\left(\frac{8}{5}\right)^{2}=2.56$

Answer: 2.56 times.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0811.jpg?height=449&width=551&top_left_y=486&top_left_x=403)

Fig. 10.135