10.320. The medians of one triangle are equal to the sides of another triangle. Find the ratio of the areas of these triangles.
Solution.

Extend each median of the original triangle $ABC$ by $1/3$ of its length (Fig. 10.109). The area of the resulting figure $AMBNCP$ will be $2S_{\triangle ABC} (\triangle BOC = \triangle BNC, \triangle AOB = \triangle AMB, \triangle AOC = \triangle APC)$. The lengths of the sides of each of the triangles $AOM, BOM, BON, CON, COP, AOP$ are $2m_1/3, 2m_2/3, 2m_1/3$; therefore, $S_{AMBNCP} = 6S_{\triangle AOM}$. Let $S_m$ be the area of the triangle constructed on the medians of $\triangle ABC$. Then $S_{\triangle AOM} = 4S_m / 9$ and $S_{AMBNCP} = 2S_{\triangle ABC} = 6 \cdot 4S_m / 9 = 8S_m / 3$, from which $S_{\triangle ABC} : S_m = 4 : 3$.

Answer: 4:3.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0792.jpg?height=533&width=570&top_left_y=79&top_left_x=145)

Fig. 10.110

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0792.jpg?height=474&width=458&top_left_y=110&top_left_x=740)

Fig. 10.111