10.270. Find the ratio of the sum of the squares of all medians of a triangle to the sum of the squares of all its sides.
## Solution.

Let $a, b, c$ be the sides of a triangle, and $m_{a}, m_{b}, m_{c}$ be the medians drawn to these sides. We use the formulas:

$$
m_{a}=\frac{1}{2} \sqrt{2\left(b^{2}+c^{2}\right)-a^{2}}, m_{b}=\frac{1}{2} \sqrt{2\left(a^{2}+c^{2}\right)-b^{2}}, m_{c}=\frac{1}{2} \sqrt{2\left(a^{2}+b^{2}\right)-c^{2}}
$$

Then $m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{1}{4}\left(2 b^{2}+2 c^{2}-a^{2}+2 a^{2}+2 c^{2}-b^{2}+2 a^{2}+2 b^{2}-c^{2}\right)=$ $=\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)$. Therefore, $\frac{m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}{a^{2}+b^{2}+c^{2}}=\frac{3}{4}$.

Answer: $\frac{3}{4}$.