10.266. Find the third side of an acute-angled triangle if two of its sides are equal to $a$ and $b$ and it is known that the medians of these sides intersect at a right angle.
Solution.

Given $B C=a, A C=b, A E$ and $B F$ are medians, $\angle A M F=90^{\circ}$ (Fig. 10.67). Let $A E=m_{a}, B F=m_{b}$. Then in $\triangle B M E$ and $\triangle A M F$ respectively we have $\frac{m_{a}^{2}}{9}+\frac{4 m_{b}^{2}}{9}=\frac{a^{2}}{4}, \frac{4 m_{a}^{2}}{9}+\frac{m_{b}^{2}}{9}=\frac{b^{2}}{4}$. Adding these equations, we get $\frac{5\left(m_{a}^{2}+m_{b}^{2}\right)}{9}=\frac{a^{2}+b^{2}}{4}$ or $m_{a}^{2}+m_{b}^{2}=\frac{9\left(a^{2}+b^{2}\right)}{20}$. From $\triangle A M B$ we have $A B^{2}=\frac{4\left(m_{a}^{2}+m_{b}^{2}\right)}{9}=\frac{a^{2}+b^{2}}{5}$, from which $A B=\sqrt{\frac{a^{2}+b^{2}}{5}}$.

Answer: $\sqrt{\frac{a^{2}+b^{2}}{5}}$.