1. Given are the lengths of two sides $a$ and $b$ of triangle $\triangle A B C$. The medians dropped to these sides are mutually perpendicular. Calculate the length of the third side using the lengths of sides $a$ and $b$.
Solution. Let $T$ be the centroid of $\triangle A B C$, and let the medians to sides $\overline{B C}=a$ and $\overline{A C}=b$ be $\overline{A A_{1}}=t_{a}$ and $\overline{B B_{1}}=t_{b}$ respectively. Since $\overline{A_{1} T}=\frac{1}{3} t_{a}$ and $\overline{T B}=\frac{2}{3} t_{b}$, from the right triangle $\triangle B A_{1} T$ we have

![](https://cdn.mathpix.com/cropped/2024_06_05_3cde5d119ebb42d699c4g-5.jpg?height=409&width=514&top_left_y=1841&top_left_x=947)

$$
\left(\frac{a}{2}\right)^{2}=\left(\frac{1}{3} t_{a}\right)^{2}+\left(\frac{2}{3} t_{b}\right)^{2}
$$

Since $\overline{A T}=\frac{2}{3} t_{a}$ and $\overline{T B_{1}}=\frac{1}{3} t_{b}$, from the right triangle $\triangle B_{1} A T$ we have

$$
\left(\frac{b}{2}\right)^{2}=\left(\frac{2}{3} t_{a}\right)^{2}+\left(\frac{1}{3} t_{b}\right)^{2}
$$

By adding the two equations, we get

$$
\frac{a^{2}+b^{2}}{4}=\frac{5}{9}\left(t_{a}^{2}+t_{b}^{2}\right)
$$

which simplifies to

$$
t_{a}^{2}+t_{b}^{2}=\frac{9}{20}\left(a^{2}+b^{2}\right)
$$

Finally, from the right triangle $\triangle A B T$ we have

$$
c^{2}=\left(\frac{2}{3} t_{a}\right)^{2}+\left(\frac{2}{3} t_{b}\right)^{2}=\frac{4}{9}\left(t_{a}^{2}+t_{b}^{2}\right)=\frac{a^{2}+b^{2}}{5}
$$

from which $c=\sqrt{\frac{a^{2}+b^{2}}{5}}$.