A triangle's two sides' lengths are given, and we know that the medians corresponding to these sides are perpendicular to each other. Calculate the length of the third side.
Solution. Let the sides $a$ and $c$ be given, $A F_{1}=x, C F_{2}=y$ (where $F_{1}$ is the midpoint of $B C$ and $F_{2}$ is the midpoint of $A B$).

![](https://cdn.mathpix.com/cropped/2024_05_02_ca506f6b7401d7d81a74g-1.jpg?height=377&width=545&top_left_y=229&top_left_x=779)

Write down the Pythagorean theorem for the right triangles $A S C, A S F_{2}, C S F_{1}$:

$$
\begin{aligned}
b^{2} & =\left(\frac{2}{3} x\right)^{2}+\left(\frac{2}{3} y\right)^{2} \\
\left(\frac{c}{2}\right)^{2} & =\left(\frac{2}{3} x\right)^{2}+\left(\frac{1}{3} y\right)^{2} \\
\left(\frac{a}{2}\right)^{2} & =\left(\frac{1}{3} x\right)^{2}+\left(\frac{2}{3} y\right)^{2}
\end{aligned}
$$

Solve the system of equations. Subtract the (3) equation from the 4 times the (2) equation:

$$
c^{2}=\frac{16}{9} x^{2}+\frac{4}{9} y^{2}, \quad-\frac{a^{2}}{4}=-\frac{1}{9} x^{2}-\frac{4}{9} y^{2}, \quad \text { hence } \quad x^{2}=\frac{36 c^{2}-9 a^{2}}{60}
$$

Similarly, if we multiply the (3) equation by 4 and subtract the (2) equation from it, we get $y^{2}=\frac{36 a^{2}-9 c^{2}}{60}$. Substitute the obtained values into (1):

$$
b^{2}=\left(\frac{2}{3}\right)^{2} \frac{36 c^{2}-9 a^{2}}{60}+\left(\frac{2}{3}\right)^{2} \frac{36 a^{2}-9 c^{2}}{60}
$$

After rearranging, we get $b=\sqrt{\frac{a^{2}+c^{2}}{5}}$.