## Task B-1.5.

In a right-angled triangle, the sum of the lengths of the medians drawn from the vertices to the hypotenuse is $5 \sqrt{5}$, and their product is 30. What is the length of the hypotenuse?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6dc7744662d888032875g-05.jpg?height=439&width=645&top_left_y=543&top_left_x=634)

According to the conditions of the problem, and using the labels as in the figure, we have $t_{a}+t_{b}=5 \sqrt{5}, t_{a} \cdot t_{b}=30$. Apply the Pythagorean theorem to triangles $A Q C$ and $B C P$.

$$
\begin{aligned}
& t_{a}^{2}=b^{2}+\left(\frac{a}{2}\right)^{2} \\
& t_{b}^{2}=a^{2}+\left(\frac{b}{2}\right)^{2}
\end{aligned}
$$

If we add these equations, we get

$$
t_{a}^{2}+t_{b}^{2}=a^{2}+b^{2}+\frac{a^{2}+b^{2}}{4}=\frac{5}{4} c^{2}
$$

From $\left(t_{a}+t_{b}\right)^{2}=t_{a}^{2}+2 t_{a} \cdot t_{b}+t_{b}^{2}$, it follows that

$$
\begin{aligned}
t_{a}^{2}+t_{b}^{2} & =\left(t_{a}+t_{b}\right)^{2}-2 t_{a} \cdot t_{b}=125-60=65 \\
c^{2} & =\frac{4}{5}\left(t_{a}^{2}+t_{b}^{2}\right)=52 \Longrightarrow c=2 \sqrt{13}
\end{aligned}
$$