## Task 26/74

Let $A B C$ be a right-angled triangle at $A$ with sides $a, b$, and $c$ (where the sides are opposite to the correspondingly named vertices). What values can the ratio of the medians $s_{c}: s_{b}$ take?
According to the Pythagorean theorem, the lengths of the medians are

$$
s_{b}=\sqrt{\frac{1}{4} b^{2}+c^{2}} \quad ; \quad s_{c}=\sqrt{\frac{1}{4} c^{2}+b^{2}}
$$

The ratio $s_{c}: s_{b}$ is then

$$
\frac{s_{c}}{s_{b}}=\sqrt{\frac{4 b^{2}+c^{2}}{b^{2}+4 c^{2}}}
$$

If we simplify the radicand by $c^{2}$ and introduce the ratio $\lambda=\frac{b}{c}$ (where $00, c>0$), we get

$$
\frac{s_{c}}{s_{b}}=\sqrt{\frac{4 \lambda^{2}+1}{\lambda^{2}+4}}=\sqrt{4-\frac{15}{\lambda^{2}+4}}
$$

For $\lambda \rightarrow 0$, we get $s_{c}: s_{b} \rightarrow \sqrt{4-\frac{15}{4}}=\frac{1}{2}$. For $\lambda \rightarrow \infty$, we get $s_{c}: s_{b} \rightarrow \sqrt{4}=2$. Therefore, $\frac{1}{2}<s_{c}: s_{b}<2$.