In a right-angled triangle, $s_{a}$ and $s_{b}$ are the medians to the legs, and $s_{c}$ is the median to the hypotenuse. Determine the maximum value of the expression $\frac{s_{a}+s_{b}}{s_{c}}$.
Based on the notation in the figure, in the right-angled triangle $ABC$, according to the Pythagorean theorem, $c^{2}=a^{2}+b^{2}$. Applying the Pythagorean theorem to the right-angled triangles $AF_{a}C$ and $BF_{B}C$:

$$
s_{a}^{2}=\left(\frac{a}{2}\right)^{2}+b^{2} \quad \text{and} \quad s_{b}^{2}=\left(\frac{b}{2}\right)^{2}+a^{2}
$$

By Thales' theorem, $s_{c}=\frac{c}{2}=\sqrt{\frac{a^{2}+b^{2}}{4}}$, thus

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

applying the arithmetic and quadratic mean inequality to the positive numbers $\sqrt{\frac{a^{2}+4 b^{2}}{a^{2}+b^{2}}}$ and $\sqrt{\frac{4 a^{2}+b^{2}}{a^{2}+b^{2}}}$. We can divide by $a^{2}+b^{2}>0$, so

$$
\frac{s_{a}+s_{b}}{s_{c}} \leq 2 \cdot \sqrt{\frac{5}{2}}=\sqrt{10}
$$

Equality holds if $a^{2}+4 b^{2}=4 a^{2}+b^{2}$, which in our case means if the right-angled triangle is isosceles, $a=b$. Therefore, the maximum value of the expression is $\sqrt{10}$.

Noémi Varga (Budapest, Szent István Gymnasium, 10th grade)

Consensus. Common mistakes:

1. Many did not even mention the theorems they applied in the solution.
2. Some only provided the location of the maximum, not the maximum itself.
3. Those who tried to solve the extremum problem using differential calculus often only concluded that the first derivative of the expression is 0, which does not guarantee that there is an extremum at that point!

![](https://cdn.mathpix.com/cropped/2024_05_02_d7a246517abe4fbf515fg-1.jpg?height=295&width=455&top_left_y=1316&top_left_x=824)