Let's calculate the lengths of the medians $\left(s_{a}, s_{b}, s_{c}\right)$ of a triangle, given the sides of the triangle $(a, b, c)$.
I. Solution: Let the feet of $s_{a}$ and $m_{a}$ be $A_{1}$ and $A_{2}$ respectively (Figure 1) and let $A_{1} A_{2}=d$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5719292bd8f741b141bdg-1.jpg?height=371&width=458&top_left_y=224&top_left_x=820)

Figure 1

For the unknowns $s_{a}, m_{a}$, and $d$, the following three equations can be established based on the Pythagorean theorem:

$$
\begin{aligned}
s_{a}^{2} & =m_{a}^{2}+d^{2} \\
b^{2} & =m_{a}^{2}+\left(\frac{a}{2}+d\right)^{2} \\
c^{2} & =m_{a}^{2}+\left(\frac{a}{2}-d\right)^{2}
\end{aligned}
$$

From (1), $m_{a}^{2}=s_{a}^{2}-d^{2}$. Substituting this value into (2) and (3),

$$
\begin{aligned}
& b^{2}=s_{a}^{2}-d^{2}+\frac{a^{2}}{4}+a d+d^{4} \\
& c^{2}=s_{a}^{2}-d^{2}+\frac{a^{2}}{4}-a d+d^{2}
\end{aligned}
$$

The sum of (4) and (5) is

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

from which

$$
2 s_{a}^{2}=b^{2}+c^{2}-\frac{a^{2}}{2}=\frac{2 b^{2}+2 c^{2}-a^{2}}{2}
$$

and thus

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

Similarly,

$$
s_{b}=\frac{\sqrt{2 a^{2}+2 c^{2}-b^{2}}}{2} \quad \text { and } \quad s_{c}=\frac{\sqrt{2 a^{2}+2 b^{2}-c^{2}}}{2}
$$

George Bali (Gyöngyös, Vak Bottyán g. IV. o. t.)

II. Solution: Let the reflection of $A$ over the midpoint $A_{1}$ of side $BC$ be $A^{\prime}$. (Figure 2.)

![](https://cdn.mathpix.com/cropped/2024_05_02_5719292bd8f741b141bdg-1.jpg?height=391&width=480&top_left_y=2057&top_left_x=801)

Figure 2

Applying the well-known theorem to the parallelogram $A B A^{\prime} C$, which states that the sum of the squares of the four sides is equal to the sum of the squares of the diagonals (L. II. oszt. textbook),

$$
2\left(b^{2}+c^{2}\right)=\left(2 s_{a}\right)^{2}+a^{2}
$$

from which

$$
4 s_{a}^{2}=2 b^{2}+2 c^{2}-a^{2}
$$

and thus

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

George Vastag (Bp. XXI., Jedlik A. g. III. o. t.)