A triangle has side lengths of $\sqrt{3}, \sqrt{4}(=2), \sqrt{5}$.

In what ratio does the altitude perpendicular to the middle side divide it?
We denote the height corresponding to the middle side by $m$ and this height divides the middle side into segments of $x$ and $2-x$. (See the figure.)

![](https://cdn.mathpix.com/cropped/2024_05_02_fc276e2fe64def6ed9d2g-1.jpg?height=403&width=439&top_left_y=243&top_left_x=824)

By the Pythagorean theorem, on one hand,

$$
m^{2}=3-x^{2}
$$

and on the other hand,

$$
m^{2}=5-(2-x)^{2}
$$

Thus,

$$
3-x^{2}=5-(2-x)^{2}
$$

which simplifies to

$$
4 x=2
$$

or

$$
x=\frac{1}{2}, \quad 2-x=\frac{3}{2}
$$

Therefore, the height divides the middle side in the ratio $1:3$.

Kardos Erzzébet (Kaposvár, Munkácsy M. lg. II. o. t.) then

Remark: If we perform the calculation generally with sides $a, b, c$ and seek the ratio $\frac{x}{a-x}$ on side $a$,

$$
\begin{gathered}
m_{a}^{2}=b^{2}-x^{2}=c^{2}-(a-x)^{2} \\
x^{2}-(a-x)^{2}=a(2 x-a)=b^{2}-c^{2}
\end{gathered}
$$

from which

$$
\begin{gathered}
x=\frac{b^{2}-c^{2}}{2 a}+\frac{a}{2}=\frac{a^{2}+b^{2}-c^{2}}{2 a} \\
a-x=\frac{a}{2}-\frac{b^{2}-c^{2}}{2 a}=\frac{a^{2}-\left(b^{2}-c^{2}\right)}{2 a}
\end{gathered}
$$

and thus the sought ratio is

$$
\frac{x}{a-x}=\frac{a^{2}+\left(b^{2}-c^{2}\right)}{a^{2}-\left(b^{2}-c^{2}\right)}
$$

Note that this ratio is certainly a rational number if $a^{2}, b^{2}, c^{2}$ are rational, and indeed if $a, b, c$ are rational lengths, then $x$ and $a-x$ are certainly rational individually as well.