4A. The height divides the right-angled triangle into two triangles which have perimeters $m$ and $n$. Determine the perimeter of the original triangle.
Solution. Let $ABC$ be a right triangle (with the right angle at point $C$) and let $CH$ be its altitude. From the problem statement, we have

$$
L_{1}=L_{AHC}=m \quad \text{ and } \quad L_{2}=L_{CHB}=n
$$

The triangles $AHC$, $BHC$, and $ABC$ are similar, with the similarity coefficients being

$$
k_{1}=\frac{\overline{AC}}{\overline{AB}} \text{ and } k_{2}=\frac{\overline{BC}}{\overline{AB}}.
$$

Furthermore,

$$
\frac{L_{1}}{L}=k_{1} \text{ and } \frac{L_{2}}{L}=k_{2}
$$

From equations (1), we have $\overline{AC}=k_{1} \overline{AB}$ and $\overline{BC}=k_{2} \overline{AB}$. According to the Pythagorean theorem,

$$
\overline{AC}^{2}+\overline{BC}^{2}=\overline{AB}^{2}
$$

from which we obtain the equation $k_{1}^{2}+k_{2}^{2}=1$. If we substitute (2) into the previous equation, we get $\left(\frac{L_{1}}{L}\right)^{2}+\left(\frac{L_{2}}{L}\right)^{2}=1$, i.e., $L=\sqrt{L_{1}^{2}+L_{2}^{2}}=\sqrt{m^{2}+n^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_dd8561913dbd5a53296cg-07.jpg?height=368&width=565&top_left_y=941&top_left_x=914)