What are the conditions for a triangle constructed from the altitudes of a right triangle to also be a right triangle?
The figure shows the construction.

![](https://cdn.mathpix.com/cropped/2024_05_02_e2af7dee5ccb8ef2b8e6g-1.jpg?height=518&width=631&top_left_y=221&top_left_x=736)

We can assume that $a \leq b$. Since $m < a \leq b$, if a right triangle can be constructed from the heights, then the hypotenuse in that triangle will be $b$. However, the hypotenuse $b$ and the leg $m$ uniquely determine the right triangle that can be constructed from them, and thus the leg $AD = c_1$ in the $\triangle ACD$.

Therefore, a right triangle can be constructed from $a$, $b$, and $m$ if and only if the shorter leg of the right triangle is equal to the segment of the hypotenuse adjacent to the other leg. This can be restated using the geometric mean theorem in the right triangle: the longer leg is the geometric mean of the shorter leg and the hypotenuse.

We can also determine the angles of this special triangle. Indeed, $\operatorname{tg} \alpha = \frac{a}{b}$, and in the triangle with sides $m$, $c_1$, and $b$, $\cos \alpha = \frac{c_1}{a} = \frac{a}{b}$, so

\[
\begin{aligned}
\cos \alpha & = \operatorname{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} \\
\cos^2 \alpha & = \sin \alpha, \quad 1 - \sin^2 \alpha = \sin \alpha
\end{aligned}
\]

From the resulting quadratic equation, only the positive root for $\sin \alpha$ is valid:

\[
\begin{gathered}
\sin \alpha = \frac{-1 + \sqrt{5}}{2} \approx 0.6180 \\
\alpha \approx 38^\circ 10'
\end{gathered}
\]

The other angle is

\[
\beta \approx 51^\circ 50'
\]

Therefore, a right triangle can be constructed from the heights if the sine of the angle $\alpha$ is the above value, i.e., the angles of the triangle are of the above magnitudes.

Balázs János (Bp. V., Eötvös g. II. o. t.)