Let's construct a right-angled triangle, given the hypotenuse; the difference of the squares of the legs is equal to four times the square of the height corresponding to the hypotenuse.
Let $c$ be the hypotenuse, $a$ and $b$ the two legs, $m$ the altitude to the hypotenuse, and finally $p$ and $q$ the segments of the hypotenuse. Then

$$
a^{2}-b^{2}=p^{2}+m^{2}-\left(q^{2}+m^{2}\right)=p^{2}-q^{2}=4 m^{2}
$$

But

$$
m^{2}=p q \quad \text { and } \quad q=c-p
$$

so

$$
p^{2}-(c-p)^{2}=4 p(c-p)
$$

or

$$
4 p^{2}-2 c p-c^{2}=0
$$

from which $p=\frac{c}{4}(1+\sqrt{5})=c \sin 54^{\circ}$.

Therefore, $p$ is one of the legs of a right triangle whose hypotenuse is $c$ and the opposite angle is $54^{\circ}$.

(Tandlich Emil, Körmöczbánya.)

The problem was also solved by: Ádámffy E., Blum J., Csada I., Dömény I., Epstein K., Földes R., Heimlich P., Jánosy Gy., Merse P., Miklóssy K., Morvai O., Rosenberg J., Ruvald S., Sárközy P., Schöffer I., Székely J., Tóth J.