What are the angles of a right-angled triangle whose area is divided in a $1: 5$ ratio by the height corresponding to the hypotenuse?
In the right-angled triangle $ABC$, let the height corresponding to the hypotenuse $A D$ be $m$, and $B D=p, D C=q$. Then

$$
\frac{p m}{q m}=\frac{p}{q}=\frac{1}{5}
$$

But also

$$
a p=c^{2} \text { and } a q=b^{2} \text {, }
$$

so

$$
\frac{p}{q}=\left(\frac{c}{b}\right)^{2}=\operatorname{tg}^{2} \gamma
$$

and thus, considering (1)

$$
\operatorname{tg}^{2} \gamma=\frac{1}{5}
$$

from which

$$
\gamma=24^{\circ} 5^{\prime} 41^{\prime \prime} \text { and } \beta=65^{\circ} 54^{\prime} 19^{\prime \prime}
$$

(József Vámos, Szeged.)

Number of solutions: 56.