In a right-angled triangle with hypotenuse $c$ given, six times the tangent of one acute angle is 1 greater than the tangent of the other acute angle. What are the lengths of the legs?
Let the legs opposite to the acute angles $\alpha$ and $\beta$ be denoted by $a$ and $b$ respectively, then $\operatorname{tg} \alpha=\frac{a}{b}, \operatorname{tg} \beta=\frac{b}{a}$.

According to the problem,

$$
6 \operatorname{tg} \beta=\operatorname{tg} \alpha+1
$$

which means

$$
6 \cdot \frac{b}{a}=\frac{a}{b}+1
$$

Multiplying both sides by $a b$,

$$
6 b^{2}=a^{2}+a b
$$

thus,

$$
a^{2}+b a-6 b^{2}=0
$$

from which

$$
a_{1}=2 b, \quad\left[a_{2}=-3 b\right]
$$

Only the positive root makes sense.

By the Pythagorean theorem,

$$
(2 b)^{2}+b^{2}=c^{2}
$$

from which

$$
5 b^{2}=c^{2}
$$

thus

$$
b=\frac{c}{\sqrt{5}}=\frac{c \sqrt{5}}{5}, \quad \text { and hence } \quad a=2 b=\frac{2 c \sqrt{5}}{5}
$$

Tibor Behringer (Bp., III., Árpád g. I. o. t.)