For a triangle inscribed in a circle of radius $r$, we know that

$$
\operatorname{tg} \alpha=\frac{1}{3}, \quad \sin \alpha \sin \gamma=\frac{1}{\sqrt{10}}
$$

Express the radius of the inscribed circle (@) as a function of $r$.
# I. Solution:

$$
\sin \alpha=\frac{\operatorname{tg} \alpha}{\sqrt{1+\operatorname{tg}^{2} \alpha}}=\frac{\frac{1}{3}}{\sqrt{1+\frac{1}{9}}}=\frac{1}{\sqrt{10}}
$$

Substituting this value into the equation

$$
\sin \alpha \sin \gamma=\frac{1}{\sqrt{10}}
$$

we obtain

$$
\sin \gamma=1 \quad \text { or } \quad \gamma=90^{\circ}
$$

Thus, the triangle is a right triangle. Let the hypotenuse be $c$, and the two legs be $a$ and $b$, then $2 r=c, \operatorname{tg} \alpha=\frac{a}{b}=\frac{1}{3}$, so $b=3 a$, and by the Pythagorean theorem

$$
c^{2}=4 r^{2}=a^{2}+b^{2}=a^{2}+9 a^{2}=10 a^{2}
$$

from which

$$
a=r \sqrt{\frac{4}{10}}=r \frac{\sqrt{10}}{5}
$$

and thus

$$
b=3 a=r \frac{3 \sqrt{10}}{5}
$$

It is also known that

$$
\begin{gathered}
\varrho=\frac{t}{s}=\frac{2 t}{2 s}=\frac{a b}{a+b+c}=\frac{r \frac{\sqrt{10}}{5} \cdot r \frac{3 \sqrt{10}}{5}}{\frac{r \sqrt{10}}{5}+\frac{3 r \sqrt{10}}{5}+2 r}=\frac{\frac{30}{25}}{\frac{4 \sqrt{10}}{5}+2} r= \\
=\frac{30}{20 \sqrt{10}+50} r=\frac{3}{2 \sqrt{10}+5} r .
\end{gathered}
$$

Rationalizing the denominator,

$$
\varrho=\frac{2 \sqrt{10}-5}{5} r
$$

Bálint Benkő (Sárospatak, Rákóczi g. IV. o. t.)

II. Solution: After calculating the sides, we can reach the goal more simply by using the well-known relationship in a right triangle

$$
a+b=c+2 \varrho
$$

From this,

$$
\varrho=\frac{1}{2}(a+b-c)=\frac{1}{2}\left(r \frac{\sqrt{10}}{5}+r \frac{3 \sqrt{10}}{5}-2 r\right)=\frac{r}{2} \cdot \frac{4 \sqrt{10}-10}{5}=\frac{2 \sqrt{10}-5}{5} r
$$

Tibor Harza (Székesfehérvár, József A. g. IV. o. t.)