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

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

Express the radius of the inscribed circle $(g)$ as a function of $r$. (Be careful! This is not the same as problem 737.)
Express the area of a triangle with sides $a, b, c$ first using $\varrho$ and the semiperimeter, then using two sides and the included angle:

$$
T=\varrho \frac{a+b+c}{2}=\frac{b c \sin \alpha}{2}
$$

From this,

$$
\varrho=\frac{b c \sin \alpha}{a+b+c}
$$

It is known that the sides can be expressed using the angles and the circumradius as follows:

$$
a=2 r \sin \alpha, \quad b=2 r \sin \beta, \quad c=2 r \sin \gamma
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_c02f8d9ff39d72d650a9g-1.jpg?height=507&width=588&top_left_y=603&top_left_x=758)

Substituting this into the formula for $\varrho$ and simplifying by $2 r$:

$$
\varrho=\frac{2 r \sin \alpha \sin \beta \sin \gamma}{\sin \alpha+\sin \beta+\sin \gamma}
$$

Using the given values in the problem, calculate the sines (since $\alpha<180^{\circ}$, take the positive value for $\sin \alpha$):

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

$\sin \gamma=\sin \left[180^{\circ}-(\alpha+\beta)\right]=\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$, multiplying through by $\sin \beta$ and substituting $\sin \beta \sin \gamma$:

$$
\left.\frac{1}{\sqrt{10}}=\sin \beta\left(\sin \alpha \cos \beta+\sqrt{1-\sin ^{2} \alpha} \sin \beta\right)\right)=\frac{1}{\sqrt{10}} \sin \beta \cos \beta+\frac{3}{\sqrt{10}} \sin ^{2} \beta
$$

(Expressing $\cos \alpha$ in terms of $\sin \alpha$, it was sufficient to take the positive root, since $\operatorname{tg} \alpha$ is positive, and thus $\alpha<90^{\circ}$.)

Multiply through by $\sqrt{10}$, express $\cos \beta$ in terms of $\sin \beta$. Squaring the resulting equation, we get:

$$
\sin ^{2} \beta\left(1-\sin ^{2} \beta\right)=1-\sin ^{2} \beta+9 \sin ^{4} \beta
$$

Expanding the parentheses and rearranging:

$$
10 \sin ^{4} \beta-7 \sin ^{2} \beta+1=0
$$

From this,

$$
\sin ^{2} \beta_{1}=\frac{1}{2} \quad \text { and } \quad \sin ^{2} \beta_{2}=\frac{1}{5}
$$

For $\sin \beta$, take only the positive value:

$$
\sin \beta_{1}=\frac{1}{\sqrt{2}}, \quad \sin \beta_{2}=\frac{1}{\sqrt{5}}
$$

From the given $\sin \beta \sin \gamma=\frac{1}{\sqrt{10}}$ relationship:

$$
\sin \gamma_{1}=\frac{1}{\sqrt{5}}, \quad \sin \gamma_{2}=\frac{1}{\sqrt{2}}
$$

Since $\varrho$ in the expression (1) can have $\sin \beta$ and $\sin \gamma$ interchanged, it is sufficient to substitute only one of the obtained value pairs. Thus, using the calculated trigonometric function values:

$$
\varrho=2 r \frac{\frac{1}{\sqrt{10}} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{5}}}{\sqrt{10}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{5}}}=\frac{\sqrt{10}}{(1+\sqrt{2}+\sqrt{5}) 5} r(\sim 0.136 r)
$$

Katalin Fanta (Szombathely, Kanizsai Dorottya High School, 4th grade)