2. In triangle $A B C$, the following holds: $|B C|=4,|C A|=5, \angle B C A=2 \angle C A B$. Calculate the radii of the circumscribed and inscribed circles of this triangle.
Solution.

The conditions can be written as $a=4, b=5, \gamma=2 \alpha$.

By the Law of Sines, we have

$$
\frac{c}{a}=\frac{\sin \gamma}{\sin \alpha}=\frac{\sin 2 \alpha}{\sin \alpha}=2 \cos \alpha
$$

so $\cos \alpha=\frac{c}{8}$. On the other hand, the Law of Cosines gives

$$
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{c^{2}+9}{10 c}
$$

so by comparing, we get the equation

$$
\frac{c}{8}=\frac{c^{2}+9}{10 c}
$$

from which $c=6$.

Now $\cos \alpha=\frac{c}{8}=\frac{3}{4}$, which gives $\sin \alpha=\frac{\sqrt{7}}{4}$. From the formula $a=2 R \sin \alpha$, we can calculate the radius of the circumscribed circle $R$.

$$
R=\frac{a}{2 \sin \alpha}=\frac{8}{\sqrt{7}}=\frac{8 \sqrt{7}}{7}
$$

The area of the triangle is $P=\frac{1}{2} b c \sin \alpha=\frac{1}{2} \cdot 5 \cdot 6 \cdot \frac{\sqrt{7}}{4}=\frac{15 \sqrt{7}}{4}$, and the semiperimeter is $s=\frac{a+b+c}{2}=\frac{4+5+6}{2}=\frac{15}{2}$, so we can finally calculate the radius of the inscribed circle.

$$
r=\frac{P}{s}=\frac{\frac{15 \sqrt{7}}{4}}{\frac{15}{2}}=\frac{\sqrt{7}}{2}
$$