A triangle has exradii $r_{a}, r_{b}$, and $r_{c}$, and the radius of its circumcircle is $R$. Determine the angles of the triangle if

$$
r_{a}+r_{b}=3 R \quad \text { and } \quad r_{b}+r_{c}=2 R
$$
Solution. First, we show that the triangle is a right triangle.

Let the sides of the triangle be $a, b$, and $c$, the perimeter be $2s$, and the area be $T$. According to the known formulas for the radii of the circles,

$$
r_{a}=\frac{T}{s-a}, \quad r_{b}=\frac{T}{s-b}, \quad r_{c}=\frac{T}{s-c} \quad \text { and } \quad R=\frac{a b c}{4 T}
$$

Substituting these into the condition $r_{b}+r_{c}=2 R$, rearranging, and then applying Heron's formula, which states

$$
T^{2}=s(s-a)(s-b)(s-c)
$$

and finally using the relationship $(s-c)+(s-b)=a$, we get

$$
\begin{gathered}
\frac{T}{s-b}+\frac{T}{s-c}=\frac{a b c}{2 T} \\
\frac{4 T^{2}}{s-b}+\frac{4 T^{2}}{s-c}=2 a b c \\
4 s(s-a)((s-c)+(s-b))=2 a b c \\
4 s(s-a)=2 b c, \quad \text { that is } \quad 2 s(2 s-2 a)-2 b c=0
\end{gathered}
$$

Here, $2 s-2 a=b+c-a$, so

$$
\begin{aligned}
0 & =2 s(2 s-2 a)-2 b c=(b+c+a)(b+c-a)-2 b c= \\
& =(b+c)^{2}-a^{2}-2 b c=b^{2}+c^{2}-a^{2}
\end{aligned}
$$

Thus, by the converse of the Pythagorean theorem, the angle opposite the side $a$ of the triangle is a right angle.

Starting from the condition $r_{a}+r_{b}=3 R$ and performing the same transformations, we get

$$
\begin{gathered}
\frac{T}{s-a}+\frac{T}{s-b}=\frac{3 a b c}{4 T} \\
\frac{4 T^{2}}{s-a}+\frac{4 T^{2}}{s-b}=3 a b c \\
4 s(s-c)((s-b)+(s-a))=3 a b c \\
4 s(s-c)=3 a b, \quad \text { that is } \quad a^{2}+b^{2}-c^{2}-a b=0
\end{gathered}
$$

Since $a^{2}=b^{2}+c^{2}$, rearranging the last equality gives

$$
2 b^{2}-a b=0, \quad \text { that is } \quad b(2 b-a)=0
$$

From $b>0$, it follows that $b=a / 2$. Thus, the acute angle opposite the side $b$ of the triangle has a sine of $1 / 2$, which means this angle is $30^{\circ}$.

Therefore, the angles of the triangle are $30^{\circ}, 60^{\circ}$, and $90^{\circ}$. It can be easily verified by calculation that for triangles with these angles, the sides satisfy $a=2 b$ and $c=\sqrt{3} b$, and the radii of the circles are

$$
r_{a}=\frac{(3+\sqrt{3}) b}{2}, \quad r_{b}=\frac{(3-\sqrt{3}) b}{2}, \quad r_{c}=\frac{(1+\sqrt{3}) b}{2} \quad \text { and } \quad R=b
$$

which satisfy the conditions of the problem.