The radii of the excircles of a triangle are $r_{a}, r_{b}$, and $r_{c}$, and the radius of its circumcircle is $R$. We know that $r_{a}+r_{b}=3 R$ and $r_{b}+r_{c}=2 R$. What are the angles of the triangle?
Solution. Let the sides of the triangle be denoted by $a, b, c$, the angles by $\alpha, \beta, \gamma$, the area by $T$, and the perimeter by $2 s$. First, we prove a lemma:

In any triangle, if we multiply the value of one more than the cosine of an angle by twice the radius of the circumcircle, we get the sum of the radii of the excircles opposite to the sides enclosing the angle.

Proof: Since the roles of the angles in the triangle are symmetric, it is sufficient to show that

$$
2 R(1+\cos \gamma)=r_{a}+r_{b}
$$

It is known (see, for example, Kiss Gy.: What is Good to Know About Triangles, KöMaL 2002/3, pp. 130-139) that

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

From the cosine rule, $\cos \gamma$ can be expressed in terms of the sides as

$$
\cos \gamma=\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$

To prove our statement, it is sufficient to show that

$$
2 \cdot \frac{a b c}{4 T} \cdot\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}+1\right)=\frac{T}{s-a}+\frac{T}{s-b}
$$

We get an equivalent statement if we multiply both sides by $4 T$ and use Heron's formula, which states that $T^{2}=s(s-a)(s-b)(s-c)$.

$$
2 a b c \cdot\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}+1\right)=4 \cdot(s(s-b)(s-c)+s(s-a)(s-c))
$$

Simplifying both sides:

$$
c \cdot\left(a^{2}+b^{2}-c^{2}+2 a b\right)=4 s(s-c)((s-b)+(s-a))
$$

This is clearly true because, by the definition of $s$, $(s-b)+(s-a)=c$ and $4 s(s-c)=2 s(2 s-2 c)=(a+b)^{2}-c^{2}$. This completes the proof of the lemma.

Using the lemma, we can easily answer the question of the problem, as the two given equations can be written as:

$$
2 R(1+\cos \gamma)=3 R \quad \text { and } \quad 2 R(1+\cos \alpha)=2 R
$$

Dividing both equations by $2 R$ and rearranging, we get:

$$
\cos \gamma=\frac{1}{2} \quad \text { and } \quad \cos \alpha=0
$$

from which $\gamma=60^{\circ}$ and $\alpha=90^{\circ}$.

Thus, the angles of the triangle can be $90^{\circ}, 60^{\circ}$, and $180^{\circ}-\left(90^{\circ}+60^{\circ}\right)=30^{\circ}$. It can be verified that in triangles with these angles, the desired relationships for the radii of the excircles hold.