Let the width of the circular ring be $a$. Furthermore, the ratio of the area of the circular ring to the area of the circle inscribed within it is given, and the circumference of this circle is the arithmetic mean of the circumferences of the circles bounding the circular ring. Determine the radii of the circles $r$ and $R$.
The arithmetic mean of the circumferences of the outer and inner circles of an annulus is:

$$
\frac{K+k}{2}=(R+r) \pi
$$

where $K$ is the circumference of the outer circle and $k$ is the circumference of the inner circle.

If the circumference of the circle in question is equal to $(R+r) \pi$, then its radius is

$$
R_{1}=\frac{R+r}{2}
$$

and its area is

$$
T=\frac{(R+r)^{2}}{4} \pi
$$

The area of the annulus is

$$
t=\left(R^{2}-r^{2}\right) \pi
$$

and the given ratio is

$$
t: T=c
$$

so

$$
c=\frac{4\left(R^{2}-r^{2}\right)}{(R+r)^{2}}=\frac{4(R-r)}{R+r}
$$

and from this

$$
\begin{array}{r}
R+r=\frac{4(R-r)}{c} \\
\text { and } R-r=a
\end{array}
$$

and thus (1)

$$
R+r=\frac{4 a}{c}
$$

Therefore, the radii of the circles forming the annulus are:

$$
\begin{aligned}
& r=\frac{a}{2 c}(4-c)=\frac{a}{2}\left(\frac{4}{c}-1\right) \\
& R=\frac{a}{2 c}(4+c)=\frac{a}{2}\left(\frac{4}{c}+1\right)
\end{aligned}
$$

(János Zalányi, high school teacher, Cluj-Napoca).

The problem was also solved by: Gergely Jorga, 8th grade student, main real school, Arad.