The height, surface area, and volume of a truncated cone are given. Calculate the radii of the bounding circles of the bases.

Let's denote the height of the truncated cone as \( h \), the surface area as \( A \), and the volume as \( V \). The radii of the two bases are \( R \) (larger base) and \( r \) (smaller base).

The formulas for the surface area and volume of a truncated cone are:
1. Surface Area: \( A = \pi (R + r) \sqrt{(R - r)^2 + h^2} + \pi (R^2 + r^2) \)
2. Volume: \( V = \frac{1}{3} \pi h (R^2 + Rr + r^2) \)

To find the radii \( R \) and \( r \), we need to solve these equations simultaneously. This can be a complex process, often requiring numerical methods or iterative solutions. However, if specific values for \( h \), \( A \), and \( V \) are provided, we can proceed with the calculations. 

If you have specific values for \( h \), \( A \), and \( V \), please provide them, and I can help you solve for \( R \) and \( r \).
In the known formula for the volume $V$ of a frustum, only the sought radii $R$ and $r (R \geq r)$ are unknown, hence

$$
R^{2}+r^{2}+R r=\frac{3 V}{\pi m}=A
$$

In the formula for the surface area $F$, in addition to these, the slant height $a$ of the lateral surface also appears:

$$
R^{2}+r^{2}+(R+r) a=\frac{F}{\pi}=B
$$

but this can be expressed with the radii and the height $m$:

$$
a=\sqrt{(R-r)^{2}+m^{2}}
$$

Substituting this, isolating the term containing the square root on the left side, and squaring the equation,

$$
\left(R^{2}+r^{2}+2 R r\right)\left(R^{2}+r^{2}-2 R r+m^{2}\right)=\left[B-\left(R^{2}+r^{2}\right)\right]^{2}
$$

Considering (1) and (2) as a system of equations in the unknowns

$$
R^{2}+r^{2}=u \quad \text { and } \quad R r=v
$$

Thus,

$$
\begin{gathered}
u+v=A \\
(u+2 v)\left(u-2 v+m^{2}\right)=B^{2}-2 B u+u^{2}
\end{gathered}
$$

Eliminating $u$ and rearranging,

$$
\begin{gathered}
(A+v)\left(A-3 v+m^{2}\right)=B^{2}-2 B(A-v)+(A-v)^{2} \\
4 v^{2}-\left(m^{2}-2 B\right) v+\left(B^{2}-2 A B-A m^{2}\right)=0
\end{gathered}
$$

The discriminant of this equation can be factored:

$$
\begin{gathered}
\left(m^{2}-2 B\right)^{2}-16\left(B^{2}-2 A B-A m^{2}\right)=m^{4}-4 B m^{2}-12 B^{2}+16 A\left(m^{2}+2 B\right)= \\
=\left(m^{2}+2 B\right)\left(m^{2}-6 B+16 A\right)
\end{gathered}
$$

Thus, the roots of (3) are

$$
v=\frac{1}{8}\left[m^{2}-2 B \pm \sqrt{\left(m^{2}+2 B\right)\left(m^{2}+16 A-6 B\right)}\right]
$$

Finally, according to (1 $1^{\prime}$),

$$
u=A-v=\frac{1}{8}\left[8 A+2 B-m^{2} \mp \sqrt{\left(m^{2}+2 B\right)\left(m^{2}+16 A-6 B\right)}\right]
$$

Now, denoting the square root expression by $C$,

$$
\begin{gathered}
(R+r)^{2}=R^{2}+r^{2}+2 R r=u+2 v=A+v=A+\frac{1}{8}\left(m^{2}-2 B \pm C\right) \\
(R-r)^{2}=u-2 v=A-3 v=A-\frac{3}{8}\left(m^{2}-2 B \pm C\right)
\end{gathered}
$$

Finally, by taking the square root and adding and subtracting the resulting equations:

$$
\begin{aligned}
& R=\frac{1}{2}\left[\sqrt{A+\frac{1}{8}\left(m^{2}-2 B \pm C\right)}+\sqrt{A-\frac{3}{8}\left(m^{2}-2 B \pm C\right)}\right] \\
& r=\frac{1}{2}\left[\sqrt{A+\frac{1}{8}\left(m^{2}-2 B \pm C\right)}-\sqrt{A-\frac{3}{8}\left(m^{2}-2 B \pm C\right)}\right]
\end{aligned}
$$

$u$ and $v$ are real if and only if in the discriminant of (3),

$$
m^{2}+16 A \geq 6 B
$$

and they can lead to a solution of the geometric problem if and only if both are positive and $u-2 v \geq 0$. Depending on this, the number of $R, r$ value pairs that satisfy the condition is 2, 1, or 0. The detailed discussion is complex, considering the mutual magnitude relationships of the 3 parameters.

Tattay Emóke (Budapest, Kaffka Margit lg. IV. o. t.) Szoboszlai Levente (Hódmezővásárhely, Bethlen G. g. IV. o. t.)

Remark. Several others solved the system by introducing new unknowns, such as $R+r$, in place of $u$.