A frustum of a cone has such dimensions that the sphere whose center is the center of the larger base circle and which touches the plane of the smaller base circle, also touches the lateral surface. We also know that the volume of the hemisphere that fits inside the frustum is $\frac{6}{7}$ of the volume of the frustum.

What is the ratio of the lateral surface area to the surface area of the hemisphere?
Let the radii of the bases of the frustum of a cone be $R$ and $r (R > r)$, the distance between their planes, the height $m$, is also the radius of the inscribed hemisphere.

# $1985-11-372-1 . \mathrm{eps}$ 

Any plane passing through the axis of rotation cuts a symmetrical trapezoid from the frustum of the cone and a semicircle from the hemisphere, and the latter touches the legs of the trapezoid. With the notation of the figure, the right triangles $O A E$ and $B A D$ are congruent, $A E = A D = R - r$, and due to the tangents, $B E = B C = r$, so the slant height of the frustum $A B = R$, thus

$$
m^{2} = R^{2} - (R - r)^{2} = 2 R r - r^{2}
$$

From the ratio of the volumes,

$$
\begin{gathered}
\frac{2 \pi}{3} m^{3} = \frac{6}{7} \cdot \frac{m \pi}{3} \left(R^{2} + R r + r^{2}\right) \quad (m \neq 0) \\
3 \left(R^{2} + R r + r^{2}\right) = 7 m^{2} = 14 R r - 7 r^{2} \\
3 R^{2} - 11 R r + 10 r^{2} = 0
\end{gathered}
$$

from which we get two positive values for $R / r$:

$$
\frac{R}{r} = 2 \quad \text{and} \quad \frac{R}{r} = \frac{5}{3}
$$

Since our calculations can be reversed, the two types of frustums of cones meet the given conditions, with $R = 2 r$ and $R = 5 r / 3$, respectively, and the radius of the hemisphere is

$$
m = \sqrt{3} r, \quad \text{or} \quad \sqrt{7 / 3} r
$$

For the ratio of the lateral surface area of the frustum and the surface area of the hemisphere, substituting into the appropriate formulas for the two different scalings, 1 and 20/21 are obtained, respectively.