$24 \cdot 14$ A frustum of a cone with height 8 contains a sphere $O_{1}$ with radius 2, whose center $O_{1}$ is on the axis of the frustum. The sphere $O_{2}$ is tangent to the upper base and the lateral surface of the frustum. The frustum can also contain another sphere $O_{2}$ with radius 3, which is tangent to the sphere $O_{1}$, the lower base, and the lateral surface of the frustum. Apart from the sphere $O_{2}$, the maximum number of additional spheres with radius 3 that can be placed in the frustum is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(China High School Mathematics League, 1996)
[Solution] Make an axial section through $\mathrm{O}_{2}$ of the frustum, as shown in the figure below. Then make a section through $\mathrm{O}_{2}$ perpendicular to the axis of the frustum, intersecting the axis at circle $O$. From the figure below, it is easy to find that $O O_{2}=4$.

This problem is equivalent to: On a circle with center $O$ and radius 4, how many more points can be placed, excluding $O_{2}$, such that the circles with these points and $O_{2}$ as centers and 3 as the radius have at most one common point. From the figure on the right,
$$
\sin 45^{\circ}<\sin \theta=\frac{3}{4}<\sin 60^{\circ},
$$

we have $45^{\circ}<\theta<60^{\circ}$.
Therefore, the maximum number of additional points that can be placed is
$$
\left[\frac{360^{\circ}}{2 \theta}\right]-1=3-1=2
$$

points, satisfying the above requirements.
Thus, the frustum can contain at most 2 more spheres with a radius of 3.
Therefore, the answer is (B).