Problem:
A solid in the shape of a right circular cone is inches tall and its base has a -inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid and a frustum-shaped solid , in such a way that the ratio between the areas of the painted surfaces of and and the ratio between the volumes of and are both equal to . Given that , where and are relatively prime positive integers, find .
Solution:
The lateral surface area of a cone with radius and slant height can be found by cutting the cone along a slant height and then unrolling it to form a sector of a circle. The sector's arc has length and its radius is , so its area, and the cone's lateral surface area, is . Let , and represent the radius, height, and slant height of the smaller cone formed by the cut. Then
Thus . Because , let , and substitute to find that , and then that . The ratio of the volume of to that of the large cone is therefore , so the ratio of the volumes of and is . Thus .
The problems on this page are the property of the MAA's American Mathematics Competitions