# 2004 AIME I Problems/Problem 11

## Problem

A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-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

Our original solid has volume equal to and has surface area , where is the slant height of the cone. Using the Pythagorean Theorem, we get and .

Let denote the radius of the small cone. Let and denote the area of the painted surface on cone and frustum , respectively, and let and denote the volume of cone and frustum , respectively. Because the plane cut is parallel to the base of our solid, is similar to the uncut solid and so the height and slant height of cone are and , respectively. Using the formula for lateral surface area of a cone, we find that . By subtracting from the surface area of the original solid, we find that .

Next, we can calculate . Finally, we subtract from the volume of the original cone to find that . We know that Plugging in our values for , , , and , we obtain the equation . We can take reciprocals of both sides to simplify this equation to and so . Then so the answer is .