2004 AIME I Problems/Problem 11
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
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 .
and are similar cones, because the plane that cut out was parallel to the base of . Let be the scale factor between the original cone and the small cone in one dimension. Because the scale factor is uniform in all dimensions, relates corresponding areas of and , and relates corresponding volumes. Then, the ratio of the painted areas is and the ratio of the volumes is . Since both ratios are equal to , they are equal to each other. Therefore, .
Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives . Dividing both sides by and distributing the on the right, we have , and so and . Substituting back into the easier ratio, we have . And so we have .
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