2004 AIME I Problems/Problem 11
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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 
.
See also
| 2004 AIME I (Problems • Resources) | ||
| Preceded by Problem 10 | Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
