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
.