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
Solution 1
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
.
Solution 2
Our original solid has surface area
, where
is the slant height of the cone. Using the Pythagorean Theorem or Pythagorean Triple knowledge, we obtain
and lateral area
. The area of the base 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
to
is
and the ratio of the volumes
to
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
.
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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