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 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 .
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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