2000 AIME I Problems/Problem 8
Problem
A container in the shape of a right circular cone is inches tall and its base has a -inch radius. The liquid that is sealed inside is inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is from the base where and are positive integers and is not divisible by the cube of any prime number. Find .
Solution
Solution 1
The scale factor is uniform in all dimensions, so the volume of the liquid is of the container. The remaining section of the volume is of the volume, and therefore of the height when the vertex is at the top.
So, the liquid occupies of the height, or . Thus .
Solution 2
(Computational) The volume of a cone can be found by . In the second container, if we let represent the height, radius (respectively) of the air (so is the height of the liquid), then the volume of the liquid can be found by .
By similar triangles, we find that the dimensions of the liquid in the first cone to the entire cone is , and that ; equating,
Thus the answer is , and .
Solution 3
From the formula , we can find that the volume of the container is . The cone formed by the liquid is similar to the original, but scaled down by in all directions, so its volume is . The volume of the air in the container is the volume of the container minus the volume of the liquid, which is , which is of the volume of the container. When the point faces upwards, the air forms a cone at the top of the container. This cone must have of the height of the container. This means that the height of the liquid is inches, so our answer is . Solution by Zeroman
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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