Difference between revisions of "2000 AIME I Problems/Problem 8"
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== Solution == | == Solution == | ||
− | {{ | + | The scale factor is uniform in all dimensions, so the volume of the liquid is <math>\left(\frac{3}{4}\right)^{3}</math> of the container. |
+ | |||
+ | The remaining section of the volume is <math>\frac{1-\left(\frac{3}{4}\right)^{3}}{1}</math> of the volume, and therefore <math>\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}</math> of the height when it is consolidated at the tip. | ||
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+ | So, the liquid occupies <math>\frac{1-\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}</math> of the height, or <math>12-12\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}=12-3\left(37^{1/3}\right)</math> | ||
+ | |||
+ | <math>\boxed{m+n+p=052}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=I|num-b=7|num-a=9}} | {{AIME box|year=2000|n=I|num-b=7|num-a=9}} |
Revision as of 17:24, 31 December 2007
Problem
A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 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 liquid is where and are positive integers and is not divisible by the cube of any prime number. Find .
Solution
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 it is consolidated at the tip.
So, the liquid occupies of the height, or
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 | ||
All AIME Problems and Solutions |