Difference between revisions of "2006 AIME I Problems/Problem 14"
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We also have the [[area ratios|area ratio]] <math>\frac {[SBC]}{[ABC]} = \frac {[SM]}{[AM]}</math>. | We also have the [[area ratios|area ratio]] <math>\frac {[SBC]}{[ABC]} = \frac {[SM]}{[AM]}</math>. | ||
− | The triangle <math>TOA</math> is a <math>3-4-5</math> [[right triangle]] so <math>[ | + | The triangle <math>TOA</math> is a <math>3-4-5</math> [[right triangle]] so <math>[AM] = \frac {3}{2}\cdot[AO] = \frac {9}{2}</math> and <math>\cos{\angle{TAO}} = \frac {3}{5}</math>. |
Applying [[Law of Cosines]] to the triangle <math>SAM</math> with <math>[SA] = 1</math>, <math>[AM] = \frac {9}{2}</math> and <math>\cos{\angle{SAM}} = \frac {3}{5}</math>, we find: | Applying [[Law of Cosines]] to the triangle <math>SAM</math> with <math>[SA] = 1</math>, <math>[AM] = \frac {9}{2}</math> and <math>\cos{\angle{SAM}} = \frac {3}{5}</math>, we find: |
Revision as of 14:10, 3 August 2013
Problem
A tripod has three legs each of length feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then can be written in the form where and are positive integers and is not divisible by the square of any prime. Find (The notation denotes the greatest integer that is less than or equal to )
Solution
We will use to denote volume (four letters), area (three letters) or length (two letters).
Let be the top of the tripod, are end points of three legs. Let be the point on such that and . Let be the center of the base equilateral triangle . Let be the midpoint of segment . Let be the distance from to the triangle ( is what we want to find).
We have the volume ratio .
So .
We also have the area ratio .
The triangle is a right triangle so and .
Applying Law of Cosines to the triangle with , and , we find:
Putting it all together, we find .
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.