Difference between revisions of "2012 AMC 12B Problems/Problem 20"
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So <math>r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5</math>, which rounds down to <math>\boxed{\textbf{(D)}\ 63}</math>. | So <math>r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5</math>, which rounds down to <math>\boxed{\textbf{(D)}\ 63}</math>. | ||
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+ | == Video Solution == | ||
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+ | https://youtu.be/8w1vrsD2urs ~Math Problem Solving Skills | ||
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== See Also == | == See Also == |
Latest revision as of 17:29, 20 June 2022
Contents
Problem 20
A trapezoid has side lengths 3, 5, 7, and 11. The sum of all the possible areas of the trapezoid can be written in the form of , where , , and are rational numbers and and are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to ?
Solution
Name the trapezoid , where is parallel to , , and . Draw a line through parallel to , crossing the side at . Then , . One needs to guarantee that , so there are only three possible trapezoids:
In the first case, by Law of Cosines, , so . Therefore the area of this trapezoid is .
In the second case, , so . Therefore the area of this trapezoid is .
In the third case, , therefore the area of this trapezoid is .
So , which rounds down to .
Video Solution
https://youtu.be/8w1vrsD2urs ~Math Problem Solving Skills
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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