2012 AMC 12B Problems/Problem 20
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 1
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 .
Solution 2
Let the area of the trapezoid be , the area of the triangle be , the area of the parallelogram be .
If , , ,
If , , ,
, which is impossible as
If , , ,
, which is impossible as
If , , ,
, which is impossible as
If , , ,
If , , ,
Thus the answer is , which rounds down to
Note: the three invalid cases can also be determined by the triangle inequality.
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 |
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