Difference between revisions of "2012 AMC 12B Problems/Problem 21"
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+ | == See Also == | ||
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+ | {{AMC12 box|year=2012|ab=B|num-b=20|num-a=22}} |
Revision as of 22:22, 12 January 2013
Problem
Square is inscribed in equiangular hexagon with on , on , and on . Suppose that , and . What is the side-length of the square?
Solution (Long)
Extend and so that they meet at . Then , so and therefore is parallel to . Also, since is parallel and equal to , we get , hence is congruent to . We now get .
Let , , and .
Drop a perpendicular line from to the line of that meets line at , and a perpendicular line from to the line of that meets at , then is congruent to since is complementary to . Then we have the following equations:
The sum of these two yields that
So, we can now use the law of cosines in :
Therefore
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 |