Difference between revisions of "2012 AMC 12B Problems/Problem 21"
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<math> \textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}</math> | <math> \textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}</math> | ||
− | ==Solution | + | ==Solution== |
Extend <math>AF</math> and <math>YE</math> so that they meet at <math>G</math>. Then <math>\angle FEG=\angle GFE=60^{\circ}</math>, so <math>\angle FGE=60^{\circ}</math> and therefore <math>AB</math> is parallel to <math>YE</math>. Also, since <math>AX</math> is parallel and equal to <math>YZ</math>, we get <math>\angle BAX = \angle ZYE</math>, hence <math>\triangle ABX</math> is congruent to <math>\triangle YEZ</math>. We now get <math>YE=AB=40</math>. | Extend <math>AF</math> and <math>YE</math> so that they meet at <math>G</math>. Then <math>\angle FEG=\angle GFE=60^{\circ}</math>, so <math>\angle FGE=60^{\circ}</math> and therefore <math>AB</math> is parallel to <math>YE</math>. Also, since <math>AX</math> is parallel and equal to <math>YZ</math>, we get <math>\angle BAX = \angle ZYE</math>, hence <math>\triangle ABX</math> is congruent to <math>\triangle YEZ</math>. We now get <math>YE=AB=40</math>. |
Revision as of 19:31, 4 December 2016
Problem 21
Square is inscribed in equiangular hexagon with on , on , and on . Suppose that , and . What is the side-length of the square?
Solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.