Difference between revisions of "2015 AMC 10B Problems/Problem 22"
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Therefore, <math>FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math> | Therefore, <math>FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math> | ||
− | ==Solution 2(Trigonometry)== | + | ==Solution 2 (Trigonometry)== |
Note that since <math>ABCDE</math> is a regular pentagon, all of its interior angles are <math>108^\circ</math>. We can say that pentagon <math>FGHIJ</math> is also regular by symmetry. So, all of the interior angles of <math>FGHIJ</math> are <math>108^\circ</math>. Now, we can angle chase and use trigonometry to get that <math>FG=2\sin18^\circ</math>, <math>JH=2\sin18^\circ*(2\sin18^\circ+1)</math>, and <math>DC=2\sin18^\circ*(2\sin18^\circ+2)</math>. Adding these together, we get that <math>FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)</math>. Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to <math>8\sin18^\circ*(1+\sin18^\circ)</math>, but we can find that this is closest to <math>\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>. | Note that since <math>ABCDE</math> is a regular pentagon, all of its interior angles are <math>108^\circ</math>. We can say that pentagon <math>FGHIJ</math> is also regular by symmetry. So, all of the interior angles of <math>FGHIJ</math> are <math>108^\circ</math>. Now, we can angle chase and use trigonometry to get that <math>FG=2\sin18^\circ</math>, <math>JH=2\sin18^\circ*(2\sin18^\circ+1)</math>, and <math>DC=2\sin18^\circ*(2\sin18^\circ+2)</math>. Adding these together, we get that <math>FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)</math>. Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to <math>8\sin18^\circ*(1+\sin18^\circ)</math>, but we can find that this is closest to <math>\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math>. | ||
Revision as of 16:58, 1 January 2018
Problem
In the figure shown below, is a regular pentagon and . What is ?
Solution 1
Triangle is isosceles, so . since is also isosceles. Using the symmetry of pentagon , notice that . Therefore, .
Since , .
However, since must be greater than 0.
Notice that .
Therefore,
Solution 2 (Trigonometry)
Note that since is a regular pentagon, all of its interior angles are . We can say that pentagon is also regular by symmetry. So, all of the interior angles of are . Now, we can angle chase and use trigonometry to get that , , and . Adding these together, we get that . Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to , but we can find that this is closest to .
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.