Difference between revisions of "2015 AMC 10B Problems/Problem 22"
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<math>\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} </math> | <math>\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} </math> | ||
− | ==Solution== | + | ==Solution 1== |
Triangle <math>AFG</math> is isosceles, so <math>AG=AF=1</math>. <math>FJ = FG</math> since <math>\triangle FGJ</math> is also isosceles. Using the symmetry of pentagon <math>FGHIJ</math>, notice that <math>\triangle JHG \cong \triangle AFG</math>. Therefore, <math>JH=AF=1</math>. | Triangle <math>AFG</math> is isosceles, so <math>AG=AF=1</math>. <math>FJ = FG</math> since <math>\triangle FGJ</math> is also isosceles. Using the symmetry of pentagon <math>FGHIJ</math>, notice that <math>\triangle JHG \cong \triangle AFG</math>. Therefore, <math>JH=AF=1</math>. | ||
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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> | ||
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+ | ==Solution 2== | ||
+ | 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>. | ||
==See Also== | ==See Also== |
Revision as of 14:54, 29 December 2017
Contents
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
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 .
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AMC 10 Problems and Solutions |
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