Difference between revisions of "2015 AMC 10B Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | Triangle <math>AFG</math> is isosceles, so <math>AG=AF=1</math>. Using the symmetry of pentagon <math>FGHIJ</math>, notice that <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>. |
− | Since <math>\triangle AJH \sim \triangle AFG</math>, < | + | Since <math>\triangle AJH \sim \triangle AFG</math>, |
+ | <cmath>\frac{JH}{AF+FJ}=\frac{FG}{FA}</cmath>. | ||
+ | <cmath>\frac{1}{1+FG} = \frac{FG}1</cmath> | ||
+ | <cmath>1 = FG^2 + FG</cmath> | ||
+ | <cmath>FG^2+FG-1 = 0</cmath> | ||
+ | <cmath>FG = \frac{-1 \pm \sqrt{5} }{2}</cmath> | ||
− | + | However, <math>FG=\frac{-1 + \sqrt{5}}{2}</math> since <math>FG</math> must be greater than 0. | |
− | Therefore, <math>FG+JH+CD=\frac{\sqrt5 | + | Notice that <math>CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}</math>. |
+ | |||
+ | Therefore, <math>FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math> | ||
==See Also== | ==See Also== |
Revision as of 18:17, 8 October 2017
Problem
In the figure shown below, is a regular pentagon and . What is ?
Solution
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,
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 |
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