Difference between revisions of "2015 AMC 10B Problems/Problem 22"
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Since <math>\triangle AJH \sim \triangle AFG</math>, <math>\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1</math>. From this, we get <math>FG=\frac{\sqrt{5} -1}{2}</math>. | Since <math>\triangle AJH \sim \triangle AFG</math>, <math>\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1</math>. From this, we get <math>FG=\frac{\sqrt{5} -1}{2}</math>. | ||
− | Since <math>\triangle DIJ \cong \triangle AFG</math>, <math>DJ=DI=AF=1</math>. Since <math>\triangle AFG \sim ADC</math>, <math> \frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac{\frac{\sqrt{5} - 1}{2}}{CD}</math>. Thus, <math>CD = (\sqrt | + | Since <math>\triangle DIJ \cong \triangle AFG</math>, <math>DJ=DI=AF=1</math>. Since <math>\triangle AFG \sim ADC</math>, <math> \frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac{\frac{\sqrt{5} - 1}{2}}{CD}</math>. Thus, <math>CD = (\sqrt{5} - 1) + (\frac{\sqrt{5} - 1}{2})^2 = \frac{\sqrt{5} +1}{2}</math> |
Therefore, <math>FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math> | Therefore, <math>FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math> |
Revision as of 08:37, 4 October 2015
Problem
In the figure shown below, is a regular pentagon and . What is ?
Solution
Triangle is isosceles, so . Using the symmetry of pentagon , notice that . Therefore, .
Since , . From this, we get .
Since , . Since , . Thus,
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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