Difference between revisions of "2002 AMC 10A Problems/Problem 20"
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==Solution 2== | ==Solution 2== | ||
− | As | + | As angle F is clearly congruent to itself, we get from AA similarity, <math>\triangle AGF \sim \triangle EJF</math>; hence <math>\frac {AG}{JE} =5</math>. Similarly, <math>\frac {AG}{HC} = 3</math>. Thus, <math>\frac {HC}{JE}=\left(\frac{AG}{JE}\right)\left(\frac{HC}{AG}\right) = \boxed{\frac {5}{3}\Rightarrow \text{(D)}}</math>. |
==See Also== | ==See Also== |
Revision as of 13:03, 4 January 2016
Contents
Problem
Points and lie, in that order, on , dividing it into five segments, each of length 1. Point is not on line . Point lies on , and point lies on . The line segments and are parallel. Find .
Solution 1
Since and are parallel, triangles and are similar. Hence, .
Since and are parallel, triangles and are similar. Hence, . Therefore, . The answer is .
Solution 2
As angle F is clearly congruent to itself, we get from AA similarity, ; hence . Similarly, . Thus, .
See Also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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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