2002 AMC 10A Problems/Problem 20
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
First we can draw an image.
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, .
Solution 3
Assume an arbitrary value of WLOG. and are parallel, so and are similar. So, which means . By the same logic, , so .
Video Solution
https://www.youtube.com/watch?v=AU2PJeMZ7R0 ~David
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.