2002 AMC 12B Problems/Problem 20
- The following problem is from both the 2002 AMC 12B #20 and 2002 AMC 10B #22, so both problems redirect to this page.
Contents
Problem
Let be a right-angled triangle with . Let and be the midpoints of legs and , respectively. Given that and , find .
Solution 1
Let , . By the Pythagorean Theorem on respectively,
Summing these gives .
By the Pythagorean Theorem again, we have
Alternatively, we could note that since we found , segment . Right triangles and are similar by Leg-Leg with a ratio of , so
Solution 2
Let and Then,
Since and
Adding these up:
Then, we substitute:
Solution 3 (Solution 1 but shorter)
Refer to the diagram in solution 1. and , so add them: and divide by 5: so and so , or answer .
Solution 4
Use the diagram in solution 1. Get and , and multiply the second equation by 4 to get and then subtract the first from the second. Get and . Repeat for the other variable to get and . Now XY is equal to the square root of four times these quantities, so , and
Video Solution by OmegaLearn
https://youtu.be/BIyhEjVp0iM?t=174
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=7wj6RupkO90 ~David
See also
2002 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 |
2002 AMC 12B (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.