2021 AMC 12A Problems/Problem 17
- The following problem is from both the 2021 AMC 10A #17 and 2021 AMC 12A #17, so both problems redirect to this page.
Contents
[hide]Problem
Trapezoid has , and . Let be the intersection of the diagonals and , and let be the midpoint of . Given that , the length of can be written in the form , where and are positive integers and is not divisible by the square of any prime. What is ?
Solution 1
Angle chasing reveals that , therefore Additional angle chasing shows that , therefore Since is right, the Pythagorean theorem implies that
~mn28407
Solution 2 (One Pair of Similar Triangles, then Areas)
Since is isosceles with legs and it follows that the median is also an altitude of Let and We have
Since by AA, we have
Let the brackets denote areas. Notice that (By the same base/height, Subtracting from both sides gives ). Doubling both sides, we have
In we have and Finally,
~MRENTHUSIASM
Video Solution (Using Similar Triangles, Pythagorean Theorem)
~ pi_is_3.14
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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.