2022 AMC 12B Problems/Problem 19
Contents
Problem
In medians and intersect at and is equilateral. Then can be written as , where and are relatively prime positive integers and is a positive integer not divisible by the square of any prime. What is ?
Diagram
Solution 1: Law of Cosines
Let . Since is the midpoint of , must also be .
Since the centroid splits the median in a ratio, must be equal to and must be equal to .
Applying Law of Cosines on and yields and . Finally, applying Law of Cosines on yields . The requested sum is .
Solution 2 (Also Law of Cosines, but with one less computation)
Let . Since (as is the centroid), . Also, and . By the law of cosines, .
Applying the law of cosines again on gives , so the answer is .
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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.