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 , we must have .
Since the centroid splits the median in a ratio, and .
Applying Law of Cosines on and yields and . Finally, applying Law of Cosines on yields . The requested sum is .
Solution 2 (Law of Cosines: One Fewer Step)
Let . Since (as is the centroid), . Also, and . By the law of cosines (applied on ), .
Applying the law of cosines again on gives , so the answer is .
Solution 3 (Law of Cosine)
Let . Since is the centroid, , .
By the Law of Cosine in
By the Law of Cosine in
Solution 4 (Barycentric Coordinates)
Using reference triangle , we can let If we move each over by , leaving unchanged, we have The angle between vectors and satisfies giving the answer, .
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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 |
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