2002 AMC 12B Problems/Problem 23
Problem
In , we have and . Side and the median from to have the same length. What is ?
Solution
Solution 1
Let be the foot of the median from to , and we let . Then by the Law of Cosines on , we have
Since , we can add these two equations and get
Hence and .
Solution 2
From Stewart's Theorem, we have Simplifying, we get
Solution 3
Let be the foot of the altitude from to extended past . Let and . Using the Pythagorean Theorem, we obtain the equations
Subtracting the first equation from the second and third equations, we get
Then, subtracting two times the first equation from the second and rearranging, we get , so
~greenturtle 11/26/2017
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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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