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
\begin{align*} x^2 + y^2 = 1 \t(1)\\ x^2 + y^2 + 2ya + a^2 = 4a^2 (2)\\ x^2 + y^2 + 4ya + 4a^2 = 4 (3) \end{align*} (Error compiling LaTeX. Unknown error_msg)
Subtracting (1) equation from (2) and (3), we get
Then, subtracting two times (4) from (5) 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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