2002 AMC 12B Problems/Problem 23
Contents
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 altitude from to extended past . Let and . Using the Pythagorean Theorem, we obtain the equations
Subtracting equation from and , we get
Then, subtracting from and rearranging, we get , so
~greenturtle 11/28/2017
Solution 2
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 3
From Stewart's Theorem, we have Simplifying, we get - awu2014
Solution 4 [Pappus's Median Theorem]
There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have triangle ABC, and you draw a median from point A to side BC (label this as Ma), then: (Ma)^2 = (1/4)(2(b^2) + 2(c^2) - (a^2)). Note that b is side AC, c is side AB, and a is side BC. For this specific problem, call where the median from A hits side BC point M. Let MB = MC = x. Then AM = 2x. Now, we can plug into the formula given above: Ma = 2x, b = 2, c = 1, and a = 2x. After some simple algebra, we find x = sqrt(2)/2. Then, BC = a = sqrt(2).
-Flames
Video Solution by TheBeautyofMath
~IceMatrix
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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