Difference between revisions of "2002 AMC 12B Problems/Problem 23"
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Hence <math>a = \frac{1}{\sqrt{2}}</math> and <math>BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}</math>. | Hence <math>a = \frac{1}{\sqrt{2}}</math> and <math>BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}</math>. | ||
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+ | == Alternate Solution == | ||
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+ | From [[Stewart's Theorem]], we have <math>(2)(1/2a)(2) + (1)(1/2a)(1) = (a)(a)(a) + (1/2a)a(1/2a).</math> Simplifying, we get <math>(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.</math> | ||
== See also == | == See also == |
Revision as of 22:07, 3 January 2010
Problem
In , we have and . Side and the median from to have the same length. What is ?
Solution
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 .
Alternate Solution
From Stewart's Theorem, we have Simplifying, we get
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 |