Difference between revisions of "2002 AMC 12B Problems/Problem 23"
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Revision as of 11:06, 13 January 2012
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
![\[5 = 10a^2\]](http://latex.artofproblemsolving.com/b/7/e/b7ef5bf2a35df8f95e354145fc0ffc141e39ae7a.png)
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 | |
