Difference between revisions of "2002 AMC 12B Problems/Problem 23"
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− | Then, subtracting two times the first equation from the second and rearranging, we get <math>2a^2 = 1</math>, so <math>BC = 2a = \sqrt{2}</math> | + | Then, subtracting two times the first equation from the second and rearranging, we get <math>2a^2 = 1</math>, so <math>BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}</math> |
== See also == | == See also == |
Revision as of 23:13, 26 November 2017
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.