2002 AMC 10B Problems/Problem 20
Problem
Let a, b, and c be real numbers such that and . Then is
Solution
Solution 1
Rearranging, we get and
Squaring both, and are obtained.
Adding the two equations and dividing by gives , so .
Solution 2
The easiest way is to assume a value for and then solve the system of equations. For , we get the equations and Multiplying the second equation by , we have Adding up the two equations yields , so We obtain after plugging in the value for . Therefore, which corresponds to . This time-saving trick works only because we know that for any value of , will always be constant (it's a contest), so any value of will work. This is also called without loss of generality or WLOG.
See Also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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