2002 AMC 10B Problems/Problem 20
Contents
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.
Solution 3 (fakesolve)
Notice that the coefficients of and are pretty similar (15s for reading and noticing), so let gives , and (10s writing). Since the desired quantity simplifies to , the term of the quadratics after squaring gets canceled by adding up the squares of the two equations because they have the same coefficients but opposite sign (15s mind-binom). This simplifies to , or (15s writing and addition and fraction simplification and (B) circling and submission)
Video Solution
https://www.youtube.com/watch?v=3Oq21r5OezA ~David
See Also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AMC 10 Problems and Solutions |
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