Difference between revisions of "2002 AMC 10B Problems/Problem 20"
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This time-saving trick works only because we know that for any value of <math>a</math>, <math>a^2-b^2+c^2</math> will always be constant (it's a contest), so any value of <math>a</math> will work. | This time-saving trick works only because we know that for any value of <math>a</math>, <math>a^2-b^2+c^2</math> will always be constant (it's a contest), so any value of <math>a</math> will work. | ||
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==See Also== | ==See Also== |
Revision as of 13:31, 30 November 2019
Contents
Problem
Let a, b, and c be real numbers such that and
. Then
is
Solution
Rearranging, we get and
Squaring both, and
are obtained.
Adding the two equations and dividing by gives
, so
.
Easiest Solution
The easiest way is to assume a value for and then solving 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.
See Also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.