Difference between revisions of "2011 AMC 12A Problems/Problem 23"
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Both solutions give the same lower bound, <math>1</math>. So the range is <math>\sqrt{3}-1=\boxed{\textbf{(C) }\sqrt{3}-1}</math>. | Both solutions give the same lower bound, <math>1</math>. So the range is <math>\sqrt{3}-1=\boxed{\textbf{(C) }\sqrt{3}-1}</math>. | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/FU18x_LsTeQ | ||
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+ | ~MathProblemSolvingSkills.com | ||
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== See also == | == See also == |
Revision as of 15:15, 18 August 2023
Contents
[hide]Problem
Let and , where and are complex numbers. Suppose that and for all for which is defined. What is the difference between the largest and smallest possible values of ?
Solution
By algebraic manipulations, we obtain where In order for , we must have , , and .
implies or .
implies , , or .
implies or .
Since , in order to satisfy all 3 conditions we must have either or . In the first case .
For the latter case note that and hence, . On the other hand, so, . Thus . Hence the maximum value for is while the minimum is (which can be achieved for instance when or respectively). Therefore the answer is .
Shortcut
We only need in .
Set : . Since , either or .
so .
. This is a circle in the complex plane centered at with radius since . The maximum distance from the origin is at and similarly the minimum distance is at . So .
Both solutions give the same lower bound, . So the range is .
Video Solution
~MathProblemSolvingSkills.com
See also
2011 AMC 12A (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.