Difference between revisions of "2008 AMC 12B Problems/Problem 19"
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Let <math>p=\Im(\gamma)</math> and <math>q=\Re{(\gamma)},</math> then we know <math>\Im(\alpha)=-p-1</math> and <math>\Re(\alpha)=1-p.</math> Therefore <cmath>|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},</cmath> which reaches its minimum <math>\sqrt 2</math> when <math>p=q=0</math> by the Trivial Inequality. Thus, the answer is <math>\boxed B.</math> | Let <math>p=\Im(\gamma)</math> and <math>q=\Re{(\gamma)},</math> then we know <math>\Im(\alpha)=-p-1</math> and <math>\Re(\alpha)=1-p.</math> Therefore <cmath>|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},</cmath> which reaches its minimum <math>\sqrt 2</math> when <math>p=q=0</math> by the Trivial Inequality. Thus, the answer is <math>\boxed B.</math> | ||
+ | |||
+ | [b]Solution 2:[/b] | ||
+ | |||
+ | <math>f(1)=4+i+\alpha+\gamma</math> | ||
+ | <math>f(i)=-4-i+\alpha \cdot i +\gamma</math> | ||
+ | |||
+ | Since <math>f(1)</math> and <math>f(i)</math> are both real we get, | ||
+ | <cmath>\alpha+\gamma=-i</cmath> | ||
+ | <cmath>\alpha \cdot i+\gamma=i</cmath> | ||
+ | |||
+ | Solving, we get <math>\alpha=1-i</math>, <math>\gamma</math> can be anything, to minimize the value we set <math>\gamma=0</math>, so then the answer is <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. Thus, the answer is <math>\boxed{B}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2008|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:34, 30 December 2017
Problem 19
A function is defined by for all complex numbers , where and are complex numbers and . Suppose that and are both real. What is the smallest possible value of ?
Solution
We need only concern ourselves with the imaginary portions of and (both of which must be 0). These are:
Let and then we know and Therefore which reaches its minimum when by the Trivial Inequality. Thus, the answer is
[b]Solution 2:[/b]
Since and are both real we get,
Solving, we get , can be anything, to minimize the value we set , so then the answer is . Thus, the answer is
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |
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