Difference between revisions of "2008 AMC 12B Problems/Problem 19"
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| − | Let <math>p=\textrm{Im}(\gamma)</math> and <math>q=\textrm{Re}{\gamma},</math> then we know <math>\textrm{Im}(\alpha)=-p-1</math> and <math>\textrm{Re}(\alpha)=1-p.</math> Therefore < | + | Let <math>p=\textrm{Im}(\gamma)</math> and <math>q=\textrm{Re}{\gamma},</math> then we know <math>\textrm{Im}(\alpha)=-p-1</math> and <math>\textrm{Re}(\alpha)=1-p.</math> Therefore <math>|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},</math> which reaches its minimum <math>\sqrt 2</math> when <math>p=q=0.</math> So 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 12:42, 9 September 2016
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:
\begin{align*} \textrm{Im}(f(1)) & = i+i\textrm{Im}(\alpha)+i\textrm{Im}(\gamma) \\ \textrm{Im}(f(i)) & = -i+i\textrm{Re}(\alpha)+i\textrm{Im}(\gamma) \end{align*}
Let 
and 
then we know 
and 
Therefore $|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},$ (Error compiling LaTeX. Unknown error_msg) which reaches its minimum 
when 
So 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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