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Difference between revisions of "2008 AMC 12B Problems/Problem 19"

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We need only concern ourselves with the imaginary portions of <math>f(1)</math> and <math>f(i)</math> (both of which must be 0). These are:
 
We need only concern ourselves with the imaginary portions of <math>f(1)</math> and <math>f(i)</math> (both of which must be 0). These are:
  
−
\begin{align*}
+
<cmath>\begin{align*}
 
\textrm{Im}(f(1)) & = i+i\textrm{Im}(\alpha)+i\textrm{Im}(\gamma) \\
 
\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)
 
\textrm{Im}(f(i)) & = -i+i\textrm{Re}(\alpha)+i\textrm{Im}(\gamma)
−
\end{align*}
+
\end{align*}</cmath>
  
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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 <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>
+
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^2}=\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 12:44, 9 September 2016

Problem 19

A function $f$ is defined by $f(z) = (4 + i) z^2 + \alpha z + \gamma$ for all complex numbers $z$, where $\alpha$ and $\gamma$ are complex numbers and $i^2 = - 1$. Suppose that $f(1)$ and $f(i)$ are both real. What is the smallest possible value of $| \alpha | + |\gamma |$ ?

$\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad$

Solution

We need only concern ourselves with the imaginary portions of $f(1)$ and $f(i)$ (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 $p=\textrm{Im}(\gamma)$ and $q=\textrm{Re}{\gamma},$ then we know $\textrm{Im}(\alpha)=-p-1$ and $\textrm{Re}(\alpha)=1-p.$ Therefore $|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},$ which reaches its minimum $\sqrt 2$ when $p=q=0.$ So the answer is $\boxed B.$

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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