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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>
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[b]Solution 2:[/b]
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 +
<math>f(1)=4+i+\alpha+\gamma</math>
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<math>f(i)=-4-i+\alpha \cdot i +\gamma</math>
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 +
Since <math>f(1)</math> and <math>f(i)</math> are both real we get,
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<cmath>\alpha+\gamma=-i</cmath>
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<cmath>\alpha \cdot i+\gamma=i</cmath>
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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 00:34, 30 December 2017

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*} \Im(f(1)) & = i+i\Im(\alpha)+i\Im(\gamma) \\ \Im(f(i)) & = -i+i\Re(\alpha)+i\Im(\gamma) \end{align*}

Let $p=\Im(\gamma)$ and $q=\Re{(\gamma)},$ then we know $\Im(\alpha)=-p-1$ and $\Re(\alpha)=1-p.$ 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 $p=q=0$ by the Trivial Inequality. Thus, the answer is $\boxed B.$

[b]Solution 2:[/b]

$f(1)=4+i+\alpha+\gamma$ $f(i)=-4-i+\alpha \cdot i +\gamma$

Since $f(1)$ and $f(i)$ are both real we get, \[\alpha+\gamma=-i\] \[\alpha \cdot i+\gamma=i\]

Solving, we get $\alpha=1-i$, $\gamma$ can be anything, to minimize the value we set $\gamma=0$, so then the answer is $\sqrt{1^2+1^2}=\sqrt{2}$. Thus, 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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