Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 1:)
 
(12 intermediate revisions by 7 users not shown)
Line 1: Line 1:
−
==Problem 19==
+
==Problem==
 
A function <math>f</math> is defined by <math>f(z) = (4 + i) z^2 + \alpha z + \gamma</math> for all complex numbers <math>z</math>, where <math>\alpha</math> and <math>\gamma</math> are complex numbers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alpha | + |\gamma |</math> ?
 
A function <math>f</math> is defined by <math>f(z) = (4 + i) z^2 + \alpha z + \gamma</math> for all complex numbers <math>z</math>, where <math>\alpha</math> and <math>\gamma</math> are complex numbers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alpha | + |\gamma |</math> ?
  
 
<math>\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad</math>
 
<math>\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad</math>
  
−
==Solution==
+
==Solution 1:==
 
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:
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\textrm{Im}(f(1)) & = i+i\textrm{Im}(\alpha)+i\textrm{Im}(\gamma) \\
+
\text{Im}(f(1)) & = i+i\text{Im}(\alpha)+i\text{Im}(\gamma) \\
−
\textrm{Im}(f(i)) & = -i+i\textrm{Re}(\alpha)+i\textrm{Im}(\gamma)
+
\text{Im}(f(i)) & = -i+i\text{Re}(\alpha)+i\text{Im}(\gamma)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
−
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>
+
Let <math>p=\text{Im}(\gamma)</math> and <math>q=\text{Re}{(\gamma)},</math> then we know <math>\text{Im}(\alpha)=-p-1</math> and <math>\text{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>
 +
 
 +
==Solution 2:==
 +
 
 +
<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>
 +
 
 +
By: Quaratinium
  
 
==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}}

Latest revision as of 18:28, 26 September 2023

Problem

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 1:

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

Let $p=\text{Im}(\gamma)$ and $q=\text{Re}{(\gamma)},$ then we know $\text{Im}(\alpha)=-p-1$ and $\text{Re}(\alpha)=1-p.$ 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 $\boxed B.$

Solution 2:

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

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

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 $\boxed{B}$

By: Quaratinium

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_19&oldid=198627"