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 "2011 AMC 12A Problems/Problem 23"

m (→‎Solution)
(→‎Solution)
Line 34: Line 34:
 
<cmath>2=|b^2+1|\geq|b^2|-1</cmath>
 
<cmath>2=|b^2+1|\geq|b^2|-1</cmath>
 
so,  
 
so,  
−
<cmath>|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}</cmath>. Thus <math>1\leq |b|\leq \sqrt{3}</math>. Hence the maximum value for <math>|b|</math> is <math>\sqrt{3}</math> while the minimum is <math>1</math> (which can be achieved for instance when <math>|a|=1,|b|=\sqrt{3}</math> or <math>|a|=1,|b|=1</math> respectively). Therefore the answer is <math>\boxed{\textbf{(C)}\ 2\sqrt{6}}</math>.
+
<cmath>|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}</cmath>. Thus <math>1\leq |b|\leq \sqrt{3}</math>. Hence the maximum value for <math>|b|</math> is <math>\sqrt{3}</math> while the minimum is <math>1</math> (which can be achieved for instance when <math>|a|=1,|b|=\sqrt{3}</math> or <math>|a|=1,|b|=1</math> respectively). Therefore the answer is <math>\boxed{\textbf{(C)}\ \sqrt{3}-1}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 20:20, 23 November 2013

Problem

Let $f(z)= \frac{z+a}{z+b}$ and $g(z)=f(f(z))$, where $a$ and $b$ are complex numbers. Suppose that $\left| a \right| = 1$ and $g(g(z))=z$ for all $z$ for which $g(g(z))$ is defined. What is the difference between the largest and smallest possible values of $\left| b \right|$?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \sqrt{2}-1 \qquad \textbf{(C)}\ \sqrt{3}-1 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$

Solution

By algebraic manipulations, we obtain \[h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}\] where \[P=(a+1)^2+a(b+1)^2\] \[Q=a(b+1)(b^2+2a+1)\] \[R=(b+1)(b^2+2a+1)\] \[S=a(b+1)^2+(a+b^2)^2\] In order for $h(z)=z$, we must have $R=0$, $Q=0$, and $P=S$.

$R=0$ implies $b=-1$ or $b^2+2a+1=0$.

$Q=0$ implies $a=0$, $b=-1$, or $b^2+2a+1=0$.

$P=S$ implies $b=\pm1$ or $b^2+2a+1=0$.

Since $|a|=1\neq 0$, in order to satisfy all 3 conditions we must have either $b=1$ or $b^2+2a+1=0$. In the first case $|b|=1$.

For the latter case note that \[|b^2+1|=|-2a|=2\] \[2=|b^2+1|\leq |b^2|+1\] and hence, \[1\leq|b|^2\Rightarrow1\leq |b|\]. On the other hand, \[2=|b^2+1|\geq|b^2|-1\] so, \[|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}\]. Thus $1\leq |b|\leq \sqrt{3}$. Hence the maximum value for $|b|$ is $\sqrt{3}$ while the minimum is $1$ (which can be achieved for instance when $|a|=1,|b|=\sqrt{3}$ or $|a|=1,|b|=1$ respectively). Therefore the answer is $\boxed{\textbf{(C)}\ \sqrt{3}-1}$.

See also

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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=2011_AMC_12A_Problems/Problem_23&oldid=57898"