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"

(→‎See also)
(→‎See also)
Line 72: Line 72:
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 20:52, 3 July 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

Answer: (C) $\sqrt{3} - 1$

Lemma) if $f(z) = \frac{z + a}{z + b}$, then $f(f(f(f(\frac{-a(1+b)}{a+1})))) = \frac{a ( 1 + b)}{ a+b^2}$

$f(f(f(f(\frac{-a(1+b)}{a+1})))) = f(f(f(-a))) = f(f(0)) = f(\frac{a}{b}) = \frac{a + ab}{a + b^2}$

The tedious algebra is left to the reader. (it is not bad at all)

Well, let us consider the cases where each of those step is definite ($f(-b)$ is never evaluate).

So, we have $\frac{-a(1+b)}{a+1} \neq - b$,

$-a - ab \neq - ab - b$

$a \neq b$ --- (exception -> case 2)

$-a \neq -b$ --- (exception -> case 3)

$0 \neq -b$ --- (exception -> case 4)

$\frac{a}{b} \neq -b$

$a \neq -b ^2$ --- (exception -> case 5)

If it is not any of the above 5 cases, then

$\frac{-a(1+b)}{a+1} = \frac{a + ab}{a + b^2}$

if $a(1+b) \neq 0$ (--- exception -> case 6), then $-(a+b^2) = a+1$, $b^2 = -2a - 1$, $1 \le |b^2| \le 3$

Hence, it is possible maximum of $|b| = \sqrt{3}$ and minimum is 1.

2 possible combination of $(a,b)$ are $(1, 3i)$ and $(-1, 1)$. Verification is left upto the reader. Right now, (C) is the most possible answer out of those 5.

Case 2) $a = b$, then $|b| = 1$

Case 3) $|b|$ = 1, which is in the range.

Case 5) $b^2 = -a$, hence $|b| = 1$

Case 6) Since $a \neq 0$, $(1+b) = 0$, $|b| = 1$

Case 4) $b = 0$, this is quite an annoying special case. In this case, $f(z) = \frac{z+a}{z}$, $f(0)$ is not define.

In this case, $f(f(f(f(\frac{a}{a-1})))) = f(f(f(a)))= f(f(2)) = f( \frac{2+a}{2}) = \frac{2+3a}{2+a}$ and $f(f(f(f(\frac{a}{a-2})))) = f(f(f(2a)))= f(f(\frac{3}{2})) = f( \frac{3+2a}{3}) = \frac{3+5a}{3+2a}$

Hence, $(2+a)a = (a-1)(2+3a)$ and $(a)(3+2a) = (a-2)(3+5a)$. Once, you work out this system, you will get no solution with $|a| = 1$.

Thus, answer is (C).


SOLUTION 2:


After a bit of tedius algebra (that isn't too bad, but a little lengthy) we obtain $h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}$ where $P=(a+1)^2+a(b+1)^2$, $Q=a(b+1)(b^2+2a+1)$, $R=(b+1)(b^2+2a+1)$, and $S=a(b+1)^2+(a+b^2)^2$. In order for $h(z)=z$, we must have $R=0$, $Q=0$, and $P=S$. The first implies $b=-1$ or $b^2+2a+1=0$, the second implies $a=0$, $b=-1$, or $b^2+2a+1=0$, and the third 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 $|b^2+1|=|-2a|=2$ so that $2=|b^2+1|\leq |b^2|+1$ and hence $1\leq|b|^2\Rightarrow1\leq |b|$. On the other hand $2=|b^2+1|\geq|b^2|-1$ so that $|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}$. Thus $1\leq |b|\leq \sqrt{3}$. Hence in any case 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). Hence the answer is $\sqrt{3}-1$.\\ \

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=53845"