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Difference between revisions of "2011 AMC 12A Problems/Problem 23"

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== Solution ==
 
== Solution ==
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Answer: (C) <math>\sqrt{3} - 1</math>
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Lemma) if <math>f(z) = \frac{z + a}{z + b}</math>, then <math>f(f(f(f(\frac{-a(1+b)}{a+1})))) = \frac{a ( 1 + b)}{ a+b^2}</math>
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<math>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}</math>
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The tedious algebra is left to the reader. (it is not bad at all)
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 +
Well, let us consider the cases where each of those step is definite (<math>f(-b)</math> is never evaluate).
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So, we have <math> \frac{-a(1+b)}{a+1} \neq - b</math>,
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<math> -a - ab \neq - ab - b</math>
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<math> a \neq b</math> --- (exception -> case 2)
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<math>-a \neq -b</math> --- (exception -> case 3)
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<math>0 \neq -b</math> --- (exception -> case 4)
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<math>\frac{a}{b} \neq -b</math>
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<math>a \neq -b ^2</math> --- (exception -> case 5)
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If it is not any of the above 5 cases, then
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<math> \frac{-a(1+b)}{a+1} = \frac{a + ab}{a + b^2}</math>
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if <math>a(1+b) \neq 0</math> (--- exception -> case 6), then <math>-(a+b^2) = a+1</math>, <math>b^2 = -2a - 1</math>, <math>1 \le |b^2| \le 3</math>
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Hence, it is possible maximum of <math>|b| = \sqrt{3}</math> and minimum is 1.
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2 possible combination of <math>(a,b)</math> are <math>(1, 3i)</math> and <math>(-1, 1)</math>. Verification is left upto the reader. Right now, (C) is the most possible answer out of those 5.
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Case 2) <math>a = b</math>, then <math>|b| = 1</math>
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Case 3) <math>|b|</math> = 1, which is in the range.
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Case 5) <math>b^2 = -a</math>, hence <math>|b| = 1</math>
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Case 6) Since <math>a \neq 0</math>, <math>(1+b) = 0</math>, <math>|b| = 1</math>
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Case 4) <math>b = 0</math>, this is quite an annoying special case. In this case, <math>f(z) = \frac{z+a}{z}</math>, <math>f(0)</math> is not define.
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In this case, <math>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}</math> and <math>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}</math>
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Hence, <math>(2+a)a = (a-1)(2+3a)</math> and <math>(a)(3+2a) = (a-2)(3+5a)</math>. Once, you work out this system, you will get no solution with <math>|a| = 1</math>.
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Thus, answer is (C).
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=22|num-a=24|ab=A}}
 
{{AMC12 box|year=2011|num-b=22|num-a=24|ab=A}}

Revision as of 23:26, 14 February 2011

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).

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
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