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

Latest revision as of 15:36, 10 June 2025

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 1 (plug and chug)

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 $b=\pm1$ 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 $\boxed{\textbf{(C)}\ \sqrt{3}-1}$ .

Shortcut

We only need $Q$ in $f^4(z)=g^2(z)=\frac{Pz+\textcolor{red}{Q}}{Rz+S}$ .

Set $Q=0$ : $a(b+1)\left(b^2+2a+1\right)=0$ . Since $|a|=1$ , either $b+1=0$ or $b^2+2a+1=0$ .

$b+1=0\rightarrow b=-1$ so $|b|=1$ .

$b^2+2a+1=0\rightarrow b^2=-1-2a$ . This is a circle in the complex plane centered at $(-1,0)$ with radius $2$ since $|a|=1$ . The maximum distance from the origin is $3$ at $(-3,0)$ and similarly the minimum distance is $1$ at $(1,0)$ . So $1\le |b^2|\le 3\rightarrow 1\le |b|\le \sqrt{3}$ .

Both solutions give the same lower bound, $1$ . So the range is $\sqrt{3}-1=\boxed{\textbf{(C) }\sqrt{3}-1}$ .

Solution 2

note: the "begin{align*}" environment is weird on aops, so if the two things that use it below are duplicated in front of each other, just reload the page or something.

$\textbf{Lemma.}$ Let $h(x)=\frac{ax+b}{cx+d}$ and $h\left (h(x)\right )=x$ for all such $x$ where $h(h(x))$ is defined. Then either $d=-a$ or $h(x)=x$ .

This lemma is one that is typically known by people who could make it this far into this AMC 12 anyway (USAMO qualifiers likely).

$\textit{Proof.}$ We plug in $h\left (h(x)\right )$ , and equating it to $x$ , we see \begin{align*} h\left (h(x)\right ) &=h\left (\frac{ax+b}{cx+d}\right ) \\ &= \frac{a\cdot \frac{ax+b}{cx+d}+b}{c\cdot \frac{ax+b}{cx+d}+d} \\ &= \frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)} \\ &= \frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)}=x. \end{align*} Then if we clear the denominator, it rearranges into $(a^2+bc)x+(ab+bd)=(ac+cd)x^2+(bc+d^2)x\implies (ac+cd)x^2+(d^2-a^2)x-(ab+bd)=0.$ Since this equation is true for all $x$ (since now the fractions are gone), this implies that the coefficients of this quadratic are all equal to $0$ . Focus on the condition $d^2-a^2=0$ .

$\textbf{Case 1:}$ $d+a=0$

Then all the other coefficients automatically become zero, so $d+a=0\implies d=-a$ is a sufficient condition.

$\textbf{Case 2:}$ $d-a=0$ (and $d+a\neq 0$ )

Then in $ab+bd=0$ , we divide both sides by $a+d$ to see that $b=0$ . Similarly, in $ac+cd=0$ , we divide to also see that $c=0$ . Then since $d=a$ , $h(x)=\frac{ax+b}{cx+d}=\frac{ax+0}{0x+a}=x.$ Therefore, in this case, it is required that $h(x)=x$ , which now works. This proves our desired claim. $\qquad \qquad \square$

Note that we are asking when $g(g(z))=z$ . So either $g(z)=z$ , or in the expansion of $g(z)=f(f(z))$ , the two corresponding coefficients are negations of each other.

$\textbf{Case 1:}$ $g(z)=z$

Then $f(f(z))=z$ , so either $f(z)=z$ or the two coefficients are negative of each other. Clearly $f(z)\neq z$ , since the numerator and denominator are both degree $1$ polynomials. So we would require the coefficients, namely $1$ and $b$ , to be negatives of each other. Then $b=-1$ is a possibility here, giving $|b|=1$ .

$\textbf{Case 2:}$ The two coefficients are negative of each other.

It can be found that \begin{align*} f(f(x))&=f\left (\frac{z+a}{z+b}\right ) \\ &=\frac{\frac{z+a}{z+b}+a}{\frac{z+a}{z+b}+b} \\ &=\frac{z+a+a(z+b)}{z+a+b(z+b)} \\ &= \frac{(a+1)z+(ab+a)}{(b+1)z+(b^2+a)}.\end{align*} By our lemma, we would require $a+1=-(b^2+a)$ , which rearranges into $b^2=-1-2a$ . But we are given that $|a|=1$ . The triangle inequality for complex numbers states that $|a|+|b|\geq |a+b|$ , where equality happens when the arguments of $a$ and $b$ are equal. Here, we can see that $|-1-2a|\leq |-1|+|-2a|=1+2=3,$ but we can also find that $|-1-2a|+1=|-1-2a|+|1|\geq |-2a|=2\implies |-1-2a|\geq 1.$ Therefore, because $|b^2|=|-1-2a|$ , then $1\leq |b|^2\leq 3$ , giving us $1\leq |b|\leq \sqrt{3}$ .

When we combine the two cases, we find that the possible range of $|b|$ is $|b|\in [1,\sqrt{3}]$ , so the answer is $\boxed{\textbf{(C) }\sqrt{3}-1}$ .

~ethanzhang1001

Video Solution

https://youtu.be/FU18x_LsTeQ

~MathProblemSolvingSkills.com

Note

This problem is kinda similar to 2002 AIME I Problems/Problem 12

@@ Line 4: / Line 4: @@
 <math>
 \textbf{(A)}\ 0 \qquad
-\textbf{(B)}\ \sqrt{2}1 \qquad
+\textbf{(B)}\ \sqrt{2}-1 \qquad
 \textbf{(C)}\ \sqrt{3}-1 \qquad
 \textbf{(D)}\ 1 \qquad
 \textbf{(E)}\ 2 </math>
-== Solution ==
+== Solution 1 (plug and chug)==
+By algebraic manipulations, we obtain
+<cmath>h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}</cmath> where
+<cmath>P=(a+1)^2+a(b+1)^2</cmath>
+<cmath>Q=a(b+1)(b^2+2a+1)</cmath>
+<cmath>R=(b+1)(b^2+2a+1)</cmath>
+<cmath>S=a(b+1)^2+(a+b^2)^2</cmath>
+In order for <math>h(z)=z</math>, we must have <math>R=0</math>, <math>Q=0</math>, and <math>P=S</math>.
+<math>R=0</math> implies <math>b=-1</math> or <math>b^2+2a+1=0</math>.
+<math>Q=0</math> implies <math>a=0</math>, <math>b=-1</math>, or <math>b^2+2a+1=0</math>.
+<math>P=S</math> implies <math>b=\pm1</math> or <math>b^2+2a+1=0</math>.
+Since <math>|a|=1\neq 0</math>, in order to satisfy all 3 conditions we must have either <math>b=\pm1</math> or <math>b^2+2a+1=0</math>. In the first case <math>|b|=1</math>.
+For the latter case note that
+<cmath>|b^2+1|=|-2a|=2</cmath>
+<cmath>2=|b^2+1|\leq |b^2|+1</cmath>
+and hence,
+<cmath>1\leq|b|^2\Rightarrow1\leq |b|</cmath>.
+On the other hand,
+<cmath>2=|b^2+1|\geq|b^2|-1</cmath>
+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)}\ \sqrt{3}-1}</math>.
+==Shortcut==
+We only need <math>Q</math> in <math>f^4(z)=g^2(z)=\frac{Pz+\textcolor{red}{Q}}{Rz+S}</math>.
+Set <math>Q=0</math>: <math>a(b+1)\left(b^2+2a+1\right)=0</math>. Since <math>|a|=1</math>, either <math>b+1=0</math> or <math>b^2+2a+1=0</math>.
+<math>b+1=0\rightarrow b=-1</math> so <math>|b|=1</math>.
+<math>b^2+2a+1=0\rightarrow b^2=-1-2a</math>. This is a circle in the complex plane centered at <math>(-1,0)</math> with radius <math>2</math> since <math>|a|=1</math>. The maximum distance from the origin is <math>3</math> at <math>(-3,0)</math> and similarly the minimum distance is <math>1</math> at <math>(1,0)</math>. So <math>1\le |b^2|\le 3\rightarrow 1\le |b|\le \sqrt{3}</math>.
+Both solutions give the same lower bound, <math>1</math>. So the range is <math>\sqrt{3}-1=\boxed{\textbf{(C) }\sqrt{3}-1}</math>.
+== Solution 2 ==
+note: the "begin{align*}" environment is weird on aops, so if the two things that use it below are duplicated in front of each other, just reload the page or something.
+<math>\textbf{Lemma.}</math> Let <math>h(x)=\frac{ax+b}{cx+d}</math> and <math>h\left (h(x)\right )=x</math> for all such <math>x</math> where <math>h(h(x))</math> is defined. Then either <math>d=-a</math> or <math>h(x)=x</math>.
+This lemma is one that is typically known by people who could make it this far into this AMC 12 anyway (USAMO qualifiers likely).
+<math>\textit{Proof.}</math> We plug in <math>h\left (h(x)\right )</math>, and equating it to <math>x</math>, we see
+\begin{align*}
+h\left (h(x)\right ) &=h\left (\frac{ax+b}{cx+d}\right ) \\ &= \frac{a\cdot \frac{ax+b}{cx+d}+b}{c\cdot \frac{ax+b}{cx+d}+d} \\ &= \frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)} \\ &= \frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)}=x.
+\end{align*}
+Then if we clear the denominator, it rearranges into <cmath>(a^2+bc)x+(ab+bd)=(ac+cd)x^2+(bc+d^2)x\implies (ac+cd)x^2+(d^2-a^2)x-(ab+bd)=0.</cmath> Since this equation is true for all <math>x</math> (since now the fractions are gone), this implies that the coefficients of this quadratic are all equal to <math>0</math>. Focus on the condition <math>d^2-a^2=0</math>.
+<math>\textbf{Case 1:}</math> <math>d+a=0</math>
+Then all the other coefficients automatically become zero, so <math>d+a=0\implies d=-a</math> is a sufficient condition.
+<math>\textbf{Case 2:}</math> <math>d-a=0</math> (and <math>d+a\neq 0</math>)
+Then in <math>ab+bd=0</math>, we divide both sides by <math>a+d</math> to see that <math>b=0</math>. Similarly, in <math>ac+cd=0</math>, we divide to also see that <math>c=0</math>. Then since <math>d=a</math>, <cmath>h(x)=\frac{ax+b}{cx+d}=\frac{ax+0}{0x+a}=x.</cmath> Therefore, in this case, it is required that <math>h(x)=x</math>, which now works. This proves our desired claim. <math>\qquad \qquad \square</math>
+Note that we are asking when <math>g(g(z))=z</math>. So either <math>g(z)=z</math>, or in the expansion of <math>g(z)=f(f(z))</math>, the two corresponding coefficients are negations of each other.
+<math>\textbf{Case 1:}</math> <math>g(z)=z</math>
+Then <math>f(f(z))=z</math>, so either <math>f(z)=z</math> or the two coefficients are negative of each other. Clearly <math>f(z)\neq z</math>, since the numerator and denominator are both degree <math>1</math> polynomials. So we would require the coefficients, namely <math>1</math> and <math>b</math>, to be negatives of each other. Then <math>b=-1</math> is a possibility here, giving <math>|b|=1</math>.
+<math>\textbf{Case 2:}</math> The two coefficients are negative of each other.
+It can be found that \begin{align*}
+f(f(x))&=f\left (\frac{z+a}{z+b}\right ) \\ &=\frac{\frac{z+a}{z+b}+a}{\frac{z+a}{z+b}+b} \\ &=\frac{z+a+a(z+b)}{z+a+b(z+b)} \\ &= \frac{(a+1)z+(ab+a)}{(b+1)z+(b^2+a)}.\end{align*} By our lemma, we would require <math>a+1=-(b^2+a)</math>, which rearranges into <math>b^2=-1-2a</math>. But we are given that <math>|a|=1</math>. The triangle inequality for complex numbers states that <math>|a|+|b|\geq |a+b|</math>, where equality happens when the arguments of <math>a</math> and <math>b</math> are equal. Here, we can see that <cmath>|-1-2a|\leq |-1|+|-2a|=1+2=3,</cmath> but we can also find that <cmath>|-1-2a|+1=|-1-2a|+|1|\geq |-2a|=2\implies |-1-2a|\geq 1.</cmath> Therefore, because <math>|b^2|=|-1-2a|</math>, then <math>1\leq |b|^2\leq 3</math>, giving us <math>1\leq |b|\leq \sqrt{3}</math>.
+When we combine the two cases, we find that the possible range of <math>|b|</math> is <math>|b|\in [1,\sqrt{3}]</math>, so the answer is <math>\boxed{\textbf{(C) }\sqrt{3}-1}</math>.
+~ethanzhang1001
+== Video Solution ==
+https://youtu.be/FU18x_LsTeQ
+~MathProblemSolvingSkills.com
+==Note==
+This problem is kinda similar to [[2002 AIME I Problems/Problem 12]]
 == See also ==
 {{AMC12 box|year=2011|num-b=22|num-a=24|ab=A}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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

Page

Toolbox

Search