Difference between revisions of "2011 AMC 12A Problems/Problem 23"
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== Solution == | == Solution == | ||
+ | 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>. | ||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/FU18x_LsTeQ | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | ==Note== | ||
+ | This problem is kinda similar to [[2002 AIME I Problems/Problem 12]] | ||
+ | |||
== 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}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:49, 5 November 2023
Problem
Let and
, where
and
are complex numbers. Suppose that
and
for all
for which
is defined. What is the difference between the largest and smallest possible values of
?
Solution
By algebraic manipulations, we obtain
where
In order for
, we must have
,
, and
.
implies
or
.
implies
,
, or
.
implies
or
.
Since , in order to satisfy all 3 conditions we must have either
or
. In the first case
.
For the latter case note that
and hence,
.
On the other hand,
so,
. Thus
. Hence the maximum value for
is
while the minimum is
(which can be achieved for instance when
or
respectively). Therefore the answer is
.
Shortcut
We only need in
.
Set :
. Since
, either
or
.
so
.
. This is a circle in the complex plane centered at
with radius
since
. The maximum distance from the origin is
at
and similarly the minimum distance is
at
. So
.
Both solutions give the same lower bound, . So the range is
.
Video Solution
~MathProblemSolvingSkills.com
Note
This problem is kinda similar to 2002 AIME I Problems/Problem 12
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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