Difference between revisions of "2011 AMC 12A Problems/Problem 23"
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+ | After a bit of tedius algebra (that isn't too bad, but a little lengthy) we obtain | ||
+ | <math>h(z)=g(g(z))=f(f(f(f(z))))=\frac{Pz+Q}{Rz+S}</math> where <math>P=(a+1)^2+a(b+1)^2</math>, <math>Q=a(b+1)(b^2+2a+1)</math>, <math>R=(b+1)(b^2+2a+1)</math>, and <math>S=a(b+1)^2+(a+b^2)^2</math>. In order for <math>h(z)=z</math>, we must have <math>R=0</math>, <math>Q=0</math>, and <math>P=S</math>. The first implies <math>b=-1</math> or <math>b^2+2a+1=0</math>, the second implies <math>a=0</math>, <math>b=-1</math>, or <math>b^2+2a+1=0</math>, and the third 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=1</math> or <math>b^2+2a+1=0</math>. In the first case <math>|b|=1</math>. For the latter case note that <math>|b^2+1|=|-2a|=2</math> so that <math>2=|b^2+1|\leq |b^2|+1</math> and hence <math>1\leq|b|^2\Rightarrow1\leq |b|</math>. On the other hand <math>2=|b^2+1|\geq|b^2|-1</math> so that <math>|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}</math>. Thus <math>1\leq |b|\leq \sqrt{3}</math>. Hence in any case 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). Hence the answer is <math>\sqrt{3}-1</math>.\\ | ||
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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 05:00, 20 February 2011
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
Answer: (C)
Lemma) if , then
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 ( is never evaluate).
So, we have ,
--- (exception -> case 2)
--- (exception -> case 3)
--- (exception -> case 4)
--- (exception -> case 5)
If it is not any of the above 5 cases, then
if (--- exception -> case 6), then
,
,
Hence, it is possible maximum of and minimum is 1.
2 possible combination of are
and
. Verification is left upto the reader. Right now, (C) is the most possible answer out of those 5.
Case 2) , then
Case 3) = 1, which is in the range.
Case 5) , hence
Case 6) Since ,
,
Case 4) , this is quite an annoying special case. In this case,
,
is not define.
In this case, and
Hence, and
. Once, you work out this system, you will get no solution with
.
Thus, answer is (C).
SOLUTION 2:
After a bit of tedius algebra (that isn't too bad, but a little lengthy) we obtain
where
,
,
, and
. In order for
, we must have
,
, and
. The first implies
or
, the second implies
,
, or
, and the third 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
so that
and hence
. On the other hand
so that
. Thus
. Hence in any case the maximum value for
is
while the minimum is
(which can be achieved for instance when
or
respectively). Hence the answer is
.\\
\
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 |