Difference between revisions of "2011 AMC 12A Problems/Problem 23"
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== Solution == | == Solution == | ||
+ | Answer: (C) <math>\sqrt{3} - 1</math> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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 (<math>f(-b)</math> is never evaluate). | ||
+ | |||
+ | So, we have <math> \frac{-a(1+b)}{a+1} \neq - b</math>, | ||
+ | |||
+ | <math> -a - ab \neq - ab - b</math> | ||
+ | |||
+ | <math> a \neq b</math> --- (exception -> case 2) | ||
+ | |||
+ | <math>-a \neq -b</math> --- (exception -> case 3) | ||
+ | |||
+ | <math>0 \neq -b</math> --- (exception -> case 4) | ||
+ | |||
+ | <math>\frac{a}{b} \neq -b</math> | ||
+ | |||
+ | <math>a \neq -b ^2</math> --- (exception -> case 5) | ||
+ | |||
+ | If it is not any of the above 5 cases, then | ||
+ | |||
+ | <math> \frac{-a(1+b)}{a+1} = \frac{a + ab}{a + b^2}</math> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Hence, it is possible maximum of <math>|b| = \sqrt{3}</math> and minimum is 1. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Case 2) <math>a = b</math>, then <math>|b| = 1</math> | ||
+ | |||
+ | Case 3) <math>|b|</math> = 1, which is in the range. | ||
+ | |||
+ | Case 5) <math>b^2 = -a</math>, hence <math>|b| = 1</math> | ||
+ | |||
+ | Case 6) Since <math>a \neq 0</math>, <math>(1+b) = 0</math>, <math>|b| = 1</math> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Thus, answer is (C). | ||
+ | |||
== 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 00:26, 15 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).
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 |