2011 AMC 12A Problems/Problem 23
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 .\\
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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 |