Difference between revisions of "2002 AMC 12A Problems/Problem 24"
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== Solution 2 == | == Solution 2 == | ||
− | As in the other solution, split the problem into when <math>s=0</math> and when <math>s=1</math>. When <math>s=1</math> and <math>a+bi=\cos\theta+\sin\theta</math>, <math>(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta)= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta) | + | As in the other solution, split the problem into when <math>s=0</math> and when <math>s=1</math>. When <math>s=1</math> and <math>a+bi=\cos\theta+i\sin\theta</math>, |
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+ | <math>(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta) </math> <math> =a-bi= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta)</math> | ||
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+ | so we must have <math>2002\theta=-\theta+2\pi k</math> and hence <math>\theta=\frac{2\pi k}{2003}</math>. Since <math>\theta</math> is restricted to <math>[0,2\pi)</math>, <math>k</math> can range from <math>0</math> to <math>2002</math> inclusive, which is <math>2002-0+1=2003</math> values. Thus the total is <math>1+2003 = \boxed{\textbf{(E)}\ 2004}</math>. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2002|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2002|ab=A|num-b=23|num-a=25}} |
Revision as of 12:04, 21 February 2012
Contents
Problem
Find the number of ordered pairs of real numbers such that .
Solution
Let be the magnitude of . Then the magnitude of is , while the magnitude of is . We get that , hence either or .
For we get a single solution .
Let's now assume that . Multiply both sides by . The left hand side becomes , the right hand side becomes . Hence the solutions for this case are precisely all the rd complex roots of unity, and there are of those.
The total number of solutions is therefore .
Solution 2
As in the other solution, split the problem into when and when . When and ,
so we must have and hence . Since is restricted to , can range from to inclusive, which is values. Thus the total is .
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |