Difference between revisions of "2002 AMC 12A Problems/Problem 24"
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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>. | 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>. | ||
+ | == Solution 3 == | ||
+ | |||
+ | Notice that r=0 or r=1 for this to be true. We know this because we are taking magnitude to the 2003rd power, and if the magnitude of a+bi is larger than 1, it will increase and if it is smaller than 1 it will decrease. However, the magnitude on the RHS is still r, so this is not possible. Again, only r=0 and r=1 satisfy. | ||
+ | |||
+ | Now if r=0, we must have (0,0) for (a,b). No exceptions. | ||
+ | |||
+ | However if r=1, we then have: | ||
+ | |||
+ | <math>cis(2002 \theta) = cis(-\theta)</math>. This has solution of <math>\theta = 0</math>. This would represent the number 1+0i, with conjugate 1-0i. This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by 2002 due to DeMoivre's Theorem and also we do <math>-\theta</math> because it is a reflection, angles therefore is negative. | ||
+ | |||
+ | We then write: | ||
+ | |||
+ | <math>cis(2002 \theta) = cis(360-\theta)</math> which has solution of <math>\theta = \frac{360}{2003}</math>. | ||
+ | |||
+ | We can also write: | ||
+ | |||
+ | <math>cis(2002 \theta) = cis(720-\theta)</math> which has solution <math>\theta = \frac{720}{2003}</math>. | ||
+ | |||
+ | We notice that it is simply headed upwards and the answer is of the form <math>\frac{720}{2003} n</math>, where n is some integer from 0 to infinity inclusive. | ||
+ | |||
+ | Well wait, it repeats itself n=2003, that is 360 which is also 0! Hence we only have n=0 to 2002 as original solutions, or 2003 solutions. | ||
+ | |||
+ | 1+2003 = <math>\boxed{2004}</math>. | ||
+ | |||
+ | Solution by Blackhawk 9-10-17 | ||
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:45, 10 September 2017
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
.
Solution 3
Notice that r=0 or r=1 for this to be true. We know this because we are taking magnitude to the 2003rd power, and if the magnitude of a+bi is larger than 1, it will increase and if it is smaller than 1 it will decrease. However, the magnitude on the RHS is still r, so this is not possible. Again, only r=0 and r=1 satisfy.
Now if r=0, we must have (0,0) for (a,b). No exceptions.
However if r=1, we then have:
. This has solution of
. This would represent the number 1+0i, with conjugate 1-0i. This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by 2002 due to DeMoivre's Theorem and also we do
because it is a reflection, angles therefore is negative.
We then write:
which has solution of
.
We can also write:
which has solution
.
We notice that it is simply headed upwards and the answer is of the form , where n is some integer from 0 to infinity inclusive.
Well wait, it repeats itself n=2003, that is 360 which is also 0! Hence we only have n=0 to 2002 as original solutions, or 2003 solutions.
1+2003 = .
Solution by Blackhawk 9-10-17
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.