2002 AMC 12A Problems/Problem 24
Find the number of ordered pairs of real numbers such that .
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 .
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 .
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
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