2002 AMC 12A Problems/Problem 24
Problem
Find the number of ordered pairs of real numbers such that .
Solution 1
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
Let be the magnitude of . Notice that must be either or for this to be true, as shown in the above solutions. We know this because we are taking magnitude to the rd power, and if the magnitude of is larger than , it will increase and if it is smaller than it will decrease. However, the magnitude on the RHS is still , so this is not possible. Again, only and satisfy the equation.
Now if , then must be .
However if , we then have:
. This has solution of . This would represent the number , with conjugate . This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by through De Moivre'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 to infinity, inclusive.
Well wait, it repeats itself , that is which is also ! Hence we only have to as original solutions, or solutions.
.
Solution by Blackhawk 9-10-17
ed (with some edits) by PhunsukhWangdu 7/27/22
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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