Difference between revisions of "2002 AMC 12A Problems/Problem 24"
(→Solution 3) |
(→Solution) |
||
Line 15: | Line 15: | ||
</math> | </math> | ||
− | == Solution == | + | == Solution 1 == |
Let <math>s=\sqrt{a^2+b^2}</math> be the magnitude of <math>a+bi</math>. Then the magnitude of <math>(a+bi)^{2002}</math> is <math>s^{2002}</math>, while the magnitude of <math>a-bi</math> is <math>s</math>. We get that <math>s^{2002}=s</math>, hence either <math>s=0</math> or <math>s=1</math>. | Let <math>s=\sqrt{a^2+b^2}</math> be the magnitude of <math>a+bi</math>. Then the magnitude of <math>(a+bi)^{2002}</math> is <math>s^{2002}</math>, while the magnitude of <math>a-bi</math> is <math>s</math>. We get that <math>s^{2002}=s</math>, hence either <math>s=0</math> or <math>s=1</math>. | ||
Line 24: | Line 24: | ||
The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>. | The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>. | ||
− | |||
== Solution 2 == | == Solution 2 == |
Revision as of 23:59, 29 January 2019
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
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.