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| | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math> | | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5</math> |
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| − | [[2020 AMC 12B Problems/Problem 23|Solution]]
| + | == Solution == |
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| − | == Solutions ==
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| − | === Solution 1 ===
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| | For <math>n=2</math>, we see that if <math>z_{1}+z_{2}=0</math>, then <math>z_{1}=-z_{2}</math>, so they are evenly spaced along the unit circle. | | For <math>n=2</math>, we see that if <math>z_{1}+z_{2}=0</math>, then <math>z_{1}=-z_{2}</math>, so they are evenly spaced along the unit circle. |
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| − | For <math>n=3</math>, WLOG, we can set <math>z_{1}=1</math>. Notice that now <math>\Re(z_{2}+z_{3})=-1</math> and <math>\Im\{z_{2}\}=-\Im\{z_{3}\}</math>. This forces <math>z_{2}</math> and <math>z_{3}</math> to be equal to <math>e^{i\frac{2\pi}{3}}</math> and <math>e^{-i\frac{2\pi}{3}}</math>, meaning that all three are equally spaced along the unit circle. | + | For <math>n=3</math>, WLOG, we can set <math>z_{1}=1</math>. Notice that now Re<math>(z_{2}+z_{3})=-1</math> and Im<math>\{z_{2}\}</math> = <math>-</math>Im<math>\{z_{3}\}</math>. This forces <math>z_{2}</math> and <math>z_{3}</math> to be equal to <math>e^{i\frac{2\pi}{3}}</math> and <math>e^{-i\frac{2\pi}{3}}</math>, meaning that all three are equally spaced along the unit circle. |
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| | We can now show that we can construct complex numbers when <math>n\geq 4</math> that do not satisfy the conditions in the problem. | | We can now show that we can construct complex numbers when <math>n\geq 4</math> that do not satisfy the conditions in the problem. |
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| | -Solution by Qqqwerw | | -Solution by Qqqwerw |
| | + | -Minor edit made by HappySharks |
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| − | === Solution 2 === | + | == Video Solution by On The Spot STEM == |
| − | (for people not used to notation)
| + | https://www.youtube.com/watch?v=JOgSOni5HhM |
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| − | Note: I personally don't think the question was very well worded; it should explicitly state that if the first two conditions/equations are satisfied, then the final condition MUST be satisfied as well. The question can be misinterpreted as "how many different sizes of a list of complex numbers exist such that at least one satisfies the conditions" (in which case the answer would be infinitely many).
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| − | The problem asks for the number of list lengths that satisfy the three conditions given; i.e. how many different sizes can the list of complex numbers be such that the elements in the list always satisfy the conditions. This means that for a list of length 3, for example, all possible lists have to satisfy the conditions. If even one variation of a list of length 3 does not satisfy the conditions, then 3 is not an acceptable value of n.
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| − | A complex number can be written in the form <math>a + bi</math> where <math>a</math> represents the real part and <math>b</math> represents the coefficient of the imaginary part. If a complex number is plotted on the complex plane, then <math>a</math> becomes the x-coordinate, and <math>b</math> becomes the y-coordinate.
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| − | The first condition states that the absolute value of each complex number is equal to 1; i.e. the point (a, b) is a distance of one from the origin. If all of the complex numbers are a distance of 1 from the origin, they must all be on the unit circle. This condition simply allows us to be able to check condition 3, think of it as a sort of "given".
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| − | The second condition states that the sum of all the complex numbers is equal to 0. Therefore, the sum of all the x-coordinates and the y-coordinates will yield 0. However, we know that real numbers and complex numbers cannot be simplified together (assuming that <math>a</math> and <math>b</math> are real numbers). This means that the real parts of the complex numbers, i.e. the x-coordinates, must sum to 0 by themselves. The same is true for the imaginary parts (the y-coordinates).
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| − | The third condition states that the complex numbers must be evenly spaced along the unit circle. This condition can be used to determine acceptable values of x.
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| − | Start with <math>n = 2</math>. If we have 2 complex numbers, in order to satisfy the second condition, the x coordinates must be opposites. The same is true for the y-coordinates. After drawing (or imagining) we find that the two points are directly across from one another on the unit circle. Upon further thought, we find that whenever we have an even number of points, if they are paired up in this way, they will always satisfy the first and second conditions. However, it is possible to arrange these pairs of points (whenever we have at least 2 pairs) such as the points are not evenly spaced out. Therefore, any even value of <math>n</math> greater than 2 will not work (since we proved that not all variations satisfy the last condition).
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| − | Now we consider <math>n = 3</math>. It is much easier to consider 2 numbers instead of 3, so we will assume that <math>z_{1}=1</math>, or <math>(1, 0)</math> (we can do this because a circle has rotational symmetry). Since the y-coordinate of the first number is 0, the y-coordinates of the other 2 numbers must be opposites. Since the x coordinate of the first number is 1, the x-coordinates of the other 2 numbers must sum to -1. Using this knowledge, we find that the coordinates of the other 2 numbers are
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| − | <math>-\frac{1}{2}</math> <math>+</math> <math>\sqrt{3}</math> and <math>-\frac{1}{2}</math> <math>-</math> <math>\sqrt{3}</math>
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| − | After calculating angle measurements, we see that the 3 points are, indeed, equally spaced apart. Therefore 3 is an acceptable value of <math>n</math> (all variations of <math>n = 3</math> are simply rotations of this specific case, and therefore satisfy all of the conditions).
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| − | Lastly we consider the odd numbers. If we start with the same arrangement of points as above, we see that adding pairs of opposite points will continue to satisfy conditions 1 and 2, but NOT ALWAYS condition 3. Therefore, any odd number greater than 3 is not
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| − | an acceptable value of <math>n</math>.
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| − | So we have <math>n = 2</math> and <math>n = 3</math> for a total of 2 possibilities <math>\rightarrow</math> <math>\boxed{B}</math>
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| − | == Video Solution ==
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| − | On The Spot STEM: https://www.youtube.com/watch?v=JOgSOni5HhM
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| | ==See Also== | | ==See Also== |
Problem
How many integers 
are there such that whenever 
are complex numbers such that
![\[|z_1| = |z_2| = ... = |z_n| = 1 \text{ and } z_1 + z_2 + ... + z_n = 0,\]](//latex.artofproblemsolving.com/3/b/5/3b5ec1e5f9204ea20f9438cf2c48b0c9464f27ac.png)
then the numbers are equally spaced on the unit circle in the complex plane?
Solution
For 
, we see that if 
, then 
, so they are evenly spaced along the unit circle.
For 
, WLOG, we can set 
. Notice that now Re
and Im
= 
Im
. This forces 
and 
to be equal to 
and 
, meaning that all three are equally spaced along the unit circle.
We can now show that we can construct complex numbers when 
that do not satisfy the conditions in the problem.
Suppose that the condition in the problem holds for some 
. We can now add two points 
and 
anywhere on the unit circle such that 
, which will break the condition. Now that we have shown that and works, by this construction, any does not work, making the answer 
.
-Solution by Qqqwerw
-Minor edit made by HappySharks
Video Solution by On The Spot STEM
https://www.youtube.com/watch?v=JOgSOni5HhM
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
are there such that whenever 
are complex numbers such that
![\[|z_1| = |z_2| = ... = |z_n| = 1 \text{ and } z_1 + z_2 + ... + z_n = 0,\]](http://latex.artofproblemsolving.com/3/b/5/3b5ec1e5f9204ea20f9438cf2c48b0c9464f27ac.png)
then the numbers 
, we see that if 
, then 
, so they are evenly spaced along the unit circle.
, WLOG, we can set 
. Notice that now Re
and Im
= 
Im
. This forces 
and 
to be equal to 
and 
, meaning that all three are equally spaced along the unit circle.
that do not satisfy the conditions in the problem.
. We can now add two points 
and 
anywhere on the unit circle such that 
, which will break the condition. Now that we have shown that 
.
