Difference between revisions of "2020 AMC 12B Problems/Problem 23"
m (→Solutions) |
Happysharks (talk | contribs) (→Solution) |
||
(9 intermediate revisions by 3 users not shown) | |||
Line 7: | Line 7: | ||
<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> | ||
− | == | + | == Solution == |
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. | ||
− | For <math>n=3</math>, WLOG, we can set <math>z_{1}=1</math>. Notice that now <math> | + | 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. |
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. | ||
Line 18: | Line 18: | ||
-Solution by Qqqwerw | -Solution by Qqqwerw | ||
+ | -Minor edit made by HappySharks | ||
− | == Video Solution | + | == Video Solution by On The Spot STEM == |
− | On The Spot STEM | + | https://www.youtube.com/watch?v=JOgSOni5HhM |
==See Also== | ==See Also== |
Latest revision as of 16:45, 28 January 2024
Problem
How many integers are there such that whenever are complex numbers such that
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
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.