Difference between revisions of "2022 AMC 12B Problems/Problem 11"
(Created page with "==Problem == Let <math> f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n </math>, where <math>i = \sqrt{-1}</math>. What is <math>f(20...") |
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~ <math>\color{magenta} zoomanTV</math> | ~ <math>\color{magenta} zoomanTV</math> | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2022|ab=B|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:44, 17 November 2022
Problem
Let 
, where 
. What is 
?
Solution
Converting both summands to exponential form, ![\[-1 + i\sqrt{3} = 2e^{\frac{2\pi i}{3}}\]](http://latex.artofproblemsolving.com/a/e/8/ae8dc98d141e6f5ec1c2b4262b67c493f3b367ef.png)
![\[-1 - i\sqrt{3} = 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}\]](http://latex.artofproblemsolving.com/4/1/8/41861437526de79f11bf244993fa6d83b8891183.png)
Notice that both are scaled copies of the third roots of unity.
When we replace the summands with their exponential form, we get
![\[f(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n\]](http://latex.artofproblemsolving.com/0/1/5/015ae4eb6e2858fec3e03a6db0efc0d2859ac86d.png)
When we substitute 
, we get
![\[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{2022} + \left(e^{\frac{4\pi i}{3}}\right)^{2022}\]](http://latex.artofproblemsolving.com/b/e/1/be1131d2f981a4ed1e479b928dfdf22ecbaa5b53.png)
We can rewrite 
as 
, how does that help?
![\[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{3 \cdot 674} + \left(e^{\frac{4\pi i}{3}}\right)^{3 \cdot 674} =\]](http://latex.artofproblemsolving.com/7/9/6/79684ce25c7c0c5569d5b1e3a83b1037922f81d9.png)
![\[\left(\left(e^{\frac{2\pi i}{3}}\right)^{3}\right)^{674} + \left(\left(e^{\frac{4\pi i}{3}}\right)^{3}\right)^{674} =\]](http://latex.artofproblemsolving.com/d/6/d/d6dd534625a9ff88a686bb379eec1a610899d01f.png)
![\[1^{674} + 1^{674} = \boxed{\textbf{(E)} \ 2}\]](http://latex.artofproblemsolving.com/c/1/9/c195dbf9de51d64f9f519083d170cca4023954c7.png)
Since any third root of unity must cube to 
.
~ 
See Also
| 2022 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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. 
