2022 AMC 12B Problems/Problem 11
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Eisenstein Units)
- 4 Solution 3 (Quick and Easy)
- 5 Solution 4 (Third-order Homogeneous Linear Recurrence Relation)
- 6 Solution 5 (Polynomial + Recursion)
- 7 Solution 6 (SO FAST)
- 8 Video Solution by mop 2024
- 9 Video Solution (Under 2 min!)
- 10 Video Solution(1-16)
- 11 See Also
Problem
Let , where . What is ?
Solution 1
Converting both summands to exponential form, Notice that both are scaled copies of the third roots of unity. When we replace the summands with their exponential form, we get When we substitute , we get We can rewrite as , how does that help? Since any third root of unity must cube to .
~ zoomanTV
Solution 2 (Eisenstein Units)
The numbers and are both (along with ), denoted as and , respectively. They have the property that when they are cubed, they equal to . Thus, we can immediately solve: ~mathboy100
Solution 3 (Quick and Easy)
We begin by recognizing this form looks similar to the definition of cosine: We can convert our two terms into exponential form to find This simplifies nicely: Thus,
~Indiiiigo
Solution 4 (Third-order Homogeneous Linear Recurrence Relation)
Notice how this looks like the closed form of the Fibonacci sequence except different roots. This is motivation to turn this closed formula into a recurrence relation. The base of the exponents are the roots of the characteristic equation . So we have Every time is multiple of as is true when , ~lopkiloinm
Solution 5 (Polynomial + Recursion)
Let and . We know that and . Therefore, a and b are the roots of . By the factor theorem, and . Multiply the first equation by and the second equation by . This gives us and . Adding both equations together we get . This is the same as . Therefore, . Plugging in , and , we get . Therefore we know that if is a multiple of , then is . Since is a multiple of , our answers is . ~vpeddi18
Solution 6 (SO FAST)
Converting the two terms into rectangular form,
By DeMoivre's Theorem,
Note that if is even and if is odd, and that for all integers .
All arguments are even in the second equation for , so the two terms are equal to , and the two terms are equal to .
Therefore the answer is
-Benedict T (countmath1)
Video Solution by mop 2024
https://youtu.be/ezGvZgBLe8k&t=70s
~r00tsOfUnity
Video Solution (Under 2 min!)
https://youtu.be/ifPUOy_uctM ~ Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
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.