Difference between revisions of "2022 AMC 12B Problems/Problem 11"

Revision as of 21:17, 27 October 2023

Problem

Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$ ?

$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$

Solution 1

Converting both summands to exponential form, $\begin{align*} -1 + i\sqrt{3} &= 2e^{\frac{2\pi i}{3}}, \\ -1 - i\sqrt{3} &= 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}. \end{align*}$ Notice that both are scaled copies of the third roots of unity. When we replace the summands with their exponential form, we get $f(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n.$ When we substitute $n = 2022$ , we get $f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{2022} + \left(e^{\frac{4\pi i}{3}}\right)^{2022}.$ We can rewrite $2022$ as $3 \cdot 674$ , how does that help? $f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{3 \cdot 674} + \left(e^{\frac{4\pi i}{3}}\right)^{3 \cdot 674} = \left(\left(e^{\frac{2\pi i}{3}}\right)^{3}\right)^{674} + \left(\left(e^{\frac{4\pi i}{3}}\right)^{3}\right)^{674} = 1^{674} + 1^{674} = \boxed{\textbf{(E)} \ 2}.$ Since any third root of unity must cube to $1$ .

~ zoomanTV

Solution 2 (Eisenstein Units)

The numbers $\frac{-1+i\sqrt{3}}{2}$ and $\frac{-1-i\sqrt{3}}{2}$ are both $\textbf{Eisenstein Units}$ (along with $1$ ), denoted as $\omega$ and $\omega^2$ , respectively. They have the property that when they are cubed, they equal to $1$ . Thus, we can immediately solve: $\omega^{2022} + \omega^{2 \cdot 2022} = \omega^{3 \cdot 674} + \omega^{3 \cdot 2 \cdot 674} = 1^{674} + 1^{2 \cdot 674} = \boxed{\textbf{(E)} \ 2}.$ ~mathboy100

Solution 3 (Quick and Easy)

We begin by recognizing this form looks similar to the definition of cosine: $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.$ We can convert our two terms into exponential form to find $f(n) = \left( e^{\frac{2\pi i}{3}} \right ) ^n + \left ( e^{-\frac{2\pi i}{3}} \right ) ^n=e^{\frac{2 \pi i n}{3}} + e^{-\frac{2\pi i n}{3}}.$ This simplifies nicely: $f(n)=2\cos\left( \frac{2\pi n}{3} \right).$ Thus, $f(2022)=2\cos \left ( \frac{2\pi (2022) }{3} \right) = 2\cos(1348 \pi) = \boxed{\textbf{(E)}\ 2}.$

~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 $r^3-1=0$ . So we have $\begin{align*} a_0&=2\\ a_1&=-1\\ a_2&=-1\\ a_n&=a_{n-3} \end{align*}$ Every time $n$ is multiple of $3$ as is true when $n=2022$ , $a_n= \boxed{\textbf{(E)} \ 2}$ ~lopkiloinm

Solution 5 (Polynomial + Recursion)

Let $a = \frac{-1+i\sqrt{3}}{2}$ and $b = \frac{-1-i\sqrt{3}}{2}$ . We know that $a + b = -1$ and $a \cdot b = 1$ . Therefore, a and b are the roots of $x^2 + x + 1 = 0$ . By the factor theorem, $a^2 + a + 1 = 0$ and $b^2 + b + 1 = 0$ . Multiply the first equation by $a^{n-2}$ and the second equation by $b^{n-2}$ . This gives us $a^n + a^{n-1} + a^{n-2} = 0$ and $b^n + b^{n-1} + b^{n-2} = 0$ . Adding both equations together we get $a^n + b^n + a^{n-1} + b^{n-1}+ a^{n-2} + b^{n-2} = 0$ . This is the same as $f(n) + f(n-1) + f(n-2) = 0$ . Therefore, $f(n) = -f(n-1) - f(n-2)$ . Plugging in $n=1,2,3,4,5$ , and $6$ , we get $f(n) = -1, -1, 2, -1, -1, 2$ . Therefore we know that if $n$ is a multiple of $3$ , then $f(n)$ is $2$ . Since $2022$ is a multiple of $3$ , our answers is $E) 2$ . ~vpeddi18

Solution 6 (SO FAST)

Converting the two terms into rectangular form,

$f(2022)=\left(\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}\right)^{2022}+\left(\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}\right)^{2022}.$

By DeMoivre's Theorem,

$f(2022)=\left(\cos{\left(\frac{2\pi}{3}\cdot{2022}\right)}+i\sin{\left(\frac{2\pi}{3}\cdot{2022}\right)}\right)+\left(\cos{\left(\frac{4\pi}{3}\cdot{2022}\right)}+i\sin{\left(\frac{4\pi}{3}\cdot{2022}\right)}\right).$

Note that $\cos{\pi\cdot{k}}=1$ if $k$ is even and $-1$ if $k$ is odd, and that $\sin{\pi\cdot{k}}=0$ for all integers $k$ .

All arguments are even in the second equation for $f(2022)$ , so the two $\cos$ terms are equal to $1$ , and the two $\sin$ terms are equal to $0$ .

Therefore the answer is $1+1=\boxed{\textbf{(E) } 2}.$

-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)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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