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

Revision as of 04:04, 19 November 2022

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, $-1 + i\sqrt{3} = 2e^{\frac{2\pi i}{3}}$ $-1 - i\sqrt{3} = 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}$

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 $1$ .

~ $\color{magenta} zoomanTV$

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:

$\omega^{2022} + \omega^{2 \cdot 2022}$ $= \omega^{3 * 674} + \omega^{3 \cdot 2 \cdot 674}$ $= 1^{674} + 1^{2 \cdot 674}$ $= \boxed{\textbf{(E)} \ 2}$

~mathboy100

Solution 3 (Linear Second-order Homogeneous Difference Equation)

Notice how this looks like the closed form of the Fibonacci sequence. The base of the exponents are the roots of the characteristic equation $r^2+r+1=0$ . So we have $a_0=2,a_1=-1,a_n=-a_{n-1}-a_{n-2}$ . This recurrence relationship goes like $2,-1,-1,2,-1,-1,2,\ldots$ . Every time $n$ is multiple of $3$ as is true when $n=2022$ , $a_n= \boxed{\textbf{(E)} \ 2}$ ~lopkiloinm

@@ Line 36: / Line 36: @@
 ~mathboy100
 == Solution 3 (Linear Second-order Homogeneous Difference Equation) ==
-Notice that this is a recurrence relationship where <math>a_0=2,a_1=-1,a_n=-a_{n-1}-a_{n-2}</math>. This recurrence relationship goes like <math>2,-1,-1,2,-1,-1,2,\ldots</math>. Every time <math>n</math> is multiple of <math>3</math> as is true when <math>n=2022</math>, <math>a_n= \boxed{\textbf{(E)} \ 2}</math>
+Notice how this looks like the closed form of the Fibonacci sequence. The base of the exponents are the roots of the characteristic equation <math>r^2+r+1=0</math>. So we have <math>a_0=2,a_1=-1,a_n=-a_{n-1}-a_{n-2}</math>. This recurrence relationship goes like <math>2,-1,-1,2,-1,-1,2,\ldots</math>. Every time <math>n</math> is multiple of <math>3</math> as is true when <math>n=2022</math>, <math>a_n= \boxed{\textbf{(E)} \ 2}</math>
 ~lopkiloinm

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