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

Revision as of 04:34, 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 Third-order Homogeneous Difference Equation)

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

@@ Line 35: / Line 35: @@
 ~mathboy100
-== Solution 3 (Linear Second-order Homogeneous Difference Equation) ==
+== Solution 3 (Linear Third-order Homogeneous Difference Equation) ==
 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 <math>r^3-1=0</math>. So we have
 <cmath>\begin{align*}

2022 AMC 12B (Problems • Answer Key • Resources)
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