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

Latest revision as of 22:57, 1 November 2024

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 the two terms in the problem are two of the third roots of unity (that is, both of them equal $1$ when raised to the power of $3$ ). 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

@@ Line 10: / Line 10: @@
 == Solution 1 ==
-Converting both summands to exponential form, <cmath>-1 + i\sqrt{3} = 2e^{\frac{2\pi i}{3}}</cmath>
+Converting both summands to exponential form,
-<cmath>-1 - i\sqrt{3} = 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}</cmath>
+<cmath>\begin{align*}
+-1 + i\sqrt{3} &= 2e^{\frac{2\pi i}{3}}, \\
-Notice that both are scaled copies of the third roots of unity.
+-1 - i\sqrt{3} &= 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}.
-When we replace the summands with their exponential form, we get
+\end{align*}</cmath>
-<cmath>f(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n</cmath>
+Notice that the two terms in the problem are two of the third roots of unity (that is, both of them equal <math>1</math> when raised to the power of <math>3</math>).
-When we substitute <math>n = 2022</math>, we get
+When we replace the summands with their exponential form, we get <cmath>f(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n.</cmath>
-<cmath>f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{2022} + \left(e^{\frac{4\pi i}{3}}\right)^{2022}</cmath>
+When we substitute <math>n = 2022</math>, we get <cmath>f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{2022} + \left(e^{\frac{4\pi i}{3}}\right)^{2022}.</cmath>
 We can rewrite <math>2022</math> as <math>3 \cdot 674</math>, how does that help?
-<cmath>f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{3 \cdot 674} + \left(e^{\frac{4\pi i}{3}}\right)^{3 \cdot 674} = </cmath>
+<cmath>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} =
-<cmath>\left(\left(e^{\frac{2\pi i}{3}}\right)^{3}\right)^{674} + \left(\left(e^{\frac{4\pi i}{3}}\right)^{3}\right)^{674} = </cmath>
+^{674} + 1^{674} = \boxed{\textbf{(E)} \ 2}.</cmath>
-<cmath>1^{674} + 1^{674} = \boxed{\textbf{(E)} \ 2}</cmath>
 Since any third root of unity must cube to <math>1</math>.
-~ <math>\color{magenta} zoomanTV</math>
+~ zoomanTV
 == Solution 2 (Eisenstein Units) ==
 The numbers <math>\frac{-1+i\sqrt{3}}{2}</math> and <math>\frac{-1-i\sqrt{3}}{2}</math> are both <math>\textbf{Eisenstein Units}</math> (along with <math>1</math>), denoted as <math>\omega</math> and <math>\omega^2</math>, respectively. They have the property that when they are cubed, they equal to <math>1</math>. Thus, we can immediately solve:
+<cmath>\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}.</cmath>
+~mathboy100
-<cmath>\omega^{2022} + \omega^{2 \cdot 2022}</cmath>
+== Solution 3 (Quick and Easy) ==
-<cmath> = \omega^{3 * 674} + \omega^{3 \cdot 2 \cdot 674}</cmath>
-<cmath> = 1^{674} + 1^{2 \cdot 674}</cmath>
-<cmath> = \boxed{\textbf{(E)} \ 2}</cmath>
-~mathboy100
+We begin by recognizing this form looks similar to the definition of cosine: <cmath>\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.</cmath> We can convert our two terms into exponential form to find <cmath>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}}.</cmath> This simplifies nicely: <cmath>f(n)=2\cos\left( \frac{2\pi n}{3} \right).</cmath> Thus, <cmath>f(2022)=2\cos \left ( \frac{2\pi (2022) }{3} \right) = 2\cos(1348 \pi) = \boxed{\textbf{(E)}\ 2}.</cmath>
-== Solution 3 (Linear Second-order 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>
+~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 <math>r^3-1=0</math>. So we have
+<cmath>\begin{align*}
+a_0&=2\\
+a_1&=-1\\
+a_2&=-1\\
+a_n&=a_{n-3}
+\end{align*}</cmath>
+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
+== Solution 5 (Polynomial + Recursion) ==
+Let <math>a = \frac{-1+i\sqrt{3}}{2}</math> and <math>b = \frac{-1-i\sqrt{3}}{2}</math>.
+We know that <math>a + b = -1</math> and <math>a \cdot b = 1</math>.
+Therefore, a and b are the roots of <math>x^2 + x + 1 = 0</math>.
+By the factor theorem, <math>a^2 + a + 1 = 0</math> and <math>b^2 + b + 1 = 0</math>.
+Multiply the first equation by <math>a^{n-2}</math> and the second equation by <math>b^{n-2}</math>.
+This gives us <math>a^n + a^{n-1} + a^{n-2} = 0</math> and <math>b^n + b^{n-1} + b^{n-2} = 0</math>.
+Adding both equations together we get <math>a^n + b^n + a^{n-1} + b^{n-1}+ a^{n-2} + b^{n-2} = 0</math>.
+This is the same as <math>f(n) + f(n-1) + f(n-2) = 0</math>.
+Therefore, <math>f(n) = -f(n-1) - f(n-2)</math>.
+Plugging in <math>n=1,2,3,4,5</math>, and <math>6</math>, we get <math>f(n) = -1, -1, 2, -1, -1, 2</math>. Therefore we know that if <math>n</math> is a multiple of <math>3</math>, then <math>f(n)</math> is <math>2</math>.
+Since <math>2022</math> is a multiple of <math>3</math>, our answers is <math>E) 2</math>.
+~vpeddi18
+== Solution 6 (SO FAST) ==
+Converting the two terms into rectangular form,
+<cmath>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}.</cmath>
+By DeMoivre's Theorem,
+<cmath>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).</cmath>
+Note that <math>\cos{\pi\cdot{k}}=1</math> if <math>k</math> is even and <math>-1</math> if <math>k</math> is odd, and that <math>\sin{\pi\cdot{k}}=0</math> for all integers <math>k</math>.
+All arguments are even in the second equation for <math>f(2022)</math>, so the two <math>\cos</math> terms are equal to <math>1</math>, and the two <math>\sin</math> terms are equal to <math>0</math>.
+Therefore the answer is <math>1+1=\boxed{\textbf{(E) } 2}.</math>
+-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
+~<i> Education, the Study of Everything </i>
+==Video Solution(1-16)==
+https://youtu.be/SCwQ9jUfr0g
+~~Hayabusa1
 ==See Also==
 {{AMC12 box|year=2022|ab=B|num-b=10|num-a=12}}
 {{MAA Notice}}

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