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

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== Solution 1 ==
 
== 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}}, \\
 +
-1 - i\sqrt{3} &= 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}.
 +
\end{align*}</cmath>
 
Notice that both are scaled copies of the third roots of unity.
 
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 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(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n</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>
−
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?
 
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>
+
1^{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>.
 
Since any third root of unity must cube to <math>1</math>.
  
−
~ <math>\color{magenta} zoomanTV</math>
+
~ zoomanTV
  
 
== Solution 2 (Eisenstein Units) ==
 
== 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:
 
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>
−
<cmath>\omega^{2022} + \omega^{2 \cdot 2022}</cmath>
 
−
<cmath> = \omega^{3 \cdot 674} + \omega^{3 \cdot 2 \cdot 674}</cmath>
 
−
<cmath> = 1^{674} + 1^{2 \cdot 674}</cmath>
 
−
<cmath> = \boxed{\textbf{(E)} \ 2}</cmath>
 
−
 
 
 
~mathboy100
 
~mathboy100
  
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== Solution 5 (Polynomial + Recursion) ==
 
== Solution 5 (Polynomial + Recursion) ==
−
let <math>a = \frac{-1+i\sqrt{3}}{2}</math> and <math>b = \frac{-1-i\sqrt{3}}{2}</math>
+
Let <math>a = \frac{-1+i\sqrt{3}}{2}</math> and <math>b = \frac{-1-i\sqrt{3}}{2}</math>.
−
<math>a + b = -1</math>
+
We know that <math>a + b = -1</math> and <math>a \cdot b = 1</math>.
−
<math>a * b = 1</math>
+
Therefore, a and b are the roots of <math>x^2 + x + 1 = 0</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>.
−
By 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>.
−
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>.
 
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>
+
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>.
 
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>
+
Therefore, <math>f(n) = -f(n-1) - f(n-2)</math>.
−
Plugging in <math>n=1,2,3,4,5,6</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>.
+
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>
+
Since <math>2022</math> is a multiple of <math>3</math>, our answers is <math>E) 2</math>.
 
~vpeddi18
 
~vpeddi18
  
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<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>
 
<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> <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>.  
+
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>.  
 
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{(B) } 2}.</math>
+
Therefore the answer is <math>1+1=\boxed{\textbf{(E) } 2}.</math>
  
 
-Benedict T (countmath1)
 
-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==
 
==See Also==
 
{{AMC12 box|year=2022|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2022|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

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

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