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}} | + | Converting both summands to exponential form, |
− | + | <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>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}.</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>. | ||
− | ~ | + | ~ 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} | ||
− | |||
− | |||
− | |||
− | |||
~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>. | |
− | <math>a + b = -1</math> | + | We know that <math>a + b = -1</math> and <math>a \cdot b = 1</math>. |
− | <math>a | + | 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> | + | 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 | + | Since <math>2022</math> is a multiple of <math>3</math>, our answers is <math>E) 2</math>. |
~vpeddi18 | ~vpeddi18 | ||
− | == Solution 6 ( | + | == Solution 6 (SO FAST) == |
Converting the two terms into rectangular form, | Converting the two terms into rectangular form, | ||
− | <cmath>f(2022)=\left(\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}\right)^{2022}+ | + | <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, | By DeMoivre's Theorem, | ||
− | <cmath>f(2022)=\left(\cos{\frac{2\pi}{3}\cdot{2022}}+i\sin{\frac{2\pi}{3}\cdot{2022}}\right)+ | + | <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{( | + | 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}} |
Latest revision as of 22:17, 27 October 2023
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Eisenstein Units)
- 4 Solution 3 (Quick and Easy)
- 5 Solution 4 (Third-order Homogeneous Linear Recurrence Relation)
- 6 Solution 5 (Polynomial + Recursion)
- 7 Solution 6 (SO FAST)
- 8 Video Solution by mop 2024
- 9 Video Solution (Under 2 min!)
- 10 Video Solution(1-16)
- 11 See Also
Problem
Let , where
. What is
?
Solution 1
Converting both summands to exponential form,
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 substitute
, we get
We can rewrite
as
, how does that help?
Since any third root of unity must cube to
.
~ zoomanTV
Solution 2 (Eisenstein Units)
The numbers and
are both
(along with
), denoted as
and
, respectively. They have the property that when they are cubed, they equal to
. Thus, we can immediately solve:
~mathboy100
Solution 3 (Quick and Easy)
We begin by recognizing this form looks similar to the definition of cosine: We can convert our two terms into exponential form to find
This simplifies nicely:
Thus,
~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 . So we have
Every time
is multiple of
as is true when
,
~lopkiloinm
Solution 5 (Polynomial + Recursion)
Let and
.
We know that
and
.
Therefore, a and b are the roots of
.
By the factor theorem,
and
.
Multiply the first equation by
and the second equation by
.
This gives us
and
.
Adding both equations together we get
.
This is the same as
.
Therefore,
.
Plugging in
, and
, we get
. Therefore we know that if
is a multiple of
, then
is
.
Since
is a multiple of
, our answers is
.
~vpeddi18
Solution 6 (SO FAST)
Converting the two terms into rectangular form,
By DeMoivre's Theorem,
Note that if
is even and
if
is odd, and that
for all integers
.
All arguments are even in the second equation for , so the two
terms are equal to
, and the two
terms are equal to
.
Therefore the answer is
-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)
~~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.