Difference between revisions of "2021 AMC 12A Problems/Problem 22"
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==Solution 4 (Complex Numbers)== | ==Solution 4 (Complex Numbers)== | ||
+ | Using geometric series, we can show that <math>\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=0:</math> <cmath>\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=1+e^{\frac{2\pi i}{7}}+e^{\frac{4\pi i}{7}}+\cdots+e^{\frac{12\pi i}{7}}=\frac{1-1}{1-e^{\frac{2\pi i}{7}}}=0.</cmath> | ||
− | + | Alternatively, note that the <math>7</math>th roots of unity are <math>z=e^{\frac{2k\pi i}{7}}</math> for <math>i=0,1,2,\cdots,6,</math> in which <math>z^7-1=0.</math> By Vieta's Formulas, the sum of these seven roots is <math>0.</math> | |
− | < | ||
− | |||
− | |||
− | |||
− | |||
− | + | It follows that the real parts of these complex numbers must sum to <math>0,</math> so we get | |
− | + | <cmath>\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}=\left(\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}\right)-1=-1,</cmath> | |
− | |||
from which <cmath>\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac12,</cmath> | from which <cmath>\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac12,</cmath> | ||
− | as it contributes half the real part of <math>\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}.</math> Two solutions follow from here: | + | as it contributes half the real part of <math>\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}.</math> |
+ | |||
+ | Two solutions follow from here: | ||
− | ===Solution 4.1 ( | + | ===Solution 4.1 (Trigonometric Identities)=== |
− | We know that <math>\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}</math> are solutions of <cmath>\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \ (*)</cmath> as they can be verified geometrically or algebraically (by the identity <math>\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)</math>). Now, let <math>x=\cos\theta.</math> It follows that | + | We know that <math>\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}</math> are solutions of <cmath>\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \ \ \ \ \ (*)</cmath> as they can be verified geometrically or algebraically (by the identity <math>\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)</math>). Now, let <math>x=\cos\theta.</math> It follows that |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\cos(2\theta)&=2\cos^2\theta-1 \\ | \cos(2\theta)&=2\cos^2\theta-1 \\ | ||
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\cos(3\theta)&=\cos(2\theta+\theta) \\ | \cos(3\theta)&=\cos(2\theta+\theta) \\ | ||
&=\cos(2\theta)\cos\theta-\sin(2\theta)\sin\theta \\ | &=\cos(2\theta)\cos\theta-\sin(2\theta)\sin\theta \\ | ||
− | &=\ | + | &=\cos(2\theta)\cos\theta-2\sin^2\theta\cos\theta \\ |
− | + | &=\cos(2\theta)\cos\theta-2\left(1-\cos^2\theta\right)\cos\theta \\ | |
− | &=\ | ||
&=\left(2x^2-1\right)x-2\left(1-x^2\right)x \\ | &=\left(2x^2-1\right)x-2\left(1-x^2\right)x \\ | ||
&=2x^3-x-2x+2x^3 \\ | &=2x^3-x-2x+2x^3 \\ | ||
&=4x^3-3x. | &=4x^3-3x. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Rewriting <math>(*)</math> | + | Rewriting <math>(*)</math> in terms of <math>x,</math> we have |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
x+\left(2x^2-1\right)+\left(4x^3-3x\right)&=-\frac12 \\ | x+\left(2x^2-1\right)+\left(4x^3-3x\right)&=-\frac12 \\ | ||
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x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0. | x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | Therefore, we obtain <math>(a,b,c)=\left(\frac12,-\frac12,-\frac18\right),</math> from which <math>abc=\boxed{\textbf{(D) }\frac{1}{32}}.</math> | |
~MRENTHUSIASM (inspired by Peeyush Pandaya et al) | ~MRENTHUSIASM (inspired by Peeyush Pandaya et al) | ||
− | ===Solution 4.2 (Vieta's Formulas | + | ===Solution 4.2 (Vieta's Formulas)=== |
Let <math>z=e^{\frac{2\pi i}{7}}.</math> Since <math>z</math> is a <math>7</math>th root of unity, <math>z^7=1.</math> Geometrically, it follows that | Let <math>z=e^{\frac{2\pi i}{7}}.</math> Since <math>z</math> is a <math>7</math>th root of unity, <math>z^7=1.</math> Geometrically, it follows that | ||
<cmath>\begin{array}{ccccc} | <cmath>\begin{array}{ccccc} | ||
− | \cos{\frac{2\pi}{7}} &=& \frac{z+z^6}{2} &=& \frac{z+z^{-1}}{2} \\ [2ex] | + | \cos{\frac{2\pi}{7}} &=& \frac{z+z^6}{2} &=& \frac{z+z^{-1}}{2}, \\ [2ex] |
− | \cos{\frac{4\pi}{7}} &=& \frac{z^2+z^5}{2} &=& \frac{z^2+z^{-2}}{2} \\ [2ex] | + | \cos{\frac{4\pi}{7}} &=& \frac{z^2+z^5}{2} &=& \frac{z^2+z^{-2}}{2}, \\ [2ex] |
− | \cos{\frac{ | + | \cos{\frac{6\pi}{7}} &=& \frac{z^3+z^4}{2} &=& \frac{z^3+z^{-3}}{2}. |
\end{array}</cmath> | \end{array}</cmath> | ||
− | Recall that <math>\sum_{k=0}^{6}z^k=0</math> ( | + | Recall that <math>\sum_{k=0}^{6}z^k=0</math> (from which <math>\sum_{k=1}^{6}z^k=-1</math>), and let <math>(r,s,t)=\left(\cos{\frac{2\pi}{7}},\cos{\frac{4\pi}{7}},\cos{\frac{6\pi}{7}}\right).</math> By Vieta's Formulas, the answer is |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
abc&=[-(r+s+t)](rs+st+tr)(-rst) \\ | abc&=[-(r+s+t)](rs+st+tr)(-rst) \\ | ||
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~MRENTHUSIASM (inspired by Peeyush Pandaya et al) | ~MRENTHUSIASM (inspired by Peeyush Pandaya et al) | ||
+ | |||
+ | ===Remark=== | ||
+ | Graph of the <math>7</math>th roots of unity in Desmos: | ||
+ | |||
+ | * Rectangular Coordinate System: https://www.desmos.com/calculator/rzzk5geric (shown below) | ||
+ | |||
+ | * Polar Coordinate System: https://www.desmos.com/calculator/xta6sfbnot | ||
+ | |||
+ | [[File:2021 AMC 12A Problem 22 Exact Values + LaTeX.png|center]] | ||
+ | |||
+ | Geometrically, it is clear that the imaginary parts of these complex numbers sum to <math>0.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
== Video Solution by OmegaLearn (Euler's Identity + Vieta's ) == | == Video Solution by OmegaLearn (Euler's Identity + Vieta's ) == |
Latest revision as of 21:47, 3 April 2021
Contents
Problem
Suppose that the roots of the polynomial are and , where angles are in radians. What is ?
Solution 1
Part 1: solving for a
is the negation of the sum of roots
The real values of the 7th roots of unity are: and they sum to .
If we subtract 1, and condense identical terms, we get:
Therefore, we have
Part 2: solving for b
is the sum of roots two at a time by Vieta's
We know that
By plugging all the parts in we get:
Which ends up being:
Which was shown in the first part to equal , so
Part 3: solving for c
Notice that
is the negation of the product of roots by Vieta's formulas
Multiply by
Then use sine addition formula backwards:
Finally multiply or .
~Tucker
Solution 2 (Approximation)
Letting the roots be , , and , Vietas gives We use the Taylor series for , to approximate the roots. Taking the sum up to yields a close approximation, so we have Note that these approximations get worse as gets larger, but they will be fine for the purposes of this problem. We then have We further approximate these values to , , and (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have . ~ciceronii
Remark: In order to be more confident in your answer, you can go a few terms further in the Taylor series.
Solution 3 (Only using Product to Sum Identity)
Note sum of roots of unity equal zero, sum of real parts equal zero, and thus which means
By product to sum, so
By product to sum, so
~ ccx09
Solution 4 (Complex Numbers)
Using geometric series, we can show that
Alternatively, note that the th roots of unity are for in which By Vieta's Formulas, the sum of these seven roots is
It follows that the real parts of these complex numbers must sum to so we get from which as it contributes half the real part of
Two solutions follow from here:
Solution 4.1 (Trigonometric Identities)
We know that are solutions of as they can be verified geometrically or algebraically (by the identity ). Now, let It follows that Rewriting in terms of we have Therefore, we obtain from which
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Solution 4.2 (Vieta's Formulas)
Let Since is a th root of unity, Geometrically, it follows that
Recall that (from which ), and let By Vieta's Formulas, the answer is
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Remark
Graph of the th roots of unity in Desmos:
- Rectangular Coordinate System: https://www.desmos.com/calculator/rzzk5geric (shown below)
- Polar Coordinate System: https://www.desmos.com/calculator/xta6sfbnot
Geometrically, it is clear that the imaginary parts of these complex numbers sum to
~MRENTHUSIASM
Video Solution by OmegaLearn (Euler's Identity + Vieta's )
~ pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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.