Difference between revisions of "2021 AMC 12A Problems/Problem 22"
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Suppose that the roots of the polynomial <math>P(x)=x^3+ax^2+bx+c</math> are <math>\cos \frac{2\pi}7,\cos \frac{4\pi}7,</math> and <math>\cos \frac{6\pi}7</math>, where angles are in radians. What is <math>abc</math>? | Suppose that the roots of the polynomial <math>P(x)=x^3+ax^2+bx+c</math> are <math>\cos \frac{2\pi}7,\cos \frac{4\pi}7,</math> and <math>\cos \frac{6\pi}7</math>, where angles are in radians. What is <math>abc</math>? | ||
− | <math>\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{\sqrt[3]7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}</math> | + | <math>\textbf{(A) }{-}\frac{3}{49} \qquad \textbf{(B) }{-}\frac{1}{28} \qquad \textbf{(C) }\frac{\sqrt[3]7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}</math> |
− | ==Solution 1== | + | ==Solution 1 (Complex Numbers: 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, we have <math>z^7=1.</math> For all integers <math>k,</math> note that <math>\cos\frac{2k\pi}{7}=\operatorname{Re}\left(z^k\right)=\operatorname{Re}\left(z^{-k}\right)</math> and <math>\sin\frac{2k\pi}{7}=\operatorname{Im}\left(z^k\right)=-\operatorname{Im}\left(z^{-k}\right).</math> It follows that | ||
+ | <cmath>\begin{alignat*}{4} | ||
+ | \cos\frac{2\pi}{7} &= \frac{z+z^{-1}}{2} &&= \frac{z+z^6}{2}, \\ | ||
+ | \cos\frac{4\pi}{7} &= \frac{z^2+z^{-2}}{2} &&= \frac{z^2+z^5}{2}, \\ | ||
+ | \cos\frac{6\pi}{7} &= \frac{z^3+z^{-3}}{2} &&= \frac{z^3+z^4}{2}. | ||
+ | \end{alignat*}</cmath> | ||
+ | By geometric series, we conclude that <cmath>\sum_{k=0}^{6}z^k=\frac{1-1}{1-z}=0.</cmath> | ||
+ | Alternatively, recall that the <math>7</math>th roots of unity satisfy the equation <math>z^7-1=0.</math> By Vieta's Formulas, the sum of these seven roots is <math>0.</math> | ||
− | + | As a result, we get <cmath>\sum_{k=1}^{6}z^k=-1.</cmath> | |
+ | 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*} | ||
+ | abc&=[-(r+s+t)](rs+st+tr)(-rst) \\ | ||
+ | &=(r+s+t)(rs+st+tr)(rst) \\ | ||
+ | &=\left(\frac{\sum_{k=1}^{6}z^k}{2}\right)\left(\frac{2\sum_{k=1}^{6}z^k}{4}\right)\left(\frac{1+\sum_{k=0}^{6}z^k}{8}\right) \\ | ||
+ | &=\frac{1}{32}\left(\sum_{k=1}^{6}z^k\right)\left(\sum_{k=1}^{6}z^k\right)\left(1+\sum_{k=0}^{6}z^k\right) \\ | ||
+ | &=\frac{1}{32}(-1)(-1)(1) \\ | ||
+ | &=\boxed{\textbf{(D) }\frac{1}{32}}. | ||
+ | \end{align*}</cmath> | ||
+ | ~MRENTHUSIASM (inspired by Peeyush Pandaya et al) | ||
− | + | ==Solution 2 (Complex Numbers: Trigonometric Identities)== | |
− | + | Let <math>z=e^{\frac{2\pi i}{7}}.</math> In Solution 1, we conclude that <math>\sum_{k=1}^{6}z^k=-1,</math> so <cmath>\sum_{k=1}^{6}\operatorname{Re}\left(z^k\right)=\sum_{k=1}^{6}\cos\frac{2k\pi}{7}=-1.</cmath> | |
− | <math> | + | Since <math>\cos\theta=\cos(2\pi-\theta)</math> holds for all <math>\theta,</math> this sum becomes |
− | + | <cmath>\begin{align*} | |
− | + | 2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)&=-1\\ | |
− | + | \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}&=-\frac12. | |
− | + | \end{align*}</cmath> | |
− | + | Note that <math>\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}</math> are roots of <cmath>\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \hspace{15mm} (\bigstar)</cmath> as they can be verified either algebraically (by the identity <math>\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)</math>) or geometrically (by the graph below). | |
− | + | <asy> | |
− | + | /* Made by MRENTHUSIASM */ | |
− | + | size(200); | |
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− | + | int xMin = -2; | |
+ | int xMax = 2; | ||
+ | int yMin = -2; | ||
+ | int yMax = 2; | ||
+ | int numRays = 24; | ||
− | < | + | //Draws a polar grid that goes out to a number of circles |
+ | //equal to big, with numRays specifying the number of rays: | ||
+ | void polarGrid(int big, int numRays) | ||
+ | { | ||
+ | for (int i = 1; i < big+1; ++i) | ||
+ | { | ||
+ | draw(Circle((0,0),i), gray+linewidth(0.4)); | ||
+ | } | ||
+ | for(int i=0;i<numRays;++i) | ||
+ | draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); | ||
+ | } | ||
− | < | + | //Draws the horizontal gridlines |
+ | void horizontalLines() | ||
+ | { | ||
+ | for (int i = yMin+1; i < yMax; ++i) | ||
+ | { | ||
+ | draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); | ||
+ | } | ||
+ | } | ||
− | < | + | //Draws the vertical gridlines |
+ | void verticalLines() | ||
+ | { | ||
+ | for (int i = xMin+1; i < xMax; ++i) | ||
+ | { | ||
+ | draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); | ||
+ | } | ||
+ | } | ||
− | + | horizontalLines(); | |
+ | verticalLines(); | ||
+ | polarGrid(xMax,numRays); | ||
+ | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("Re",(xMax,0),(2,0)); | ||
+ | label("Im",(0,yMax),(0,2)); | ||
− | + | //The n such that we're taking the nth roots of unity | |
+ | int n = 7; | ||
+ | pair A[]; | ||
+ | for(int i = 0; i <= n-1; i+=1) { | ||
+ | A[i] = rotate(360*i/n)*(1,0); | ||
+ | } | ||
+ | label("$1$",A[0],NE, UnFill); | ||
+ | for(int i =1; i < n; ++i) | ||
+ | { | ||
+ | label("$e^{\frac{" +string(2i)+"\pi i}{" + string(n) + "}}$",A[i],dir(A[i]), UnFill); | ||
+ | } | ||
− | + | draw(Circle((0,0),1),red); | |
− | + | for(int i = 0; i< n; ++i) dot(A[i],linewidth(3.5)); | |
− | + | </asy> | |
− | + | Let <math>x=\cos\theta.</math> It follows that | |
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\cos(2\theta)&=2\cos^2\theta-1 \\ | \cos(2\theta)&=2\cos^2\theta-1 \\ | ||
Line 128: | Line 106: | ||
\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 \\ | ||
− | |||
&=4x^3-3x. | &=4x^3-3x. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Rewriting <math>( | + | Rewriting <math>(\bigstar)</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 \\ | ||
4x^3+2x^2-2x-\frac12&=0 \\ | 4x^3+2x^2-2x-\frac12&=0 \\ | ||
− | x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0 | + | x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0, |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | in which the roots are <math>x=\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}.</math> | |
+ | |||
+ | 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 3 (Trigonometric Identities)== |
− | + | We solve for <math>a,b,</math> and <math>c</math> separately: | |
− | < | + | <ol style="margin-left: 1.5em;"> |
− | \cos | + | <li>Solve for <math>a:</math> By Vieta's Formulas, we have <math>a = - \left( \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 \right).</math><p> |
− | \cos | + | The real parts of the <math>7</math>th roots of unity are <math>1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7</math> and they sum to <math>0.</math> <p> |
− | \ | + | Note that <math>\cos\theta=\cos(2\pi-\theta)</math> for all <math>\theta.</math> Excluding <math>1,</math> the other six roots add to <cmath>2\left(\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7\right) = -1,</cmath> from which <cmath>\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 = -\frac12.</cmath> |
− | \end{ | + | Therefore, we get <math>a = -\left(-\frac12\right) = \frac12.</math></li><p> |
+ | <li>Solve for <math>b:</math> By Vieta's Formulas, we have <math>b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7.</math><p> | ||
+ | Note that <math>\cos \alpha \cos \beta = \frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}</math> for all <math>\alpha</math> and <math>\beta.</math> Therefore, we get <cmath>b=\frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{2\pi}7}2=\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7=-\frac12.</cmath></li> | ||
+ | <li>Solve for <math>c:</math> By Vieta's Formulas, we have <math>c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{6\pi}7=-\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7.</math> <p> | ||
+ | We multiply both sides by <math>8 \sin{\frac{2\pi}{7}},</math> then repeatedly apply the angle addition formula for sine: | ||
+ | <cmath>\begin{align*} | ||
+ | c \cdot 8 \sin{\frac{2\pi}{7}} &= -8 \sin{\frac{2\pi}{7}} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7 \\ | ||
+ | &= -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7 \\ | ||
+ | &= -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7 \\ | ||
+ | &= -\sin \frac{16\pi}7 \\ | ||
+ | &= -\sin \frac{2\pi}7. | ||
+ | \end{align*}</cmath> | ||
+ | Therefore, we get <math>c = -\frac18.</math><p> | ||
+ | </li> | ||
+ | </ol> | ||
+ | Finally, the answer is <math>abc=\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac18\right)=\boxed{\textbf{(D) }\frac{1}{32}}.</math> | ||
+ | |||
+ | ~Tucker | ||
− | + | == Solution 4 (Product-to-Sum Identity) == | |
+ | Note that the sum of the roots of unity equal zero, so the sum of their real parts equal zero, and <math>\operatorname{Re}\left(\omega^{m}\right) = \operatorname{Re}\left(\omega^{-m}\right).</math> We have <cmath>\cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} = \frac12(0 - \cos 0) = -\frac12,</cmath> so <math>a = \frac{1}{2}.</math> | ||
+ | |||
+ | By the Product-to-Sum Identity, we have | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | + | \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} + \cos \frac{2 \pi}{7} \cos \frac{6 \pi}{7} + \cos \frac{4 \pi}{7} \cos \frac{6 \pi}{7} &= \frac{1}{2} \left(2 \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{10 \pi}{7}\right) \\ | |
− | &=( | + | &= \frac{1}{2}\left(2 \cos \frac{2 \pi}{7} + 2 \cos \frac{4 \pi}{7} + 2 \cos \frac{6 \pi}{7}\right) \\ |
− | &= | + | &= \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \\ |
− | &=\frac{1}{ | + | &= -\frac{1}{2}, |
− | &=\frac{1}{ | + | \end{align*}</cmath> |
− | + | so <math>b = -\frac{1}{2}.</math> | |
+ | |||
+ | By the Product-to-Sum Identity, we have | ||
+ | <cmath>\begin{align*} | ||
+ | \cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} \cos \frac{6 \pi}{7} &= \frac{1}{2}\cos \frac{6 \pi}{7}\left(\cos \frac{2 \pi}{7} + \cos \frac{6 \pi}{7}\right) \\ | ||
+ | &= \frac{1}{4}\left(\cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7}\right) + \frac{1}{4}\left(1 + \cos \frac{12 \pi}{7}\right) \\ | ||
+ | &= \frac{1}{4}\left(1 + \cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}\right) \\ | ||
+ | &= \frac{1}{8}, | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | so <math>c = -\frac{1}{8}.</math> | ||
+ | |||
+ | Finally, we get <math>abc=\boxed{\textbf{(D) }\frac{1}{32}}.</math> | ||
− | ~ | + | ~ccx09 |
− | == Video Solution by OmegaLearn (Euler's Identity + Vieta's ) == | + | == Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula) == |
https://youtu.be/Im_WTIK0tss | https://youtu.be/Im_WTIK0tss | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution by MRENTHUSIASM (English & Chinese) == | ||
+ | https://youtu.be/X6oqEpFAJBk | ||
+ | |||
+ | ~MRENTHUSIASM | ||
==See also== | ==See also== |
Latest revision as of 02:25, 21 September 2023
Contents
- 1 Problem
- 2 Solution 1 (Complex Numbers: Vieta's Formulas)
- 3 Solution 2 (Complex Numbers: Trigonometric Identities)
- 4 Solution 3 (Trigonometric Identities)
- 5 Solution 4 (Product-to-Sum Identity)
- 6 Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula)
- 7 Video Solution by MRENTHUSIASM (English & Chinese)
- 8 See also
Problem
Suppose that the roots of the polynomial are
and
, where angles are in radians. What is
?
Solution 1 (Complex Numbers: Vieta's Formulas)
Let Since
is a
th root of unity, we have
For all integers
note that
and
It follows that
By geometric series, we conclude that
Alternatively, recall that the
th roots of unity satisfy the equation
By Vieta's Formulas, the sum of these seven roots is
As a result, we get
Let
By Vieta's Formulas, the answer is
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Solution 2 (Complex Numbers: Trigonometric Identities)
Let In Solution 1, we conclude that
so
Since
holds for all
this sum becomes
Note that
are roots of
as they can be verified either algebraically (by the identity
) or geometrically (by the graph below).
Let
It follows that
Rewriting
in terms of
we have
in which the roots are
Therefore, we obtain from which
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Solution 3 (Trigonometric Identities)
We solve for and
separately:
- Solve for
By Vieta's Formulas, we have
The real parts of the
th roots of unity are
and they sum to
Note that
for all
Excluding
the other six roots add to
from which
Therefore, we get
- Solve for
By Vieta's Formulas, we have
Note that
for all
and
Therefore, we get
- Solve for
By Vieta's Formulas, we have
We multiply both sides by
then repeatedly apply the angle addition formula for sine:
Therefore, we get
Finally, the answer is
~Tucker
Solution 4 (Product-to-Sum Identity)
Note that the sum of the roots of unity equal zero, so the sum of their real parts equal zero, and We have
so
By the Product-to-Sum Identity, we have
so
By the Product-to-Sum Identity, we have
so
Finally, we get
~ccx09
Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula)
~ pi_is_3.14
Video Solution by MRENTHUSIASM (English & Chinese)
~MRENTHUSIASM
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.