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) | ||
− | <math> | + | ==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> | ||
+ | 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); | ||
− | + | 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 109: | ||
&=\cos(2\theta)\cos\theta-2\left(1-\cos^2\theta\right)\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> | 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{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\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{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) | ||
+ | |||
+ | ~MRENTHUSIASM (Reformatting) | ||
− | + | == 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 (Solution) | ||
− | ~MRENTHUSIASM ( | + | ~MRENTHUSIASM (Reformatting) |
− | === | + | == Solution 5 (Approximations) == |
− | + | Letting the roots be <math>p=\cos\frac{2\pi}{7},q=\cos\frac{4\pi}{7},</math> and <math>r=\cos\frac{6\pi}{7}.</math> Vieta gives | |
+ | <cmath>\begin{align*} | ||
+ | p + q + r &= a, \\ | ||
+ | pq + qr + rp &= -b, \\ | ||
+ | pqr &= c. | ||
+ | \end{align*}</cmath> | ||
+ | We use the Taylor series <cmath>\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}</cmath> | ||
+ | to approximate the roots. | ||
− | * | + | Taking the sum up to <math>k = 3</math> yields a close approximation, so we have |
+ | <cmath>\begin{alignat*}{8} | ||
+ | \cos\left(\frac{2\pi}{7}\right) &\approx 1-\frac{\left(\frac{2\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{2\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{2\pi}{7}\right)^{6}}{720} &&\approx 0.623, \\ | ||
+ | \cos\left(\frac{4\pi}{7}\right) &\approx 1-\frac{\left(\frac{4\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{4\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{4\pi}{7}\right)^{6}}{720} &&\approx -0.225, \\ | ||
+ | \cos\left(\frac{6\pi}{7}\right) &\approx 1-\frac{\left(\frac{6\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{6\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{6\pi}{7}\right)^{6}}{720} &&\approx -0.964. | ||
+ | \end{alignat*}</cmath> | ||
+ | Note that these approximations get worse as <math>x</math> gets larger, but they will be fine for the purposes of this problem. We then have | ||
+ | <cmath>\begin{alignat*}{8} | ||
+ | p + q + r &= a &&\approx -0.56, \\ | ||
+ | pq + qr + rq &= -b &&\approx -0.524, \\ | ||
+ | pqr &= c &&\approx 0.135. | ||
+ | \end{alignat*}</cmath> | ||
+ | We further approximate these values to <math>a \approx -0.5,b \approx 0.5,</math> and <math>c \approx 0.125</math> (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have <math>abc \approx \boxed{\textbf{(D) }\frac{1}{32}}.</math> | ||
− | + | <u><b>Remark</b></u> | |
− | + | In order to be more confident in your answer, you can go a few terms further in the Taylor series. | |
− | + | ~ciceronii (Solution) | |
− | ~MRENTHUSIASM | + | ~MRENTHUSIASM (Reformatting) |
− | == 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 | ||
Revision as of 17:41, 4 December 2021
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 Solution 5 (Approximations)
- 7 Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula)
- 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)
~MRENTHUSIASM (Reformatting)
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 (Solution)
~MRENTHUSIASM (Reformatting)
Solution 5 (Approximations)
Letting the roots be and Vieta gives We use the Taylor series 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
Remark
In order to be more confident in your answer, you can go a few terms further in the Taylor series.
~ciceronii (Solution)
~MRENTHUSIASM (Reformatting)
Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula)
~ 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.