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Difference between revisions of "2021 AMC 12A Problems/Problem 22"

(→‎Solution 1 (Complex Numbers: Vieta's Formulas))
(→‎Solution 2 (Complex Numbers and Trigonometric Identities))
 
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~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
 
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
  
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==Solution 2 (Complex Numbers: Trigonometric Identities)==
+
==Solution 2 (Complex Numbers and 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>
 
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
 
Since <math>\cos\theta=\cos(2\pi-\theta)</math> holds for all <math>\theta,</math> this sum becomes
Line 101: Line 101:
 
for(int i = 0; i< n; ++i) dot(A[i],linewidth(3.5));  
 
for(int i = 0; i< n; ++i) dot(A[i],linewidth(3.5));  
 
</asy>
 
</asy>
−
Let <math>x=\cos\theta.</math> We apply the Double-Angle and Triple-Angle Identities:
+
Let <math>x=\cos\theta.</math> We apply the Double-Angle and Triple-Angle Identities to rewrite <math>(\bigstar)</math> in terms of <math>x,</math> and then rearrange:
−
<cmath>\begin{align*}
 
−
\cos(2\theta) &= 2\cos^2\theta-1 = 2x^2-1, \\
 
−
\cos(3\theta) &= 4\cos^3\theta-3\cos\theta =4x^3-3x.
 
−
\end{align*}</cmath>
 
−
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>
+
Recall that 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>

Latest revision as of 02:27, 27 October 2025

Problem

Suppose that the roots of the polynomial $P(x)=x^3+ax^2+bx+c$ are $\cos \frac{2\pi}7,\cos \frac{4\pi}7,$ and $\cos \frac{6\pi}7$, where angles are in radians. What is $abc$?

$\textbf{(A) }{-}\frac{3}{49} \qquad \textbf{(B) }{-}\frac{1}{28} \qquad \textbf{(C) }\frac{\sqrt[3]7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}$

Solution 1 (Complex Numbers and Vieta's Formulas)

Let $z=e^{\frac{2\pi i}{7}}.$ Since $z$ is a $7$th root of unity, we have $z^7=1.$ For all integers $k,$ note that \[\cos\frac{2k\pi}{7}=\operatorname{Re}\left(z^k\right)=\operatorname{Re}\left(z^{-k}\right).\] It follows that \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*} By geometric series, we conclude that \[\sum_{k=0}^{6}z^k=\frac{1-1}{1-z}=0.\] Alternatively, recall that the $7$th roots of unity satisfy the equation $z^7-1=0.$ By Vieta's Formulas, the sum of these seven roots is $0.$

As a result, we get \[\sum_{k=1}^{6}z^k=-1.\] Let $(r,s,t)=\left(\cos{\frac{2\pi}{7}},\cos{\frac{4\pi}{7}},\cos{\frac{6\pi}{7}}\right).$ By Vieta's Formulas, the answer is \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*} ~MRENTHUSIASM (inspired by Peeyush Pandaya et al)

Solution 2 (Complex Numbers and Trigonometric Identities)

Let $z=e^{\frac{2\pi i}{7}}.$ In Solution 1, we conclude that $\sum_{k=1}^{6}z^k=-1,$ so \[\sum_{k=1}^{6}\operatorname{Re}\left(z^k\right)=\sum_{k=1}^{6}\cos\frac{2k\pi}{7}=-1.\] Since $\cos\theta=\cos(2\pi-\theta)$ holds for all $\theta,$ this sum becomes \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*} Note that $\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}$ are roots of \[\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \hspace{15mm} (\bigstar)\] as they can be verified either algebraically (by the identity $\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)$) 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 $x=\cos\theta.$ We apply the Double-Angle and Triple-Angle Identities to rewrite $(\bigstar)$ in terms of $x,$ and then rearrange: \begin{align*} x+\left(2x^2-1\right)+\left(4x^3-3x\right)&=-\frac12 \\ 4x^3+2x^2-2x-\frac12&=0 \\ x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0. \end{align*} Recall that the roots are $x=\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}.$

Therefore, we obtain $(a,b,c)=\left(\frac12,-\frac12,-\frac18\right),$ from which $abc=\boxed{\textbf{(D) }\frac{1}{32}}.$

~MRENTHUSIASM (inspired by Peeyush Pandaya et al)

Solution 3 (Trigonometric Identities)

We solve for $a,b,$ and $c$ separately:

  1. Solve for $a:$ By Vieta's Formulas, we have $a = - \left( \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 \right).$

    The real parts of the $7$th roots of unity are $1, \cos \frac{2\pi}7, \cos \frac{4\pi}7, \cos \frac{6\pi}7, \cos \frac{8\pi}7, \cos \frac{10\pi}7, \cos \frac{12\pi}7$ and they sum to $0.$

    Note that $\cos\theta=\cos(2\pi-\theta)$ for all $\theta.$ Excluding $1,$ the other six roots add to \[2\left(\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7\right) = -1,\] from which \[\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 = -\frac12.\] Therefore, we get $a = -\left(-\frac12\right) = \frac12.$

  2. Solve for $b:$ By Vieta's Formulas, we have $b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7.$

    Note that $\cos \alpha \cos \beta = \frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}$ for all $\alpha$ and $\beta.$ Therefore, we get \[b=\frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{2\pi}7}2=\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7=-\frac12.\]

  3. Solve for $c:$ By Vieta's Formulas, we have $c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{6\pi}7=-\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7.$

    We multiply both sides by $8 \sin{\frac{2\pi}{7}},$ then repeatedly apply the angle addition formula for sine: \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*} Therefore, we get $c = -\frac18.$

Finally, the answer is $abc=\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac18\right)=\boxed{\textbf{(D) }\frac{1}{32}}.$

~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 $\operatorname{Re}\left(\omega^{m}\right) = \operatorname{Re}\left(\omega^{-m}\right).$ We have \[\cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} = \frac12(0 - \cos 0) = -\frac12,\] so $a = \frac{1}{2}.$

By the Product-to-Sum Identity, we have \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}{2}, \end{align*} so $b = -\frac{1}{2}.$

By the Product-to-Sum Identity, we have \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*} so $c = -\frac{1}{8}.$

Finally, we get $abc=\boxed{\textbf{(D) }\frac{1}{32}}.$

~ccx09

Easy Video Solution by Scholars Foundation Without Complex Numbers and Euler's Identity (Using Trigonometry + Vieta's Formula)

https://youtu.be/m4N4KN6_tA0

Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula)

https://youtu.be/Im_WTIK0tss

~ pi_is_3.14

Video Solution by MRENTHUSIASM (English & Chinese)

https://youtu.be/X6oqEpFAJBk

~MRENTHUSIASM

Video Solution by grogg007

https://www.youtube.com/watch?v=hP4cpuAszLo

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

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