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2021 AMC 12A Problems/Problem 22

Revision as of 04:17, 4 September 2021 by MRENTHUSIASM (talk | contribs)

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 (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 \[b=\frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2=\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7=-\frac12.\]

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

    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)

~MRENTHUSIASM (Reformatting)

Solution 2 (Approximations)

Letting the roots be $p=\cos\frac{2\pi}{7},q=\cos\frac{4\pi}{7},$ and $r=\cos\frac{6\pi}{7}.$ Vieta gives \begin{align*} p + q + r &= a, \\ pq + qr + rp &= -b, \\ pqr &= c. \end{align*} We use the Taylor series \[\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}\] to approximate the roots.

Taking the sum up to $k = 3$ yields a close approximation, so we have \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*} Note that these approximations get worse as $x$ gets larger, but they will be fine for the purposes of this problem. We then have \begin{alignat*}{8} p + q + r &= a &&\approx -0.56, \\ pq + qr + rq &= -b &&\approx -0.524, \\ pqr &= c &&\approx 0.135. \end{alignat*} We further approximate these values to $a \approx -0.5,b \approx 0.5,$ and $c \approx 0.125$ (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have $abc \approx \boxed{\textbf{(D) }\frac{1}{32}}.$

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)

Solution 3 (Product-to-Sum Identity)

Note sum of roots of unity equal zero, sum of 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) \\ &= -\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 (Solution)

~MRENTHUSIASM (Reformatting)

Solution 4 (Complex Numbers)

By geometric series, we have \[\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=\frac{1-1}{1-e^{\frac{2\pi i}{7}}}=0.\] Alternatively, note that the $7$th roots of unity are $z=e^{\frac{2k\pi i}{7}}$ for $k\in\{0,1,2,\cdots,6\},$ in which $z^7-1=0.$ By Vieta's Formulas, the sum of these seven roots is $0.$

It follows that the real parts of these complex numbers must sum to $0,$ so we get $\sum_{k=0}^{6}\cos\frac{2k\pi}{7}=0,$ or \[\sum_{k=1}^{6}\cos\frac{2k\pi}{7}=-1.\] Since $\cos\theta=\cos(2\pi-\theta)$ holds for all $\theta,$ we can rewrite this as \begin{align*} \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\underbrace{\cos\frac{8\pi}{7}}_{\cos\tfrac{6\pi}{7}}+\underbrace{\cos\frac{10\pi}{7}}_{\cos\tfrac{4\pi}{7}}+\underbrace{\cos\frac{12\pi}{7}}_{\cos\tfrac{2\pi}{7}}&=-1\\ 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*} Two solutions follow from here:

Solution 4.1 (Trigonometric Identities)

We know that $\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}$ are roots of \[\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \hspace{15mm} (*)\] as they can be verified algebraically (by the identity $\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)$ for all $\theta$) or geometrically (by the Remark section).

Let $x=\cos\theta.$ It follows that \begin{align*} \cos(2\theta)&=2\cos^2\theta-1 \\ &=2x^2-1, \\ \cos(3\theta)&=\cos(2\theta+\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 \\ &=2x^3-x-2x+2x^3 \\ &=4x^3-3x. \end{align*} Rewriting $(*)$ in terms of $x,$ we have \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*} in which 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 4.2 (Vieta's Formulas)

Let $z=e^{\frac{2\pi i}{7}}.$ Since $z$ is a $7$th root of unity, it follows that $z^7=1.$ Geometrically (shown in the Remark section), we get \begin{alignat*}{4} \cos{\frac{2\pi}{7}} &= \frac{z+z^6}{2} &&= \frac{z+z^{-1}}{2}, \\ \cos{\frac{4\pi}{7}} &= \frac{z^2+z^5}{2} &&= \frac{z^2+z^{-2}}{2}, \\ \cos{\frac{6\pi}{7}} &= \frac{z^3+z^4}{2} &&= \frac{z^3+z^{-3}}{2}. \end{alignat*} Recall that $\sum_{k=0}^{6}z^k=0$ (from which $\sum_{k=1}^{6}z^k=-1$), and 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)

Remark (Geometric Representations)

Graph of the $7$th roots of unity: [asy] /* Made by MRENTHUSIASM */ size(220); import TrigMacros; int big = 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)); } polarGrid(big, numRays); rr_cartesian_axes(-big,big,-big,big,complexplane=true); //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] Geometrically, it is clear that the imaginary parts of these complex numbers sum to $0.$

~MRENTHUSIASM

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

https://youtu.be/Im_WTIK0tss

~ 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

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