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

Revision as of 03:49, 4 September 2021

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:

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 $a = -\left(-\frac12\right) = \frac12.$

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 $b=\frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2=\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7=-\frac12.$
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 (Only Using Product to Sum Identity)

Note sum of roots of unity equal zero, sum of real parts equal zero, and $\text{Re} \omega^{m} = \text{Re} \omega^{-m},$ thus $\cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} = 1/2(0 - \cos 0) = -1/2$ which means $A = \frac{1}{2}.$

By product to sum, $\cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} + \cos \frac{2 \pi}{7} \cos \frac{6 \pi}{7} + \cos \frac{4 \pi}{7} \cos \frac{6 \pi}{7} = \frac{1}{2} (2 \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{10 \pi}{7})$ $= \frac{1}{2}(2 \cos \frac{2 \pi}{7} + 2 \cos \frac{4 \pi}{7} + 2 \cos \frac{6 \pi}{7}) = -1/2,$ so $B = - \frac{1}{2}.$

By product to sum, $\cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} \cos \frac{6 \pi}{7} = \frac{1}{2}(\cos \frac{2 \pi}{7} + \cos \frac{6 \pi}{7}) \cos \frac{6 \pi}{7} = \frac{1}{4}(\cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7}) + \frac{1}{4}(1 + \cos \frac{12 \pi}{7})$ $= \frac{1}{4}(1 + \cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}) = 1/8,$ so $C = -1/8.$

$ABC =\boxed{ \frac{1}{32}}.$

~ ccx09

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

@@ Line 31: / Line 31: @@
 ~MRENTHUSIASM (Reformatting)
-== Solution 2 (Approximation) ==
+== Solution 2 (Approximations) ==
-Letting the roots be <math>p</math>, <math>q</math>, and <math>r</math>, Vietas gives
+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>p + q + r = a</cmath>
+<cmath>\begin{align*}
-<cmath>pq + qr + pq = -b</cmath>
+p + q + r &= a, \\
-<cmath>pqr = c</cmath>
+pq + qr + rp &= -b, \\
-We use the Taylor series for <math>\cos x</math>,
+pqr &= c.
-<cmath>\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}</cmath>
+\end{align*}</cmath>
-to approximate the roots. Taking the sum up to <math>k = 3</math> yields a close approximation, so we have
+We use the Taylor series <cmath>\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}</cmath>
-<cmath>\cos\left(\frac{2\pi}{7}\right) \simeq 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} \simeq 0.623</cmath>
+to approximate the roots.
-<cmath>\cos\left(\frac{4\pi}{7}\right) \simeq 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} \simeq -0.225</cmath>
-<cmath>\cos\left(\frac{6\pi}{7}\right) \simeq 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} \simeq -0.964.</cmath>
+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>p + q + r = a \simeq -0.56</cmath>
+<cmath>\begin{alignat*}{8}
-<cmath>pq + qr + pr = -b \simeq -0.524</cmath>
+p + q + r &= a &&\approx -0.56, \\
-<cmath>pqr = c \simeq 0.135</cmath>
+pq + qr + rq &= -b &&\approx -0.524, \\
-We further approximate these values to <math>a \simeq -0.5</math>, <math>b \simeq 0.5</math>, and <math>c \simeq 0.125</math> (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have <math>abc \simeq \boxed{\textbf{(D) } \frac{1}{32}}</math>. ~ciceronii
+pqr &= c &&\approx 0.135.
+\end{alignat*}</cmath>
+We further approximate these values to <math>a \approx -0.5</math>, <math>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)
-Remark: In order to be more confident in your answer, you can go a few terms further in the Taylor series.
+~MRENTHUSIASM (Reformatting)
 == Solution 3 (Only Using Product to Sum Identity) ==

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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