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

Revision as of 21:47, 3 April 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

Part 1: solving for a

$a$ is the negation of the sum of roots

$a = - \left( \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 \right)$

The real values of the 7th 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 $0$ .

If we subtract 1, and condense identical terms, we get:

$2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1$

Therefore, we have $a = -\left(-\frac{1}2\right) = \frac{1}2$

Part 2: solving for b

$b$ is the sum of roots two at a time by Vieta's

$b = \cos \frac{2\pi}7 \cos \frac{4\pi}7 + \cos \frac{2\pi}7 \cos \frac{6\pi}7 + \cos \frac{4\pi}7 \cos \frac{6\pi}7$

We know that $\cos \alpha \cos \beta = \frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}$

By plugging all the parts in we get:

$\frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2$

Which ends up being:

$\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7$

Which was shown in the first part to equal $-\frac{1}2$ , so $b = -\frac{1}2$

Part 3: solving for c

Notice that $\cos \frac{6\pi}7 = \cos \frac{8\pi}7$

$c$ is the negation of the product of roots by Vieta's formulas

$c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$

Multiply by $8 \sin{\frac{2\pi}{7}}$

$c \cdot 8 \sin{\frac{2\pi}{7}} = -8 \sin{\frac{2\pi}{7}} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$

Then use sine addition formula backwards:

$2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7$

$c \cdot 8 \sin{\frac{2\pi}{7}} = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7$

$c \cdot 8 \sin{\frac{2\pi}{7}} = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7$

$c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{16\pi}7$

$c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{2\pi}7$

$c = -\frac{1}8$

Finally multiply $abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}{32}$ or $\boxed{D) \frac{1}{32}}$ .

~Tucker

Solution 2 (Approximation)

Letting the roots be $p$ , $q$ , and $r$ , Vietas gives $p + q + r = a$ $pq + qr + pq = -b$ $pqr = c$ We use the Taylor series for $\cos x$ , $\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}$ to approximate the roots. Taking the sum up to $k = 3$ yields a close approximation, so we have $\cos\left(\frac{2\pi}{7}\right) \simeq 1-\frac{\left(\frac{2\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{2\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{2\pi}{7}\right)^{6}}{720} \simeq 0.623$ $\cos\left(\frac{4\pi}{7}\right) \simeq 1-\frac{\left(\frac{4\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{4\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{4\pi}{7}\right)^{6}}{720} \simeq -0.225$ $\cos\left(\frac{6\pi}{7}\right) \simeq 1-\frac{\left(\frac{6\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{6\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{6\pi}{7}\right)^{6}}{720} \simeq -0.964.$ Note that these approximations get worse as $x$ gets larger, but they will be fine for the purposes of this problem. We then have $p + q + r = a \simeq -0.56$ $pq + qr + pr = -b \simeq -0.524$ $pqr = c \simeq 0.135$ We further approximate these values to $a \simeq -0.5$ , $b \simeq 0.5$ , and $c \simeq 0.125$ (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have $abc \simeq \boxed{\textbf{(D) } \frac{1}{32}}$ . ~ciceronii

Remark: In order to be more confident in your answer, you can go a few terms further in the Taylor series.

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)

Using geometric series, we can show that $\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=0:$ $\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=1+e^{\frac{2\pi i}{7}}+e^{\frac{4\pi i}{7}}+\cdots+e^{\frac{12\pi i}{7}}=\frac{1-1}{1-e^{\frac{2\pi i}{7}}}=0.$

Alternatively, note that the $7$ th roots of unity are $z=e^{\frac{2k\pi i}{7}}$ for $i=0,1,2,\cdots,6,$ 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 $\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}=\left(\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}\right)-1=-1,$ from which $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac12,$ as it contributes half the real part of $\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}.$

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 solutions of $\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \ \ \ \ \ (*)$ as they can be verified geometrically or algebraically (by the identity $\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)$ ). Now, 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 $\begin{align*} x+\left(2x^2-1\right)+\left(4x^3-3x\right)&=-\frac12 \\ 4x^3+2x^2-2x-\frac12&=0 \\ x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0. \end{align*}$ 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, $z^7=1.$ Geometrically, it follows that $\begin{array}{ccccc} \cos{\frac{2\pi}{7}} &=& \frac{z+z^6}{2} &=& \frac{z+z^{-1}}{2}, \\ [2ex] \cos{\frac{4\pi}{7}} &=& \frac{z^2+z^5}{2} &=& \frac{z^2+z^{-2}}{2}, \\ [2ex] \cos{\frac{6\pi}{7}} &=& \frac{z^3+z^4}{2} &=& \frac{z^3+z^{-3}}{2}. \end{array}$

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

Graph of the $7$ th roots of unity in Desmos:

Rectangular Coordinate System: https://www.desmos.com/calculator/rzzk5geric (shown below)

Polar Coordinate System: https://www.desmos.com/calculator/xta6sfbnot

File:2021 AMC 12A Problem 22 Exact Values + LaTeX.png

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 2: / Line 2: @@
 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{^3\sqrt7}{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==
-Part 1: solving for c
+Part 1: solving for a
-Notice that <math>\cos \frac{6\pi}7 = \cos \frac{8\pi}7</math>
+<math>a</math> is the negation of the sum of roots
-<math>c</math> is the negation of the product of roots by Vieta's formulas
+<math>a = - \left( \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 \right)</math>
-<math>c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math>
+The real values of the 7th 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>.
-Multiply by <math>8 \sin{\frac{2\pi}{7}}</math>
+If we subtract 1, and condense identical terms, we get:
-<math>c 8 \sin{2\pi}7 = -8 \sin{\frac{2\pi}{7}} \cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math>
+<math>2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1</math>
-Then use sine addition formula backwards:
+Therefore, we have <math>a = -\left(-\frac{1}2\right) = \frac{1}2</math>
-<math>2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7</math>
-<math>c \cdot 8 \sin{\frac{2\pi}{7}} = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math>
+Part 2: solving for b
-<math>c \cdot 8 \sin{\frac{2\pi}{7}} = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7</math>
-<math>c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{16\pi}7</math>
-<math>c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{2\pi}7</math>
-<math>c = -\frac{1}8</math>
-Part 2: starting to solve for b
 <math>b</math> is the sum of roots two at a time by Vieta's
@@ Line 49: / Line 37: @@
 <math> \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 </math>
-Which is shown in the next part to equal <math>-\frac{1}2</math>, so <math>b = -\frac{1}2</math>
+Which was shown in the first part to equal <math>-\frac{1}2</math>, so <math>b = -\frac{1}2</math>
-Part 3: solving for a and b as the sum of roots
-<math>a</math> is the negation of the sum of roots
+Part 3: solving for c
+Notice that <math>\cos \frac{6\pi}7 = \cos \frac{8\pi}7</math>
+<math>c</math> is the negation of the product of roots by Vieta's formulas
+<math>c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math>
+Multiply by <math>8 \sin{\frac{2\pi}{7}}</math>
+<math>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</math>
+Then use sine addition formula backwards:
+<math>2 \sin \frac{2\pi}7 \cos \frac{2\pi}7 = \sin \frac{4\pi}7</math>
+<math>c \cdot 8 \sin{\frac{2\pi}{7}} = -4 \sin \frac{4\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7</math>
+<math>c \cdot 8 \sin{\frac{2\pi}{7}} = -2 \sin \frac{8\pi}7 \cos \frac{8\pi}7</math>
-<math>a = - \left( \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 \right)</math>
+<math>c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{16\pi}7</math>
-The real values of the 7th 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>.
+<math>c \cdot 8 \sin{\frac{2\pi}{7}} = -\sin \frac{2\pi}7</math>
-If we subtract 1, and condense identical terms, we get:
+<math>c = -\frac{1}8</math>
-<math>2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1</math>
-Therefore, we have <math>a = -\left(-\frac{1}2\right) = \frac{1}2</math>
 Finally multiply <math>abc = \frac{1}2 * - \frac{1}2 * -\frac{1}8 = \frac{1}{32}</math> or <math>\boxed{D) \frac{1}{32}}</math>.
@@ Line 88: / Line 91: @@
 Remark: In order to be more confident in your answer, you can go a few terms further in the Taylor series.
+== Solution 3 (Only using Product to Sum Identity) ==
+Note sum of roots of unity equal zero, sum of real parts equal zero, and <math>\text{Re} \omega^{m} = \text{Re} \omega^{-m},</math>
+thus <math>\cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} = 1/2(0 - \cos 0) = -1/2</math> which means <math>A = \frac{1}{2}.</math>
+By product to sum, <math>\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}) </math>
+<math>= \frac{1}{2}(2 \cos \frac{2 \pi}{7} + 2 \cos \frac{4 \pi}{7} + 2 \cos \frac{6 \pi}{7}) = -1/2,</math> so <math>B = - \frac{1}{2}.</math>
+By product to sum, <math>\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})</math>
+<math>= \frac{1}{4}(1 + \cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}) = 1/8,</math> so <math>C = -1/8.</math>
+<math>ABC =\boxed{ \frac{1}{32}}.</math>
+~ ccx09
+==Solution 4 (Complex Numbers)==
+Using geometric series, we can show that <math>\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=0:</math> <cmath>\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=1+e^{\frac{2\pi i}{7}}+e^{\frac{4\pi i}{7}}+\cdots+e^{\frac{12\pi i}{7}}=\frac{1-1}{1-e^{\frac{2\pi i}{7}}}=0.</cmath>
+Alternatively, note that the <math>7</math>th roots of unity are <math>z=e^{\frac{2k\pi i}{7}}</math> for <math>i=0,1,2,\cdots,6,</math> in which <math>z^7-1=0.</math> By Vieta's Formulas, the sum of these seven roots is <math>0.</math>
+It follows that the real parts of these complex numbers must sum to <math>0,</math> so we get
+<cmath>\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}=\left(\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}\right)-1=-1,</cmath>
+from which <cmath>\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac12,</cmath>
+as it contributes half the real part of <math>\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}.</math>
+Two solutions follow from here:
+===Solution 4.1 (Trigonometric Identities)===
+We know that <math>\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}</math> are solutions of <cmath>\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \ \ \ \ \ (*)</cmath> as they can be verified geometrically or algebraically (by the identity <math>\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)</math>). Now, let <math>x=\cos\theta.</math> It follows that
+<cmath>\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*}</cmath>
+Rewriting <math>(*)</math> in terms of <math>x,</math> we have
+<cmath>\begin{align*}
+x+\left(2x^2-1\right)+\left(4x^3-3x\right)&=-\frac12 \\
+x^3+2x^2-2x-\frac12&=0 \\
+x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0.
+\end{align*}</cmath>
+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)
+===Solution 4.2 (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, <math>z^7=1.</math> Geometrically, it follows that
+<cmath>\begin{array}{ccccc}
+\cos{\frac{2\pi}{7}} &=& \frac{z+z^6}{2} &=& \frac{z+z^{-1}}{2}, \\ [2ex]
+\cos{\frac{4\pi}{7}} &=& \frac{z^2+z^5}{2} &=& \frac{z^2+z^{-2}}{2}, \\ [2ex]
+\cos{\frac{6\pi}{7}} &=& \frac{z^3+z^4}{2} &=& \frac{z^3+z^{-3}}{2}.
+\end{array}</cmath>
+Recall that <math>\sum_{k=0}^{6}z^k=0</math> (from which <math>\sum_{k=1}^{6}z^k=-1</math>), and 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)
+===Remark===
+Graph of the <math>7</math>th roots of unity in Desmos:
+* Rectangular Coordinate System: https://www.desmos.com/calculator/rzzk5geric (shown below)
+* Polar Coordinate System: https://www.desmos.com/calculator/xta6sfbnot
+[[File:2021 AMC 12A Problem 22 Exact Values + LaTeX.png|center]]
+Geometrically, it is clear that the imaginary parts of these complex numbers sum to <math>0.</math>
+~MRENTHUSIASM
 == Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==
@@ Line 96: / Line 181: @@
 ==See also==
 {{AMC12 box|year=2021|ab=A|num-b=21|num-a=23}}
+[[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

2021 AMC 12A (Problems • Answer Key • Resources)
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