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

m (Solution 4 (Complex Numbers): Graphically -> Geometrically?)
m (Solution 4.1 (Function Composition): Auto-sizing parentheses.)
Line 128: Line 128:
 
\cos(3\theta)&=\cos(2\theta+\theta) \\
 
\cos(3\theta)&=\cos(2\theta+\theta) \\
 
&=\cos(2\theta)\cos\theta-\sin(2\theta)\sin\theta \\
 
&=\cos(2\theta)\cos\theta-\sin(2\theta)\sin\theta \\
&=(2x^2-1)x-\sin(2\theta)\sin\theta \\
+
&=\left(2x^2-1\right)x-\sin(2\theta)\sin\theta \\
&=(2x^2-1)x-2\sin^2\theta\cos\theta \\
+
&=\left(2x^2-1\right)x-2\sin^2\theta\cos\theta \\
&=(2x^2-1)x-2(1-\cos^2\theta)\cos\theta \\
+
&=\left(2x^2-1\right)x-2\left(1-\cos^2\theta\right)\cos\theta \\
&=(2x^2-1)x-2(1-x^2)x \\
+
&=\left(2x^2-1\right)x-2\left(1-x^2\right)x \\
 
&=2x^3-x-2x+2x^3 \\
 
&=2x^3-x-2x+2x^3 \\
 
&=4x^3-3x.
 
&=4x^3-3x.
Line 137: Line 137:
 
Rewriting <math>(*)</math> from above in terms of <math>x,</math> we have
 
Rewriting <math>(*)</math> from above in terms of <math>x,</math> we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
x+(2x^2-1)+(4x^3-3x)&=-\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.

Revision as of 05:12, 2 March 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 $8 \sin{\frac{2\pi}{7}}$

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

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 $\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=0:$ \begin{align*} \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. \end{align*}

Desmos graph of $e^{\frac{2k\pi i}{7}},$ where $k=0,1,2,\cdots,6$ (the $7$th roots of unity): https://www.desmos.com/calculator/kjnnkhgq6u

Geometrically, the imaginary parts of these complex numbers sum to $0.$ Using the above result, the real parts of these complex numbers sum to $0$ too. It follows that \[\sum_{k=1}^{6}e^{\frac{2k\pi i}{7}}=\left(\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}\right)-1=-1,\] 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 (Function Composition)

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 \\ &=\left(2x^2-1\right)x-\sin(2\theta)\sin\theta \\ &=\left(2x^2-1\right)x-2\sin^2\theta\cos\theta \\ &=\left(2x^2-1\right)x-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 $(*)$ from above 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*} It follows that $(a,b,c)=\left(\frac12,-\frac12,-\frac18\right),$ and $abc=\boxed{\textbf{(D) }\frac{1}{32}}.$

~MRENTHUSIASM (inspired by Peeyush Pandaya et al)

Solution 4.2 (Vieta's Formulas--Explains Solution 1 Using De Moivre's Theorem)

Let $z=e^{\frac{2\pi i}{7}}.$ Since $z$ is a $7$th root of unity, $z^7=1.$ Geometrically, it follows that \[\begin{array}{ccccc} \cos{\frac{2\pi}{7}} &=& \frac{z+z^6}{2} &=& \frac{z+z^{-1}}{2} \\ [2ex] \cos{\frac{4\pi}{7}} &=& \frac{z^2+z^5}{2} &=& \frac{z^2+z^{-2}}{2} \\ [2ex] \cos{\frac{7\pi}{7}} &=& \frac{z^3+z^4}{2} &=& \frac{z^3+z^{-3}}{2} \end{array}\]

Recall that $\sum_{k=0}^{6}z^k=0$ (so that $\sum_{k=1}^{6}z^k=-1$), 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 and the results above, 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)

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

https://youtu.be/Im_WTIK0tss

~ pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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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