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

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<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{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}</math>
  
==Solution==
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==Solution 1==
  
 
Part 1: solving for c  
 
Part 1: solving for c  
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~Tucker
 
~Tucker
 +
 +
== Solution 2 (Approximation) ==
 +
Letting the roots be <math>p</math>, <math>q</math>, and <math>r</math>, Vietas gives
 +
\begin{align*}
 +
    p + q + r &= a \\
 +
    pq + qr + pq &= -b \\
 +
    pqr &= c.
 +
\end{align*}
 +
We use the Taylor series for <math>\cos x</math>,
 +
<cmath>\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}</cmath>
 +
to approximate the roots. Taking the sum up to <math>k = 3</math> yields a close approximation, so we have
 +
<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>
 +
<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>
 +
    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
 +
\begin{align*}
 +
    p + q + r &= a \simeq -0.56 \\
 +
    pq + qr + pr &= -b \simeq -0.524 \\
 +
    pqr &= c \simeq 0.135.
 +
\end{align*}
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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
  
 
== Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==
 
== Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==

Revision as of 14:57, 12 February 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 $abc$?

$\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}$

Solution 1

Part 1: 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:

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

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

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

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

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

$c = -\frac{1}8$


Part 2: starting to solve 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 is shown in the next part to equal $-\frac{1}2$, so $b = -\frac{1}2$


Part 3: solving for a and b as the sum of roots

$a$ 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 $a = -\left(-\frac{1}2\right) = \frac{1}2$

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 \begin{align*}

   p + q + r &= a \\
   pq + qr + pq &= -b \\
   pqr &= c.

\end{align*} 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

\begin{align*}

   p + q + r &= a \simeq -0.56 \\
   pq + qr + pr &= -b \simeq -0.524 \\
   pqr &= c \simeq 0.135.

\end{align*} 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

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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