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

m (→‎Solution 4 (Complex Numbers): Added in the polar coordinates system view.)
(→‎Solution 3 (Trigonometric Identities))
 
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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>?
 
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{\sqrt[3]7}{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==
+
==Solution 1 (Complex Numbers: 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, we have <math>z^7=1.</math> For all integers <math>k,</math> note that <math>\cos\frac{2k\pi}{7}=\operatorname{Re}\left(z^k\right)=\operatorname{Re}\left(z^{-k}\right)</math> and <math>\sin\frac{2k\pi}{7}=\operatorname{Im}\left(z^k\right)=-\operatorname{Im}\left(z^{-k}\right).</math> It follows that
 +
<cmath>\begin{alignat*}{4}
 +
\cos\frac{2\pi}{7} &= \frac{z+z^{-1}}{2} &&= \frac{z+z^6}{2}, \\
 +
\cos\frac{4\pi}{7} &= \frac{z^2+z^{-2}}{2} &&= \frac{z^2+z^5}{2}, \\
 +
\cos\frac{6\pi}{7} &= \frac{z^3+z^{-3}}{2} &&= \frac{z^3+z^4}{2}.
 +
\end{alignat*}</cmath>
 +
By geometric series, we conclude that <cmath>\sum_{k=0}^{6}z^k=\frac{1-1}{1-z}=0.</cmath>
 +
Alternatively, recall that the <math>7</math>th roots of unity satisfy the equation <math>z^7-1=0.</math> By Vieta's Formulas, the sum of these seven roots is <math>0.</math>
  
−
Part 1: solving for a
+
As a result, we get <cmath>\sum_{k=1}^{6}z^k=-1.</cmath>
 +
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)
  
−
<math>a</math> is the negation of the sum of roots
+
==Solution 2 (Complex Numbers: Trigonometric Identities)==
−
 
+
Let <math>z=e^{\frac{2\pi i}{7}}.</math> In Solution 1, we conclude that <math>\sum_{k=1}^{6}z^k=-1,</math> so <cmath>\sum_{k=1}^{6}\operatorname{Re}\left(z^k\right)=\sum_{k=1}^{6}\cos\frac{2k\pi}{7}=-1.</cmath>
−
<math>a = - \left( \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 \right)</math>
+
Since <math>\cos\theta=\cos(2\pi-\theta)</math> holds for all <math>\theta,</math> this sum becomes
−
 
+
<cmath>\begin{align*}
−
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>.
+
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.
−
If we subtract 1, and condense identical terms, we get:
+
\end{align*}</cmath>
−
 
+
Note that <math>\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}</math> are roots of <cmath>\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \hspace{15mm} (\bigstar)</cmath> as they can be verified either algebraically (by the identity <math>\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)</math>) or geometrically (by the graph below).
−
<math>2\cos \frac{2\pi}7 + 2\cos \frac{4\pi}7 + 2\cos \frac{6\pi}7 = -1</math>
+
<asy>
−
 
+
/* Made by MRENTHUSIASM */
−
Therefore, we have <math>a = -\left(-\frac{1}2\right) = \frac{1}2</math>
+
size(200);
−
 
 
−
 
 
−
Part 2: solving for b
 
−
 
 
−
<math>b</math> is the sum of roots two at a time by Vieta's
 
−
 
 
−
<math>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</math>
 
−
 
 
−
We know that <math>\cos \alpha \cos \beta = \frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}</math>
 
−
 
 
−
By plugging all the parts in we get:
 
−
 
 
−
<math> \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 </math>
 
−
 
 
−
Which ends up being:
 
−
 
 
−
<math> \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 </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 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>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>
 
−
 
 
−
 
 
−
 
 
−
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>.
 
  
−
~Tucker
+
int xMin = -2;
 +
int xMax = 2;
 +
int yMin = -2;
 +
int yMax = 2;
 +
int numRays = 24;
  
−
== Solution 2 (Approximation) ==
+
//Draws a polar grid that goes out to a number of circles
−
Letting the roots be <math>p</math>, <math>q</math>, and <math>r</math>, Vietas gives
+
//equal to big, with numRays specifying the number of rays:
−
<cmath>p + q + r = a</cmath>
+
void polarGrid(int big, int numRays)  
−
<cmath>pq + qr + pq = -b</cmath>
+
{
−
<cmath>pqr = c</cmath>
+
  for (int i = 1; i < big+1; ++i)
−
We use the Taylor series for <math>\cos x</math>,
+
  {
−
<cmath>\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}</cmath>
+
    draw(Circle((0,0),i), gray+linewidth(0.4));
−
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>
+
  for(int i=0;i<numRays;++i)
−
<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>
+
  draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
−
<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
 
−
<cmath>p + q + r = a \simeq -0.56</cmath>
 
−
<cmath>pq + qr + pr = -b \simeq -0.524</cmath>
 
−
<cmath>pqr = c \simeq 0.135</cmath>
 
−
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
 
  
−
Remark: In order to be more confident in your answer, you can go a few terms further in the Taylor series.
+
//Draws the horizontal gridlines
 +
void horizontalLines()
 +
{
 +
  for (int i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
 +
  }
 +
}
  
−
== Solution 3 (Only using Product to Sum Identity) ==
+
//Draws the vertical gridlines
−
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>
+
void verticalLines()
−
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>
+
{
 +
  for (int i = xMin+1; i < xMax; ++i)
 +
  {
 +
    draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
 +
  }
 +
}
  
−
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>
+
horizontalLines();
−
<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>
+
verticalLines();
 +
polarGrid(xMax,numRays);
 +
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
 +
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
 +
label("Re",(xMax,0),(2,0));
 +
label("Im",(0,yMax),(0,2));
  
−
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>
+
//The n such that we're taking the nth roots of unity
−
<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>
+
int n = 7;
  
−
<math>ABC =\boxed{ \frac{1}{32}}.</math>
+
pair A[];
 +
for(int i = 0; i <= n-1; i+=1) {
 +
  A[i] = rotate(360*i/n)*(1,0);
 +
}
  
−
~ ccx09
+
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);
 +
}
  
−
==Solution 4 (Complex Numbers)==
+
draw(Circle((0,0),1),red);
  
−
Using geometric series, we can show that <math>\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=0:</math>
+
for(int i = 0; i< n; ++i) dot(A[i],linewidth(3.5));
−
<cmath>\begin{align*}
+
</asy>
−
\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}} \\
+
Let <math>x=\cos\theta.</math> It follows that
−
&=\frac{1-1}{1-e^{\frac{2\pi i}{7}}} \\
 
−
&=0.
 
−
\end{align*}</cmath>
 
−
 
 
−
<b>Graph of <math>\boldsymbol{e^{\frac{2k\pi i}{7}}}</math> in Desmos, where <math>\boldsymbol{k=0,1,2,\cdots,6}</math> (the <math>\boldsymbol{7}</math>th roots of unity)</b>
 
−
 
 
−
<math>\quad\bullet</math> Rectangular Coordinate System: https://www.desmos.com/calculator/rzzk5geric (shown below)
 
−
 
 
−
<math>\quad\bullet</math> Polar Coordinate System: https://www.desmos.com/calculator/xta6sfbnot
 
−
[[File:2021 AMC 12A Problem 22 Exact Values + LaTeX.png|center]]
 
−
Geometrically, the imaginary parts of these complex numbers sum to <math>0.</math> Using the above result, the real parts of these complex numbers sum to <math>0</math> too. It follows that <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 (Function Composition)===
 
−
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*}
 
<cmath>\begin{align*}
 
\cos(2\theta)&=2\cos^2\theta-1 \\
 
\cos(2\theta)&=2\cos^2\theta-1 \\
Line 132: Line 106:
 
\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 \\
−
&=\left(2x^2-1\right)x-\sin(2\theta)\sin\theta \\
+
&=\cos(2\theta)\cos\theta-2\sin^2\theta\cos\theta \\
−
&=\left(2x^2-1\right)x-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-\cos^2\theta\right)\cos\theta \\
 
 
&=\left(2x^2-1\right)x-2\left(1-x^2\right)x \\
 
&=\left(2x^2-1\right)x-2\left(1-x^2\right)x \\
−
&=2x^3-x-2x+2x^3 \\
 
 
&=4x^3-3x.
 
&=4x^3-3x.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Rewriting <math>(*)</math> from above in terms of <math>x,</math> we have
+
Rewriting <math>(\bigstar)</math> in terms of <math>x,</math> we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
x+\left(2x^2-1\right)+\left(4x^3-3x\right)&=-\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,
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
It follows that <math>(a,b,c)=\left(\frac12,-\frac12,-\frac18\right),</math> and <math>abc=\boxed{\textbf{(D) }\frac{1}{32}}.</math>
+
in which the roots are <math>x=\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}.</math>
 +
 
 +
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)
 
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
  
−
===Solution 4.2 (Vieta's Formulas--Explains Solution 1 Using De Moivre's Theorem)===
+
==Solution 3 (Trigonometric Identities)==
−
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
+
We solve for <math>a,b,</math> and <math>c</math> separately:
−
<cmath>\begin{array}{ccccc}
+
<ol style="margin-left: 1.5em;">
−
\cos{\frac{2\pi}{7}} &=& \frac{z+z^6}{2} &=& \frac{z+z^{-1}}{2} \\ [2ex]
+
  <li>Solve for <math>a:</math> By Vieta's Formulas, we have <math>a = - \left( \cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 \right).</math><p>
−
\cos{\frac{4\pi}{7}} &=& \frac{z^2+z^5}{2} &=& \frac{z^2+z^{-2}}{2} \\ [2ex]
+
The real parts of the <math>7</math>th 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> <p>
−
\cos{\frac{7\pi}{7}} &=& \frac{z^3+z^4}{2} &=& \frac{z^3+z^{-3}}{2}
+
Note that <math>\cos\theta=\cos(2\pi-\theta)</math> for all <math>\theta.</math> Excluding <math>1,</math> the other six roots add to <cmath>2\left(\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7\right) = -1,</cmath> from which <cmath>\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7 = -\frac12.</cmath>
−
\end{array}</cmath>
+
Therefore, we get <math>a = -\left(-\frac12\right) = \frac12.</math></li><p>
 +
  <li>Solve for <math>b:</math> By Vieta's Formulas, we have <math>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.</math><p>
 +
Note that <math>\cos \alpha \cos \beta = \frac{ \cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) }{2}</math> for all <math>\alpha</math> and <math>\beta.</math> Therefore, we get <cmath>b=\frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{2\pi}7}2=\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7=-\frac12.</cmath></li>
 +
  <li>Solve for <math>c:</math> By Vieta's Formulas, we have <math>c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{6\pi}7=-\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7.</math> <p>
 +
We multiply both sides by <math>8 \sin{\frac{2\pi}{7}},</math> then repeatedly apply the angle addition formula for sine:
 +
<cmath>\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*}</cmath>
 +
Therefore, we get <math>c = -\frac18.</math><p>
 +
</li>
 +
</ol>
 +
Finally, the answer is <math>abc=\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac18\right)=\boxed{\textbf{(D) }\frac{1}{32}}.</math>
  
−
Recall that <math>\sum_{k=0}^{6}z^k=0</math> (so that <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 and the results above, the answer is
+
~Tucker
 +
 
 +
== Solution 4 (Product-to-Sum Identity) ==
 +
Note that the sum of the roots of unity equal zero, so the sum of their real parts equal zero, and <math>\operatorname{Re}\left(\omega^{m}\right) = \operatorname{Re}\left(\omega^{-m}\right).</math> We have <cmath>\cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} = \frac12(0 - \cos 0) = -\frac12,</cmath> so <math>a = \frac{1}{2}.</math>
 +
 
 +
By the Product-to-Sum Identity, we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
abc&=[-(r+s+t)](rs+st+tr)(-rst) \\
+
\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) \\
−
&=(r+s+t)(rs+st+tr)(rst) \\
+
&= \frac{1}{2}\left(2 \cos \frac{2 \pi}{7} + 2 \cos \frac{4 \pi}{7} + 2 \cos \frac{6 \pi}{7}\right) \\
−
&=\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) \\
+
&= \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \\
−
&=\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}{2},
−
&=\frac{1}{32}(-1)(-1)(1) \\
+
\end{align*}</cmath>
−
&=\boxed{\textbf{(D) }\frac{1}{32}}.
+
so <math>b = -\frac{1}{2}.</math>
 +
 
 +
By the Product-to-Sum Identity, we have
 +
<cmath>\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*}</cmath>
 
\end{align*}</cmath>
 +
so <math>c = -\frac{1}{8}.</math>
 +
 +
Finally, we get <math>abc=\boxed{\textbf{(D) }\frac{1}{32}}.</math>
  
−
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
+
~ccx09
  
−
== Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==
+
== Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula) ==
 
https://youtu.be/Im_WTIK0tss
 
https://youtu.be/Im_WTIK0tss
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
== Video Solution by MRENTHUSIASM (English & Chinese) ==
 +
https://youtu.be/X6oqEpFAJBk
 +
 +
~MRENTHUSIASM
  
 
==See also==
 
==See also==

Latest revision as of 01:25, 21 September 2023

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 (Complex Numbers: Vieta's Formulas)

Let $z=e^{\frac{2\pi i}{7}}.$ Since $z$ is a $7$th root of unity, we have $z^7=1.$ For all integers $k,$ note that $\cos\frac{2k\pi}{7}=\operatorname{Re}\left(z^k\right)=\operatorname{Re}\left(z^{-k}\right)$ and $\sin\frac{2k\pi}{7}=\operatorname{Im}\left(z^k\right)=-\operatorname{Im}\left(z^{-k}\right).$ It follows that \begin{alignat*}{4} \cos\frac{2\pi}{7} &= \frac{z+z^{-1}}{2} &&= \frac{z+z^6}{2}, \\ \cos\frac{4\pi}{7} &= \frac{z^2+z^{-2}}{2} &&= \frac{z^2+z^5}{2}, \\ \cos\frac{6\pi}{7} &= \frac{z^3+z^{-3}}{2} &&= \frac{z^3+z^4}{2}. \end{alignat*} By geometric series, we conclude that \[\sum_{k=0}^{6}z^k=\frac{1-1}{1-z}=0.\] Alternatively, recall that the $7$th roots of unity satisfy the equation $z^7-1=0.$ By Vieta's Formulas, the sum of these seven roots is $0.$

As a result, we get \[\sum_{k=1}^{6}z^k=-1.\] 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)

Solution 2 (Complex Numbers: Trigonometric Identities)

Let $z=e^{\frac{2\pi i}{7}}.$ In Solution 1, we conclude that $\sum_{k=1}^{6}z^k=-1,$ so \[\sum_{k=1}^{6}\operatorname{Re}\left(z^k\right)=\sum_{k=1}^{6}\cos\frac{2k\pi}{7}=-1.\] Since $\cos\theta=\cos(2\pi-\theta)$ holds for all $\theta,$ this sum becomes \begin{align*} 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*} Note that $\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}$ are roots of \[\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \hspace{15mm} (\bigstar)\] as they can be verified either algebraically (by the identity $\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)$) or geometrically (by the graph below). [asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 2; int yMin = -2; int yMax = 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)); } //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } horizontalLines(); verticalLines(); polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),(2,0)); label("Im",(0,yMax),(0,2)); //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] 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 \\ &=4x^3-3x. \end{align*} Rewriting $(\bigstar)$ 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 3 (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 \[b=\frac{\cos \frac{6\pi}7 + \cos \frac{2\pi}7}2 + \frac{\cos \frac{6\pi}7 + \cos \frac{4\pi}7}2 + \frac{\cos \frac{4\pi}7 + \cos \frac{2\pi}7}2=\cos \frac{2\pi}7 + \cos \frac{4\pi}7 + \cos \frac{6\pi}7=-\frac12.\]

  3. Solve for $c:$ By Vieta's Formulas, we have $c = -\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{6\pi}7=-\cos \frac{2\pi}7 \cos \frac{4\pi}7 \cos \frac{8\pi}7.$

    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 4 (Product-to-Sum Identity)

Note that the sum of the roots of unity equal zero, so the sum of their 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) \\ &= \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \\ &= -\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

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

https://youtu.be/Im_WTIK0tss

~ pi_is_3.14

Video Solution by MRENTHUSIASM (English & Chinese)

https://youtu.be/X6oqEpFAJBk

~MRENTHUSIASM

See also

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

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