Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Reformatted all solutions, and arranged solutions based on thoroughness. Let me know if anyone is unhappy with this.)
(→‎Easy Video Solution by Scholars Foundation Without complex number and Euler's identity (Using Trigonometry + Vieta's Formula))
 
(20 intermediate revisions by 4 users not shown)
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>?
 
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 4 (Complex Numbers)==
+
==Solution 1 (Complex Numbers: Vieta's Formulas)==
−
By geometric series, we have <cmath>\sum_{k=0}^{6}e^{\frac{2k\pi i}{7}}=\frac{1-1}{1-e^{\frac{2\pi i}{7}}}=0.</cmath>
+
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
−
Alternatively, note that the <math>7</math>th roots of unity are <math>z=e^{\frac{2k\pi i}{7}}</math> for <math>k\in\{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 <math>\sum_{k=0}^{6}\cos\frac{2k\pi}{7}=0,</math> or <cmath>\sum_{k=1}^{6}\cos\frac{2k\pi}{7}=-1.</cmath>
 
−
Since <math>\cos\theta=\cos(2\pi-\theta)</math> holds for all <math>\theta,</math> we can rewrite this as
 
−
<cmath>\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*}</cmath>
 
−
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 roots of <cmath>\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \hspace{15mm} (*)</cmath> as they can be verified algebraically (by the identity <math>\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)</math> for all <math>\theta</math>) or geometrically (by the <b>Remark</b> section).
 
−
 
 
−
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 \\
 
−
4x^3+2x^2-2x-\frac12&=0 \\
 
−
x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0,
 
−
\end{align*}</cmath>
 
−
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)
 
−
 
 
−
===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, it follows that <math>z^7=1.</math> Geometrically (shown in the <b>Remark</b> section), we get
 
 
<cmath>\begin{alignat*}{4}
 
<cmath>\begin{alignat*}{4}
−
\cos{\frac{2\pi}{7}} &= \frac{z+z^6}{2} &&= \frac{z+z^{-1}}{2}, \\
+
\cos\frac{2\pi}{7} &= \frac{z+z^{-1}}{2} &&= \frac{z+z^6}{2}, \\
−
\cos{\frac{4\pi}{7}} &= \frac{z^2+z^5}{2} &&= \frac{z^2+z^{-2}}{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^4}{2} &&= \frac{z^3+z^{-3}}{2}.
+
\cos\frac{6\pi}{7} &= \frac{z^3+z^{-3}}{2} &&= \frac{z^3+z^4}{2}.
 
\end{alignat*}</cmath>
 
\end{alignat*}</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
+
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>
 +
 
 +
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*}
 
<cmath>\begin{align*}
 
abc&=[-(r+s+t)](rs+st+tr)(-rst) \\
 
abc&=[-(r+s+t)](rs+st+tr)(-rst) \\
Line 62: Line 26:
 
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
 
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
  
−
===Remark (Geometric Representations)===
+
==Solution 2 (Complex Numbers: Trigonometric Identities)==
−
Graph of the <math>7</math>th roots of unity:
+
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>
 +
Since <math>\cos\theta=\cos(2\pi-\theta)</math> holds for all <math>\theta,</math> this sum becomes
 +
<cmath>\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*}</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).
 
<asy>
 
<asy>
 
/* Made by MRENTHUSIASM */
 
/* Made by MRENTHUSIASM */
−
size(220);
+
size(200);  
−
import TrigMacros;
 
  
−
int big = 2;
+
int xMin = -2;
 +
int xMax = 2;
 +
int yMin = -2;
 +
int yMax = 2;
 
int numRays = 24;
 
int numRays = 24;
  
Line 78: Line 50:
 
   for (int i = 1; i < big+1; ++i)
 
   for (int i = 1; i < big+1; ++i)
 
   {
 
   {
−
     draw(Circle((0,0),i), gray+ linewidth(0.4));
+
     draw(Circle((0,0),i), gray+linewidth(0.4));
 
   }
 
   }
 
   for(int i=0;i<numRays;++i)  
 
   for(int i=0;i<numRays;++i)  
−
   draw(rotate(i*360/numRays)*((-big,0)--(big,0)),gray+ linewidth(0.4));
+
   draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
 
}
 
}
  
−
polarGrid(big, numRays);
+
//Draws the horizontal gridlines
−
rr_cartesian_axes(-big,big,-big,big,complexplane=true);
+
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
 
//The n such that we're taking the nth roots of unity
Line 105: Line 100:
 
for(int i = 0; i< n; ++i) dot(A[i],linewidth(3.5));  
 
for(int i = 0; i< n; ++i) dot(A[i],linewidth(3.5));  
 
</asy>
 
</asy>
−
Geometrically, it is clear that the imaginary parts of these complex numbers sum to <math>0.</math>  
+
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 \\
 +
&=4x^3-3x.
 +
\end{align*}</cmath>
 +
Rewriting <math>(\bigstar)</math> in terms of <math>x,</math> we have
 +
<cmath>\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*}</cmath>
 +
in which the roots are <math>x=\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}.</math>
  
−
~MRENTHUSIASM
+
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 3 (Trigonometric Identities)==
 
==Solution 3 (Trigonometric Identities)==
Line 117: Line 131:
 
Therefore, we get <math>a = -\left(-\frac12\right) = \frac12.</math></li><p>
 
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>
 
   <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{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.</cmath></li>
+
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{8\pi}7.</math> <p>
+
   <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:
 
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*}
 
<cmath>\begin{align*}
Line 132: Line 146:
 
Finally, the answer is <math>abc=\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac18\right)=\boxed{\textbf{(D) }\frac{1}{32}}.</math>
 
Finally, the answer is <math>abc=\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac18\right)=\boxed{\textbf{(D) }\frac{1}{32}}.</math>
  
−
~Tucker (Solution)
+
~Tucker
−
 
 
−
~MRENTHUSIASM (Reformatting)
 
  
 
== Solution 4 (Product-to-Sum Identity) ==
 
== Solution 4 (Product-to-Sum Identity) ==
−
Note sum of roots of unity equal zero, sum of 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>
+
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  
 
By the Product-to-Sum Identity, we have  
Line 143: Line 155:
 
\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) \\
 
\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) \\
 
&= \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},
 
&= -\frac{1}{2},
 
\end{align*}</cmath>
 
\end{align*}</cmath>
Line 158: Line 171:
 
Finally, we get <math>abc=\boxed{\textbf{(D) }\frac{1}{32}}.</math>
 
Finally, we get <math>abc=\boxed{\textbf{(D) }\frac{1}{32}}.</math>
  
−
~ ccx09 (Solution)
+
~ccx09
  
−
~MRENTHUSIASM (Reformatting)
+
== Easy Video Solution by Scholars Foundation Without Complex Numbers and Euler's Identity (Using Trigonometry + Vieta's Formula) ==
 +
 +
https://youtu.be/m4N4KN6_tA0
  
−
== Solution 5 (Approximations) ==
+
== Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula) ==
−
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
+
https://youtu.be/Im_WTIK0tss
−
<cmath>\begin{align*}
 
−
p + q + r &= a, \\
 
−
pq + qr + rp &= -b, \\
 
−
pqr &= c.
 
−
\end{align*}</cmath>
 
−
We use the Taylor series <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
+
~ pi_is_3.14
−
<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>\begin{alignat*}{8}
 
−
p + q + r &= a &&\approx -0.56, \\
 
−
pq + qr + rq &= -b &&\approx -0.524, \\
 
−
pqr &= c &&\approx 0.135.
 
−
\end{alignat*}</cmath>
 
−
We further approximate these values to <math>a \approx -0.5,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>
+
== Video Solution by MRENTHUSIASM (English & Chinese) ==
 +
https://youtu.be/X6oqEpFAJBk
  
−
In order to be more confident in your answer, you can go a few terms further in the Taylor series.
+
~MRENTHUSIASM
−
 
 
−
~ciceronii (Solution)
 
−
 
 
−
~MRENTHUSIASM (Reformatting)
 
−
 
 
−
== Video Solution by OmegaLearn (Euler's Identity + Vieta's ) ==
 
−
https://youtu.be/Im_WTIK0tss
 
−
 
 
−
~ pi_is_3.14
 
  
 
==See also==
 
==See also==

Latest revision as of 12:52, 21 November 2024

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

Easy Video Solution by Scholars Foundation Without Complex Numbers and Euler's Identity (Using Trigonometry + Vieta's Formula)

https://youtu.be/m4N4KN6_tA0

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_22&oldid=235241"