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 "2023 AIME II Problems/Problem 8"

(→‎Solution 3)
m (Corrected a small mistake)
 
(18 intermediate revisions by 6 users not shown)
Line 31: Line 31:
 
\left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\
 
\left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\
 
& = 3 \cdot 2^3 \\
 
& = 3 \cdot 2^3 \\
−
& = \boxed{\textbf{(024) }}.
+
& = \boxed{\textbf{024}}.
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
Line 45: Line 45:
 
<cmath>|z_0||z_1||z_2||z_3||z_4||z_5||z_6|.</cmath>
 
<cmath>|z_0||z_1||z_2||z_3||z_4||z_5||z_6|.</cmath>
  
−
<math>z_0 = 0</math>, and we may observe that <math>z_x</math> and <math>z_{7-x}</math> are conjugates for any <math>x</math>, meaning that their magnitudes are the same. Thus, our product is
+
<math>z_0 = 3</math>, and we may observe that <math>z_x</math> and <math>z_{7-x}</math> are conjugates for any <math>x</math>, meaning that their magnitudes are the same. Thus, our product is
  
 
<cmath>3|z_1|^2|z_2|^2|z_3|^2</cmath>
 
<cmath>3|z_1|^2|z_2|^2|z_3|^2</cmath>
Line 85: Line 85:
 
~mathboy100
 
~mathboy100
  
−
 
+
==Solution 3 (Inspecting the exponents of powers of <math>\omega</math>)==
−
==Solution 3==
 
 
We write out the product in terms of <math>\omega</math>:
 
We write out the product in terms of <math>\omega</math>:
 
<cmath>\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).</cmath>
 
<cmath>\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).</cmath>
Line 113: Line 112:
  
  
−
Note that <math>\omega=e^{\frac{2\pi i}{7}}</math>. When raising <math>\omega</math> to a power, the numerator of the fraction is <math>2</math> times whatever power <math>\omega</math> is raised to, multiplied by <math>\pi i</math>. Since the period of <math>\omega</math> is <math>2\pi,</math> we multiply each array by <math>2</math> then reduce each entry <math>\mod{14}.</math>
+
Note that <math>\omega=e^{\frac{2\pi i}{7}}</math>. When raising <math>\omega</math> to a power, the numerator of the fraction is <math>2</math> times whatever power <math>\omega</math> is raised to, multiplied by <math>\pi i</math>. Since the period of <math>\omega</math> is <math>2\pi,</math> we multiply each array by <math>2</math> then reduce each entry <math>\mod{14},</math> as each entry in an array represents an exponent which <math>\omega</math> is raised to.
  
  
Line 133: Line 132:
 
To obtain the correct exponents, we seperately add each element of the lower row to one element of the top row.  
 
To obtain the correct exponents, we seperately add each element of the lower row to one element of the top row.  
  
−
Therefore (after reducing <math>\mod 14</math> again), we get the following sums:  
+
Therefore (after reducing <math>\mod 14</math> again), we get the following sets:  
  
 
<cmath>\textbf{(A)}\ \{0, 4, 6, 10, 0, 2, 8, 12, 0\}</cmath>
 
<cmath>\textbf{(A)}\ \{0, 4, 6, 10, 0, 2, 8, 12, 0\}</cmath>
Line 139: Line 138:
 
<cmath>\textbf{(C)}\ \{0, 12, 4, 2, 0, 8, 10, 8, 0\}.</cmath>
 
<cmath>\textbf{(C)}\ \{0, 12, 4, 2, 0, 8, 10, 8, 0\}.</cmath>
  
−
Raising <math>\omega</math> to the power of each of these, then multiplying over <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}</math> yields  
+
Raising <math>\omega</math> to the power of each element in every set then multiplying over <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}</math> yields  
 +
 
 +
<cmath>\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)\left(\sum_{b\in \textbf{(B)}} \omega^b\right)\left(\sum_{c\in \textbf{(C)}} \omega^c\right)</cmath>
 +
 
 +
<cmath>=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)^3</cmath>
  
−
<cmath>\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(\omega^0+\omega^4+\omega^6+\omega^{10}+\omega^0+\omega^2+\omega^8+\omega^{12}+\omega^0\right)^3</cmath>
+
<cmath>=\left(\omega^0+\omega^4+\omega^6+\omega^{10}+\omega^0+\omega^2+\omega^8+\omega^{12}+\omega^0\right)^3</cmath>
  
 
<cmath>=\left(3+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}+\omega^{12}\right)^3,</cmath>
 
<cmath>=\left(3+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}+\omega^{12}\right)^3,</cmath>
Line 163: Line 166:
  
 
-Benedict T (countmath1)
 
-Benedict T (countmath1)
 +
 +
 +
==Solution 4==
 +
 +
The product can be factored into <math>-(r-1)(s-1)(t-1)(r-w)(s-w)(t-w)(r-w^2)(s-w^2)(t-w^2)....(r-w^6)(s-w^6)(t-w^6)</math>,
 +
 +
 +
where <math>r,s,t</math> are the roots of the polynomial <math>x^3+x+1=0</math>.
 +
 +
 +
This is then <math>-(r^7-1)(s^7-1)(t^7-1)</math> because <math>(r^7-1)</math> and <math>(r-1)(r-w)(r-w^2)...(r-w^6)</math> share the same roots.
 +
 +
 +
To find <math>-(r^7-1)(s^7-1)(t^7-1)</math>,
 +
 +
 +
Notice that <math>(r^7-1)=(r-1)(r^6+r^5+r^4+r^3+r^2+r+1)</math>. Since r satisfies <math>x^3+x+1=0</math>, <math>r^6+r^4+r^3=0</math>
 +
 +
 +
Substituting, you are left with <math>r^5+r^2+r+1</math>. This is <math>r^2(r^3+1)+r+1</math>, and after repeatedly substituting <math>r^3+x+1=0</math> you are left with <math>-2r^3</math>.
 +
 +
 +
So now the problem is reduced to finding <math>-(r-1)(s-1)(t-1)(-2r^3)(-2s^3)(-2t^3)=8(rst)^3(r-1)(s-1)(t-1)</math>, and vietas gives you the result of <math>\boxed{24}</math> -resources
 +
 +
 +
==Video Solution by The Power of Logic==
 +
https://youtu.be/o6w9t43GpJs?si=aoe-uM3m5AIwpz_H
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2023|num-b=7|num-a=9|n=II}}
 
{{AIME box|year=2023|num-b=7|num-a=9|n=II}}
 +
 +
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:25, 14 October 2024

Problem

Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product\[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\]

Solution 1

For any $k\in Z$, we have, \begin{align*} & \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = \omega^{3 \cdot 7} + \omega^{2k + 7} + \omega^{3k} + \omega^{-2k + 3 \cdot 7} + \omega^7 + \omega^k + \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \\ & = 1 + \omega^{2k} + \omega^{3k} + \omega^{-2k} + 1 + \omega^k + \omega^{-3k} + \omega^{-k} + 1 \\ & = 2 + \omega^{-3k} \sum_{j=0}^6 \omega^{j k} \\ & = 2 + \omega^{-3k} \frac{1 - \omega^{7 k}}{1 - \omega^k} \\ & = 2 . \end{align*} The second and the fifth equalities follow from the property that $\omega^7 = 1$.

Therefore, \begin{align*} \Pi_{k=0}^6 \left( \omega^{3k} + \omega^k + 1 \right) & = \left( \omega^{3 \cdot 0} + \omega^0 + 1 \right) \Pi_{k=1}^3 \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = 3 \cdot 2^3 \\ & = \boxed{\textbf{024}}. \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2 (Moduli)

Because the answer must be a positive integer, it is just equal to the modulus of the product. Define $z_n = \left(\textrm{cis }\frac{2n\pi}{7}\right)^3 + \textrm{cis }\frac{2n\pi}{7} + 1$.

Then, our product is equal to

\[|z_0||z_1||z_2||z_3||z_4||z_5||z_6|.\]

$z_0 = 3$, and we may observe that $z_x$ and $z_{7-x}$ are conjugates for any $x$, meaning that their magnitudes are the same. Thus, our product is

\[3|z_1|^2|z_2|^2|z_3|^2\] \[= 3\left((\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7} + 1)^2 + (\sin \frac{6\pi}{7} + \sin \frac{2\pi}{7})^2\right) \left((\cos \frac{12\pi}{7} + \cos \frac{4\pi}{7} + 1)^2 + (\sin \frac{12\pi}{7} + \sin \frac{4\pi}{7})^2\right) \left((\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} + 1)^2 + (\sin \frac{4\pi}{7} + \sin \frac{6\pi}{7})^2\right)\]

Let us simplify the first term. Expanding, we obtain

\[\cos^2 \frac{6\pi}{7} + \cos^2 \frac{2\pi}{7} + 1 + 2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\cos \frac{6\pi}{7} + 2\cos \frac{2\pi}{7} + \sin^2 \frac{6\pi}{7} + \sin^2 \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7}.\]

Rearranging and cancelling, we obtain

\[3 + 2\cos \frac{6\pi}{7} + 2\cos \frac{2\pi}{7} + 2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7}.\]

By the cosine subtraction formula, we have $2\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} + 2\sin \frac{6\pi}{7}\sin \frac{2\pi}{7} = \cos \frac{6\pi - 2\pi}{7} = \cos \frac{4\pi}{7}$.

Thus, the first term is equivalent to

\[3 + 2(\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}).\]

Similarly, the second and third terms are, respectively,

\[3 + 2(\cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} + \cos \frac{12\pi}{7}),\textrm{ and}\] \[3 + 2(\cos \frac{6\pi}{7} + \cos \frac{12\pi}{7} + \cos \frac{4\pi}{7}).\]

Next, we have $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$. This is because

\[\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = \frac{1}{2}(\textrm{cis }\frac{2\pi}{7} + \textrm{cis }\frac{4\pi}{7} + \textrm{cis }\frac{6\pi}{7} + \textrm{cis }\frac{8\pi}{7} + \textrm{cis }\frac{10\pi}{7} + \textrm{cis }\frac{12\pi}{7})\]

\[= \frac{1}{2}(-1)\] \[= -\frac{1}{2}.\]

Therefore, the first term is simply $2$. We have $\cos x = \cos 2\pi - x$, so therefore the second and third terms can both also be simplified to $3 + 2(\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}) = 2$. Thus, our answer is simply

\[3 \cdot 2 \cdot 2 \cdot 2\] \[= \boxed{\mathbf{024}}.\]

~mathboy100

Solution 3 (Inspecting the exponents of powers of $\omega$)

We write out the product in terms of $\omega$: \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3(\omega^3+\omega+1)(\omega^6+\omega^2+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1)(\omega^{15}+\omega^5+1)(\omega^{18}+\omega^6+1).\]

Grouping the terms in the following way exploits the fact that $\omega^{7k}=1$ for an integer $k$, when multiplying out two adjacent products from left to right:

\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=(\omega^3+\omega+1)(\omega^{18}+\omega^6+1)(\omega^6+\omega^2+1)(\omega^{15}+\omega^5+1)(\omega^9+\omega^3+1)(\omega^{12}+\omega^4+1).\]


When multiplying two numbers with like bases, we add the exponents. We can now rewrite the exponents of each product (two at a time, where $1$ is treated as the identity) as a series of arrays:

\[\textbf{(A)}\begin{bmatrix} 3&1 &0 \\ 18&6&0\\ \end{bmatrix}\]

\[\textbf{(B)}\begin{bmatrix} 6&2 &0 \\ 15&5&0\\ \end{bmatrix}\]

\[\textbf{(C)}\begin{bmatrix} 9&3 &0 \\ 12&4&0\\ \end{bmatrix}.\]


Note that $\omega=e^{\frac{2\pi i}{7}}$. When raising $\omega$ to a power, the numerator of the fraction is $2$ times whatever power $\omega$ is raised to, multiplied by $\pi i$. Since the period of $\omega$ is $2\pi,$ we multiply each array by $2$ then reduce each entry $\mod{14},$ as each entry in an array represents an exponent which $\omega$ is raised to.


\[\textbf{(A)}\begin{bmatrix} 6&2 &0 \\ 8&12&0\\ \end{bmatrix}\]

\[\textbf{(B)}\begin{bmatrix} 12&4 &0 \\ 2&10&0\\ \end{bmatrix}\]

\[\textbf{(C)}\begin{bmatrix} 4&6 &0 \\ 10&8&0\\ \end{bmatrix}.\]

To obtain the correct exponents, we seperately add each element of the lower row to one element of the top row.

Therefore (after reducing $\mod 14$ again), we get the following sets:

\[\textbf{(A)}\ \{0, 4, 6, 10, 0, 2, 8, 12, 0\}\] \[\textbf{(B)}\ \{0, 8, 12, 6, 0, 4, 2, 10, 0\}\] \[\textbf{(C)}\ \{0, 12, 4, 2, 0, 8, 10, 8, 0\}.\]

Raising $\omega$ to the power of each element in every set then multiplying over $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$ yields

\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)\left(\sum_{b\in \textbf{(B)}} \omega^b\right)\left(\sum_{c\in \textbf{(C)}} \omega^c\right)\]

\[=\left(\sum_{a\in \textbf{(A)}} \omega^a\right)^3\]

\[=\left(\omega^0+\omega^4+\omega^6+\omega^{10}+\omega^0+\omega^2+\omega^8+\omega^{12}+\omega^0\right)^3\]

\[=\left(3+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}+\omega^{12}\right)^3,\] as these sets are all identical.

Summing as a geometric series,

\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(3+\frac{\omega^2(\omega^{12}-1)}{\omega^2-1}\right)^3\]

\[=\left(3+\frac{\omega^{14}-\omega^2}{\omega^2-1}\right)^3\]

\[=\left(3+\frac{1-\omega^2}{\omega^2-1}\right)^3\]

\[=(3-1)^3=8.\]

Therefore,

\[\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=8,\] and \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=3\cdot8=\boxed{\textbf{(024)}}.\]

-Benedict T (countmath1)


Solution 4

The product can be factored into $-(r-1)(s-1)(t-1)(r-w)(s-w)(t-w)(r-w^2)(s-w^2)(t-w^2)....(r-w^6)(s-w^6)(t-w^6)$,


where $r,s,t$ are the roots of the polynomial $x^3+x+1=0$.


This is then $-(r^7-1)(s^7-1)(t^7-1)$ because $(r^7-1)$ and $(r-1)(r-w)(r-w^2)...(r-w^6)$ share the same roots.


To find $-(r^7-1)(s^7-1)(t^7-1)$,


Notice that $(r^7-1)=(r-1)(r^6+r^5+r^4+r^3+r^2+r+1)$. Since r satisfies $x^3+x+1=0$, $r^6+r^4+r^3=0$


Substituting, you are left with $r^5+r^2+r+1$. This is $r^2(r^3+1)+r+1$, and after repeatedly substituting $r^3+x+1=0$ you are left with $-2r^3$.


So now the problem is reduced to finding $-(r-1)(s-1)(t-1)(-2r^3)(-2s^3)(-2t^3)=8(rst)^3(r-1)(s-1)(t-1)$, and vietas gives you the result of $\boxed{24}$ -resources


Video Solution by The Power of Logic

https://youtu.be/o6w9t43GpJs?si=aoe-uM3m5AIwpz_H

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2023_AIME_II_Problems/Problem_8&oldid=229782"