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2023 AIME II Problems/Problem 8

Revision as of 10:03, 1 March 2023 by Tylerjdnkri (talk | contribs)

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

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

$z_0 = 0$, 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

Expand the pi notation for the problem: ∏_(k=0)^6▒〖(ω^3k 〗+ω^k+1)=(ω^(3*0)+ω^0+1)(ω^(3*1)+ω^1+1)(ω^(3*2)+ω^2+1)(ω^(3*3)+ω^3+1)(ω^(3*4)+ω^4+1)(ω^(3*5)+ω^5+1)(ω^(3*6)+ω^6+1)=(ω^0+ω^0+1)(ω^3+ω^1+1) (ω^6+ω^2+1)(ω^9+ω^3+1)(ω^12+ω^4+1)(ω^15+ω^5+1)(ω^18+ω^6 +1) =3(ω^3+ω^1+1) (ω^6+ω^2+1)(ω^9+ω^3+1)(ω^12+ω^4+1)(ω^15+ω^5+1)(ω^18+ω^6 +1) (Eq.1)

Notice that ω= cos⁡〖(2π/7)+〖i sin〗⁡〖(2π/7)〗 〗

cos⁡〖(x)+〖i sin〗⁡〖(x)=cis(x)=e^ix 〗 〗 ω=cis(2π/7)= e^(i*2π/7) Notice that 1,ω^1,ω^2,ω^3,ω^4,ω^5,ω^6 are the 7th roots of unity and the roots for ω^7-1=(ω-1)(ω^1+ω^2+ω^3+ω^4+ω^5+ω^6 )=0.

So ω^7=1 and 〖1+ω〗^1+ω^2+ω^3+ω^4+ω^5+ω^6=0

ω^(7x+y)=〖ω^7x*ω〗^y=(〖ω^7)〗^x*ω^y=1^x* ω^y=1* ω^y= ω^y where x,y ∈ Z

In Eq.1, we can simplify all w^x terms where x >= 7. ω^9=ω^(7+2)= ω^2 ω^12=ω^(7+5)= ω^5 ω^15=ω^(7*2+1)= ω^1 ω^18=ω^(7*2+4)= ω^4 Eq.1 simplifies to: 3(ω^3+ω^1+1) (ω^6+ω^2+1)(ω^2+ω^3+1)(ω^5+ω^4+1)(ω^1+ω^5+1)(ω^4+ω^6 +1) = 3(ω^3+ω^1+1) (ω^6+ω^2+1)(ω^3+ω^2+1)(ω^5+ω^4+1)(ω^5+ω^1+1)(ω^6+ω^4 +1) (Eq.2)

Rearrange the polynomials in Eq.2 to get: 3(ω^3+ω^1+1) (ω^3+ω^2+1)(ω^5+ω^4+1)(ω^5+ω^1+1)(ω^6+ω^2+1)(ω^6+ω^4 +1)

Now multiply each pair of polynomials: (ω^3+ω^1+1) (ω^3+ω^2+1)=ω^6+ω^5+ω^3+ω^4+ω^3+ω^1+ω^3+ω^2+1= ω^6+ω^5+ω^4+3ω^3+ω^2+ω^1+1=(ω^6+ω^5+ω^4+ω^3+ω^2+ω^1+1)+2ω^3=0+ 2ω^3=2ω^3 (Eq.3)


(ω^5+ω^4+1)(ω^5+ω^1+1)= ω^10+ω^6+ω^5+ω^9+ω^5+ω^4+ω^5+ω^1+1=ω^10+ω^9+ω^6+〖3ω〗^5+ω^4+ω^1+1=ω^3+ω^2+ω^6+〖3ω〗^5+ω^4+ω^1+1=〖(ω〗^6+ω^5+ω^4+ω^3+ω^2+ω^1+1)+2ω^5=0+ 2ω^5= 2ω^5 (Eq.4)


(ω^6+ω^2+1)(ω^6+ω^4 +1)= ω^12+ ω^10+ ω^6+ ω^8+ ω^6+ ω^2+ ω^6+ ω^4+ 1 

   = ω^12+ ω^10+  ω^8+  〖3ω〗^6+  ω^4+  ω^2+  1

= ω^5+ ω^3+ ω^1+ 〖3ω〗^6+ ω^4+ ω^2+ 1 = 〖(ω〗^6+ ω^5+ ω^4+ ω^3+ ω^2+ ω^1+ 1)+2ω^6=0+2ω^6= 2ω^6 (Eq.5)

Substituting Eq.3-5 into Eq.2, our final result is: 3(2ω^3 )(2ω^5 )(2ω^6 )=3*2*2*2*ω^(3+5+6)=24*ω^14=24*〖〖(w〗^7)〗^2= 24*1^2= 24*1= 24. Answer: 024

-TylerJdnkri


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

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