Difference between revisions of "2023 AIME II Problems/Problem 8"

Revision as of 18:00, 17 February 2023

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

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. Since the period of $\omega$ is $2\pi,$ we multiply each array by $2$ then reduce each entry $\mod{14}.$

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

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

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

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 sums:

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

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

$\frac{1}{3} \prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right)=\left(\omega^0+\omega^4+\omega^6+\omega^{10}+\omega^0+\omega^2+\omega^8+\omega^{12}+\omega^0\right)^3$

$=\left(3+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}+\omega^{12}\right)^3,$ 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)

@@ Line 146: / Line 146: @@
 as these sets are all identical.
-Summing as a geometric sereis,
+Summing as a geometric series,
 <cmath>\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</cmath>
-<cmath>=\left(3+\frac{\omega^14-\omega^2}{\omega^2-1}\right)^3</cmath>
+<cmath>=\left(3+\frac{\omega^{14}-\omega^2}{\omega^2-1}\right)^3</cmath>
 <cmath>=\left(3+\frac{1-\omega^2}{\omega^2-1}\right)^3</cmath>

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

Page

Toolbox

Search