Difference between revisions of "2023 AIME II Problems/Problem 8"
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<cmath>3|z_1|^2|z_2|^2|z_3|^2</cmath> | <cmath>3|z_1|^2|z_2|^2|z_3|^2</cmath> | ||
<cmath> = 3((\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7} + 1)^2 + (\sin \frac{6\pi}{7} + \sin \frac{2\pi}{7})^2) | <cmath> = 3((\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7} + 1)^2 + (\sin \frac{6\pi}{7} + \sin \frac{2\pi}{7})^2) | ||
− | ((\cos \frac{12\pi}{7} + \cos \frac{4\pi}{7} + 1)^2 + (\sin \frac{12\pi}{7} + \sin \frac{4\pi}{7})^2) | + | ((\cos \frac{12\pi}{7} + \cos \frac{4\pi}{7} + 1)^2 + (\sin \frac{12\pi}{7} + \sin \frac{4\pi}{7})^2) |
− | ((\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} + 1)^2 + (\sin \frac{4\pi}{7} + \sin \frac{6\pi}{7})^2)<cmath> | + | ((\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} + 1)^2 + (\sin \frac{4\pi}{7} + \sin \frac{6\pi}{7})^2)</cmath> |
Let us simplify the first term. Expanding, we obtain | Let us simplify the first term. Expanding, we obtain | ||
− | < | + | <cmath>\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}.</cmath> |
Rearranging and cancelling, we obtain | Rearranging and cancelling, we obtain | ||
− | < | + | <cmath>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}.</cmath> |
By the cosine subtraction formula, we have <math>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}</math>. | By the cosine subtraction formula, we have <math>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}</math>. |
Revision as of 13:28, 17 February 2023
Problem
Let where
Find the value of the product
Solution 1
For any , we have,
The second and the fifth equalities follow from the property that
.
Therefore,
~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 .
Then, our product is equal to
, and we may observe that
and
are conjugates for any
, meaning that their magnitudes are the same. Thus, our product is
Let us simplify the first term. Expanding, we obtain
Rearranging and cancelling, we obtain
By the cosine subtraction formula, we have .
Thus, the first term is equivalent to
Similarly, the second and third terms are, respectively,
Now, we have . This is because
Therefore, the first term is simply . We have
, so therefore the second and third terms can both also be simplified to
. Thus, our answer is simply
~mathboy100
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.