Difference between revisions of "2023 AIME II Problems/Problem 8"
Mathboy100 (talk | contribs) (→Solution 2 (Moduli)) |
Mathboy100 (talk | contribs) (→Solution 2 (Moduli)) |
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<cmath>3z_1^2z_2^2z_3^2</cmath> | <cmath>3z_1^2z_2^2z_3^2</cmath> | ||
− | <cmath> = 3((\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7} + 1)^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)((\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> |
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:21, 17 February 2023
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
\[= 3((\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7} + 1)^2 + (\sin \frac{6\pi}{7] + \sin \frac{2\pi}{7})^2)((\cos \frac{12\pi}{7} + \cos \frac{4\pi}{7} + 1)^2 + (\sin \frac{12\pi}{7] + \sin \frac{4\pi}{7})^2)((\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} + 1)^2 + (\sin \frac{4\pi}{7] + \sin \frac{6\pi}{7})^2)\] (Error compiling LaTeX. Unknown error_msg)
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.