Difference between revisions of "2003 AIME II Problems/Problem 15"
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And we can proceed as above. | And we can proceed as above. | ||
+ | ==Solution 3== | ||
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+ | As in Solution 1, we find that the roots of <math>P(x)</math> we care about are the 24th roots of unity except <math>1</math>. Therefore, the squares of these roots are the 12th roots of unity. In particular, every 12th root of unity is counted twice, except for <math>1</math>, which is only counted once. | ||
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+ | The possible imaginary parts of the 12th roots of unity are <math>0</math>, <math>\pm\frac{1}{2}</math>, <math>\pm\frac{\sqrt{3}}{2}</math>, and <math>\pm 1</math>. We can disregard <math>0</math> because it doesn't affect the sum. | ||
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+ | <math>8</math> squares of roots have an imaginary part of <math>\pm\frac{1}{2}</math>, <math>8</math> squares of roots have an imaginary part of <math>\pm\frac{\sqrt{3}}{2}</math>, and <math>4</math> squares of roots have an imaginary part of <math>\pm 1</math>. Therefore, the sum equals <math>8\left(\frac{1}{2}\right) + 8\left(\frac{\sqrt{3}}{2}\right) + 4(1) = 8 + 4\sqrt{3}</math>. | ||
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+ | The answer is <math>8+4+3=\boxed{015}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2003|n=II|num-b=14|after=Last Question}} |
Latest revision as of 19:47, 31 January 2021
Problem
Let Let be the distinct zeros of and let for where and are real numbers. Let
where and are integers and is not divisible by the square of any prime. Find
Solution
This can be factored as:
Note that . So the roots of are exactly all -th complex roots of , except for the root .
Let . Then the distinct zeros of are .
We can clearly ignore the root as it does not contribute to the value that we need to compute.
The squares of the other roots are .
Hence we need to compute the following sum:
Using basic properties of the sine function, we can simplify this to
The five-element sum is just . We know that , , and . Hence our sum evaluates to:
Therefore the answer is .
Solution 2
Note that . Our sum can be reformed as
So
And we can proceed as above.
Solution 3
As in Solution 1, we find that the roots of we care about are the 24th roots of unity except . Therefore, the squares of these roots are the 12th roots of unity. In particular, every 12th root of unity is counted twice, except for , which is only counted once.
The possible imaginary parts of the 12th roots of unity are , , , and . We can disregard because it doesn't affect the sum.
squares of roots have an imaginary part of , squares of roots have an imaginary part of , and squares of roots have an imaginary part of . Therefore, the sum equals .
The answer is .
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.