Difference between revisions of "2003 AIME II Problems/Problem 15"
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== Problem == | == Problem == | ||
− | Let < | + | Let <cmath>P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).</cmath> Let <math>z_{1},z_{2},\ldots,z_{r}</math> be the distinct zeros of <math>P(x),</math> and let <math>z_{k}^{2} = a_{k} + b_{k}i</math> for <math>k = 1,2,\ldots,r,</math> where <math>a_{k}</math> and <math>b_{k}</math> are real numbers. Let |
<center><math>\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},</math></center> | <center><math>\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},</math></center> | ||
where <math>m, n,</math> and <math>p</math> are integers and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p.</math> | where <math>m, n,</math> and <math>p</math> are integers and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p.</math> |
Latest revision as of 18:00, 20 September 2020
Contents
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.
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.