Difference between revisions of "2003 AIME II Problems/Problem 15"
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== Problem == | == Problem == | ||
− | Let <math>P(x) = x+2x^{2}+3x^{3} \dots 24x^{24}+23x^{25}+22x^{26} \dots x^{47}</math>. 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 | + | [quote=2003 AIME II #15]Let <math>P(x) = x+2x^{2}+3x^{3} \dots 24x^{24}+23x^{25}+22x^{26} \dots x^{47}</math>. 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>[/quote] |
== Solution == | == Solution == |
Revision as of 18:48, 31 May 2017
Contents
Problem
[quote=2003 AIME II #15]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 [/quote]
Solution
We can rewrite the definition of as follows:
This can quite obviously 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.