Difference between revisions of "2003 AIME II Problems/Problem 15"
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== Solution == | == Solution == | ||
− | {{ | + | We can rewrite the definition of <math>P(x)</math> as follows: |
+ | |||
+ | <cmath> P(x) = x^{47} + 2x^{46} + \cdots + 23x^{25} + 24x^{24} + 23x^{23} + \cdots + 2x^2 + x </cmath> | ||
+ | |||
+ | This can quite obviously be factored as: | ||
+ | |||
+ | <cmath> P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2 </cmath> | ||
+ | |||
+ | Note that <math> \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 </math>. | ||
+ | So the roots of <math>x^{23} + x^{22} + \cdots + x^2 + x + 1</math> are exactly all <math>24</math>-th complex roots of <math>1</math>, except for the root <math>x=1</math>. | ||
+ | |||
+ | Let <math>\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}</math>. Then the distinct zeros of <math>P</math> are <math>0,\omega,\omega^2,\dots,\omega^{23}</math>. | ||
+ | |||
+ | We can clearly ignore the root <math>x=0</math> as it does not contribute to the value that we need to compute. | ||
+ | |||
+ | The squares of the other roots are <math>\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}</math>. | ||
+ | |||
+ | Hence we need to compute the following sum: | ||
+ | |||
+ | <cmath>R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|</cmath> | ||
+ | |||
+ | Using basic properties of the sine function, we can simplify this to | ||
+ | |||
+ | <cmath>R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)</cmath> | ||
+ | |||
+ | The five-element sum is just <math>\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ</math>. | ||
+ | We know that <math>\sin 30^\circ = \sin 150^\circ = \frac 12</math>, <math>\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2</math>, and <math>\sin 90^\circ = 1</math>. | ||
+ | Hence our sum evaluates to: | ||
+ | |||
+ | <cmath>R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3</cmath> | ||
+ | |||
+ | Therefore the answer is <math>8+4+3 = \boxed{15}</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}} |
Revision as of 23:03, 29 January 2009
Problem
Let
Let be the distinct zeros of and let for where and and are real numbers. Let
where and are integers and is not divisible by the square of any prime. Find
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 .
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 |