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 00:03, 30 January 2009
Problem
Let
![$P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).$](http://latex.artofproblemsolving.com/a/d/b/adb0a8bb705a8514b88ca50f80f549d9059435eb.png)
Let be the distinct zeros of
and let
for
where
and
and
are real numbers. Let
![$\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},$](http://latex.artofproblemsolving.com/b/6/9/b6959c0d9b67d1a2d6914af2b95338ccf226924b.png)
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 |