Difference between revisions of "2003 AIME II Problems/Problem 15"

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== Solution ==
 
== Solution ==
{{solution}}
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We can rewrite the definition of <math>P(x)</math> as follows:
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<cmath> P(x) = x^{47} + 2x^{46} + \cdots + 23x^{25} + 24x^{24} + 23x^{23} + \cdots + 2x^2 + x </cmath>
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This can quite obviously be factored as:
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<cmath> P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2 </cmath>
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Note that <math> \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 </math>.
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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>.
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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>.
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We can clearly ignore the root <math>x=0</math> as it does not contribute to the value that we need to compute.
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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>.
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Hence we need to compute the following sum:
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<cmath>R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|</cmath>
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 +
Using basic properties of the sine function, we can simplify this to
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<cmath>R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)</cmath>
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The five-element sum is just <math>\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ</math>.
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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>.
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Hence our sum evaluates to:
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<cmath>R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3</cmath>
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Therefore the answer is <math>8+4+3 = \boxed{15}</math>.
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== 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

$P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).$

Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2} = a_{k} + b_{k}i$ for $k = 1,2,\ldots,r,$ where $i = \sqrt { - 1},$ and $a_{k}$ and $b_{k}$ are real numbers. Let

$\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},$

where $m,$ $n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m + n + p.$

Solution

We can rewrite the definition of $P(x)$ as follows:

\[P(x) = x^{47} + 2x^{46} + \cdots + 23x^{25} + 24x^{24} + 23x^{23} + \cdots + 2x^2 + x\]

This can quite obviously be factored as:

\[P(x) = x\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right)^2\]

Note that $\left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1$. So the roots of $x^{23} + x^{22} + \cdots + x^2 + x + 1$ are exactly all $24$-th complex roots of $1$, except for the root $x=1$.

Let $\omega=\cos \frac{360^\circ}{24} + i\sin \frac{360^\circ}{24}$. Then the distinct zeros of $P$ are $0,\omega,\omega^2,\dots,\omega^{23}$.

We can clearly ignore the root $x=0$ as it does not contribute to the value that we need to compute.

The squares of the other roots are $\omega^2,~\omega^4,~\dots,~\omega^{24}=1,~\omega^{26}=\omega^2,~\dots,~\omega^{46}=\omega^{22}$.

Hence we need to compute the following sum:

\[R = \sum_{k = 1}^{23} \left|\, \sin \left( k\cdot \frac{360^\circ}{12} \right) \right|\]

Using basic properties of the sine function, we can simplify this to

\[R = 4 \cdot \sum_{k = 1}^{5} \sin \left( k\cdot \frac{360^\circ}{12} \right)\]

The five-element sum is just $\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ + \sin 150^\circ$. We know that $\sin 30^\circ = \sin 150^\circ = \frac 12$, $\sin 60^\circ = \sin 120^\circ = \frac{\sqrt 3}2$, and $\sin 90^\circ = 1$. Hence our sum evaluates to:

\[R = 4 \cdot \left( 2\cdot \frac 12 + 2\cdot \frac{\sqrt 3}2 + 1 \right) = 8 + 4\sqrt 3\]

Therefore the answer is $8+4+3 = \boxed{15}$.


See also

2003 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
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All AIME Problems and Solutions
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