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

Revision as of 23:04, 29 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 $8+4+3 = \boxed{015}$ .

@@ Line 38: / Line 38: @@
 <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>.
+Therefore the answer is <math>8+4+3 = \boxed{015}</math>.
 == See also ==
 {{AIME box|year=2003|n=II|num-b=14|after=Last Question}}

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

Revision as of 23:04, 29 January 2009

Problem

Solution

See also