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

m (Solution)
 
Line 1: Line 1:
 
== 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
+
Let <cmath>P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).</cmath> 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>

Latest revision as of 18:00, 20 September 2020

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 $a_{k}$ 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

This can 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 $8+4+3 = \boxed{015}$.

Solution 2

Note that $x^k + x^{k-1} + \dots + x + 1 = \frac{x^{k+1} - 1}{x - 1}$. Our sum can be reformed as \[\frac{x(x^{47} - 1) + x^2(x^{45} - 1) + \dots + x^{24}(x - 1)}{x-1}\]

So \[\frac{x^{48} + x^{47} + x^{46} + \dots + x^{25} - x^{24} - x^{23} - \dots - x}{x-1} = 0\]

$x(x^{47} + x^{46} + \dots - x - 1) = 0$

$x^{47} + x^{46} + \dots - x - 1 = 0$

$x^{47} + x^{46} + \dots + x + 1 = 2(x^{23} + x^{22} + \dots + x + 1)$

$\frac{x^{48} - 1}{x - 1} = 2\frac{x^{24} - 1}{x - 1}$

$x^{48} - 1 - 2x^{24} + 2 = 0$

$(x^{24} - 1)^2 = 0$

And we can proceed as above.

See also

2003 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS