Difference between revisions of "2003 AIME II Problems/Problem 15"
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== Problem == | == Problem == | ||
− | 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 |
− | < | ||
− | Let <math>z_{1},z_{2},\ldots,z_{r}</math> be the distinct zeros of <math>P(x),</math> and let <math>z_{ | ||
<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, | + | 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> |
== Solution == | == Solution == | ||
− | {{ | + | This can 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{015}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Note that <math>x^k + x^{k-1} + \dots + x + 1 = \frac{x^{k+1} - 1}{x - 1}</math>. Our sum can be reformed as <cmath>\frac{x(x^{47} - 1) + x^2(x^{45} - 1) + \dots + x^{24}(x - 1)}{x-1}</cmath> | ||
+ | |||
+ | So <cmath>\frac{x^{48} + x^{47} + x^{46} + \dots + x^{25} - x^{24} - x^{23} - \dots - x}{x-1} = 0</cmath> | ||
+ | |||
+ | <math>x(x^{47} + x^{46} + \dots - x - 1) = 0</math> | ||
+ | |||
+ | <math>x^{47} + x^{46} + \dots - x - 1 = 0</math> | ||
+ | |||
+ | <math>x^{47} + x^{46} + \dots + x + 1 = 2(x^{23} + x^{22} + \dots + x + 1)</math> | ||
+ | |||
+ | <math>\frac{x^{48} - 1}{x - 1} = 2\frac{x^{24} - 1}{x - 1}</math> | ||
+ | |||
+ | <math>x^{48} - 1 - 2x^{24} + 2 = 0</math> | ||
+ | |||
+ | <math>(x^{24} - 1)^2 = 0</math> | ||
+ | |||
+ | And we can proceed as above. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | As in Solution 1, we find that the roots of <math>P(x)</math> we care about are the 24th roots of unity except <math>1</math>. Therefore, the squares of these roots are the 12th roots of unity. In particular, every 12th root of unity is counted twice, except for <math>1</math>, which is only counted once. | ||
+ | |||
+ | The possible imaginary parts of the 12th roots of unity are <math>0</math>, <math>\pm\frac{1}{2}</math>, <math>\pm\frac{\sqrt{3}}{2}</math>, and <math>\pm 1</math>. We can disregard <math>0</math> because it doesn't affect the sum. | ||
+ | |||
+ | <math>8</math> squares of roots have an imaginary part of <math>\pm\frac{1}{2}</math>, <math>8</math> squares of roots have an imaginary part of <math>\pm\frac{\sqrt{3}}{2}</math>, and <math>4</math> squares of roots have an imaginary part of <math>\pm 1</math>. Therefore, the sum equals <math>8\left(\frac{1}{2}\right) + 8\left(\frac{\sqrt{3}}{2}\right) + 4(1) = 8 + 4\sqrt{3}</math>. | ||
+ | |||
+ | The answer is <math>8+4+3=\boxed{015}</math>. | ||
+ | |||
+ | ~rayfish | ||
+ | |||
+ | ==Video Solution by Sal Khan== | ||
+ | Part 1: https://www.youtube.com/watch?v=2eLAEMRrR7Q&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=3 | ||
+ | |||
+ | Part 2: https://www.youtube.com/watch?v=TljVBB7gxbE | ||
+ | |||
+ | Part 3: https://www.youtube.com/watch?v=JTpXK2mENH4 | ||
+ | |||
+ | - AMBRIGGS | ||
== 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}} | ||
+ | |||
+ | [[Category: Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 17:23, 30 July 2022
Problem
Let Let
be the distinct zeros of
and let
for
where
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
This can 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 .
Solution 2
Note that . Our sum can be reformed as
So
And we can proceed as above.
Solution 3
As in Solution 1, we find that the roots of we care about are the 24th roots of unity except
. Therefore, the squares of these roots are the 12th roots of unity. In particular, every 12th root of unity is counted twice, except for
, which is only counted once.
The possible imaginary parts of the 12th roots of unity are ,
,
, and
. We can disregard
because it doesn't affect the sum.
squares of roots have an imaginary part of
,
squares of roots have an imaginary part of
, and
squares of roots have an imaginary part of
. Therefore, the sum equals
.
The answer is .
~rayfish
Video Solution by Sal Khan
Part 1: https://www.youtube.com/watch?v=2eLAEMRrR7Q&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=3
Part 2: https://www.youtube.com/watch?v=TljVBB7gxbE
Part 3: https://www.youtube.com/watch?v=JTpXK2mENH4
- AMBRIGGS
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 |
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