Difference between revisions of "2003 AIME II Problems/Problem 15"
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And we can proceed as above. | And we can proceed as above. | ||
+ | ==Solution 3== | ||
+ | |||
+ | As in Solution 1, we find that the roots 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> roots have an imaginary part of <math>\pm\frac{1}{2}</math>, <math>8</math> roots have an imaginary part of <math>\pm\frac{\sqrt{3}}{2}</math>, and <math>4</math> 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>. | ||
== 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 19:46, 31 January 2021
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 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.
roots have an imaginary part of
,
roots have an imaginary part of
, and
roots have an imaginary part of
. Therefore, the sum equals
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.