Difference between revisions of "2003 AIME II Problems/Problem 15"
m |
|||
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> | <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{ | + | Therefore 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 00:04, 30 January 2009
Problem
Let
![$P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).$](http://latex.artofproblemsolving.com/a/d/b/adb0a8bb705a8514b88ca50f80f549d9059435eb.png)
Let be the distinct zeros of
and let
for
where
and
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
We can rewrite the definition of as follows:
This can quite obviously 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 .
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 |