2003 AIME II Problems/Problem 15
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Problem
Let
Let 
be the distinct zeros of 
and let 
for 
where 
and 
and 
are real numbers. Let
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 • Resources) | ||
| Preceded by Problem 14 | Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
