2013 AIME I Problems/Problem 10
Problem
There are nonzero integers ,
,
, and
such that the complex number
is a zero of the polynomial
. For each possible combination of
and
, let
be the sum of the zeros of
. Find the sum of the
's for all possible combinations of
and
.
Solution 1
Since is a root, by the Complex Conjugate Root Theorem,
must be the other imaginary root. Using
to represent the rational root, we have
Applying difference of squares, and regrouping, we have
So matching coefficients, we obtain
By Vieta's each so we just need to find the values of
in each pair.
We proceed by determining possible values for
,
, and
and using these to determine
and
.
If ,
so (r, s) =
Similarly, for ,
so the pairs
are
For ,
so the pairs
are
Then, since only but not
appears in the equations for
and
, we can ignore the plus minus sign for
.
The positive and negative values of r will cancel, so the sum of the
for
is
times the number of distinct
values (as each value of
generates a pair
).
Our answer is then
.
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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