Difference between revisions of "2019 AIME I Problems/Problem 10"
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Revision as of 15:39, 16 October 2020
Problem 10
For distinct complex numbers , the polynomial
can be expressed as
, where
is a polynomial with complex coefficients and with degree at most
. The sum
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1
In order to begin this problem, we must first understand what it is asking for. The notation
simply asks for the absolute value of the sum of the product of the distinct unique roots of the polynomial taken two at a time or
Call this sum
.
Now we can begin the problem. Rewrite the polynomial as . Then we have that the roots of
are
.
By Vieta's formulas, we have that the sum of the roots of is
. Thus,
Similarly, we also have that the the sum of the roots of taken two at a time is
This is equal to
Now we need to find and expression for in terms of
. We note that
Thus,
.
Plugging this into our other Vieta equation, we have . This gives
. Since 343 is relatively prime to 9,
.
Solution 2
This is a quick fake solve using where
and only
.
By Vieta's, and
Rearranging gives
and
giving
.
Substituting gives which simplifies to
.
So, ,
,
,
Solution 3
Let . By Vieta's,
Then, consider the
term. To produce the product of two roots, the two roots can either be either
for some
, or
for some
. In the former case, this can happen in
ways, and in the latter case, this can happen in
ways. Hence,
and the requested sum is
.
Solution 4
Let Therefore,
. This is also equivalent to
for some real coefficients
and
and some polynomial
with degree
. We can see that the big summation expression is simply summing the product of the roots of
taken two at a time. By Vieta's, this is just the coefficient
. The first three terms of
can be bashed in terms of
and
to get
Thus,
and
. That is
.
See Also
2019 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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.