2015 AIME I Problems/Problem 10
Contents
[hide]Problem
Let be a third-degree polynomial with real coefficients satisfying
Find
.
Solution
Let =
.
Since
is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up.
By drawing a coordinate axis, and two lines representing 12 and -12, it is easy to see that f(1)=f(5)=f(6), and f(2)=f(3)=f(7); otherwise more bends would be required in the graph. Since only the absolute value of f(0) is required, there is no loss of generalization by stating that f(1)=12, and f(2)=-12. This provides the following system of equations.
Using any four of these functions as a system of equations yields
Solution 2
Express in terms of powers of
:
By the same argument as in the first Solution, we see that
is an odd function about the line
, so its coefficients
and
are 0. From there it is relatively simple to solve
(as in the above solution, but with a smaller system of equations):
and
Solution 3
Without loss of generality, let . (If
, then take
as the polynomial, which leaves
unchanged.) Because
is third-degree, write
where
clearly must be a permutation of
from the given condition. Thus
However, subtracting the two equations gives
, so comparing
coefficients gives
and thus both values equal to
. As a result,
. As a result,
and so
. Now, we easily deduce that
and so removing the without loss of generality gives
, which is our answer.
See Also
2015 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 |
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