Difference between revisions of "2015 AIME I Problems/Problem 10"
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==Solution 6 (Finite differences)== | ==Solution 6 (Finite differences)== | ||
Because a cubic must come in a "wave form" with two "humps", we can see that <math>f(1)=f(5)=f(6)</math>, and <math>f(2)=f(3)=f(7)</math>. By symmetry, <math>f(4)=0</math>. Now, WLOG let <math>f(1)=12</math>, and <math>f(2)=f(3)=-12</math>. Then, we can use finite differences to get that the third (constant) difference is <math>12</math>, and therefore <math>f(0)=12+(24+(24+12))=\boxed{72}</math>. | Because a cubic must come in a "wave form" with two "humps", we can see that <math>f(1)=f(5)=f(6)</math>, and <math>f(2)=f(3)=f(7)</math>. By symmetry, <math>f(4)=0</math>. Now, WLOG let <math>f(1)=12</math>, and <math>f(2)=f(3)=-12</math>. Then, we can use finite differences to get that the third (constant) difference is <math>12</math>, and therefore <math>f(0)=12+(24+(24+12))=\boxed{72}</math>. | ||
+ | -Stormersyle | ||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=9|num-a=11}} | {{AIME box|year=2015|n=I|num-b=9|num-a=11}} |
Revision as of 14:12, 25 November 2018
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.
Solution 4
The following solution is similar to solution 3, but assumes nothing. Let . Since
has degree 3,
has degree 6 and has roots 1,2,3,5,6, and 7. Therefore,
for some
. Hence
. Note that
. Since
has degree 3, so do
and
; and both have the same leading coefficient. Hence
and
for some
(else
is not cubic) where
is the same as the set
. Subtracting the second equation from the first, expanding, and collecting like terms, we have that
which must hold for all
. Since
we have that (1)
, (2)
and (3)
. Since
is the sum of 1,2,3,5,6, and 7, we have
so that by (1) we have
and
. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be
with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be
and
. Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that
. Since
is the leading coefficient of
, the leading coefficient of
is
. Thus the leading coefficient of
is 4, i.e.
. Then from earlier,
so that the answer is
.
Solution 5
By drawing the function, WLOG let . Then,
. Realize that if we shift
down 12, then this function
has roots
with leading coefficient
because
. Therefore
, and then
.
Solution 6 (Finite differences)
Because a cubic must come in a "wave form" with two "humps", we can see that , and
. By symmetry,
. Now, WLOG let
, and
. Then, we can use finite differences to get that the third (constant) difference is
, and therefore
.
-Stormersyle
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.