Difference between revisions of "2015 AIME I Problems/Problem 10"
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Let <math>f(x)</math> = <math>ax^3+bx^2+cx+d</math>. | Let <math>f(x)</math> = <math>ax^3+bx^2+cx+d</math>. | ||
Since <math>f(x)</math> 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. | Since <math>f(x)</math> 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 <math>12</math> and <math>-12</math>, it is easy to see that <math>f(1)=f(5)=f(6)</math>, and <math>f(2)=f(3)=f(7)</math>; 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 <math>f(1)=12</math>, and <math>f(2)=-12</math>. This provides the following system of equations. | + | By drawing a coordinate axis, and two lines representing <math>12</math> and <math>-12</math>, it is easy to see that <math>f(1)=f(5)=f(6)</math>, and <math>f(2)=f(3)=f(7)</math>; otherwise more bends would be required in the graph. Since only the absolute value of <math>f(0)</math> is required, there is no loss of generalization by stating that <math>f(1)=12</math>, and <math>f(2)=-12</math>. This provides the following system of equations. |
<cmath> a + b + c + d = 12</cmath> | <cmath> a + b + c + d = 12</cmath> | ||
<cmath>8a + 4b + 2c + d = -12</cmath> | <cmath>8a + 4b + 2c + d = -12</cmath> | ||
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− | ==Solution 8(First few steps of solution 1)== | + | ==Solution 8 (First few steps of solution 1)== |
We can rewrite the function as <math>f(x) - 12 = z(x-1)(x-5)(x-6)</math> and <math>f(x) + 12 = z(x-2)(x-3)(x-7)</math>. Since we need to find <math>|f(0)|</math>, substitute 0 for x in these two equations. | We can rewrite the function as <math>f(x) - 12 = z(x-1)(x-5)(x-6)</math> and <math>f(x) + 12 = z(x-2)(x-3)(x-7)</math>. Since we need to find <math>|f(0)|</math>, substitute 0 for x in these two equations. | ||
Line 62: | Line 62: | ||
==Solution 9 (Cheese Solution)== | ==Solution 9 (Cheese Solution)== | ||
− | Let the leading coefficient of | + | Let the leading coefficient of <math>f(x)</math> be <math>a</math>. Then, it is obvious that <math>(f(x) - 12)(f(x) + 12) = a^2(x - 1)(x - 2)(x - 3)(x - 5)(x - 6)(x - 7)</math>. Let us now, let <math>x = 0</math>. We then have, after cleaning it up nicely, <math>[f(0)]^2 = 2^2 * 3^2 * (35a^2 + 4)</math>. We now take the square root of both sides, to obtain <math>|f(0)|</math>. So, <math>|f(0)| = 6 * \sqrt{35a^2 + 4}</math>. Now, this is the cheese part. Since this is aime, we know that the answer must be an integer, so we assume <math>a = 2</math>. Thus, we get <math>|f(0)| = 6 * \sqrt{144} = 6 * 12 = 072</math>. |
~~triggod | ~~triggod |
Latest revision as of 11:02, 15 August 2024
Contents
Problem
Let be a third-degree polynomial with real coefficients satisfying
Find
.
Solution 1
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
and
, it is easy to see that
, and
; otherwise more bends would be required in the graph. Since only the absolute value of
is required, there is no loss of generalization by stating that
, and
. This provides the following system of equations.
Using any four of these functions as a system of equations yields
Solution 2
By drawing the function, and similar to Solution 1, WLOG let . Then,
. Set
. Then the roots of
are
. So,
. Plug in
to find a. We know
.
So,
. Thus,
, and then
.
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
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 6 (Finite differences)
Because a cubic must come in a "wave form" with two points of inflection, 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
.
Solution 7 (Like solution 1 without annoying systems)
We can rewrite our function as two different cubics, . Note that
is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that
and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing
,
,
,
,
,
yields:
For the constant terms to have a difference of 24 (
), we must have
, so the constant term of our polynomial is
, the absolute value of which is
. -- Solution by eiis1000
Solution 8 (First few steps of solution 1)
We can rewrite the function as and
. Since we need to find
, substitute 0 for x in these two equations.
Isolating
in both of the equations,
Equating the two and solving for
, we see
.
~YBSuburbanTea
Solution 9 (Cheese Solution)
Let the leading coefficient of be
. Then, it is obvious that
. Let us now, let
. We then have, after cleaning it up nicely,
. We now take the square root of both sides, to obtain
. So,
. Now, this is the cheese part. Since this is aime, we know that the answer must be an integer, so we assume
. Thus, we get
.
~~triggod
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.