Difference between revisions of "2008 AIME II Problems/Problem 7"
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=== Solution 7 === | === Solution 7 === | ||
− | Let | + | Let's construct a polynomial with the roots <math>(r+s), (s+t),</math> and <math>(t+r)</math>. |
sum of the roots: | sum of the roots: | ||
Line 79: | Line 79: | ||
<math>(t+r)^3+\frac{1001}{8}(t+r)-\frac{2008}{8}=0</math> | <math>(t+r)^3+\frac{1001}{8}(t+r)-\frac{2008}{8}=0</math> | ||
adding all of the equations up, we see that | adding all of the equations up, we see that | ||
− | <math>(r+s)^3+(s+t)^3+(t+r)^3=3\cdot\frac{2008}{8}-\frac{1001}{8}(2r+2s+2t)= | + | <math>(r+s)^3+(s+t)^3+(t+r)^3=3\cdot\frac{2008}{8}-\frac{1001}{8}(2r+2s+2t)=251(3)+0=\boxed{753}</math> |
== See also == | == See also == |
Revision as of 01:05, 22 June 2020
Problem
Let ,
, and
be the three roots of the equation
Find
.
Contents
[hide]Solution
Solution 1
By Vieta's formulas, we have so
Substituting this into our problem statement, our desired quantity is
Also by Vieta's formulas we have
so negating both sides and multiplying through by 3 gives our answer of
Solution 2
By Vieta's formulas, we have , and so the desired answer is
. Additionally, using the factorization
we have that
. By Vieta's again,
Solution 3
Vieta's formulas gives . Since
is a root of the polynomial,
, and the same can be done with
. Therefore, we have
yielding the answer
.
Also, Newton's Sums yields an answer through the application. http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums
Solution 4
Expanding, you get:
This looks similar to
Substituting:
Since
,
Substituting, we get
or,
We are trying to find
.
Substituting:
Solution 5
Write and let
. Then
Solving for
and negating the result yields the answer
Solution 6
Here by Vieta's formulas:
--(1)
--(2)
By the factorisation formula:
Let ,
,
,
(By (1))
So
Solution 7
Let's construct a polynomial with the roots and
.
sum of the roots:
pairwise product of the roots:
product of the roots:
thus, the polynomial we get is
as
and
are roots of this polynomial, we know that (using power reduction)
adding all of the equations up, we see that
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.