Difference between revisions of "1989 AIME Problems/Problem 8"
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===Video Solution=== | ===Video Solution=== | ||
− | https://www.youtube.com/watch?v=4mOROTEkvWI | + | https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:06, 28 June 2020
Problem
Assume that are real numbers such that
Find the value of .
Contents
Solution
Solution 1
Notice that because we are given a system of equations with
unknowns, the values
are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three.
Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of in the first equation be
; then its coefficients in the second equation is
and the third as
. We need to find a way to sum these to make
[this is in fact a specific approach generalized by the next solution below].
Thus, we hope to find constants satisfying
. FOILing out all of the terms, we get
![$[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.$](http://latex.artofproblemsolving.com/d/e/0/de0a9256131a0d09b604f6d5b019ae307eff232d.png)
Comparing coefficents gives us the three equation system:
Subtracting the second and third equations yields that , so
and
. It follows that the desired expression is
.
Solution 2
Notice that we may rewrite the equations in the more compact form as:
and
where and
is what we're trying to find.
Now consider the polynomial given by (we are only treating the
as coefficients).
Notice that is in fact a quadratic. We are given
as
and are asked to find
. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find
.
Alternatively, applying finite differences, one obtains .
Solution 3
Notice that
I'll number the equations for convenience
Let the coefficient of in
be
. Then the coefficient of
in
is
etc.
Therefore,
So
Solution 4
Notice subracting the first equation from the second yields Then, repeating for the 2nd and 3rd equations, and then subtracting the result from the first obtained equation, we get
. Adding this twice to the first obtained equation gives difference of the desired equation and 3rd equation, which is 211. Adding to the 3rd equation, we get
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
Video Solution
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.