1989 AIME Problems/Problem 8
Contents
Problem
Assume that are real numbers such that
Find the value of
.
Solution 1 (Quadratic Function)
Note that each equation is of the form for some
When we expand and combine like terms, we obtain a quadratic function of
where
and
are linear combinations of
and
We are given that
and we wish to find
We eliminate by subtracting the first equation from the second, then subtracting the second equation from the third:
By either substitution or elimination, we get
and
Substituting these back produces
Finally, the answer is
~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2 (Linear Combination)
For simplicity purposes, we number the given equations and
in that order. Let
Subtracting
from
subtracting
from
and subtracting
from
we obtain the following equations, respectively:
Subtracting
from
and subtracting
from
we obtain the following equations, respectively:
Finally, applying the Transitive Property to
and
gives
from which
~Duohead (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 3 (Finite Differences by Arithmetic)
Note that the second differences of all quadratic sequences must be constant (but nonzero).
One example, the perfect square sequence, is shown below:
Label equations and
as Solution 2 does. Since the coefficients of
or
respectively, all form quadratic sequences with second differences
we conclude that the second differences of equations
must be constant.
It follows that the second differences of must be constant, as shown below:
NO EDIT PLEASE. DIAGRAM WILL BE READY SOON
Finally, the answer is
~MRENTHUSIASM
Solution 4 (Finite Differences by Algebra)
Notice that we may rewrite the equations in the more compact form as:
where
and
is what we are 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
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 5 (Very Cheap: Not Recommended)
We let . Thus, we have
Grinding this out, we have
which gives
as our final answer.
-Pleaseletmewin
Video Solution
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx
See also
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