1989 AIME Problems/Problem 8
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)
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)
Solution 3 (Generalized)
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 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 4 (Very Cheap: Not Recommended)
We let . Thus, we have Grinding this out, we have which gives as our final answer.
|1989 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|