1989 AIME Problems/Problem 8
Problem
Assume that are real numbers such that
Find the value of .
Solution
Solution 1
Let us try to derive a way to find the last expression in terms of the three given expressions. The coefficients of in the first equation can be denoted as , making its coefficients in the second equation as and the third as . We need to find a way to sum them up to make .
Thus, we can write that . FOILing out all of the terms, we get . We can set up the three equation system:
Subtracting the second and third equations yields that , so and . Thus, we have to add .
Solution 2
Notice that we may rewrite the equations in the more compact form:
and
, where and and is what we're trying to find.
Now undergo a paradigm shift: consider the polynomial in (we are only treating the as coefficients). Notice that the degree of must be ; it is 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 .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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All AIME Problems and Solutions |