2015 AIME I Problems/Problem 10

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Problem

Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$.

Solution

Let $f(x)$ = ax^3+bx^2+cx+d. Since f(x) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing 12 and -12, it is easy to see that f(1)=f(5)=f(6), and f(2)=f(3)=f(7); otherwise more bends would be required in the graph. Since only the absolute value of f(0) is required, there is no loss of generalization by stating that f(1)=12, and f(2)=-12. This provides the following system of equations.

     a +     b +   c +   d =  12
   8a +   4b + 2c +   d = -12
 27a +   9b + 3c +   d = -12

125a + 25b + 5c + d = 12 216a + 36b + 6c + d = 12 343a + 49b + 7c + d = -12 Using any four of these functions as a system of equations yields f(0) = 072

See Also

2015 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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