AIME 2015 I, Problem 10

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 $d = |f(0)| = \fbox{72}$

Answer: 72