2015 AIME I Problems/Problem 10

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 $f(0) = 072$

Solution 2

Express $f(x)$ in terms of powers of $(x-4)$ : $f(x) = a(x-4)^3 + b(x-4)^2 + c(x-4) + d$ By the same argument as in the first Solution, we see that $f(x)$ is an odd function about the line $x=4$ , so its coefficients $b$ and $d$ are 0. From there it is relatively simple to solve $f(2)=f(3)=-12$ (as in the above solution, but with a smaller system of equations): $a(1)^3 + c(1) = -12$ $a(2)^3 + c(2) = -12$ $a=2$ and $c=-14$ $|f(0)| = |2(-4)^3 - 14(-4)| = 072$

2015 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2015 AIME I Problems/Problem 10

Contents

Problem

Solution

Solution 2

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