Mock AIME 1 2007-2008 Problems/Problem 13

Problem

Let $F(x)$ be a polynomial such that $F(6) = 15$ and $\frac{F(3x)}{F(x+3)} = 9-\frac{48x+54}{x^2+5x+6}$ for $x \in \mathbb{R}$ such that both sides are defined. Find $F(12)$ .

Solution

Combining denominators and simplifying, $\frac{F(3x)}{F(x+3)} = \frac{9(x^2+5x+6)-48x-54}{x^2+5x+6} = \frac{9x^2 - 3x}{x^2 + 5x + 6}= \frac{3x(3x-1)}{(x+3)(x+2)}$ It becomes obvious that $F(x) = ax(x-1)$ , for some constant $a$ , matches the definition of the polynomial. To prove that $F(x)$ must have this form, note that $(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)$

Since $3x$ and $3x-1$ divides the right side of the equation, $3x$ and $3x-1$ divides the left side of the equation. Thus $3x(3x-1)$ divides $F(3x)$ , so $x(x-1)$ divides $F(x)$ .

It is easy to see that $F(x)$ is a quadratic, thus $F(x)=ax(x-1)$ as desired.

By the given, $F(6) = a(6)(5) = 15 \Longrightarrow a = \frac 12$ . Thus, $F(12) = \frac{1}{2}(12)(11) = \boxed{066}$ .

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Mock AIME 1 2007-2008 Problems/Problem 13

Problem

Solution

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