Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it Philippines

Problem:
Let $\mathbb{R}^{\star}$ be the set of all real numbers, except $1$ . Find all functions $f: \mathbb{R}^{\star} \rightarrow \mathbb{R}$ that satisfy the functional equation
$x + f(x) + 2 f\left(\frac{x+2009}{x-1}\right) = 2010$

Solution

Solution:
Let $g(x) = \frac{x+2009}{x-1}$ . Then the given functional equation becomes
$x + f(x) + 2 f(g(x)) = 2010$
Replacing $x$ with $g(x)$ in $(1\star)$ , and after noting that $g(g(x)) = x$ , we get
$g(x) + f(g(x)) + 2 f(x) = 2010$
Eliminating $f(g(x))$ in $(1\star)$ and $(2\star)$ , we obtain
$x - 3 f(x) - 2 g(x) = -2010$
Solving for $f(x)$ and using $g(x) = \frac{x+2009}{x-1}$ , we have
$f(x) = \frac{x^{2} + 2007 x - 6028}{3(x-1)}$
It is not difficult to verify that this function satisfies the given functional equation.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.