2001 AIME II Problems/Problem 8

Problem

A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$ , and that $f(x) = 1 - \mid x - 2 \mid$ for $1\leq x \leq 3$ . Find the smallest $x$ for which $f(x) = f(2001)$ .

Solution

Iterating the condition $f(3x) = 3f(x)$ , we find that $f(x) = 3^kf\left(\frac{x}{3^k}\right)$ for positive integers $k$ . We know the definition of $f(x)$ from $1 \le x \le 3$ , so we would like to express $f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6$ . Indeed,

$f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.$

We now need the smallest $x$ such that $f(x) = 3^kf\left(\frac{x}{3^k}\right) = 186$ . The range of $f(x),\ 1 \le x \le 3$ , is $0 \le f(x) \le 1$ . Then $0 \le 186 = 3^kf\left(\frac{x}{3^k}\right) \le 3^k$ , and the smallest value of $k$ is $k = 5$ . Then,

$186 = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 \cdot 243$

We want the smaller value of $x = \boxed{429}$ .

An alternative approach is to consider the graph of $f(x)$ , which repeats every power of $3$ , and resembles the section from $1 \le x \le 3$ expanded by a factor of $3$ .

2001 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2001 AIME II Problems/Problem 8

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