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2001 AIME II Problems/Problem 8

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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.

See also

2001 AIME II (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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