2001 AIME II Problem 8

Attempt Problem 8 of the 2001 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AIME II solutions, or check the answer key.

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8.

A certain function ff has the properties that f(3x)=3f(x)f(3x) = 3f(x) for all positive real values of x,x, and that f(x)=1x2f(x) = 1 - |x - 2| for 1x3.1 \le x \le 3. Find the smallest xx for which f(x)=f(2001).f(x) = f(2001).

Answer: 429
Concepts:functional equationabsolute value
Difficulty rating: 2560
Solution:

Applying f(3x)=3f(x)f(3x) = 3f(x) six times gives f(2001)=36f(2001729),f(2001) = 3^6 f\left(\frac{2001}{729}\right), and 2001729\frac{2001}{729} lies in [1,3],[1, 3], so f(2001)=729(120017292)=72920011458=729543=186. \begin{aligned} f(2001) &= 729\left(1 - \left|\tfrac{2001}{729} - 2\right|\right) \\ &= 729 - |2001 - 1458| \\ &= 729 - 543 = 186. \end{aligned}

For x[3k,3k+1],x \in [3^k, 3^{k+1}], we have f(x)=3kf(x3k)f(x) = 3^k f\left(\frac{x}{3^k}\right) =3k(1x3k2),= 3^k\left(1 - \left|\frac{x}{3^k} - 2\right|\right), a tent whose maximum value is 3k.3^k. To achieve 186186 we need 3k186,3^k \ge 186, so k5,k \ge 5, and the smallest solutions lie in [243,729],[243, 729], where f(x)=243x486.f(x) = 243 - |x - 486|.

Setting 243x486=186243 - |x - 486| = 186 gives x486=57,|x - 486| = 57, so x=429x = 429 or x=543.x = 543. The smallest xx is 429.429.

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