2001 AIME II Problems/Problem 8


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


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 $0 \le f(x) \le 1$. So when $1 \le \frac{x}{3^k} \le 3$, we have $0 \le f\left(\frac{x}{3^k}\right) = \frac{186}{3^k} \le 1$. Multiplying by $3^k$: $0 \le 186 \le 3^k$, so the smallest value of $k$ is $k = 5$. Then,

\[186 = {3^5}f\left(\frac{x}{3^5}\right).\]

Because we forced $1 \le \frac{x}{3^5} \le 3$, so

\[186 = {3^5}f\left(\frac{x}{3^5}\right) = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2  \cdot 243.\]

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

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

Solution 2 (Graphing)

Screenshot 2023-06-14 194739.png

First, we start by graphing the function when $1\leq{x}\leq3$, which consists of the lines $y=x-1$ and $y=3-x$ that intersect at $(2,1)$. Similarly, using $f(3x)=3f(x)$, we get a dilation of our initial figure by a factor of 3 for the next interval and so on. Observe that the intersection of two lines always has coordinates $(2y,y)$ where $y=3^a$ for some $a$. First, we compute $f(2001)$. The nearest intersection point is $(1458,729)$ when $a=7$. Therefore, we can safely assume that $f(2001)$ is somewhere on the line with a slope of $-1$ that intersects at that nearest point. Using the fact that the slope of the line is $-1$, we compute $f(2001)=729-543=186$. However, we want the minimum value such that $f(x)=186$ and we see that there is another intersection point on the left which has a $y>186$, namely $(486,243)$. Therefore, we want the point that lies on the line with slope $1$ that intersects this point. Once again, since the slope of the line is $1$, we get $x=486-57=\boxed{429}$.


See also

2001 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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