2001 AIME II Problems/Problem 8
Contents
Problem
A certain function 
has the properties that 
for all positive real values of 
, and that 
for 
. Find the smallest for which 
.
Solution
Iterating the condition , we find that 
for positive integers 
. We know the definition of 
from 
, so we would like to express 
. Indeed,
![\[f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.\]](http://latex.artofproblemsolving.com/e/1/2/e129ad258113f974ab9e842efa3433c64a365217.png)
We now need the smallest such that 
. The range of 
, is 
. So when 
, we have 
. Multiplying by 
: 
, so the smallest value of is 
. Then,
![\[186 = {3^5}f\left(\frac{x}{3^5}\right).\]](http://latex.artofproblemsolving.com/0/6/f/06f2b37921ee653205a539b6c9e2ab0be7c9a924.png)
Because we forced 
, so
![\[186 = {3^5}f\left(\frac{x}{3^5}\right) = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2 \cdot 243.\]](http://latex.artofproblemsolving.com/9/1/c/91c437fd96b60bcc9002ced02a93d102b03ce54a.png)
We want the smaller value of 
.
An alternative approach is to consider the graph of , which iterates every power of 
, and resembles the section from dilated by a factor of at each iteration.
Solution 2 (Graphing)
First, we start by graphing the function when 
, which consists of the lines 
and 
that intersect at 
. Similarly, using 
, 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 
where 
for some 
. First, we compute 
. The nearest intersection point is 
when 
. Therefore, we can safely assume that is somewhere on the line with a slope of 
that intersects at that nearest point. Using the fact that the slope of the line is , we compute 
. However, we want the minimum value such that 
and we see that there is another intersection point on the left which has a 
, namely 
. Therefore, we want the point that lies on the line with slope 
that intersects this point. Once again, since the slope of the line is , we get 
.
Solution 3 (Complete Bash but FAST)
We evaluate the first few terms of f(x) to try to find a pattern.
F(1)=0
F(2)=1
F(3)=0
F(4)=1
F(5) = 3(F(
)) = 2
That doesn‘t seem to be getting us anywhere. We notice what we did with f(5) will probably work with f(2001).
From here, we can evaluate f(2001) = 
when we plug in 
into 
. So all we need to find is the least number, let‘s call it, say y such that f(y)=186.
Repeating the same process we did before with f(2001),
Notice that we stopped at 
because 
is inside the range of 
, which is [0,1]. Now, f(y/243) = 186/243. Setting 
, we get 2 solutions for x: 
and 
.
Now, the problem asks for the smallest solution, so we obviously choose 429/243 as the solution for 243f(y/243) because it is smaller.
We found that 
, and solving this equation gives our answer 
~MathCosine
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
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