Difference between revisions of "2001 AIME II Problems/Problem 8"
(→Solution) |
m (→Problem) |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | A certain [[function]] <math>f</math> has the properties that <math>f(3x) = 3f(x)</math> for all positive real values of <math>x</math>, and that <math>f(x) = 1 - | + | A certain [[function]] <math>f</math> has the properties that <math>f(3x) = 3f(x)</math> for all positive real values of <math>x</math>, and that <math>f(x) = 1-|x-2|</math> for <math>1\le x \le 3</math>. Find the smallest <math>x</math> for which <math>f(x) = f(2001)</math>. |
== Solution == | == Solution == | ||
Revision as of 11:34, 30 October 2016
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 each iteration.
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
