Difference between revisions of "2001 AIME II Problems/Problem 8"

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== 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 - \mid x - 2 \mid</math> for <math>1\leq x \leq 3</math>. Find the smallest <math>x</math> for which <math>f(x) = f(2001)</math>.
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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 ==
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<cmath>f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.</cmath>
 
<cmath>f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.</cmath>
  
We now need the smallest <math>x</math> such that <math>f(x) = 3^kf\left(\frac{x}{3^k}\right) = 186</math>. The [[range]] of <math>f(x),\ 1 \le x \le 3</math>, is <math>0 \le f(x) \le 1</math>. Then <math>0 \le 186 = 3^kf\left(\frac{x}{3^k}\right) \le 3^k</math>, and the smallest value of <math>k</math> is <math>k = 5</math>. Then,
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We now need the smallest <math>x</math> such that <math>f(x) = 3^kf\left(\frac{x}{3^k}\right) = 186</math>. The [[range]] of <math>f(x),\ 1 \le x \le 3</math>, is <math>0 \le f(x) \le 1</math>. So when <math>1 \le \frac{x}{3^k} \le 3</math>, we have <math>0 \le f\left(\frac{x}{3^k}\right) = \frac{186}{3^k} \le 1</math>. Multiplying by <math>3^k</math>: <math>0 \le 186 \le 3^k</math>, so the smallest value of <math>k</math> is <math>k = 5</math>. Then,
  
<cmath>186 = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2  
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<cmath>186 = {3^5}f\left(\frac{x}{3^5}\right).</cmath>
\cdot 243</cmath>
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 +
Because we forced <math>1 \le \frac{x}{3^5} \le 3</math>, so
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 +
<cmath>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  
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\cdot 243.</cmath>
  
 
We want the smaller value of <math>x = \boxed{429}</math>.
 
We want the smaller value of <math>x = \boxed{429}</math>.
  
An alternative approach is to consider the graph of <math>f(x)</math>, which repeats every power of <math>3</math>, and resembles the section from <math>1 \le x \le 3</math> expanded by a factor of <math>3</math>.
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An alternative approach is to consider the graph of <math>f(x)</math>, which iterates every power of <math>3</math>, and resembles the section from <math>1 \le x \le 3</math> dilated by a factor of <math>3</math> at each iteration.
  
 
== See also ==
 
== See also ==

Revision as of 12:37, 30 October 2016

Problem

A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$, and that $f(x) = 1-|x-2|$ for $1\le x \le 3$. 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 $0 \le f(x) \le 1$. 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.

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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