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

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F(5) = 3(F(<math>\frac{5}{3}</math>)) = 2
 
F(5) = 3(F(<math>\frac{5}{3}</math>)) = 2
  
Hmm.. That doesn‘t seem to be getting us anywhere.  
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That doesn‘t seem to be getting us anywhere.  
 
We notice what we did with f(5) will probably work with f(2001).  
 
We notice what we did with f(5) will probably work with f(2001).  
  
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We found that <math>\frac{y}{243}=\frac{429}{243}</math>, and solving this equation gives our answer <math>\boxed{429}</math>
 
We found that <math>\frac{y}{243}=\frac{429}{243}</math>, and solving this equation gives our answer <math>\boxed{429}</math>
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~MathCosine
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==Video Solution==
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https://www.youtube.com/watch?v=j3hj2yNga0w
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by Coach Jay
  
 
== See also ==
 
== See also ==

Latest revision as of 14:53, 2 November 2024

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.

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 $x=486-57=\boxed{429}$.

~Magnetoninja

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($\frac{5}{3}$)) = 2

That doesn‘t seem to be getting us anywhere. We notice what we did with f(5) will probably work with f(2001).

$F(2001) = 3f(667)=9f(\frac{667}{3}) = 27f(\frac{667}{9}) = 81f(\frac{667}{27})=243f(\frac{667}{81})=729f(\frac{667}{243})$

From here, we can evaluate f(2001) = $186$ when we plug in $\frac{667}{243}$ into $1 - |x - 2|$. 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),

$188 = F(y)= 3f(\frac{y}{3}) = 9f(\frac{y}{9}) = 27f(\frac{y}{27})=81f(\frac{y}{81}) = 243f(\frac{y}{243})$

Notice that we stopped at $243f(\frac{y}{243})$ because $\frac{186}{243}$ is inside the range of $1-|x-2|$, which is [0,1]. Now, f(y/243) = 186/243. Setting $186/243 = 1-|x-2|$, we get 2 solutions for x: $\frac{543}{243}$ and $\frac{429}{243}$.

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 $\frac{y}{243}=\frac{429}{243}$, and solving this equation gives our answer $\boxed{429}$

~MathCosine

Video Solution

https://www.youtube.com/watch?v=j3hj2yNga0w

by Coach Jay

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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