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 - | + | 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>. | + | 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 | + | <cmath>186 = {3^5}f\left(\frac{x}{3^5}\right).</cmath> |
| − | \cdot 243</cmath> | + | |
| + | Because we forced <math>1 \le \frac{x}{3^5} \le 3</math>, so | ||
| + | |||
| + | <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 | ||
| + | \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 | + | 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. |
| + | |||
| + | ==Solution 2 (Graphing)== | ||
| + | |||
| + | [[File:Screenshot 2023-06-14 194739.png|center|200px]] | ||
| + | |||
| + | First, we start by graphing the function when <math>1\leq{x}\leq3</math>, which consists of the lines <math>y=x-1</math> and <math>y=3-x</math> that intersect at <math>(2,1)</math>. Similarly, using <math>f(3x)=3f(x)</math>, 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 <math>(2y,y)</math> where <math>y=3^a</math> for some <math>a</math>. First, we compute <math>f(2001)</math>. The nearest intersection point is <math>(1458,729)</math> when <math>a=7</math>. Therefore, we can safely assume that <math>f(2001)</math> is somewhere on the line with a slope of <math>-1</math> that intersects at that nearest point. Using the fact that the slope of the line is <math>-1</math>, we compute <math>f(2001)=729-543=186</math>. However, we want the minimum value such that <math>f(x)=186</math> and we see that there is another intersection point on the left which has a <math>y>186</math>, namely <math>(486,243)</math>. Therefore, we want the point that lies on the line with slope <math>1</math> that intersects this point. Once again, since the slope of the line is <math>1</math>, we get <math>x=486-57=\boxed{429}</math>. | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja 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(<math>\frac{5}{3}</math>)) = 2 | ||
| + | |||
| + | That doesn‘t seem to be getting us anywhere. | ||
| + | We notice what we did with f(5) will probably work with f(2001). | ||
| + | |||
| + | <math>F(2001) = 3f(667)=9f(\frac{667}{3}) = 27f(\frac{667}{9}) = 81f(\frac{667}{27})=243f(\frac{667}{81})=729f(\frac{667}{243})</math> | ||
| + | |||
| + | From here, we can evaluate f(2001) = <math>186</math> when we plug in <math>\frac{667}{243}</math> into <math>1 - |x - 2|</math>. 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), | ||
| + | |||
| + | <math>188 = F(y)= 3f(\frac{y}{3}) = 9f(\frac{y}{9}) = 27f(\frac{y}{27})=81f(\frac{y}{81}) = 243f(\frac{y}{243})</math> | ||
| + | |||
| + | Notice that we stopped at <math>243f(\frac{y}{243})</math> because <math>\frac{186}{243}</math> is inside the range of <math>1-|x-2|</math>, which is [0,1]. Now, f(y/243) = 186/243. Setting <math>186/243 = 1-|x-2|</math>, we get 2 solutions for x: <math>\frac{543}{243}</math> and <math>\frac{429}{243}</math>. | ||
| + | |||
| + | 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 <math>\frac{y}{243}=\frac{429}{243}</math>, and solving this equation gives our answer <math>\boxed{429}</math> | ||
| + | |||
| + | ~MathCosine | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://www.youtube.com/watch?v=j3hj2yNga0w | ||
| + | |||
| + | by Coach Jay | ||
== See also == | == See also == | ||
Latest revision as of 14:53, 2 November 2024
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
Video Solution
https://www.youtube.com/watch?v=j3hj2yNga0w
by Coach Jay
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
