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 == | ||
− | {{ | + | Iterating the condition <math>f(3x) = 3f(x)</math>, we find that <math>f(x) = 3^kf\left(\frac{x}{3^k}\right)</math> for positive integers <math>k</math>. We know the definition of <math>f(x)</math> from <math>1 \le x \le 3</math>, so we would like to express <math>f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6</math>. Indeed, |
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+ | <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>. 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, | ||
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+ | <cmath>186 = {3^5}f\left(\frac{x}{3^5}\right).</cmath> | ||
+ | |||
+ | 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 | ||
+ | \cdot 243.</cmath> | ||
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+ | 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 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] | ||
== See also == | == See also == | ||
{{AIME box|year=2001|n=II|num-b=7|num-a=9}} | {{AIME box|year=2001|n=II|num-b=7|num-a=9}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:41, 14 June 2023
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,
We now need the smallest such that . The range of , is . So when , we have . Multiplying by : , so the smallest value of is . Then,
Because we forced , so
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 .
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 |
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