Difference between revisions of "2001 AIME II Problems/Problem 8"
m (→Problem) |
Mathcosine (talk | contribs) (→Solution 3 (Complete Bash but FAST)) |
||
(10 intermediate revisions by 3 users not shown) | |||
Line 18: | Line 18: | ||
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 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> each iteration. | + | 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 | ||
== See also == | == See also == |
Latest revision as of 16:17, 1 August 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,
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
.
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
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.