Difference between revisions of "2005 AMC 12A Problems/Problem 20"
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== Problem == | == Problem == | ||
+ | For each <math>x</math> in <math>[0,1]</math>, define | ||
+ | <cmath> | ||
+ | \begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\ | ||
+ | f(x) & = 2 - 2x, & \text { if } \frac {1}{2} < x \leq 1. \end{array} | ||
+ | </cmath> | ||
+ | Let <math>f^{[2]}(x) = f(f(x))</math>, and <math>f^{[n + 1]}(x) = f^{[n]}(f(x))</math> for each integer <math>n \geq 2</math>. For how many values of <math>x</math> in <math>[0,1]</math> is <math>f^{[2005]}(x) = \frac {1}{2}</math>? | ||
+ | <cmath> | ||
+ | (\text {A}) \ 0 \qquad (\text {B}) \ 2005 \qquad (\text {C})\ 4010 \qquad (\text {D}) \ 2005^2 \qquad (\text {E})\ 2^{2005} | ||
+ | </cmath> | ||
+ | == Solution 1 == | ||
+ | For the two functions <math>f(x)=2x,0\le x\le \frac{1}{2}</math> and <math>f(x)=2-2x,\frac{1}{2}\le x\le 1</math>,as long as <math>f(x)</math> is between <math>0</math> and <math>1</math>, <math>x</math> will be in the right domain, so we don't need to worry about the domain of <math>x</math>. | ||
− | == Solution == | + | |
− | {{ | + | Also, every time we change <math>f(x)</math>, the expression for the final answer in terms of <math>x</math> will be in a different form(although they'll all satisfy the final equation), so we get a different starting value of <math>x</math>. Every time we have two choices for <math>f(x</math>) and altogether we have to choose <math>2005</math> times. Thus, <math>2^{2005}\Rightarrow\boxed{E}</math>. |
− | == | + | |
− | + | Note: the values of x that satisfy <math>f^{[n]}(x) = \frac {1}{2}</math> are <math>\frac{1}{2^{n+1}}</math>, <math>\frac{3}{2^{n+1}}</math>, <math>\frac{5}{2^{n+1}}</math>, <math>\cdots</math> ,<math>\frac{2^{n+1}-1}{2^{n+1}}</math>. | |
− | + | ||
− | + | == Solution 2== | |
+ | |||
+ | |||
+ | We are given that <math>f^{[2005]}(x) = \frac {1}{2}</math>. Thus, <math>f(f^{[2004]}(x))=\frac{1}{2}</math>. Let <math>f^{[2004]}(x)</math> be equal to <math>y</math>. Thus <math>f(y)=\frac{1}{2}</math> or <math>y=\frac{1}{4}</math> or <math>\frac{3}{4}</math>. Now we know <math>f^{[2004]}(x)</math> is equal to <math>\frac{1}{4}</math> or <math>\frac{3}{4}</math>. Now we know that <math>f(f^{[2003]}(x))=\frac{1}{4}</math> or <math>\frac{3}{4}</math>. Now we solve for <math>f^{[2003]}(x)</math> and let <math>f^{[2003]}(x)=z</math>. Thus <math>f(z)</math> is equal to <math>\frac{1}{8}</math>,<math>\frac{7}{8}</math>,<math>\frac{5}{8}</math>,and <math>\frac{3}{8}</math>. As we see, <math>f^{[2005]}(x)</math> has 1 solution, <math>f^{[2004]}(x)</math> has 2 solutions, and <math>f^{[2003]}(x)</math> has 4 solutions. Thus for each iteration we double the number of possible solutions. There are 2005 iterations and thus the number of solutions is <math>2^{2005}</math> <math>\Rightarrow\boxed{E}</math> | ||
+ | (not rigorous) | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|ab=A|num-b=19|num-a=21|year=2005}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:49, 28 September 2022
Contents
Problem
For each in
, define
Let
, and
for each integer
. For how many values of
in
is
?
Solution 1
For the two functions and
,as long as
is between
and
,
will be in the right domain, so we don't need to worry about the domain of
.
Also, every time we change , the expression for the final answer in terms of
will be in a different form(although they'll all satisfy the final equation), so we get a different starting value of
. Every time we have two choices for
) and altogether we have to choose
times. Thus,
.
Note: the values of x that satisfy are
,
,
,
,
.
Solution 2
We are given that . Thus,
. Let
be equal to
. Thus
or
or
. Now we know
is equal to
or
. Now we know that
or
. Now we solve for
and let
. Thus
is equal to
,
,
,and
. As we see,
has 1 solution,
has 2 solutions, and
has 4 solutions. Thus for each iteration we double the number of possible solutions. There are 2005 iterations and thus the number of solutions is
(not rigorous)
See Also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.