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Difference between revisions of "2012 AMC 12A Problems/Problem 25"

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Let <math>f(x)=|2\{x\}-1|</math> where <math>\{x\}</math> denotes the fractional part of <math>x</math>.  The number <math>n</math> is the smallest positive integer such that the equation <cmath>nf(xf(x))=x</cmath> has at least <math>2012</math> real solutions.  What is <math>n</math>?  '''Note:''' the fractional part of <math>x</math> is a real number <math>y=\{x\}</math> such that <math>0\le y<1</math> and <math>x-y</math> is an integer.
 
Let <math>f(x)=|2\{x\}-1|</math> where <math>\{x\}</math> denotes the fractional part of <math>x</math>.  The number <math>n</math> is the smallest positive integer such that the equation <cmath>nf(xf(x))=x</cmath> has at least <math>2012</math> real solutions.  What is <math>n</math>?  '''Note:''' the fractional part of <math>x</math> is a real number <math>y=\{x\}</math> such that <math>0\le y<1</math> and <math>x-y</math> is an integer.
  
−
<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 31\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 62\qquad\textbf{(E)}\64 </math>
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<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 31\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 62\qquad\textbf{(E)}\ 64 </math>
  
 
== Solution ==
 
== Solution ==
−
Our goal is to determine how many times intersects the line <math>y=x</math>. We begin by analyzing the behavior of <math>\{x\}</math>. It increases linearly with a slope of one, then when it reaches the next integer, it repeats itself. We can deduce that the function is like a sawtooth wave, with a period of one. We then analyze the function <math>f(x)=|2\{x\}-1|</math>. The slope of the teeth is multiplied by 2 to get 2, and the function is moved one unit downward. The function can then be described as starting at -1, moving upward with a slope of 2 to get to 1, and then repeating itself, still with a period of 1. The absolute value of the function is then taken. This results in all the negative segments becoming flipped in the Y direction. The positive slope starting at -1 of the function ranging from <math>u</math> to <math>u.5</math>, where u is any arbitrary integer, is now a negative slope starting at positive 1.  The function now looks like the letter V repeated within every square in the first row. It is now that we address the goal of this, which is to determine how many times the function intersects the line <math>y=x</math>. Since there are two line segments per box, the function has two chances to intersect the line <math>y=x</math> for every integer. If the height of the function is higher than <math>y=x</math> for every integer on an interval, then every chance within that interval intersects the line.   Returning to analyzing the function, we note that it is multiplied by <math>x</math>, and then fed into <math>f(x)</math>. Since <math>f(x)</math> is a periodic function, we can model it as multiplying the function's frequency by <math>x</math>. This gives us <math>2x</math> chances for every integer, which is then multiplied by 2 once more to get <math>4x</math> chances for every integer. The amplitude of this function is initially 1, and then it is multiplied by n, to give an amplitude of n. The function intersects the line <math>y=x</math> for every chance in the interval of <math>0\leq x \leq n</math>, since the function is n units high. The function ceases to intersect <math>y=x</math> when <math>n \less x</math>, since the height of the function is lower than <math>y=x</math>. The number of times the function intersects <math>y=x</math> is then therefore equal to <math>\int_0^n \! 4x \, \mathrm{d}x</math>. It is easy to see that this is equal to <math>2n^2</math>. The problem then simplifies to the algebraic expression <math>2n^2=2012</math>, which simplifies to, <math>n^2=1006</math>, and then to <math>n=\sqrt{1006}</math>, which rounds up to 32.
+
Our goal is to determine how many times the graph of <math>nf(xf(x))=x</math> intersects the graph of <math>y=x</math>. (Conversely, we can also divide the equation by <math>n</math> to get <math>f(xf(x))=\frac{x}{n}</math> and look at the graph <math>y=\frac{x}{n}</math>)
−
C: n = 32
 
  
 +
We begin by analyzing the behavior of <math>\{x\}</math>. It increases linearly with a slope of one, then when it reaches the next integer, it repeats itself. We can deduce that the function is like a sawtooth wave, with a period of one. We then analyze the function <math>f(x)=|2\{x\}-1|</math>. The slope of the teeth is multiplied by 2 to get 2, and the function is moved one unit downward. The function can then be described as starting at -1, moving upward with a slope of 2 to get to 1, and then repeating itself, still with a period of 1. The absolute value of the function is then taken. This results in all the negative segments becoming flipped in the Y direction. The positive slope starting at -1 of the function ranging from <math>u</math> to <math>u.5</math>, where u is any arbitrary integer, is now a negative slope starting at positive 1.  The function now looks like the letter V repeated within every square in the first row.
 +
 +
It is now that we address the goal of this, which is to determine how many times the function intersects the line <math>y=x</math>. Since there are two line segments per box, the function has two chances to intersect the line <math>y=x</math> for every integer. If the height of the function is higher than <math>y=x</math> for every integer on an interval, then every chance within that interval intersects the line. 
 +
 +
Returning to analyzing the function, we note that it is multiplied by <math>x</math>, and then fed into <math>f(x)</math>. Since <math>f(x)</math> is a periodic function, we can model it as multiplying the function's frequency by <math>x</math>. This gives us <math>2x</math> chances for every integer, which is then multiplied by 2 once more to get <math>4x</math> chances for every integer. The amplitude of this function is initially 1, and then it is multiplied by <math>n</math>, to give an amplitude of <math>n</math>. The function intersects the line <math>y=x</math> for every chance in the interval of <math>0\leq x \leq n</math>, since the function is n units high. The function ceases to intersect <math>y=x</math> when <math>n < x</math>, since the height of the function is lower than <math>y=x</math>.
 +
 +
The number of times the function intersects <math>y=x</math> is then therefore equal to <math>4+8+12...+4n</math>. We want this sum to be greater than 2012 which occurs when <math>n=32 \Rightarrow \boxed{(C)}</math> .
 +
 +
==Video Solution by Richard Rusczyk==
 +
https://artofproblemsolving.com/videos/amc/2012amc12a/256
 +
 +
~dolphin7
 +
 +
==See Also==
 
{{AMC12 box|year=2012|ab=A|num-b=24|after=Last Problem}}
 
{{AMC12 box|year=2012|ab=A|num-b=24|after=Last Problem}}
 +
{{MAA Notice}}

Latest revision as of 21:58, 23 August 2024

Problem

Let $f(x)=|2\{x\}-1|$ where $\{x\}$ denotes the fractional part of $x$. The number $n$ is the smallest positive integer such that the equation \[nf(xf(x))=x\] has at least $2012$ real solutions. What is $n$? Note: the fractional part of $x$ is a real number $y=\{x\}$ such that $0\le y<1$ and $x-y$ is an integer.

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 31\qquad\textbf{(C)}\ 32\qquad\textbf{(D)}\ 62\qquad\textbf{(E)}\ 64$

Solution

Our goal is to determine how many times the graph of $nf(xf(x))=x$ intersects the graph of $y=x$. (Conversely, we can also divide the equation by $n$ to get $f(xf(x))=\frac{x}{n}$ and look at the graph $y=\frac{x}{n}$)

We begin by analyzing the behavior of $\{x\}$. It increases linearly with a slope of one, then when it reaches the next integer, it repeats itself. We can deduce that the function is like a sawtooth wave, with a period of one. We then analyze the function $f(x)=|2\{x\}-1|$. The slope of the teeth is multiplied by 2 to get 2, and the function is moved one unit downward. The function can then be described as starting at -1, moving upward with a slope of 2 to get to 1, and then repeating itself, still with a period of 1. The absolute value of the function is then taken. This results in all the negative segments becoming flipped in the Y direction. The positive slope starting at -1 of the function ranging from $u$ to $u.5$, where u is any arbitrary integer, is now a negative slope starting at positive 1. The function now looks like the letter V repeated within every square in the first row.

It is now that we address the goal of this, which is to determine how many times the function intersects the line $y=x$. Since there are two line segments per box, the function has two chances to intersect the line $y=x$ for every integer. If the height of the function is higher than $y=x$ for every integer on an interval, then every chance within that interval intersects the line.

Returning to analyzing the function, we note that it is multiplied by $x$, and then fed into $f(x)$. Since $f(x)$ is a periodic function, we can model it as multiplying the function's frequency by $x$. This gives us $2x$ chances for every integer, which is then multiplied by 2 once more to get $4x$ chances for every integer. The amplitude of this function is initially 1, and then it is multiplied by $n$, to give an amplitude of $n$. The function intersects the line $y=x$ for every chance in the interval of $0\leq x \leq n$, since the function is n units high. The function ceases to intersect $y=x$ when $n < x$, since the height of the function is lower than $y=x$.

The number of times the function intersects $y=x$ is then therefore equal to $4+8+12...+4n$. We want this sum to be greater than 2012 which occurs when $n=32 \Rightarrow \boxed{(C)}$ .

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12a/256

~dolphin7

See Also

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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

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