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

(Solution)
(Solution)
Line 10: Line 10:
 
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.  
 
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 < 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. <math>\boxed{\text{C}}</math>.
+
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 < 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. <math>\boxed{\text{C}}</math>.
  
 
==See Also==
 
==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}}
 
{{MAA Notice}}

Revision as of 15:42, 15 January 2017

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 intersects the line $y=x$.

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 $\int_0^n \! 4x \, \mathrm{d}x$. It is easy to see that this is equal to $2n^2$. The problem then simplifies to the algebraic expression $2n^2=2012$, which simplifies to, $n^2=1006$, and then to $n=\sqrt{1006}$, which rounds up to 32. $\boxed{\text{C}}$.

See Also

2012 AMC 12A (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS