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Difference between revisions of "2010 AMC 12B Problems/Problem 24"

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== Problem 24 ==
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== Problem ==
 
The set of real numbers <math>x</math> for which  
 
The set of real numbers <math>x</math> for which  
  
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<math>\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}</math>
 
<math>\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}</math>
  
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== Solution ==
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== Solution 1==
 
Because the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. For simplicity of calculation, we will find the intervals where <cmath>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1</cmath>
 
Because the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. For simplicity of calculation, we will find the intervals where <cmath>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1</cmath>
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The left side of the inequality has three vertical asymptotes at <math>x=\{-1,0,1\}</math>. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity, and the function is decreasing at all intervals. In addition, the function has a horizontal asymptote at <math>y=0</math>. The function intersects <math>1</math> at some point from <math>x=-1</math> to <math>x=0</math>, and at some point from <math>x=0</math> to <math>x=1</math>, and at some point to the right of <math>x=1</math>. The intervals where the function is greater than 1 are between the points where the function equals <math>1</math> and the vertical asymptotes.
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We shall say that <math>f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}</math>. <math>f(x)</math> has three vertical asymptotes at <math>x=\{-1,0,1\}</math>. As the sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity. In addition, the function has a horizontal asymptote at <math>y=0</math>. The function intersects <math>1</math> at some point from <math>x=-1</math> to <math>x=0</math>, and at some point from <math>x=0</math> to <math>x=1</math>, and at some point to the right of <math>x=1</math>. The intervals where the function is greater than <math>1</math> are between the points where the function equals <math>1</math> and the vertical asymptotes.
  
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If <math>p</math>, <math>q</math>, and <math>r</math> are values of x where <math>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1</math>, then the sum of the lengths of the intervals is <math>(p-(-1))+(q-0)+(r-1)=p+q+r</math>.
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If <math>p</math>, <math>q</math>, and <math>r</math> are values of x where <math>f(x)=1</math>, then the sum of the lengths of the intervals is <math>(p-(-1))+(q-0)+(r-1)=p+q+r</math>.
  
 
<cmath>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1</cmath>
 
<cmath>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1</cmath>
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<cmath>x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)</cmath>
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<cmath>\implies x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)</cmath>
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<cmath>x^3-3x^2-x+1=0</cmath>
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<cmath>\implies x^3-3x^2-x+1=0</cmath>
  
 
And now our job is simply to find the sum of the roots of <math>x^3-3x^2-x+1</math>.
 
And now our job is simply to find the sum of the roots of <math>x^3-3x^2-x+1</math>.
 +
Using [[Vieta's formulas]], we find this to be <math>3</math> <math>\Rightarrow\boxed{C}</math>.
  
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<math>x^3-3x^2-x+1=(x-p)(x-q)(x-r)</math>
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''NOTE''': For the AMC, one may note that the transformed inequality should not yield solutions that involve big numbers like 67 or 134, and immediately choose <math>C</math>.
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<math>x^3-3x^2-x+1=x^3-(q+p+r)x^2+(qp+qr+pr)x-qpr</math>
 
  
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and the sum <math>q+p+r</math> is the negative of the coefficient on the <math>x^2</math> term, which is <math>3\Rightarrow\boxed{C}</math>
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== Solution 2==
 +
As in the first solution, note that the expression can be translated into <cmath>\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1</cmath> without affecting the interval lengths.
 +
 
 +
This simplifies into <cmath>\frac{-x^3+3x^2+x-1}{(x)(x+1)(x-1)}\ge0,</cmath> so <cmath>-x^3+3x^2+x-1\ge0.</cmath>
 +
Each interval is <math>(-1, a), (0, b), (1, c)</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are the roots of <math>-x^3+3x^2+x-1=0</math>
 +
so the total length is <math>a+b+c</math>, which is the sum of the roots, or <math>\boxed3</math>.
 +
 
 +
==Solution 3==
 +
Note that all of the answers but <math>C</math> have weird factors of 2010, but 2010 is a random number (set <math>x'=x-2010</math>). So therefore the answer is <math>\fbox{C(3)}</math>
 +
 
 +
== See also ==
 +
{{AMC12 box|year=2010|num-b=23|num-a=25|ab=B}}
 +
{{MAA Notice}}

Latest revision as of 11:58, 18 August 2021

Problem

The set of real numbers $x$ for which

\[\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1\]

is the union of intervals of the form $a<x\le b$. What is the sum of the lengths of these intervals?

$\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}$

Solution 1

Because the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. For simplicity of calculation, we will find the intervals where \[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1\] We shall say that $f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}$. $f(x)$ has three vertical asymptotes at $x=\{-1,0,1\}$. As the sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity. In addition, the function has a horizontal asymptote at $y=0$. The function intersects $1$ at some point from $x=-1$ to $x=0$, and at some point from $x=0$ to $x=1$, and at some point to the right of $x=1$. The intervals where the function is greater than $1$ are between the points where the function equals $1$ and the vertical asymptotes.

If $p$, $q$, and $r$ are values of x where $f(x)=1$, then the sum of the lengths of the intervals is $(p-(-1))+(q-0)+(r-1)=p+q+r$.

\[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1\] \[\implies x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)\] \[\implies x^3-3x^2-x+1=0\]

And now our job is simply to find the sum of the roots of $x^3-3x^2-x+1$. Using Vieta's formulas, we find this to be $3$ $\Rightarrow\boxed{C}$.

NOTE': For the AMC, one may note that the transformed inequality should not yield solutions that involve big numbers like 67 or 134, and immediately choose $C$.

Solution 2

As in the first solution, note that the expression can be translated into \[\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1\] without affecting the interval lengths.

This simplifies into \[\frac{-x^3+3x^2+x-1}{(x)(x+1)(x-1)}\ge0,\] so \[-x^3+3x^2+x-1\ge0.\] Each interval is $(-1, a), (0, b), (1, c)$, where $a$, $b$, and $c$ are the roots of $-x^3+3x^2+x-1=0$ so the total length is $a+b+c$, which is the sum of the roots, or $\boxed3$.

Solution 3

Note that all of the answers but $C$ have weird factors of 2010, but 2010 is a random number (set $x'=x-2010$). So therefore the answer is $\fbox{C(3)}$

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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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