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Difference between revisions of "2004 AMC 10A Problems/Problem 4"

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Alternatively, we can solve by casework (a method which should work for any similar problem involving [[absolute value]]s of [[real number]]s).  If <math>x \leq 1</math>, then <math>|x - 1| = 1-x</math> and <math>|x - 2| = 2 - x</math>, so we must solve <math>1 - x = 2 - x</math>, which has no solutions.  Similarly, if <math>x \geq 2</math>, then <math>|x - 1| = x - 1</math> and <math>|x - 2| = x - 2</math>, so we must solve <math>x - 1 = x- 2</math>, which also has no solutions.  Finally, if <math>1 \leq x \leq 2</math>, then <math>|x - 1| = x - 1</math> and <math>|x - 2| = 2-x</math>, so we must solve <math>x - 1 = 2 - x</math>, which has the unique solution <math>x = \frac32</math>.
 
Alternatively, we can solve by casework (a method which should work for any similar problem involving [[absolute value]]s of [[real number]]s).  If <math>x \leq 1</math>, then <math>|x - 1| = 1-x</math> and <math>|x - 2| = 2 - x</math>, so we must solve <math>1 - x = 2 - x</math>, which has no solutions.  Similarly, if <math>x \geq 2</math>, then <math>|x - 1| = x - 1</math> and <math>|x - 2| = x - 2</math>, so we must solve <math>x - 1 = x- 2</math>, which also has no solutions.  Finally, if <math>1 \leq x \leq 2</math>, then <math>|x - 1| = x - 1</math> and <math>|x - 2| = 2-x</math>, so we must solve <math>x - 1 = 2 - x</math>, which has the unique solution <math>x = \frac32</math>.
  
 +
==Solution 2==
 +
We know that either <math>x-1=x-2</math> or <math>x-1=-(x-2)</math>. The first equation simplifies to <math>-1=-2</math>, which is false, so <math>x-1=-(x-2)</math>. Solving, we get <math>x=\frac{3}{2}\Rightarrow\boxed{\mathrm{(D)}\ \frac{3}{2}}</math>.
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2004|ab=A|num-b=3|num-a=5}}
 
{{AMC10 box|year=2004|ab=A|num-b=3|num-a=5}}

Revision as of 21:23, 1 December 2019

Problem

What is the value of $x$ if $|x-1|=|x-2|$?

$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$

Solution

$|x-1|$ is the distance between $x$ and $1$; $|x-2|$ is the distance between $x$ and $2$.

Therefore, the given equation says $x$ is equidistant from $1$ and $2$, so $x=\frac{1+2}2=\frac{3}{2}\Rightarrow\boxed{\mathrm{(D)}\ \frac{3}{2}}$.

Alternatively, we can solve by casework (a method which should work for any similar problem involving absolute values of real numbers). If $x \leq 1$, then $|x - 1| = 1-x$ and $|x - 2| = 2 - x$, so we must solve $1 - x = 2 - x$, which has no solutions. Similarly, if $x \geq 2$, then $|x - 1| = x - 1$ and $|x - 2| = x - 2$, so we must solve $x - 1 = x- 2$, which also has no solutions. Finally, if $1 \leq x \leq 2$, then $|x - 1| = x - 1$ and $|x - 2| = 2-x$, so we must solve $x - 1 = 2 - x$, which has the unique solution $x = \frac32$.

Solution 2

We know that either $x-1=x-2$ or $x-1=-(x-2)$. The first equation simplifies to $-1=-2$, which is false, so $x-1=-(x-2)$. Solving, we get $x=\frac{3}{2}\Rightarrow\boxed{\mathrm{(D)}\ \frac{3}{2}}$.

See also

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AMC 10 Problems and Solutions

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