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

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==Solution==
 
==Solution==
<math>|x-1|</math> is the same thing as saying the distance between <math>x</math> and <math>1</math>; <math>|x-2|</math> is the same thing as saying the distance between <math>x</math> and <math>2</math>.
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<math>|x-1|</math> is equivalent to the distance between <math>x</math> and <math>1</math>; <math>|x-2|</math> is equivalent to the distance between <math>x</math> and <math>2</math>.
  
Therefore, <math>x</math> is the same distance from <math>1</math> and <math>2</math>, so <math>x=\frac{1+2}2=\frac32\Rightarrow\mathrm{(D)}</math>.
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Therefore, <math>x</math> is equidistant from <math>1</math> and <math>2</math>, so <math>x=\frac{1+2}2=\frac32\Rightarrow\mathrm{(D)}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 14:18, 12 November 2006

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

Therefore, $x$ is equidistant from $1$ and $2$, so $x=\frac{1+2}2=\frac32\Rightarrow\mathrm{(D)}$.

See Also

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