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

Revision as of 16:48, 20 December 2020

Video Solution

Education, the Study of Everything

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 $x=\frac{3}{2}\Rightarrow\boxed{\mathrm{(D)}\ \frac{3}{2}}$ .

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

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

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+==Video Solution==
+https://youtu.be/-qx9vAkfCKw
+Education, the Study of Everything
 ==Problem==
 What is the value of <math>x</math> if <math>|x-1|=|x-2|</math>?

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