2004 AMC 10A Problems/Problem 4

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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 $x=\frac{1+2}2=\frac32\Rightarrow\mathrm{(D)}$ .

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 $x = \frac32$ .

2004 AMC 10A (Problems)
Preceded by Problem 3	Followed by Problem 5
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2004 AMC 10A Problems/Problem 4

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Problem

Solution

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