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

Latest revision as of 17:01, 16 January 2021

Problem

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

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

Solutions

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

Video Solution

https://youtu.be/-qx9vAkfCKw

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

@@ Line 1: / Line 1: @@
 ==Problem==
 What is the value of <math>x</math> if <math>|x-1|=|x-2|</math>?
@@ Line 4: / Line 5: @@
 <math> \mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2 </math>
-==Solution==
+==Solutions==
+===Solution===
 <math>|x-1|</math> is the distance between <math>x</math> and <math>1</math>; <math>|x-2|</math> is the distance between <math>x</math> and <math>2</math>.
-Therefore, the given equation says <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>.
+Therefore, the given equation says <math>x</math> is equidistant from <math>1</math> and <math>2</math>, so <math>x=\frac{1+2}2=\frac{3}{2}\Rightarrow\boxed{\mathrm{(D)}\ \frac{3}{2}}</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 ==
 {{AMC10 box|year=2004|ab=A|num-b=3|num-a=5}}
+==Video Solution==
+https://youtu.be/-qx9vAkfCKw
+Education, the Study of Everything
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

2004 AMC 10A (Problems • Answer Key • Resources)
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