Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

 
(8 intermediate revisions by 7 users not shown)
Line 2: Line 2:
 
What is the value of <math>x</math> if <math>|x-1|=|x-2|</math>?
 
What is the value of <math>x</math> if <math>|x-1|=|x-2|</math>?
  
−
<math> \mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 {2}\qquad \mathrm{(E) \ } 2 </math>
+
<math> \mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2 </math>
  
 
==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>.
+
<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, <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>.
+
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>.
  
−
==See Also==
+
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>.
  
−
*[[2004 AMC 10A Problems]]
+
==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}}
  
−
*[[2004 AMC 10A Problems/Problem 3|Previous Problem]]
+
[[Category:Introductory Algebra Problems]]
−
 
+
{{MAA Notice}}
−
*[[2004 AMC 10A Problems/Problem 5|Next Problem]]
 

Revision as of 22:23, 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_4&oldid=112458"