Difference between revisions of "2004 AMC 10A Problems/Problem 4"
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==Problem== | ==Problem== | ||
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>? | ||
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<math> \mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \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== | + | ==Solutions== |
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+ | ===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>. | <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>. | ||
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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>. | 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== | + | ===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>. | 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 == | == See also == | ||
{{AMC10 box|year=2004|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2004|ab=A|num-b=3|num-a=5}} | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/-qx9vAkfCKw | ||
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+ | Education, the Study of Everything | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:01, 16 January 2021
Problem
What is the value of if ?
Solutions
Solution
is the distance between and ; is the distance between and .
Therefore, the given equation says is equidistant from and , so .
Alternatively, we can solve by casework (a method which should work for any similar problem involving absolute values of real numbers). If , then and , so we must solve , which has no solutions. Similarly, if , then and , so we must solve , which also has no solutions. Finally, if , then and , so we must solve , which has the unique solution .
Solution 2
We know that either or . The first equation simplifies to , which is false, so . Solving, we get .
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 |
Video Solution
Education, the Study of Everything The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.