Difference between revisions of "2000 AMC 10 Problems/Problem 10"

m (Solution)
m (Slight error in solution (7 is attainable!))
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
From the triangle inequality, <math>2<x<10</math> and <math>2<y<10</math>. The smallest positive number not possible is <math>10-2</math>, which is <math>8</math>.
+
From the triangle inequality, <math>2<x<10</math> and <math>2<y<10</math>. <math>7</math> can be attain by letting <math>x=9.1</math> and <math>y=2.1</math>. However, <math>8=10-2</math> cannot be attained. Thus, the answer is <math>\boxed{\bold{D}}</math>.
<math>\boxed{\text{D}}</math>
 
 
 
7 is the correct answer, but it is not listen here.
 
  
 
==See Also==
 
==See Also==

Revision as of 22:46, 3 March 2017

Problem

The sides of a triangle with positive area have lengths $4$, $6$, and $x$. The sides of a second triangle with positive area have lengths $4$, $6$, and $y$. What is the smallest positive number that is not a possible value of $|x-y|$?

$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 8 \qquad\mathrm{(E)}\ 10$

Solution

From the triangle inequality, $2<x<10$ and $2<y<10$. $7$ can be attain by letting $x=9.1$ and $y=2.1$. However, $8=10-2$ cannot be attained. Thus, the answer is $\boxed{\bold{D}}$.

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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