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

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==Problem==
 
==Problem==
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The sides of a triangle with positive area have lengths <math>4</math>, <math>6</math>, and <math>x</math>. The sides of a second triangle with positive area have lengths <math>4</math>, <math>6</math>, and <math>y</math>. What is the smallest positive number that is '''not''' a possible value of <math>|x-y|</math>? 
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<math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 8 \qquad\mathrm{(E)}\ 10</math>
  
 
==Solution==
 
==Solution==

Revision as of 21:52, 8 January 2009

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

The largest possible value for $x$ is $9$. The smallest is $3$.

$9-3=6$.

$8$ is the smallest that cannot be made (of the choices listed)

$\boxed{\text{D}}$

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
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