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2000 AMC 10 Problems/Problem 10

Revision as of 21:57, 8 January 2009 by Xpmath (talk | contribs) (Solution)

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 $2<y<10$. The smallest positive number not possible is 10-2, which is 8. $\boxed{\text{D}}$

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 10 Problems and Solutions
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