2000 AMC 10 Problems/Problem 10

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 $8$ . $\boxed{\text{D}}$

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

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2000 AMC 10 Problems/Problem 10

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