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# Difference between revisions of "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

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)

$\boxed{\text{D}}$