# 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

Since $6$ and $4$ are fixed sides, the smallest possible side has to be larger than $6-4=2$ and the largest possible side has to be smaller than $6+4=10$. This gives us the triangle inequality $2 and $2. $7$ can be attained by letting $x=9.1$ and $y=2.1$. However, $8=10-2$ cannot be attained. Thus, the answer is $\boxed{\bold{D}}$.