2000 AMC 10 Problem 10

1. Which lengths can $x$ be? In a real triangle, any two sides together must be longer than the third side.
- $4 + 6 > x$, so $x < 10$.
- $4 + x > 6$, so $x > 2$.

So $x$ is strictly between $2$ and $10$. The same is true for $y$.
2. How far apart can $x$ and $y$ be? The farthest apart they can be is when one is near $10$ and the other is near $2$. But neither can actually reach $10$ or $2$, so the gap is always less than $10 - 2 = 8$. The gap $8$ is impossible.
3. Every smaller gap is possible. For example:
- gap $2$: $x = 5$, $y = 3$.
- gap $4$: $x = 7$, $y = 3$.
- gap $6$: $x = 9$, $y = 3$.
- gap $7.9$: $x = 9.95$, $y = 2.05$.

Any gap below $8$ works the same way: put $y$ a little above $2$ and $x$ that much higher.
4. So $2$, $4$ and $6$ are all possible, and $8$ is the smallest positive number that is not.

The answer is $\boxed{\textbf{(D)}\ 8}$.

The rule "any two sides add to more than the third" gives a range for the unknown side. The words "positive area" mean the triangle cannot be flat, so the ends of the range are not allowed. Two numbers from a range with the ends left out can get close to the full width of the range, but never reach it. Watching "less than" versus "less than or equal to" is the whole problem.

For a triangle with positive area and sides $4$, $6$, $x$, the triangle inequality must hold strictly: $x < 4 + 6 = 10$ and $x + 4 > 6$, so $x > 2$. Thus $2 < x < 10$, and likewise $2 < y < 10$.

Two numbers from the open interval $(2, 10)$ differ by less than $10 - 2 = 8$, so $|x - y| < 8$; the value $8$ is impossible. On the other hand every value in $[0, 8)$ occurs: for a target $d < 8$, take $y$ slightly above $2$ and $x = y + d$, which is still below $10$. In particular $2$, $4$, $6$ are all attainable (for instance $x = 9, y = 3$ gives $6$).

Therefore the smallest positive number that $|x - y|$ cannot equal is $8$.

The answer is $\boxed{\textbf{(D)}\ 8}$.