2000 AMC 10 Problem 10

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 $\boxed{\textbf{(D)}\ 8}$.

The triangle inequality gives an open range for the unknown side, and "positive area" is the phrase that makes the endpoints unreachable. The largest difference between two numbers in the same open interval is its width, approached but never attained. Keep track of strict versus non-strict inequalities; that single detail is the whole problem.

• Treating the endpoints 2 and 10 as attainable (a degenerate triangle has zero area) and answering 10.
• Forgetting the lower bound $x > 2$ (only using $x < 10$) and concluding differences near $10$ are possible.
• Guessing that $6$ is unreachable without testing a concrete pair such as $x = 9, y = 3$.