2001 AMC 10 Problem 11

1. Start with the rings you can see. The center square is $1 \times 1$. With the first ring it becomes a $3 \times 3$ square; with the second ring, a $5 \times 5$ square. Each ring adds one square on every side, so the side grows by $2$.
2. So the $n$th ring is the border of a square with side $2n + 1$. For the $100$th ring that is a $201 \times 201$ square, and we only want its border.
3. Count the border in four pieces, making sure each corner is counted once:
- the top row: all $2n + 1$ squares, corners included;
- the bottom row: also $2n + 1$ squares;
- the left column without its two corners: $2n - 1$ squares;
- the right column without its two corners: $2n - 1$ squares.
4. Add: $(2n+1) + (2n+1) + (2n-1) + (2n-1) = 8n$.
5. Check with the given rings: $n = 1$ gives $8$ and $n = 2$ gives $16$. Both match.
6. For the $100$th ring: $8 \times 100 = 800$.

The answer is $\boxed{\textbf{(C)}\ 800}$.

A ring is just the edge of a square, so its size depends on the side length, not the area. The rings grow steadily ($8, 16, 24, \ldots$), adding $8$ each time, even though the full squares grow much faster. When counting the edge of a square, be careful to count each corner exactly once.

Subtract areas. The $n$th ring is the $(2n+1) \times (2n+1)$ square with the $(2n-1) \times (2n-1)$ square inside it removed. For $n = 100$: $201^2 - 199^2 = 40401 - 39601 = 800$. (A faster way to compute this: $201^2 - 199^2 = (201 - 199)(201 + 199) = 2 \times 400 = 800$.)