2001 AMC 10 Problem 20

1. Picture the square with a small triangle cut from each corner. Each triangle has two equal legs along the square's sides, and its slanted side (the hypotenuse) becomes one side of the octagon.
2. The octagon also has four sides that are the middle parts of the square's sides, left over between two cuts. The octagon is regular, so all eight sides are equal. Call that length $s$.
3. In an isosceles right triangle, the hypotenuse is $\sqrt{2}$ times a leg. The hypotenuse is $s$, so each leg is $\frac{s}{\sqrt 2}$.
4. Walk along one side of the square: a leg, then an octagon side, then another leg. So
$$\frac{s}{\sqrt 2} + s + \frac{s}{\sqrt 2} = 2000.$$
5. Two copies of $\frac{s}{\sqrt 2}$ make $\frac{2s}{\sqrt 2} = s\sqrt{2}$. So $s\sqrt{2} + s = 2000$, which is $s(\sqrt{2} + 1) = 2000$, and
$$s = \frac{2000}{\sqrt 2 + 1}.$$
6. The choices have no square root in the bottom, so rewrite it. Multiply top and bottom by $\sqrt 2 - 1$. The bottom becomes $(\sqrt2 + 1)(\sqrt 2 - 1) = 2 - 1 = 1$. So
$$s = 2000(\sqrt{2} - 1).$$

The answer is $\boxed{\textbf{(B)}\ 2000(\sqrt{2}-1)}$.

Cutting corners off a square makes two kinds of octagon sides: the slanted cuts and the leftover middles. "Regular" says these must be the same length, and that one fact gives the equation. Remember that $\frac{1}{\sqrt 2 + 1} = \sqrt 2 - 1$; it comes up often.

Estimate. Each leg is shorter than the octagon side, so on one side of the square $2000 = \text{leg} + s + \text{leg}$ is less than $3s$, and $s > 667$. Also $s = \sqrt2 \times \text{leg}$ is less than $2 \times \text{leg}$, so $2000$ is more than $2s$, and $s < 1000$. Using $\sqrt 2 \approx 1.414$: (A) is about $667$, (B) about $828$, (C) about $1172$, (D) is $1000$, (E) about $1414$. Only (B) lies between $667$ and $1000$.