2001 AMC 10 Problem 20

Let each cut-off triangle have legs of length $x$. Its hypotenuse is $x\sqrt{2}$, and that hypotenuse is a side of the octagon. The pieces of the square's side that remain between two cuts also form a side of the octagon, and all octagon sides are equal, so call every side $s$.

One side of the square consists of a leg, an octagon side, and another leg:
$$x + s + x = 2000, \qquad s = x\sqrt{2}.$$
Substituting $x = \frac{s}{\sqrt 2}$ gives $s\sqrt{2} + s = 2000$, so
$$s = \frac{2000}{\sqrt{2} + 1} = \frac{2000(\sqrt{2} - 1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = 2000(\sqrt{2} - 1).$$

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

Cutting corners from a square produces two kinds of octagon sides, the hypotenuses and the leftover middles, and "regular" forces them equal. That single equation, plus the $1 : 1 : \sqrt2$ ratio, determines everything. Rationalizing $\frac{1}{\sqrt 2 + 1} = \sqrt 2 - 1$ is a conversion worth memorizing.

Bound it: since $s = x\sqrt2 > x$ and $2x + s = 2000$, the side satisfies $\frac{2000}{3} < s < 1000$. Numerically $s \approx 828$, and only choice (B) ($\approx 828$) falls in that window; (A) is exactly $\frac{2000}{3}$, (D) is $1000$, and (C), (E) exceed $1000$.

• Using the leg x = 2000/(2 + sqrt 2) as the octagon side, or forgetting to rationalize and failing to match the answer form.
• Assuming the three pieces along a side are equal ($s = \frac{2000}{3}$, choice A), which would make the cut triangles' hypotenuses longer than the middle piece.