2011 AMC 10B Problem 16

Let the octagon have side $\sqrt{2}$, chosen so the corner pieces have nice legs. The four corner regions are isosceles right triangles whose hypotenuse is an octagon side, so each has legs $1$ and area $\frac{1}{2}$.

The dividing lines extend the sides of the center square out to the octagon, so the octagon sits inside a bounding square of side $1 + \sqrt{2} + 1 = 2 + \sqrt{2}$. The center square has side equal to the octagon's side, $\sqrt{2}$, hence area $2$.

Octagon area $=$ bounding square minus four corner triangles:
$$(2 + \sqrt{2})^2 - 4 \cdot \tfrac{1}{2} = 6 + 4\sqrt{2} - 2 = 4 + 4\sqrt{2}.$$

The probability is
$$\frac{2}{4 + 4\sqrt{2}} = \frac{1}{2(1 + \sqrt{2})} = \frac{\sqrt{2} - 1}{2(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{2}.$$

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

A regular octagon is a square with its corners cut off, and the cut corners are $45$-$45$-$90$ triangles. Choosing the side length so those triangles have integer legs makes every region's area immediate. Uniform darts turn "probability" into "area ratio."

Tally the nine regions with side $\sqrt{2}$: the center square ($2$), four $1 \times \sqrt{2}$ rectangles ($4\sqrt{2}$ total), and four triangles ($2$ total). Total $4 + 4\sqrt{2}$, same ratio.

• Forgetting to subtract the four corner triangles from the bounding square, or leaving the answer as $1/(2 + 2\sqrt{2})$ without rationalizing to match a choice.
• Using the wrong triangle leg (taking the leg equal to the side rather than side$/\sqrt{2}$), which produces choice (E) or (C).