2011 AMC 10B Problem 16

The dart is equally likely to land anywhere, so the probability is (area of the center square) divided by (area of the octagon). Any side length gives the same ratio, so pick one that makes the numbers easy.

1. Choose a size. Let each side of the octagon be $\sqrt{2}$. The four corner pieces are right triangles with two equal legs, and their longest side is an octagon side, $\sqrt{2}$. A right triangle with legs $1$ and $1$ has longest side $\sqrt{1 + 1} = \sqrt{2}$, so each leg is $1$.
2. The center square. Its sides are the same length as the octagon's sides, $\sqrt{2}$. So its area is $\sqrt{2} \times \sqrt{2} = 2$.
3. The octagon. Extend the octagon's corners into a big square. Each side of the big square is a leg, then an octagon side, then a leg: $1 + \sqrt{2} + 1 = 2 + \sqrt{2}$. Its area is
$$(2 + \sqrt{2})^2 = 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2}.$$
4. The octagon is this big square minus four corner triangles, each with area $\frac{1}{2} \times 1 \times 1 = \frac{1}{2}$:
$$6 + 4\sqrt{2} - 4 \times \tfrac{1}{2} = 4 + 4\sqrt{2}.$$
5. The probability.
$$\frac{2}{4 + 4\sqrt{2}} = \frac{1}{2 + 2\sqrt{2}} = \frac{1}{2(1 + \sqrt{2})}.$$
6. None of the choices looks like this, so remove the square root from the bottom. Multiply top and bottom by $\sqrt{2} - 1$, using $(\sqrt{2} + 1)(\sqrt{2} - 1) = 2 - 1 = 1$:
$$\frac{\sqrt{2} - 1}{2 \times 1} = \frac{\sqrt{2} - 1}{2}.$$

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

A regular octagon is a square with its four corners cut off, and the cut-off corners are right triangles with two equal legs. Choosing the side length so those legs are $1$ makes every area easy. With an evenly thrown dart, probability is just a ratio of areas.

Add up all nine regions instead. With side $\sqrt{2}$: the center square has area $2$; the four rectangles along the sides are each $\sqrt{2}$ by $1$, total $4\sqrt{2}$; the four corner triangles have total area $2$. The octagon's area is $4 + 4\sqrt{2}$, the same as before.