2011 AMC 10B Problem 16

Question

A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?

\dfrac{\sqrt{2} - 1}{2}

\dfrac{1}{4}

\dfrac{2 - \sqrt{2}}{2}

\dfrac{\sqrt{2}}{4}

2 - \sqrt{2}

Answer: A

Concepts:regular polygon geometric probability rationalizing denominator

Difficulty rating: 1730

Small hint:

Let each side of the center square be

1

Big hint:

Find the octagon area by adding the center square, four rectangles, and four corner triangles

Solution:

Let the side length be

1.

Then, the area of the center is

1.

Then, we must find the area of the octagon. It can be found as a square with

4

isosceles right triangles taken out. The side length of this square is

1 + 2\cdot \dfrac 1{\sqrt 2} = 1 + \sqrt 2 .

It has an area of

(1+\sqrt 2)^2 = 3 + 2 \sqrt 2.

Then, the side length of the right triangles is

\dfrac {1}{\sqrt 2} ,

making the area of one equal to

\dfrac {\frac {1}{\sqrt 2}^2}{2} = \dfrac 14 .

This makes them have a total combined area of

1,

so the area of the octagon is

2+ 2 \sqrt 2.

Thus, the ratio is

\begin{aligned}&\dfrac 1{2+ 2 \sqrt 2} \\&= \dfrac {-2+ 2 \sqrt 2}{(2+ 2 \sqrt 2)(-2+ 2 \sqrt 2)}\\ &=\dfrac {2(\sqrt 2-1)}{4} \\&=\dfrac {\sqrt 2-1}{2}. \end{aligned}

Thus, the correct answer is A .