2011 AMC 12B Problem 12

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:geometric probability area decomposition regular polygon

Difficulty rating: 1480

Small hint:

Set the octagon’s edge to

1

and split it into the center square, four rectangles, and four corner triangles

Big hint:

The corner triangles are right isosceles with legs

\dfrac{\sqrt2}{2}

Solution:

Assume the octagon has edge length

1.

The four corner triangles are right isosceles with legs

\dfrac{\sqrt2}{2}

and area

\dfrac14

each. The four rectangles are

1

by

\dfrac{\sqrt2}{2}

with area

\dfrac{\sqrt2}{2}

each, and the center square has area

1.

The total area is

4\cdot\dfrac14+4\cdot\dfrac{\sqrt2}{2}+1=2+2\sqrt2.

The probability of hitting the center square is

\dfrac{1}{2+2\sqrt2}=\dfrac{\sqrt2-1}{2}.

Thus, the correct answer is A.