Question

一个飞镖靶是如图所示分区的正八边形。假设飞镖落在靶上任意位置的概率相同。飞镖落在中心正方形内的概率是多少？

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?

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

$\dfrac{1}{4}$

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

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

$2 - \sqrt{2}$

答案：A

知识点：正多边形几何概率分母有理化

难度评级：1730

解答：

设中心正方形边长为 $1$ 。则中心面积为 $1$ 。

八边形可看作一个大正方形去掉 $4$ 个等腰直角三角形。大正方形边长为面积为 $1 + 2\cdot \dfrac 1{\sqrt 2} = 1 + \sqrt 2 .$ $(1+\sqrt 2)^2 = 3 + 2 \sqrt 2.$

这些直角三角形的直角边长为 $\dfrac {1}{\sqrt 2}$ ，每个面积为四个总面积为 $1$ ，所以八边形面积为 $2+ 2 \sqrt 2$ 。 $\dfrac {\frac {1}{\sqrt 2}^2}{2} = \dfrac 14 .$

因此比值为 $\begin{align*}&\dfrac 1{2+ 2 \sqrt 2} \\&= \dfrac {-2+ 2 \sqrt 2}{(2+ 2 \sqrt 2)(-2+ 2 \sqrt 2)}\\ &=\dfrac {2(\sqrt 2-1)}{4} \\&=\dfrac {\sqrt 2-1}{2}. \end{align*}$

所以正确答案是 A 。

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

Thus, the correct answer is A .

2011 AMC 10B 第 16 题

答案：A

难度评级：1730

解答：

其他年份的第 16 题