AMC12 2011 B

AMC12 2011 B · Q12

AMC12 2011 B · Q12. It mainly tests Polygons, Geometric probability (basic).

A dart board is a regular octagon divided into regions as shown below. 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?

一个飞镖盘是如下图所示的正八边形，并被分成若干区域。假设飞镖投向飞镖盘时，落在盘内任意位置的可能性相同。飞镖落在中心正方形内的概率是多少？

(A) \frac{\sqrt{2}-1}{2} \frac{\sqrt{2}-1}{2}

(B) \frac{1}{4} \frac{1}{4}

(D) \frac{\sqrt{2}}{4} \frac{\sqrt{2}}{4}

(E) 2 - \sqrt{2} 2 - \sqrt{2}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let's assume that the side length of the octagon is $x$. The area of the center square is just $x^2$. The triangles are all $45-45-90$ triangles, with a side length ratio of $1:1:\sqrt{2}$. The area of each of the $4$ identical triangles is $\left(\dfrac{x}{\sqrt{2}}\right)^2\times\dfrac{1}{2}=\dfrac{x^2}{4}$, so the total area of all of the triangles is also $x^2$. Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is $x$ and the other side length is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$, so the area of all of the rectangles is $2x^2\sqrt{2}$. The ratio of the area of the square to the area of the octagon is $\dfrac{x^2}{2x^2+2x^2\sqrt{2}}$. Cancelling $x^2$ from the fraction, the ratio becomes $\dfrac{1}{2\sqrt2+2}$. Multiplying the numerator and the denominator each by $2\sqrt{2}-2$ will cancel out the radical, so the fraction is now $\dfrac{1}{2\sqrt2+2}\times\dfrac{2\sqrt{2}-2}{2\sqrt{2}-2}=\dfrac{2\sqrt{2}-2}{4}=\boxed{\mathrm{(A)}\ \dfrac{\sqrt{2}-1}{2}}$

设八边形的边长为 $x$。中心正方形的面积为 $x^2$。这些三角形都是 $45-45-90$ 三角形，边长比为 $1:1:\sqrt{2}$。四个全等三角形中每个的面积为 $\left(\dfrac{x}{\sqrt{2}}\right)^2\times\dfrac{1}{2}=\dfrac{x^2}{4}$，所以所有三角形的总面积也是 $x^2$。现在求四个全等矩形的总面积。矩形的一条边长为 $x$，另一条边长为 $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$，所以所有矩形的总面积为 $2x^2\sqrt{2}$。正方形面积与八边形面积之比为 $\dfrac{x^2}{2x^2+2x^2\sqrt{2}}$。约去分子分母中的 $x^2$，得到 $\dfrac{1}{2\sqrt2+2}$。将分子分母同乘 $2\sqrt{2}-2$ 以消去根号，得到 $\dfrac{1}{2\sqrt2+2}\times\dfrac{2\sqrt{2}-2}{2\sqrt{2}-2}=\dfrac{2\sqrt{2}-2}{4}=\boxed{\mathrm{(A)}\ \dfrac{\sqrt{2}-1}{2}}$。

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