AMC 8 1996 Problem 25

AMC8 1996 年第 25 题

概率★★★★☆

A point is chosen at random from within a circular region. What is the probability that the point is closer to the center of the region than it is to the boundary of the region?

在一个圆形区域内随机选取一个点。该点离圆心比离边界更近的概率是多少？

A. $\displaystyle \frac{1}{4}$
B. $\displaystyle \frac{1}{3}$
C. $\displaystyle \frac{1}{2}$
D. $\displaystyle \frac{2}{3}$
E. $\displaystyle \frac{3}{4}$

正确答案 · Correct Answer

解析 · Solution

解法1

常规解法 · 定边界圆设大圆半径 2。「到圆心更近」即点到圆心距离

r

满足 r<2-r，得 r<1。合格区域是半径 1 的同心圆，概率

\dfrac{\pi\cdot1^2}{\pi\cdot2^2}=\dfrac14

。

Solution 1

Elementary · fix the boundary circle Let the region be a circle of radius 2. A point is closer to the center than to the boundary when its distance

r

to the center satisfies r<2-r, i.e. r<1. The favorable region is the concentric circle of radius 1, so the probability is

\dfrac{\pi\cdot1^2}{\pi\cdot2^2}=\dfrac14

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