AMC8 1996

AMC8 1996 · Q25

AMC8 1996 · Q25. It mainly tests Area & perimeter, Geometric probability (basic).

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?

从圆形区域内随机选择一点。该点比到区域边界的距离更靠近区域中心的概率是多少？

(A) $\frac{1}{4}$ $\frac{1}{4}$

(B) $\frac{1}{3}$ $\frac{1}{3}$

(D) $\frac{2}{3}$ $\frac{2}{3}$

(E) $\frac{3}{4}$ $\frac{3}{4}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Suppose that the circle has radius 1. Then, being closer to the center of the region than the boundary of the region (the circle) would mean the chosen point must be inside the circle of radius $\frac{1}{2}$ with the same center as the larger circle of radius 1. The area of the smaller region is $\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}$, and the area of the total region is $\pi(1)^2=\pi$. Since the area of the smallest region is $\frac{1}{4}$ of the area of the total region, the required probability is $\frac{1}{4}$. Note. The odds that the point is closer to the center are 1:3.

答案（A）：假设圆的半径为 1。那么，“离该区域的中心比离该区域的边界（圆周）更近”意味着所选点必须位于半径为 $\frac{1}{2}$ 的圆内，并且该小圆与半径为 1 的大圆同心。较小区域的面积为 $\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}$，而整个区域的面积为 $\pi(1)^2=\pi$。由于最小区域的面积是总区域面积的 $\frac{1}{4}$，所求概率为 $\frac{1}{4}$。注：该点更靠近中心的赔率为 1:3。

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