1971 AMC 12 第 17 题

Question

用

2n

条等间隔的半径

(n\gt0)

和一条割线分割一个圆盘。该圆盘最多能被分成多少个互不重叠的区域：

A circular disk is divided by

2n

equally spaced radii

(n\gt0)

and one secant line. The maximum number of nonoverlapping areas into which the disk can be divided is:

2n+1

2n+2

3n-1

3n

3n+1

从这

2n

条半径形成的扇形数开始计算

Begin with the

2n

sectors made by the radii

一条割线至多与

n

条半径相交，再数割线弦段被分成的部分

A secant can cross at most

n

of the radii, so count the pieces of the secant chord

这些半径先形成

2n

个扇形。一条不过圆心的割线在

n

对反向半径中的每一对里至多与一条相交，因此至多与

n

条半径相交。这些交点把圆内的割线弦段分成

n+1

段，每段都会增加一个区域。因此最大区域数为

2n+(n+1)=3n+1

。

因此，正确答案为 E。

The radii first make

2n

sectors. A secant not through the center can meet at most one radius in each of

n

opposite pairs, hence at most

n

radii. Those intersections divide the secant chord into

n+1

pieces, each of which adds one region. The maximum is therefore

2n+(n+1)=3n+1.

Therefore, the correct answer is E.

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