发信人: orion (M42), 信区: BrainTeaser
标 题: 一个圆上两个点
发信站: BBS 未名空间站 (Thu Feb 19 14:53:08 2009)

有一个半径为1的单位圆，在圆内任取两点，问两点间距的分布？
假设点是均匀分布的，即点出现在(r,theta)的概率密度为 P(r,theta) = r/Pi.
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发信人: pcasnik (pcasnik), 信区: BrainTeaser
标 题: Re: 一个圆上两个点
发信站: BBS 未名空间站 (Thu Feb 19 20:08:28 2009)

the pdf is: 2x/pi*[2*arccos(x/2)-x*sqrt(4-x^2)/2]*dx, 0<=x<=2.
(see wilmott.com/messageview.cfm?catid=26&threadid=64937)

one interesting observation is that the expression in the parentheses is the area of the intersection of two unit circles with their centers separated by x, and i always wonder whether there is an easy proof based on that, but i never figured out how...
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※ 修改:·pcasnik 於 Feb 19 20:13:21 2009 修改本文·[FROM: 75.34.]

发信人: hero080 (APM=080), 信区: BrainTeaser
标 题: Re: 一个圆上两个点
发信站: BBS 未名空间站 (Thu Feb 19 21:55:08 2009), 转信

可以考虑两个点AB距离为x时，原来的单位圆圆心必须位于单位圆A和单位圆B的交集中
。再加上一个旋转自由度就出来了。
【 在 pcasnik (pcasnik) 的大作中提到: 】
: the pdf is: 2x/pi*[2*arccos(x/2)-x*sqrt(4-x^2)/2]*dx, 0<=x<=2.
: (see wilmott.com/messageview.cfm?catid=26&threadid=64937)
: one interesting observation is that the expression in the parentheses is
the area of the intersection of two unit circles with their centers
separated by x, and i always wonder whether there is an easy proof based on
that, but i never figured out how



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