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圆内接三角形的选取


TriangleInscribing

在一个单位圆的圆周上随机选取三个点,并找出由这三个点确定的三角形面积的分布。

不失一般性,第一个点可以被赋予坐标 (1,0)。将从第一个点到第二个和第三个点的中心角分别称为 theta_1theta_2。由于对称性,theta_1 的范围可以限制在 [0,pi],但 theta_2 的范围可以是 [0,2pi)。然后

 A(theta_1,theta_2)=2sin(1/2theta_1)sin(1/2theta_2)sin[1/2(theta_1-theta_2)],
(1)

因此

A^_=(int_0^piint_0^(2pi)|A|dtheta_2dtheta_1)/(int_0^piint_0^(2pi)dtheta_2dtheta_1)
(2)
=1/(2pi^2)int_0^piint_0^(2pi)|A|dtheta_2dtheta_1.
(3)

因此,

A^_=2/(2pi^2)int_0^piint_0^(2pi)|sin(1/2theta_1)sin(1/2theta_2)sin[1/2(theta_1-theta_2)]|dtheta_2dtheta_1
(4)
=1/(pi^2)int_0^pisin(1/2theta_1)[int_0^(2pi)sin(1/2theta_2)|sin[1/2(theta_2-theta_1)]|dtheta_2]dtheta_1
(5)
=1/(pi^2)int_0^(pi)int_0^(2pi); theta_2-theta_1>0sin(1/2theta_1)sin(1/2theta_2)sin[1/2(theta_1-theta_2)]dtheta_2dtheta_1+1/(pi^2)int_0^(pi)int_0^(2pi); theta_2-theta_1<0sin(1/2theta_1)sin(1/2theta_2)sin[1/2(theta_1-theta_2)]dtheta_2dtheta_1
(6)
=1/(pi^2)int_0^pisin(1/2theta_1)[int_(theta_1)^(2pi)sin(1/2theta_2)sin[1/2(theta_2-theta_1)]dtheta_2]dtheta_1+1/(pi^2)int_0^pisin(1/2theta_1)[int_0^(theta_1)sin(1/2theta_2)sin[1/2(theta_2-theta_1)]dtheta_2]dtheta_1.
(7)

但是

int(1/2theta_2)sin[1/2(theta_2-theta_1)]dtheta_2=intsin(1/2theta_2)[sin(1/2theta_2)cos(1/2theta_2)-sin(1/2theta_1)cos(1/2theta_2)]dtheta_2
(8)
=cos(1/2theta_1)intsin^2(1/2theta_2)dtheta_2-sin(1/2theta_1)intsin(1/2theta_1)cos(1/2theta_2)dtheta_2
(9)
=1/2cos(1/2theta_1)int(1-costheta_2)dtheta_2-1/2sin(1/2theta_2)intsintheta_2dtheta_2
(10)
=1/2cos(1/2theta_1)(theta_2-sintheta_2)+1/2sin(1/2theta_1)cos(theta_2).
(11)

将 (10) 写成

 A^_=1/(pi^2)[int_0^pisin(1/2theta_1)I_1dtheta_1+int_0^pisin(1/2theta_1)I_2dtheta_1],
(12)

那么

 I_1=int_(theta_1)^(2pi)sin(1/2theta_2)sin[1/2(theta_2-theta_1)]dtheta_2,
(13)

并且

 I_2=int_0^(theta_1)sin(1/2theta_2)sin[1/2(theta_1-theta_2)]dtheta_2.
(14)

从 (12) 中,

I_1=1/2cos(1/2theta_2)[theta_2-sintheta_2]_(theta_1)^(2pi)+1/2sin(1/2theta_1)[costheta_2]_(theta_1)^(2pi)
(15)
=1/2cos(1/2theta_1)(2pi-theta_1+sintheta_1)+1/2sin(1/2theta_1)(1-costheta_1)
(16)
=picos(1/2theta_1)-1/2theta_1cos(1/2theta_1)+1/2[cos(1/2theta_1)sintheta_1-costheta_1sin(1/2theta_1)]+1/2sin(1/2theta_1)
(17)
=picos(1/2theta_1)-1/2theta_1cos(1/2theta_1)+1/2+1/2sin(theta_1-1/2theta_1)+1/2sin(1/2theta_1)
(18)
=picos(1/2theta_1)-1/2theta_1cos(1/2theta_1)+sin(1/2theta_1),
(19)

因此

 int_0^piI_1sin(1/2theta_1)dtheta_1=5/4pi.
(20)

此外,

I_2=1/2cos(1/2theta_1)[sintheta_2-theta_2]_0^(theta_1)-1/2sin(1/2theta_1)[costheta_2]_0^(theta_1)
(21)
=1/2cos(1/2theta_2)(sintheta_1-theta_1)-1/2sin(1/2theta_1)(costheta_1-1)
(22)
=-1/2theta_1cos(1/2theta_1)+1/2[sintheta_1cos(1/2theta_1)-costheta_1sin(1/2theta_2)]+1/2sin(1/2theta_1)
(23)
=-1/2theta_1cos(1/2theta_1)+sin(1/2theta_1),
(24)

因此

 int_0^piI_2sin(1/2theta_1)dtheta_1=1/4pi.
(25)

结合 (◇) 和 (◇) 得到平均三角形面积为

 A^_=1/(pi^2)((5pi)/4+pi/4)=3/(2pi)=0.47746...
(26)

(OEIS A093582)。

前几个矩是

mu_1^'=3/(2pi)
(27)
mu_2^'=3/8
(28)
mu_3^'=(35)/(32pi)
(29)
mu_4^'=(45)/(128)
(30)
mu_5^'=(3003)/(2560pi)
(31)
mu_6^'=(105)/(256)
(32)

(OEIS A093583A093584 和 OEIS A093585A093586)。

因此,方差由下式给出

 sigma_A^2=<A>^2-<A^2>=(3(pi^2-6))/(8pi^2) approx 0.1470.
(33)

在圆的圆周上随机选取的三个点确定的三角形内部包含原点的概率是 1/4。


另请参阅

圆线选取, 圆盘三角形选取, 直线选取, 球面点选取

使用 Wolfram|Alpha 探索

参考文献

Sloane, N. J. A. “整数序列在线百科全书” 中的序列 A093582, A093583, A093584, A093585A093586

在 Wolfram|Alpha 中被引用

圆内接三角形的选取

请引用为

Weisstein, Eric W. “圆内接三角形的选取。” 来自 MathWorld--Wolfram Web 资源。 https://mathworld.net.cn/CircleTrianglePicking.html

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