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Circle Triangle Picking


TriangleInscribing

Circle triangle picking considers the areas of triangles determined by three points chosen independently and uniformly at random on the circumference of a unit circle.

By rotational and reflection symmetry, the first point can be assigned coordinates (1,0) and the central angles from it to the second and third points, theta_1 and theta_2, can be taken as independent random variables with uniform distributions on [0,pi] and [0,2pi), respectively. The signed area is

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

and the ordinary area is A=|A_s|, so

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

Therefore,

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^piint_0^(2pi); theta_2-theta_1>0sin(1/2theta_1)sin(1/2theta_2)sin[1/2(theta_2-theta_1)]dtheta_2dtheta_1+1/(pi^2)int_0^piint_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_1-theta_2)]dtheta_2]dtheta_1.
(7)

But

intsin(1/2theta_2)sin[1/2(theta_2-theta_1)]dtheta_2=intsin(1/2theta_2)[sin(1/2theta_2)cos(1/2theta_1)-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_2)cos(1/2theta_2)dtheta_2
(9)
=1/2cos(1/2theta_1)int(1-costheta_2)dtheta_2-1/2sin(1/2theta_1)intsintheta_2dtheta_2
(10)
=1/2cos(1/2theta_1)(theta_2-sintheta_2)+1/2sin(1/2theta_1)cos(theta_2).
(11)

Write the mean area as

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

then

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

and

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

Evaluating the antiderivative above gives

I_1=1/2cos(1/2theta_1)[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)+1/2sin(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)

so

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

Also,

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_1)(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_1)]+1/2sin(1/2theta_1)
(23)
=-1/2theta_1cos(1/2theta_1)+sin(1/2theta_1),
(24)

so

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

Combining (20) and (25) gives the mean triangle area as

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

(OEIS A093582).

The first few raw moments are

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 A093583 and A093584 and OEIS A093585 and A093586).

The variance is therefore given by

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

The probability that the interior of the resulting triangle contains the origin is 1/4.


See also

Circle Line Picking, Disk Triangle Picking, Line Line Picking, Sphere Point Picking

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References

Sloane, N. J. A. Sequences A093582, A093583, A093584, A093585, and A093586 in "The On-Line Encyclopedia of Integer Sequences."

Referenced on Wolfram|Alpha

Circle Triangle Picking

Cite this as:

Weisstein, Eric W. "Circle Triangle Picking." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/CircleTrianglePicking.html

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