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Circle Triangle Picking

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Select three points at random on the circumference of a unit circle and find the distribution of areas of the resulting triangles determined by these three points.

The first point can be assigned coordinates (1,0) without loss of generality. Call the central angles from the first point to the second and third theta_1 and theta_2. The range of theta_1 can be restricted to [0,pi] because of symmetry, but theta_2 can range from [0,2pi). Then

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

so

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)

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^(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)

But

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)

Write (10) 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)

From (12),

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)

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_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)

so

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

Combining (◇) and (◇) gives the mean triangle area as

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

(OEIS A093582).

The first few 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 triangle determined by the three points picked at random on the circumference of a circle 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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