Triangle Inscribing in a Circle

Select three points at random on a unit Circle. Find the distribution of possible areas. 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 $[0,2\pi)$. Then

$$A(\theta_1,\theta_2) =2\left\vert{\sin({\textstyle{1\over 2}... ...2)\sin[{\textstyle{1\over 2}}(\theta_1-\theta_2)]}\right\vert,$$ (1)

so

$$\bar A = {\int_0^\pi\int_0^{2\pi} A(\theta_1,\theta_2)\,d\theta_2\,d\theta_1\over C},$$ (2)

where

$$C\equiv \int_0^\pi\int_0^{2\pi} \,d\theta_2\,d\theta_1 = 2\pi^2.$$ (3)

Therefore,

$$\bar A = {2\over 2\pi^2} \int_0^\pi\int_0^{2\pi} \left\vert{\sin({\textsty... ...sin[{\textstyle{1\over 2}}(\theta_1-\theta_2)]}\right\vert d\theta_2\,d\theta_1$$

$$= {1\over \pi^2} \int_0^\pi \sin({\textstyle{1\over 2}}\theta_1) \l... ...extstyle{1\over 2}}(\theta_2-\theta_1)]}\right\vert\,d\theta_2}\right]d\theta_1$$

$$= {1\over \pi^2} {\int_0^\pi\int_0^{2\pi}\atop \scriptstyle\theta_2... ...}\theta_2)\sin[{\textstyle{1\over 2}}(\theta_1-\theta_2)]\,d\theta_2\,d\theta_1$$

$$\phantom{=} + {1\over \pi^2} {\int_0^\pi\int_0^{2\pi}\atop\scriptstyle\theta_... ...}\theta_2)\sin[{\textstyle{1\over 2}}(\theta_1-\theta_2)]\,d\theta_2\,d\theta_1$$

$$= {1\over \pi^2} \int_0^\pi \sin({\textstyle{1\over 2}}\theta_1) \l... ..._2) \sin[{\textstyle{1\over 2}}(\theta_2-\theta_1)]\,d\theta_2}\right]d\theta_1$$

$$\phantom{=} +{1\over \pi^2} \int_0^\pi \sin({\textstyle{1\over 2}}\theta_1) \... ...2) \sin[{\textstyle{1\over 2}}(\theta_2-\theta_1)]\,d\theta_2}\right]d\theta_1.$$ (4)

But

$\int({\textstyle{1\over 2}}\theta_2)\sin[{\textstyle{1\over 2}}(\theta_2-\theta_1)]\,d\theta_2$
$=\int \sin({\textstyle{1\over 2}}\theta_2) \left[{\sin({\textstyle{1\over 2}}\t... ...tyle{1\over 2}}\theta_1)\cos({\textstyle{1\over 2}}\theta_2)}\right]\,d\theta_2$
$=\cos({\textstyle{1\over 2}}\theta_1)\int\sin^2({\textstyle{1\over 2}}\theta_2)... ...({\textstyle{1\over 2}}\theta_1)\cos({\textstyle{1\over 2}}\theta_2)\,d\theta_2$
$= {\textstyle{1\over 2}}\cos ({\textstyle{1\over 2}}\theta_1)\int(1-\cos\theta_... ...style{1\over 2}}\sin({\textstyle{1\over 2}}\theta_2)\int\sin\theta_2\,d\theta_2$
$= {\textstyle{1\over 2}}\cos({\textstyle{1\over 2}}\theta_1)(\theta_2-\sin\theta_2)+{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)\cos(\theta_2).$	(5)

Write (4) as

$$\bar A={1\over\pi^2}\left[{\int_0^\pi \sin({\textstyle{1\ove... ...pi \sin({\textstyle{1\over 2}}\theta_1)I_2\,d\theta_1}\right],$$ (6)

then

$$I_1 \equiv \int_{\theta_1}^{2\pi} \sin({\textstyle{1\over 2}... ..._2)\sin[{\textstyle{1\over 2}}(\theta_2-\theta_1)]\,d\theta_2,$$ (7)

and

$$I_2 \equiv \int_0^{\theta_1}\sin({\textstyle{1\over 2}}\theta_2)\sin[{\textstyle{1\over 2}}(\theta_1-\theta_2)]\,d\theta_2.$$ (8)

From (6),

$$I_1 = {\textstyle{1\over 2}}\cos({\textstyle{1\over 2}}\theta_2)[\theta... ...{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)[\cos\theta_2]^{2\pi}_{\theta_1}$$

$$= {\textstyle{1\over 2}}\cos({\textstyle{1\over 2}}\theta_1) (2\pi-... ...a_1)+{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)(1-\cos\theta_1)$$

$$= \pi\cos({\textstyle{1\over 2}}\theta_1)-{\textstyle{1\over 2}}\th... ...1\over 2}}\theta_1)]+{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)$$

$$= \pi \cos({\textstyle{1\over 2}}\theta_1)-{\textstyle{1\over 2}}\t... ...{1\over 2}}\theta_1)+{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)$$

$$= \pi \cos({\textstyle{1\over 2}}\theta_1)-{\textstyle{1\over 2}}\theta_1\cos ({\textstyle{1\over 2}}\theta_1)+\sin({\textstyle{1\over 2}}\theta_1),$$ (9)

so

$$\int_0^\pi I_1\sin({\textstyle{1\over 2}}\theta_1)\,d\theta_1 = {\textstyle{5\over 4}}\pi.$$ (10)

Also,

$$I_2 = {\textstyle{1\over 2}}\cos ({\textstyle{1\over 2}}\theta_1)[\sin ... ...style{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)[\cos\theta_2]^{\theta_1}_0$$

$$= {\textstyle{1\over 2}}\cos({\textstyle{1\over 2}}\theta_2)(\sin \... ..._1)-{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)(\cos \theta_1-1)$$

$$= -{\textstyle{1\over 2}}\theta_1\cos ({\textstyle{1\over 2}}\theta... ...1\over 2}}\theta_2)]+{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)$$

$$= -{\textstyle{1\over 2}}\theta_1\cos ({\textstyle{1\over 2}}\theta_1)+\sin ({\textstyle{1\over 2}}\theta_1),$$ (11)

so

$$\int_0^\pi I_2\sin({\textstyle{1\over 2}}\theta_1)\,d\theta_1 = {\textstyle{1\over 4}}\pi.$$ (12)

Combining (10) and (12) gives

$$\bar A={1\over \pi^2} \left({{5\pi\over 4}+{\pi\over 4}}\right)= {3\over 2\pi} \approx 0.4775.$$ (13)

The Variance is

$${\sigma_A}^2 = {1\over 2\pi^2} \int_0^\pi \int_0^{2\pi} [A(\theta_1,\theta_2)-{\textstyle{3\over 2\pi}}]^2\,d\theta_2\,d\theta_1$$

$$= {1\over 2\pi^2} \int_0^\pi \int_0^{2\pi}\left[{2\left\vert{\sin({... ...}(\theta_1-\theta_2)]}\right\vert-{3\over 2\pi}}\right]^2\,d\theta_2\,d\theta_1$$

$$= {1\over 2\pi^2}\int_0^\pi\int_0^{2\pi}\left\{{4\sin^2({\textstyle... ...le{1\over 2}}\theta_2)\sin^2[{\textstyle{1\over 2}}(\theta_2-\theta_1)]}\right.$$

$$\phantom{=} \left.{-{6\over\pi}\left\vert{\sin({\textstyle{1\over 2}}\theta_1... ...\theta_1-\theta_2)]}\right\vert +{9\over 4\pi^2}}\right\}\,d\theta_2\,d\theta_1$$

$$= {1\over 2\pi^2}\left[{\int_0^\pi \pi(2+\theta_1)\sin^2({\textstyl... ...r \pi}\left({{5\pi\over 4}+{\pi\over 4}}\right)+{9\over 4\pi^2}(2\pi^2)}\right]$$

$$= {1\over 2\pi^2} \left({{3\pi^2\over 4}-9+{9\over 2}}\right)= {1\over 2\pi^2}\left({{3\pi^2\over 4}-{9\over 2}}\right)$$

$$= {3(\pi^2-6)\over 8\pi^2}\approx 0.1470.$$ (14)

See also Point-Point Distance--1-D, Tetrahedron Inscribing