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Triangle Inscribing in a Circle

\begin{figure}\begin{center}\BoxedEPSF{TriangleInscribing.epsf}\end{center}\end{figure}

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

\begin{displaymath}
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,
\end{displaymath} (1)

so
\begin{displaymath}
\bar A = {\int_0^\pi\int_0^{2\pi} A(\theta_1,\theta_2)\,d\theta_2\,d\theta_1\over C},
\end{displaymath} (2)

where
\begin{displaymath}
C\equiv \int_0^\pi\int_0^{2\pi} \,d\theta_2\,d\theta_1 = 2\pi^2.
\end{displaymath} (3)

Therefore,


$\displaystyle \bar A$ $\textstyle =$ $\displaystyle {2\over 2\pi^2} \int_0^\pi\int_0^{2\pi}
\left\vert{\sin({\textsty...
...sin[{\textstyle{1\over 2}}(\theta_1-\theta_2)]}\right\vert d\theta_2\,d\theta_1$  
  $\textstyle =$ $\displaystyle {1\over \pi^2} \int_0^\pi \sin({\textstyle{1\over 2}}\theta_1) \l...
...extstyle{1\over 2}}(\theta_2-\theta_1)]}\right\vert\,d\theta_2}\right]d\theta_1$  
  $\textstyle =$ $\displaystyle {1\over \pi^2} {\int_0^\pi\int_0^{2\pi}\atop \scriptstyle\theta_2...
...}\theta_2)\sin[{\textstyle{1\over 2}}(\theta_1-\theta_2)]\,d\theta_2\,d\theta_1$  
  $\textstyle \phantom{=}$ $\displaystyle + {1\over \pi^2} {\int_0^\pi\int_0^{2\pi}\atop\scriptstyle\theta_...
...}\theta_2)\sin[{\textstyle{1\over 2}}(\theta_1-\theta_2)]\,d\theta_2\,d\theta_1$  
  $\textstyle =$ $\displaystyle {1\over \pi^2} \int_0^\pi \sin({\textstyle{1\over 2}}\theta_1) \l...
..._2)
\sin[{\textstyle{1\over 2}}(\theta_2-\theta_1)]\,d\theta_2}\right]d\theta_1$  
  $\textstyle \phantom{=}$ $\displaystyle +{1\over \pi^2} \int_0^\pi \sin({\textstyle{1\over 2}}\theta_1) \...
...2)
\sin[{\textstyle{1\over 2}}(\theta_2-\theta_1)]\,d\theta_2}\right]d\theta_1.$ (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

\begin{displaymath}
\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],
\end{displaymath} (6)

then
\begin{displaymath}
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,
\end{displaymath} (7)

and
\begin{displaymath}
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.
\end{displaymath} (8)

From (6),


$\displaystyle I_1$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\cos({\textstyle{1\over 2}}\theta_2)[\theta...
...{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)[\cos\theta_2]^{2\pi}_{\theta_1}$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\cos({\textstyle{1\over 2}}\theta_1) (2\pi-...
...a_1)+{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)(1-\cos\theta_1)$  
  $\textstyle =$ $\displaystyle \pi\cos({\textstyle{1\over 2}}\theta_1)-{\textstyle{1\over 2}}\th...
...1\over 2}}\theta_1)]+{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)$  
  $\textstyle =$ $\displaystyle \pi \cos({\textstyle{1\over 2}}\theta_1)-{\textstyle{1\over 2}}\t...
...{1\over 2}}\theta_1)+{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)$  
  $\textstyle =$ $\displaystyle \pi \cos({\textstyle{1\over 2}}\theta_1)-{\textstyle{1\over 2}}\theta_1\cos ({\textstyle{1\over 2}}\theta_1)+\sin({\textstyle{1\over 2}}\theta_1),$ (9)

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

Also,


$\displaystyle I_2$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\cos ({\textstyle{1\over 2}}\theta_1)[\sin ...
...style{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)[\cos\theta_2]^{\theta_1}_0$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\cos({\textstyle{1\over 2}}\theta_2)(\sin \...
..._1)-{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)(\cos \theta_1-1)$  
  $\textstyle =$ $\displaystyle -{\textstyle{1\over 2}}\theta_1\cos ({\textstyle{1\over 2}}\theta...
...1\over 2}}\theta_2)]+{\textstyle{1\over 2}}\sin({\textstyle{1\over 2}}\theta_1)$  
  $\textstyle =$ $\displaystyle -{\textstyle{1\over 2}}\theta_1\cos ({\textstyle{1\over 2}}\theta_1)+\sin ({\textstyle{1\over 2}}\theta_1),$ (11)

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

Combining (10) and (12) gives
\begin{displaymath}
\bar A={1\over \pi^2} \left({{5\pi\over 4}+{\pi\over 4}}\right)= {3\over 2\pi} \approx 0.4775.
\end{displaymath} (13)

The Variance is


$\displaystyle {\sigma_A}^2$ $\textstyle =$ $\displaystyle {1\over 2\pi^2} \int_0^\pi \int_0^{2\pi} [A(\theta_1,\theta_2)-{\textstyle{3\over 2\pi}}]^2\,d\theta_2\,d\theta_1$  
  $\textstyle =$ $\displaystyle {1\over 2\pi^2} \int_0^\pi \int_0^{2\pi}\left[{2\left\vert{\sin({...
...}(\theta_1-\theta_2)]}\right\vert-{3\over 2\pi}}\right]^2\,d\theta_2\,d\theta_1$  
  $\textstyle =$ $\displaystyle {1\over 2\pi^2}\int_0^\pi\int_0^{2\pi}\left\{{4\sin^2({\textstyle...
...le{1\over 2}}\theta_2)\sin^2[{\textstyle{1\over 2}}(\theta_2-\theta_1)]}\right.$  
  $\textstyle \phantom{=}$ $\displaystyle \left.{-{6\over\pi}\left\vert{\sin({\textstyle{1\over 2}}\theta_1...
...\theta_1-\theta_2)]}\right\vert +{9\over 4\pi^2}}\right\}\,d\theta_2\,d\theta_1$  
  $\textstyle =$ $\displaystyle {1\over 2\pi^2}\left[{\int_0^\pi \pi(2+\theta_1)\sin^2({\textstyl...
...r \pi}\left({{5\pi\over 4}+{\pi\over 4}}\right)+{9\over 4\pi^2}(2\pi^2)}\right]$  
  $\textstyle =$ $\displaystyle {1\over 2\pi^2} \left({{3\pi^2\over 4}-9+{9\over 2}}\right)= {1\over 2\pi^2}\left({{3\pi^2\over 4}-{9\over 2}}\right)$  
  $\textstyle =$ $\displaystyle {3(\pi^2-6)\over 8\pi^2}\approx 0.1470.$ (14)

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



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© 1996-9 Eric W. Weisstein
1999-05-26