*Posted by Sreenath in : potd , trackback
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Four points are placed randomly on a circle with radius 1. Find the expected value of the area of the quadrilateral formed by these points.

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The radii formed by the center and the four points form angles x, y, z, and $2\pi – x -y -z$. The area of the quadrilateral can be expressed as the sum of the areas of these triangles. Thus we have $$A = \frac{1}{2} \left ( \sin(x)+\sin(y)+\sin(z)+\sin(2\pi-x-y-z)\right )$$

The expected value of the area is $$\frac{\int _0 ^{2\pi} dx \int _0 ^{2\pi-x}dy\int_0^{2\pi-x-y}dz\ A }{\int _0 ^{2\pi} dx \int _0 ^{2\pi-x}dy\int_0^{2\pi-x-y}dz\ 1}$$

The numerator is $$\small \begin{eqnarray}&\frac{1}{2}&\int _0 ^{2\pi} dx \int _0 ^{2\pi-x}dy\int_0^{2\pi-x-y}dz\ \big[\sin(x)+\sin(y)+\sin(z)- \sin(x+y+z)\big]\\&=&\frac{1}{2} \int _0 ^{2\pi} dx \int _0 ^{2\pi-x}dy\ \big[ 2-2 \cos(x+y)+(2 \pi -x-y) \sin(x)+2 \pi \sin(y) \\ && -x \sin(y)-y \sin(y) \big]\\ &=& \frac{1}{2}\int _0 ^{2\pi} dx \ \Bigg[6 \pi -3 x+\frac{1}{2} \left(6+4 \pi ^2-4 \pi x+x^2\right) \sin(x) \Bigg] \\ &=& \frac{1}{2} \cdot 8\pi ^2 \\ &=& 4\pi ^2 \end{eqnarray}$$

The denominator is $$\begin{eqnarray} \int _0 ^{2\pi} dx \int _0 ^{2\pi-x}dy\int_0^{2\pi-x-y}dz\ 1 &=&\int _0 ^{2\pi} dx \int _0 ^{2\pi-x}dy\ \big[2\pi -x -y \big]\\ &=& \int _0 ^{2\pi} dx \ \Bigg[2 \pi ^2-2 \pi x+\frac{x^2}{2}\Bigg] \\ &=& \frac{4\pi^3}{3}

\end{eqnarray}$$

Therefore the answer is $$\frac{4\pi^2}{\frac{4\pi^3}{3}} = \boxed{\frac{3}{\pi}} $$

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