Triangle

TriangleWhat is the average area of a triangle whose vertices are randomly chosen inside the unit square? Here is a picture of 10 random triangles.

 Results of an experiment with 10\,000\,000 trials repeated 10 times:

    \begin{equation*} \begin{array}{ccccc} 0.076371294 & 0.076382215 & 0.076401898 & 0.076394862 & 0.076416114 \\ 0.076411850 & 0.076410552 & 0.076422611 & 0.076388065 & 0.076402319 \end{array}% \end{equation*}

The triangle with vertices \left( u,v\right), \left( w,x\right), and \left( y,z\right) has area

    \begin{equation*} \frac{1}{2}\left\vert \left( w-u,x-v,0\right) \times \left( y-u,z-v,0\right) \right\vert =\frac{1}{2}\left\vert \left( w-u\right) \left( z-v\right) -\left( x-v\right) \left( y-u\right) \right\vert \end{equation*}

The expected area is given by

    \begin{equation*} \frac{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \int_{0}^{1}\left\vert \left( w-u\right) \left( z-v\right) -\left( x-v\right) \left( y-u\right) \right\vert \,du\,dv\,dw\,dx\,dy\,dz}{ \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\,du \,dv\,dw\,dx\,dy\,dz} \end{equation*}

The absolute values can be removed by decomposing this cube into disjoint
regions, obtaining an average area of

    \begin{equation*} \frac{11}{144}=0.0763\bar{8}\, \end{equation*}