Random triangle in square: geometrical approach

We call our approach geometrical as instead of considering 6-fold integral in abstract space we consider random triangle (RT) inside the plane rectangle when all possible cases are explicitly apparent. Area of triangle with vertices p1=(x1,y1), p2=(x2,y2), p3=(x3,y3) is equal to s = 1 2 (x1(y2− y3) + x2 (−y1 + y3) + x3 (y1− y2)). (1) Let points p1, p2, p3 are randomly (with constant differential probability function) distributed over the rectangle with sides A, B. What is the mean area of triangles with vertices p1,p2,p3? Answer is evident: zero, as any given triangle corresponds to 6 cases of full permutation of three points at vertices of the triangle. Mean area of this 6 triangles, as given by (1), is zero. But if we take triangle as geometrical figure and if we consider an area of such a figure as positive value, then we must take absolute value of s in formula (1) and... calculation of relevant integrals become impossible even for Mathematica. So Michael Trott in his recent brilliant paper in Mathematica Journal [1] found 496 different integrals each over subregion with the same sign of s, and then used Mathematica to solve such an enormously difficult task. Needless to say that M.Trott’s stunning skill of using Mathematica is far out of scope of ordinary reader (as me, e.g.), so I’ve spend some three weeks in searching a more simple solution. The result is most easily get by the explicit geometrical approach.