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Inside a regular tetrahedron, 4 random points are selected and a new tetrahedron is built on them (let's call it the second tetrahedron). It is necessary to find the average value of the square of this second tetrahedron $\langle V^2 \rangle $ (we can assume that the edge of the original regular tetrahedron is 1).

My ideas are primitive: I expressed the volume the "Shoelace formula", also known as Gauss's area formula $$ V = \frac{1}{3!} \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end{vmatrix} , $$the volume of the second tetrahedron by its coordinates, then I calculated the integral of $V^2$ by the volume of the first tetrahedron (to find the average value) and got... first of all: the integral contains $3\cdot4=12$ variables! You can go crazy while you're integrating. Secondly, the integral is not that simple, because the limits of integration for at least the coordinates of the vertices of the first point inside the tetrahedron are not easy to imagine.

Maybe, I do not know, there are some powerful arguments of symmetry, probability theory, mathematical statistics, already solved problems in an understandable way, but similar to this one? I don't know what to do anymore, I'm completely desperate. And the task was given on a test under the pretext of solving it in 3 hours.

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  • $\begingroup$ I'm not sure that's a "pretext": what are you claiming the real intent of putting this on a test was, then? $\endgroup$ Commented Feb 20 at 21:40
  • $\begingroup$ @Misha Lavrov 1) Understand how to simplify the integral to calculate the average (if possible, of course). Maybe we can somehow use symmetry for this, or try to fix one point inside the tetrahedron, and average over the rest (only, of course, in this case we are interested in how to set the limits of integration) 2) To understand if there are more interesting and other ways to solve (suddenly you can consider the problem, for example, relative to the center of mass of the tetrahedron and it will be more convenient to integrate) $\endgroup$ Commented Feb 20 at 21:50
  • $\begingroup$ In general, I am interested in everything that can simplify calculations to find the average square of the volume of a tetrahedron inside a tetrahedron. $\endgroup$ Commented Feb 20 at 21:50
  • $\begingroup$ Sorry, I don't understand; are you saying that you put this on a test in the hope that the students taking the test will find an interesting solution? $\endgroup$ Commented Feb 20 at 21:57
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    $\begingroup$ @Kotlopou Yes - that was my thought too, though I would have said affine transformation. I found it easier to do the simulation with the tetrahedron $(0,0,0), (0,0,1), (0,1,0), (1,0,0)$. $\endgroup$ Commented Feb 22 at 0:29

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