This is inspired by this problem about randomly broken sticks that can form a triangle. It goes in a different direction than this generalization about randomly broken sticks that can form a tetrahedron.
Choose a point $P$ at random inside the unit disk. Choose two directions at random and cut the disk in four parts by the two segments through $P$ in these directions.
Question: What is the probability that the areas of those four parts can occur as areas of the four faces of a tetrahedron?
It is easy to see that the necessary condition "the sum of any three areas must be bigger than the fourth one" is also sufficient for the existence of such a tetrahedron, in an analogous way as the triangle inequalities.
Variant: Take for $P$ the point $(x,0)$, where $0<x<1$ is chosen at random. With random segments as above, what is the probability now?
Intuitively, it seems clear that for this distribution the probability will be smaller because $P$ is in average "more excentric". Note that if $P$ is close to the center $O$, both directions must be close to each other to violate the condition. If $P$ is close to the boundary, the condition is violated as soon as both directions are more or less far from the direction $OP$, which seems to be a priori more probable.
