Splitting a Tetrahedron into Smaller Tetrahedrons 
Consider $n$ points $A_1, A_2, ..., A_n$ in the interior of a
  tetrahedron $B_1B_2B_3B_4$ and decompose the tetrahedron into smaller
  tetrahedrons such that the vertices of all pieces are from the set
  $\{A_1, A_2, ..., A_n, B_1, B_2, B_3, B_4\}$ (and the union of the
  pieces is the intial tetrahedron). Prove that for any $n \geq 4$ there
  are two configurations of the points $A_1, A_2, ..., A_n$ such that
  the corresponding decompositions contain different numbers of small
  tetrahedrons.

This is a problem that I got from a friend and I am unable to solve it. One observation that I noted is that we can have an inductive procedure of forming a decomposition by first "randomly" creating tetrahedrons and if there is a point $P$ inside a tetrahedron $UVWY$, we can create the four tetrahedrons $PUVW, PUVY, PUWY, PVWY$. Other than that, I have made no progress. Please post thoughts, possible approaches, solutions, etc. 

Here's one way to do it. Or rather, here's two ways to do it, where by "do it" I mean "split $B_1B_2B_3B_4$ into small tetrahedra using only four internal vertices".

For the first approach, we'll need a gadget: a decomposition of a triangular prism into three tetrahedra. If the prism has opposite triangular faces $ABC$ and $A'B'C'$, then the three tetrahedra we use are $ABCA'$, $BCA'B'$, and $CA'B'C'$. In picture form, step-by-step:

This picture is drawn with $ABC$ and $A'B'C'$ the same size, but the gadget doesn't rely on that.
We'll use this decomposition as follows. Place $A_1, A_2, A_3, A_4$ at the vertices of a scaled copy of the tetrahedron $B_1 B_2 B_3 B_4$: say, each $A_i$ is halfway between $B_i$ and the center of the tetrahedron. One of the small tetrahedra will be $A_1 A_2 A_3 A_4$ itself. For the others, we'll take the four prisms with opposite faces $A_i A_j A_k$ and $B_i B_j B_k$ (with $\{i,j,k\} \subset \{1,2,3,4\}$) and decompose each into three tetrahedra.
This is hard to draw, but here's a picture anyway:

The first approach produces $13$ small tetrahedra: three from every prism, plus the middle one.

For the second approach, we'll place the points $A_1, A_2, A_3, A_4$ in exactly the opposite way: place $A_i$ on the line from $B_i$ to the center, such that the center is between $A_i$ and $B_i$. For example, we could take $A_i$ to be halfway between the center of the tetrahedron $B_1 B_2 B_3 B_4$, and the center of the face opposite $B_i$. This is shown in the first figure below:

Then we add:


*

*Tetrahedra $A_1A_2A_3B_4$, $A_1A_2B_3A_4$, $A_1B_2A_3A_4$, and $B_1A_2A_3A_4$ (second figure).

*Tetrahedra $A_1A_2B_3B_4$, $A_1B_2A_3B_4$, $A_1B_2B_3A_4$, $B_1A_2A_3B_4$, $B_1A_2B_3A_4$, and $B_1B_2A_3A_4$ (third figure).

*Tetrahedra $A_1B_2B_3B_4$, $B_1A_2B_3B_4$, $B_1B_2A_3B_4$, and $B_1B_2B_3A_4$ (third figure).


Essentially, we first add small tetrahedra to extend our shape to the vertices of $B_1B_2B_3B_4$, then to its edges, then finally to its faces.
This approach uses $1+4+6+4 = 15$ tetrahedra.

This is just a solution for $n=4$. But any decomposition with $n$ points and $t$ tetrahedra can be extended to a decomposition with $n+1$ points and $t+3$ tetrahedra by placing an extra point in the middle of some small tetrahedron, and replacing that small tetrahedron with four tetrahedral pieces. So we can use these two decompositions to solve the problem for all $n\ge 4$.
