At least one vertex of a tetrahedron projects to the interior of the opposite triangle How i can give a fast proof of the following fact:
Given four points on $\mathbb{R}^3$ not contained in a plane we can choose one such that its projection to the plane passing through the others is in the triangle generated by the three others points.
Thanks in advance.
 A: The statement is false. 
Just like anyone else, start from 4 points on the plane $z = 0$ which forms a square . For example, $(1,0,0)$,$(0,1,0)$,$(-1,0,0)$ and $(0,-1,0)$. Perturb them by a small amount in the $z$-direction to form a tetrahedron. More precisely, pick some small $\eta > 0$ and consider the tetrahedron formed by following 4 vertices
$$A(1,0,\eta),\quad B(0,1,-\eta),\quad C(-1,0,\eta),\quad D(0,-1,-\eta)$$
It is easy to see this tetrahedron is isohedral. The faces are not merely congruent, they are transitive. For any faces $F_1$ and $F_2$, there is a symmetry of the tetrahedron which maps the face $F_1$ into $F_2$.
For this tetrahedron, if the statement works for one vertex, then it will work for all vertices. Let's look at the vertex $D$ with respect to the face $ABC$. In order for $D$ to project into some point inside $ABC$ (either interior or boundary), a necessary condition
is the outward pointing normals for the faces $ABC$ and $CDA$ is making an angle $\ge 90^\circ$.
By direct computation, the outward pointing normals for the faces $ABC$ and $CDA$ are 
$$
\begin{cases}
\hat{n}_{ABC} &= \frac{1}{\sqrt{1+4\eta^2}}(0,2\eta,1)\\
\hat{n}_{CDA} &= \frac{1}{\sqrt{1+4\eta^2}}(0,-2\eta,1)
\end{cases}
\quad\implies\quad \hat{n}_{ABC}\cdot\hat{n}_{CDA} = \frac{1 - 4\eta^2}{1 + 4\eta^2}
$$
When $|\eta| < \frac12$, we find $\hat{n}_{ABC}\cdot\hat{n}_{CDA} > 0$. This means the two normal vectors is making an angle less than $90^\circ$. As a result, the projection of $D$ onto the plane containing $ABC$ doesn't fall inside $ABC$ (either interior or boundary). Since the tetrahedron is isohedral, same thing happens to other three vertex/face combinations.
A: Suppose the four points are the vertices of a square?
A: Here is a graphical supplement (that I cannot place into a comment) to the excellent answer by @achille hui .
I have taken the case $\eta=0.4.$ with normals in red.
The (complicated) name of this polyhedron is "tetragonal disphenoid" (https://en.wikipedia.org/wiki/Disphenoid).

