Coordinates of a tetrahedron containing a cube I have a cube with side $x$ and center of $P.$ Using this knowledge how can I find the vertices of the tetrahedron containing this cube?

I found that the side length of such a tetrahedron is $\left(\dfrac{5}{3} + \sqrt{3}\right)x$, but I'm running in trouble when I'm trying to calculate the vertices of the tetrahedron. 
I also found the height if the triangles that make the large tetrahedron, $\frac{\sqrt{34 + 30\sqrt{3}}}{3}\,x$, not sure if this is correct, next would be to find the point where the center of the cube splits this line, then I would have the lengths to find the $3$ bottom vertices.
For the upper vertex I would need to move by $\sqrt{\dfrac{2}{3}}\cdot\left(\dfrac{5}{3} + \sqrt{3}\right)x - \dfrac{x}{2}$ up and then some additional length to side. Here I don't know the side vector.
If someone could tell me how to find where the $P$ is projected on the bottom and how much to the side I need to move to calculate the top vector, I think I could manage the rest.
 A: If the edge of the cube is $1$, it is easy to find that the blue triangle has side $AB=1+2/\sqrt3$. 
On the other hand, as the altitude of the tetrahedron is 
$VH=\sqrt{2/3}VC$ and $AJ=1$, a simple proportion shows that 
$AC=\sqrt{3/2}$ and $CJ=1/\sqrt2$, which entails
$$
JK=\sqrt2/4 
\quad\hbox{and}\quad 
VC=AB+AC=1+2/\sqrt3+\sqrt{3/2}.
$$

A: Use for the top green square the coords $(\pm1, 2, 0)$ and $(\pm1, 0, 0)$. - Thus the center of that green line, which aligns with the blue line, is taken as origin, and the cube here will have an edge size of 2 units.
Add a small regular triangle on top of this square, still within the same plane of $x_3=0$, then its tip has coordinates $(0, 2+\sqrt{3}, 0)$.
The other vertices of the blue triangle then are $(\pm(1+\frac{2}{3}\sqrt{3}), 0, 0)$.
What has been used here several times is the ratio in a regular triangle between its side and its height, which is $2$ : $\sqrt{3}$.
Next consider the red base triangle. According to the grren cube that layer is 2 units below. The tip then is located at $(0, 2+\sqrt{2}+\sqrt{3}, -2)$.
What has been used here are the ratios in a regular tetrahedron between its side, its height and the radius of its face triangle, which is $\sqrt{3}$ : $\sqrt{2}$ : $1$.
The other vertices of that red bottom triangle then are $(\pm(1+\frac{2}{3}\sqrt{3}+\frac{1}{2}\sqrt{6}), -\frac{1}{2}\sqrt{2}, -2)$.
Remains just the upper tip of the red tetrahedron. That one then is $(0, \frac{2}{3}+\frac{1}{3}\sqrt{3}, \frac{2}{3}\sqrt{2}+\frac{1}{3}\sqrt{6})$.
--- rk
