Coordinates of a tetrahedron containing a cube

Question

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.

Answer 1

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 $$ JK=\sqrt2/4 \quad\hbox{and}\quad VC=AB+AC=1+2/\sqrt3+\sqrt{3/2}. $$

Answer 2

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

Answer 3

Armed with only a basic knowledge of vectors and knowing the required length proportions for an equilateral triangle and a regular tetrahedron, here is a plodding derivation of the coordinates for your circumscribing tetrahedron.

I start with a unit cube with corners at $(0,0,0)$ and $(1,1,1)$ for convenience.

As noted by previous answers, the equilateral triangle slice on top of the unit cube can be dissected, so that there is a smaller unit equilateral triangle on top, and two $30-60-90$ triangle "ears" on the side.

We know that the height of an equilateral triangle with side length $s$ is $\dfrac{\sqrt{3}}{2}s$ (which follows from using the Pythagorean theorem on a $30-60-90$ triangle).

Thus, for the ears, if the "height" of the $30-60-90$ triangle is $1$, the offset from the cube corners would be $\dfrac1{\sqrt{3}}$, giving two of the triangle points as

$$\left(-\frac1{\sqrt{3}},0,1\right),\left(1+\frac1{\sqrt{3}},0,1\right)$$

The third point can be obtained by making an offset of $\dfrac{\sqrt{3}}{2}$ off the midpoint of the furthest top edge of the unit cube, $\left(\frac12,1,1\right)$ in the $y$ direction. Thus, the three points are

$$\left(-\frac1{\sqrt{3}},0,1\right),\left(1+\frac1{\sqrt{3}},0,1\right),\left(\frac12,1+\frac{\sqrt{3}}{2},1\right)$$

At this point, we recall that the height of a regular tetrahedron with edge length $s$ is $\dfrac{\sqrt{6}}{3}s$.

We use this piece of information first to get the peak of the circumscribing tetrahedron. First, we determine the centroid of our initial equilateral triangle to be

$$\left(\frac12,\frac13+\frac{\sqrt3}{6},1\right)$$

We then make an offset of $\left(1+\dfrac2{\sqrt 3}\right)\left(\dfrac{\sqrt{6}}{3}\right)=\frac{\sqrt 2}{3}\left(2+\sqrt 3\right)$ in the $z$ direction, yielding the coordinates

$$\left(\frac12,\frac13+\frac{\sqrt3}{6},1+\frac{2\sqrt 2}{3}+\frac{\sqrt 6}{3}\right)$$

This expression is particularly convenient; from here we find that the height of the circumscribing tetrahedron ought to be $1+\dfrac{2\sqrt 2}{3}+\dfrac{\sqrt 6}{3}$ as well. We thus determine the edge length of the circumscribing tetrahedron to be

$$\frac{1+\frac{2\sqrt 2}{3}+\frac{\sqrt 6}{3}}{\frac{\sqrt{6}}{3}}=1+\frac{2}{\sqrt 3}+\sqrt{\frac32}$$

We can use this to determine the coordinates of the other three points of the circumscribing tetrahedron. We first note that the centroid of the tetrahedron's base ought to be at $\left(\dfrac12,\dfrac13+\dfrac{\sqrt3}{6},0\right)$ (why?).

From there, we can find the base points by making an offset of $\dfrac{\sqrt{\frac{3}{2}}+\frac{2}{\sqrt{3}}+1}{\sqrt{3}}=\dfrac23+\dfrac1{\sqrt{3}}+\dfrac1{\sqrt{2}}$ from the centroid, in the $-30^\circ$, $90^\circ$ and $210^\circ$ directions (why?). For example, we obtain one point as

$$\begin{align*} &\left(\frac12,\frac13+\frac{\sqrt3}{6},0\right)+\left(\frac23+\frac1{\sqrt{3}}+\frac1{\sqrt{2}}\right)\left(\cos(-30^\circ),\sin(-30^\circ),0\right)\\ &=\left(1+\sqrt{\frac38}+\frac1{\sqrt{3}},-\frac1{2\sqrt{2}},0\right) \end{align*}$$

We finally obtain the four corners of the circumscribing tetrahedron as

$$\begin{align*} &\left(1+\sqrt{\frac38}+\frac1{\sqrt{3}},-\frac1{2\sqrt{2}},0\right)\\ &\left(\frac12,1+\frac{\sqrt{2}+\sqrt{3}}{2},0\right)\\ &\left(-\frac1{\sqrt{3}}-\sqrt{\frac38},-\frac1{2\sqrt{2}},0\right)\\ &\left(\frac12,\frac13+\frac{\sqrt3}{6},1+\frac{2\sqrt 2}{3}+\frac{\sqrt 6}{3}\right) \end{align*}$$

In Mathematica:

Graphics3D[{{Opacity[2/3, White], EdgeForm[Directive[AbsoluteThickness[1/2], White]],
             Cuboid[]},
            {EdgeForm[Directive[AbsoluteThickness[5], Pink]], FaceForm[], 
             Polygon[{{1 + 1/Sqrt[3], 0, 1}, {-1/Sqrt[3], 0, 1},
                      {1/2, (2 + Sqrt[3])/2, 1}}]},
            {FaceForm[], EdgeForm[Directive[AbsoluteThickness[4], Black]], 
             Simplex[{{1 + 1/Sqrt[3] + Sqrt[3/8], -1/(2 Sqrt[2]), 0},
                      {1/2, 1 + (Sqrt[2] + Sqrt[3])/2, 0},
                      {-1/Sqrt[3] - Sqrt[3/8], -1/(2 Sqrt[2]), 0},
                      {1/2, (2 + Sqrt[3])/6, (3 + 2 Sqrt[2] + Sqrt[6])/3}}]}}, 
            Axes -> True, Boxed -> False]