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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? It looks like this

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.

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  • $\begingroup$ It's not clear how you get your expression for the length of the edge of the large tetrahedron, nor any of your other results $\endgroup$ – David Quinn Feb 28 '18 at 20:57
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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 $$ JK=\sqrt2/4 \quad\hbox{and}\quad VC=AB+AC=1+2/\sqrt3+\sqrt{3/2}. $$

enter image description here

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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

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

triangle slice

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

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

finding the peak

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]

circumscribing tetrahedron

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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

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  • $\begingroup$ So as i understand the up direction is the Z axis. And for the coordinates i just multiply by X and add P to get the size of the tetrahedron to correspond to the Cube and to make the center of the cube to be (0,0,0). Correct? $\endgroup$ – Marko Taht Mar 1 '18 at 15:12
  • $\begingroup$ yes, $Z$ or $x_3$ is the up direction. But, as now being clarified a bit more explizitely now, the edge size of the cube was taken to be 2 units. Thus you would have to multiply all coords by $\frac{X}{2}$ if you'd like to have the cube to be of edge size $X$ instead. Moreover, just as being pointed out now as well above, the origin was taken at the midpoint of one of the edges of the cube. Thus you might want to reduce all $x_2$ values each by 1 and increase all $x_3$ values each by 1, when you'd prefer to have the origin at the center of the cube instead. --- rk $\endgroup$ – Dr. Richard Klitzing Mar 2 '18 at 12:40

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