# Stacking the dual tetrahedron in an ordinary tetrahedron. Is it possible?

I'm thinking about the following problem.

## Introduction

First let me introduce the problem with a 2D example. The area of the triangle constructed by connecting the midpoints of a triangle is 1/4 of the the area of the full triangle.

Very straightforward and a picture is a nice clarification. I like how the center triangle can be found four times in the full triangle.

## Onto 3 dimensions

Consider a tetrahedron, then the volume of the dual tetrahedron is 1/27 of the full tetrahedron.

This seems a lot less obvious then the 2D example. I printed the model using my 3D printer for better visualizing, but still it is not obvious.

I would like to stack the dual tetrahedron 27 times to visualize this quantity, but thus far I'm unable to do so. My first approach to divide each side in three and then connect corresponding points doesn't seem successful as there are pyramids with a square base to be found.

I guess I should start like in the picture below, but it's unclear how to proceed. Which makes me think it's not possible to stack the smaller tetrahedron in a similar way.

Is it possible to stack the tetrahedron? And if so, could someone provide me with some pointers on how to proceed.

No, tetrahedrons cannot fill space. They can stack with regular octahedrons to fill space. Your figures justify the statement that the volume of the dual is $1/27$ the volume of the original because you find the centroid of the faces of the original tetrahedron. As the centroid is $2/3$ of the way along the bisector, the altitude of the dual is $1/3$ the altitude of the original. As the volume scales with the cube of the size, the volume is $1/3^3=1/27$ of the original. But you can't take $27$ of the duals and stack them together to make a copy of the original.

• hm, that's a bit disappointing, thanks for the info though. Any ideas on how to visualize the number 27? My best visualization yet seems to be the 11 tetrahedra and the 4 regular octahedrons. – dietervdf Jun 27 '16 at 20:41
• @dietervdf: just what I gave-the side length is $1/3$ and you cube that for volumes. 3D packing is much more constrained than 2D. In 2D, every polygon can be cut into a finite number of pieces and reassembled to make any polygon of equal area. That is not true in 3D. The Dehn invariant makes an issue – Ross Millikan Jun 27 '16 at 20:42
• Right and sorry, deleted my comment just before your above comment. Please consider deleting yours too. – Narasimham Jun 27 '16 at 20:44
• I know it's pretty easy to prove, but I find it difficult to grasp visualizing the tetrahedron. Looking at the object it's hard to imagine that the volume is 1/27 of the full tetrahedron. Math can be mind-boggling ;). – dietervdf Jun 27 '16 at 20:44

If the vertices of the tetrahedron are $A,B,C,D$, the vertices of the dual tetrahedron are $$V_D=\frac{A+B+C}{3},\quad V_C=\frac{A+B+D}{3},\quad V_B=\frac{A+C+D}{3},\quad V_A=\frac{B+C+D}{3}$$ and the centroid $G$ is the same: $$G = \frac{A+B+C+D}{4}=\frac{V_A+V_B+V_C+V_D}{4}.$$ It follows that a dilation with centre at $G$ and ratio $\lambda=-3$ brings $V_X$ to $X$ for any $X\in\{A,B,C,D\}$, so the ratio between the volume of $ABCD$ and the volume of $V_A V_B V_C V_D$
is $|\lambda|^3 = \color{red}{27}$.

As already pointed by Ross Millikan, a tetrahedron is not a space-filling solid, since neither the angle between faces nor the solid angle in a vertex are rational multiples of a radian/steradian.

You may also use a pretty elegant argument based on Dehn's invariant, solving Hilbert's third problem.

• I like the "angle" reasoning, thanks. I didn't know of the link with Hilbert's third problem, cool! – dietervdf Jun 27 '16 at 20:47