1
$\begingroup$

As was known to the ancients, two congruent regular tetrahedra can be inscribed in a cube and likewise 5 congruent regular tetrahedra can be inscribed in a regular dodecahedron. Is the converse to these two statements true as well, under the following rather stringent conditions:

Question 1 Suppose I have 8 points $\{x_1,\ldots, x_8\}$ in $\mathbb{R}^3$ such that $\|x_i-x_j\|=\|x_\ell-x_k\|$ for all $\{i,j\}\subset \{1,2,3,4\}$ and all $\{\ell,k\}\subset \{5,6,7,8\}$. In particular, $\{x_1,\ldots, x_4\}$ and $\{x_5, \ldots, x_8\}$ are the vertex set of two congruent regular tetrahedra. Are my 8 points the vertices of a cube?

Likewise,

Question 2 Suppose I have 20 points $\{x_1,\ldots, x_{20}\}$ in $\mathbb{R}^3$ such that $\|x_{i_1}-x_{i_2}\|=\| x_{i_3}-x_{i_4}\|=\cdots = \|x_{i_4}-x_{i_5}\|$ for all $\{i_1,i_2\}\in \{1,\ldots,4\}$, all $\{i_2, i_3\}\in \{5,\ldots, 8\}, \ldots,$ and all $\{i_5,i_6\}\subset \{17,\ldots, 20\}$. In particular, $\{x_1,\ldots, x_4\}$, $\{x_5, \ldots, x_8\}$, $\ldots$, $\{x_{17}, \ldots, x_{20}\}$ are the vertex set of five congruent tetrahedra. Are my 20 points the vertices of a regular dodecahedron?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ Since the conditions in question 1 say nothing about the relationship between $x_1,\dots x_4$ and $x_5,\dots,x_8$, the answer is "no". Same for Question 2. Some further conditions would be needed. (I thought of asking that the barycenters coincide, but that is still too weak.) $\endgroup$
    – Hugh Thomas
    Commented Jun 28, 2019 at 19:25
  • $\begingroup$ Since the point sets are disjoint, the answer should be no. (You can tweak the cube and do Dec a he deal configurations slightly to get near misses, for example.) It would be more interesting if a complex of tetrahedra which shared vertices would define the shape of the complex. I suspect you need a stronger condition to eliminate flexible versions and another one to rule out some alternate possible configurations. Gerhard "Easier To Stack Legos Together" Paseman, 2019.06.28. $\endgroup$
    – Gerhard Paseman
    Commented Jun 28, 2019 at 19:29
  • $\begingroup$ Thanks Hugh and Gerhard. Here's a condition to make the question less trivial $\endgroup$
    – Fred
    Commented Jun 28, 2019 at 19:54
  • $\begingroup$ Unfortunately, it is an equivalent condition. (I suspect typos in the Q2 numbering of indices.). It looks like my point about disjoint vertex sets was not considered. Gerhard "It Really Needs More Hinges" Paseman, 2019.06.28. $\endgroup$
    – Gerhard Paseman
    Commented Jun 28, 2019 at 20:26
  • $\begingroup$ As @HughThomas and others have pointed out, perhaps not emphatically enough, the hypotheses of Question 1 allow the possibility that $x_i$ is in my office for $i=1,2,3,4$ and on the moon for $i=5,6,7,8$. They say only that you have two congruent regular tetrahedra, but nothing about their relative locations. Similarly for the five tetrahedra in Question 2. $\endgroup$
    – Andreas Blass
    Commented Jun 29, 2019 at 0:32

0

Reset to default