(This question occurred to me after reading @IanAgol's reminisces of Conway's spiral tetrahedron billiard path.)
Let $T_i$ be a regular tetrahedron, and $P$ a collection of regular tetrahedra glued together face-to-face. Say that $P$ constitutes a "path of stacked regular tetrahedra" iff two conditions hold:
- The dual graph (node for each $T_i$, arc if $T_i$ shares a face with $T_j$) is a path.
- No edge of the construction is incident to more than three tetrahedra.
The first condition intuitively insists on a snake-like object. The second condition excludes too many tetrahedra circling about one edge. (Without this, $5$ dihedral angles of $70.5^\circ$ fit into $360^\circ$, but $6$ do not.)
My question is:
Q. What is the fewest number of tetrahedra in a path $P$ of stacked regular tetrahedra that self-intersects?
$P = \cup_i T_i$ self intersects if a pair of distinct tetrahedra share a point strictly interior to both. So such a self-intersecting snake might be called a tetrahedral ouroboros.
This example1 establishes an upperbound of $31$ tetrahedra (adding one more would self-intersect), but clearly this is not the minimum number of tetrahedra. (This example was aiming toward closure, not self-intersection.)
Fig.6(detail): $QH_7$: $4L+2=30$.
1 Elgersma, Michael, and Stan Wagon. "The quadrahelix: A nearly perfect loop of tetrahedra." arXiv:1610.00280 (2016).