# Self-intersecting path of stacked regular tetrahedra

(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:

1. The dual graph (node for each $$T_i$$, arc if $$T_i$$ shares a face with $$T_j$$) is a path.
2. 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).

Michael Elgersma, a coauthor on the paper I cited, provided an answer to my question: an $$11$$-tetrahedra snake suffices to self-intersect. Here is his illustration of the first $$10$$ tetrahedra:
Theorem: The shortest tetrahedal snake that has self intersections, has length $$11$$.