Does every thrackle have average degree at most 2

Definitions and mathematical statement

A thrackle drawing of a finite simple graph $G=(V,E)$ in the plane places distinct vertices at distinct points and represents every edge by a simple Jordan arc joining its endpoints. An arc contains no other vertex. Any two distinct edges meet exactly once: either at their shared endpoint, when adjacent, or at one proper crossing of their interiors, when nonadjacent. Tangencies and overlapping subarcs are not counted as proper crossings and are excluded. One may use general position, with no three edge interiors meeting at a point.

Conway's thrackle conjecture is \[ |E|\le |V| \] for every graph admitting such a drawing. The assertion concerns the existence of a topological drawing, not whether a prescribed placement of the vertices permits one.

Curved edges are essential to the scope of the question. Requiring straight segments gives a more restrictive geometric problem and does not resolve the conjecture for Jordan arcs. Adjacent edges have already used their unique allowed meeting at their common endpoint, so they may not cross again elsewhere. Isolated vertices are allowed and contribute to $|V|$; deleting them only strengthens a possible counterexample. A resolution must either establish the inequality for every finite thrackle or exhibit a fully specified eligible drawing with more edges than vertices.

Short English statement

Must every finite graph drawn as a planar thrackle have at most as many edges as vertices?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3100008 (AMR-030-0008), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Cooper - Combinatorial Problems I Like (2020 snapshot), Geometry, source-order bullet 8. Original source indexed by AMR-030-0008. https://people.math.sc.edu/cooper/combprob.html.