# Two triangulations that share only their hull

Open problem settled, formally verified in Lean 4. Vamshi Jandhyala.

> From nine points on, two triangulations can share nothing but the five outline edges, answering Rote.

Canonical: https://vamshij.com/research/two-triangulations-hull
Code and Lean proofs: https://github.com/jvvk/mathematics/tree/main/two-triangulations-hull
Paper (PDF): https://github.com/jvvk/mathematics/blob/main/two-triangulations-hull/paper/paper.pdf

Any two triangulations of a finite point set in the plane contain the edges of its convex hull. How few other edges must they share? For every n ≥ 9 there are n points in general position, with a pentagonal hull, whose two triangulations share nothing else; since five edges must always be shared once n ≥ 5, the minimum is exactly five. The construction starts from nine explicit points and inserts points one at a time, carrying along a *chain* (a face of one triangulation whose side passes through two suitable faces of the other) that makes the next insertion possible.

Started from a convex polygon, the same construction settles every hull size: n points with h hull vertices admit two triangulations sharing only the hull exactly when n = h, or h ≥ 6, or h = 5 and n ≥ 9. The cases of a pentagonal hull with six to eight points rest on an exhaustive search over the order-type database of Aichholzer, Aurenhammer and Krasser.

The question is from MathOverflow ([question 454425](https://mathoverflow.net/q/454425)). The nine-point example and the pentagonal case were first posted as the author's [MathOverflow answer](https://mathoverflow.net/a/515648) on 30 September 2026, with certificates in [triangulations-sharing-five-edges](https://github.com/jvvk/triangulations-sharing-five-edges).

Preprint, 8 October 2026. The author used AI tools (Claude, and ChatGPT for a later revision) in developing the proofs, code and text, as the paper's acknowledgements describe, and takes full responsibility for its content.