Equal-Area Drawings of Plane Triangulations

Definitions and mathematical statement

Let $G$ be a finite four-connected plane triangulation with a specified outer face. Four-connected means that deletion of fewer than four vertices leaves the graph connected; a plane triangulation has triangular faces, including its outer face. A straight-line drawing assigns distinct points $p(v)\in{\mathbb{R}}^2$ to vertices, draws every edge as its endpoint segment, has no crossings except at common endpoints, and realizes the prescribed plane embedding and outer face.

The question is whether every such $G$ admits a drawing for which there is a single $a>0$ such that every bounded triangular face $uvw$ has \[ \frac12\left|\det\left(p(v)-p(u),p(w)-p(u)\right)\right|=a. \] The positive-area requirement excludes collapsed faces. The outer face is specified but is not required to have the same area as an interior triangle. If its triangle has area $A$, the common interior-face area is necessarily $A/F$, where $F$ is the number of bounded faces, equal to $2|V(G)|-5$ in this convention.

If the three outer-vertex coordinates are additionally prescribed as any nondegenerate triangle, an invertible affine map transfers a drawing between outer triangles while multiplying every face area by the same positive factor. Thus fixing those coordinates does not change existence in this triangular-boundary formulation. No integer-grid, rational-coordinate or arbitrary individual face-area prescription is added.

The literature assessment checked on 21 August 2026 reports progress in prescribed-face-area drawing theory but no general theorem for this equal-area, four-connected, fixed-outer-face question. Arbitrary prescribed-area results for special classes must not be treated as an all-triangulations solution. This is dated supplied triage requiring independent verification.

Short English statement

Can every four-connected plane triangulation be drawn with straight edges and equal positive areas for all bounded triangular faces, preserving the chosen outer face? The outer triangle itself is not one of the equal-area faces.

Sources

• Discrete Differential Geometry (2006), Oberwolfach Reports, DOI 10.4171/owr/2006/12, pp. 653--728, https://doi.org/10.4171/owr/2006/12.
• Literature assessment checked 21 August 2026: original OWR problem, https://doi.org/10.4171/owr/2006/12; Kleist, "prescribed face areas," https://doi.org/10.20382/jocg.v9i1a9.
• Local UnsolvedMath registry, permanent ID $30000432$, exact source code catalogue entry, created 25 August 2026 and updated 16 September 2026. The exact statement is supplied as checked against the official report with source-level content unchanged.