Bracing A Polygon Without Triangles
Answer : Any rigid framework, hence all regular polygons, can be converted to a triangle-free equivalent. Simply chaining copies of the 12 12 12 -vertex triangle-free braced square shown in the question (which I discovered) along the two collinear edges gives a rigid line segment of arbitrary whole number length without triangles: Then any triangular grid can be mimicked without triangles as follows (all straight fuchsia edges are made with the graph chaining construction above, all black edges are single sticks): For example, to brace the hexagon without triangles: However, the above hexagon bracing is quite big. Another approach to triangle-free bracing is the virtual edge : in any embedding of the cubical graph with one edge removed, the distance between the two degree- 2 2 2 vertices (incident to the missing edge) must always be 1 1 1 . This leads to the following triangle-free rigid regular hexagon in 16 16 16 vertices and 29 29 29 edges (Shibuya commit proof): Th...