We are given a figure like this in the plane. Does any triangulation without addition of new vertices of such a figure have the same number of triangles? For a polygon, I know any triangulation gives n-2 triangles for n vertices.

I am thinking of the Euler characteristic. Does $\chi = V -E + F$ hold in such cases ? Determining $F$ confuses me without doing the triangulation. Even if it holds, though $V$ remains same we might seem to change $E$ and get a different $F$ which is the number of triangles.

The figure is from Delaunay Mesh Generation book Chapter 2, Section 2.10. On the left is the figure we want to triangulate and on right is the triangulation of said figure. The link for chapter 2.


  • $\begingroup$ why would you expect the same number of triangles? Any triangulation can be further subdivided via barycentric subdivision $\endgroup$ Jul 10 '18 at 0:33
  • $\begingroup$ There is no addition of vertices allowed to the figure. In barycentric subdivision, there seems to be. $\endgroup$
    – T.Harish
    Jul 10 '18 at 2:35
  • 1
    $\begingroup$ In the plane $V-E+F=b_0-b_1$ where $b_0$ is the number of connected components and $b_1$ is the number of holes (connected components of the complement). $\endgroup$ Jul 10 '18 at 22:27

At each vertex, measure the total grey angle. The sum of the angles of all the triangles is $\pi\times$ the number of triangles and is equal to the sum of the grey angles at every vertex as:

  1. Triangles do not overlap
  2. All triangle angles are on grey at a vertex
  3. All grey areas are covered by triangles

Therefore the number of triangles is fixed.

By total grey angle, I mean draw a circle around the vertex and measure the proportion (times $2\pi$) of its circumference that’s on grey. Take the limit of this measurement as the radius goes to 0.


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