Triangular grid with 4 edges per vertex I am trying to create a triangular grid/mesh for a rectangular domain in $\mathbb{R}^2$ with the property that each vertex is shared by (at most) four edges. Is this possible to accomplish?
 A: Since there is no requirement that the outer region must be a triangle, the question is not quite as trivial as indicated in earlier comments. The rectangle $[0,n] \times [0,1]$ can be triangulated by dividing it into unit squares and then inserting the SW-NE diagonal in each square. Still, this might not be the kind of grid/mesh one wants. To see the problem, it might be easier to think in terms of angles than to use Euler's polyhedron formula: If there are interior points in the triangulation, then the angles at those points have to be at least $90^\circ$ on average, while the average angle in a triangle is only $60^\circ$. It follows that most vertices of the triangulation have to be on the boundary of the region. 
A: This question is too elementary for MO, but here's a hint.  (See the FAQ for alternative sites to ask your question.)


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*Familiarize yourself with the notion of Euler characteristic

*Convince yourself that the Euler characteristic of your triangulation of a rectangle is 1.

*Deduce some inequalities on the numbers of vertices, edges and faces, assuming all faces are triangles and all vertex valences are less than or equal to 4.
