most recent 30 from http://mathoverflow.net 2013-06-20T02:47:40Z http://mathoverflow.net/feeds/user/23052 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89180/triangulations-of-lattice-polygons/94518#94518 Rick Walcott 2012-04-19T12:07:51Z 2012-04-19T12:23:49Z

<p>This is not my own formula: A = 2i + b - 1 when the area of one triangle is 1. I haven't tested this, but I think it works on any triangular grid (not just equilateral). I did test A=sqrt(3)*(i+b/2-1)/2 on an equilateral triangle grid. This works because each square can be mapped to a parallelogram comprised of 2 equilateral triangles, and the area of the parallelogram is sqrt(3)/2.</p>

http://mathoverflow.net/questions/89180/triangulations-of-lattice-polygons/94518#94518 Rick Walcott 2012-04-19T12:35:53Z 2012-04-19T12:35:53Z

I can't argue with i=0, b=3, A=1. I should have tested this.

http://mathoverflow.net/questions/89180/triangulations-of-lattice-polygons/94518#94518 Rick Walcott 2012-04-19T12:27:47Z 2012-04-19T12:27:47Z

I would think it's -2 except that this can be compared to Euler's polyhedral formula F+V=E+2, flattened to 2D, and F has to include the one face outside the bounded figure. A+1=F