Here is a mesh I computed in order to evoke the new Gridcoin logo 
https://steemitimages.com/0x0/http://i.imgur.com/0UPGfqI.png (thanks  @joshoeah).
First, I decomposed the logo into elementary shapes like lines, circles and meeting points.
Then I computed the distance function to the global shape defined as the minimum of all the distances from those elementary shapes.
Then I ran a mesher which adapts to the metrics defined by that function.
Et voilà!

https://upload.wikimedia.org/wikipedia/commons/4/4b/GridcoinMesh.png
By Zipity.grc (Own work) [<a href="https://creativecommons.org/licenses/by-sa/4.0">CC BY-SA 4.0</a>], <a href="https://commons.wikimedia.org/wiki/File%3AGridcoinMesh.png">via Wikimedia Commons</a>

The hardest part (beside writing the mesher) was to tweak half a dozen of coefficients in order to stress the 'active' locations -- the shape itself -- from the backgound by varying triangle density.
The result has T = 7202 triangles, and as you can count there are B = 12 border edges (and 12 border vertices obviously).
Fun fact : this is enough to compute E the total number of edges!
Indeed, as
<center> <a href="https://www.codecogs.com/eqnedit.php?latex=2E-3T=B" target="_blank"><img src="https://latex.codecogs.com/gif.latex?2E-3T=B" title="2E-3T=B" /></a></center>
for any triangular mesh, we have E = 10809 edges. And now, since the Euler characteristic is 1 for a planar mesh
<center> <a href="https://www.codecogs.com/eqnedit.php?latex=V-E+T=1" target="_blank"><img src="https://latex.codecogs.com/gif.latex?V-E+T=1" title="V-E+T=1" /></a></center>
so V = 3608 vertices.

(edit: hints for a proof [here](https://steemit.com/mathematics/@zipity/the-dehn-sommerville-equations))