I can create a boundary mesh region that consists of a rectangular region with two independently-rotated pentagonal regions excluded using the following code:
d = 10;
x1 = -d/2; y1 = 0; rad1 = 2; \[Theta]1 = 0 Degree;
x2 = d/2; y2 = 0; rad2 = rad1; \[Theta]2 = 20 Degree;
pts1 = CirclePoints[{x1, y1}, {rad1, (Pi/5 - Pi/2) + \[Theta]1}, 5];
pts2 = CirclePoints[{x2, y2}, {rad2, (Pi/5 - Pi/2) + \[Theta]2}, 5];
sx = 10; sy = 8;
boundpts = {{-sx, -sy}, {sx, -sy}, {sx, sy}, {-sx, sy}};
bmr = BoundaryMeshRegion[Flatten[{pts1, pts2, boundpts}, 1],
Line[{1, 2, 3, 4, 5, 1}], Line[{6, 7, 8, 9, 10, 6}],
Line[{11, 12, 13, 14, 11}]]
The graphical output is
I can then create a triangle mesh (with as large cell sizes as possible) with the following code
TriangulateMesh[bmr, MaxCellMeasure -> \[Infinity]]
which results in
My question is whether the triangular mesh can be made into a mesh of the largest possible convex polygons (with greater than 3 sides) instead, so that the number of individual cells is minimized, and, if so, how. I would greatly appreciate any help. Thank you very much!
Addendum: Furthermore, is it possible to specify that the outside boundary edges should not be at all split (or, at most, at one point -- say, the middle -- of each boundary edge)?