Partition into Isosceles Triangles

Question: What is the best way to partition a given polygon into isosceles triangles?

What is known to us: One has an upper bound of 4*(n-2) isosceles triangles for any n-gon. Indeed, the n-gon can be cut into n-2 triangles and each triangle cut into 4 isosceles triangles.

And this bound appears tight - eg. consider the following n-gon, with vertices lying randomly spaced along two broad arcs as shown below. The two long inclined sides are much closer to horizontal than shown. It appears impossible to partition it into less than 4*(n-2) isosceles traingles.



But there are also polygons which can be triangulated into only n-2 isosceles triangles; so we suspect it could be an interesting algorithmic challenge to partition a given polygon into the least number of isosceles triangles.

What follows is based on Martin Gardner's 'New Mathematical Diversions':

Further question: What about partitioning a polygon into ACUTE isosceles triangles? Can any obtuse triangle be dissected into seven acute isosceles triangles? The answer is no. Verner E. Hoggatt, Jr., and Russ Denman (American Mathematical Monthly, November 1961, pages 912-913) proved that eight such triangles are sufficient for all obtuse triangles, and Free Jamison (ibid., June-July 1962, pages 550-552) proved that eight are also necessary.

Question: So, does one have a tight upper bound of 8(n-2) for the least number of acute isosceles triangles that result from partitioning an n-gon?