Partitioning polygons into obtuse isosceles triangles

Question

Ref:

Partitioning polygons into acute isosceles triangles

Partition of polygons into 'strongly acute' and 'strongly obtuse' triangles

https://math.stackexchange.com/questions/1052063/dividing-an-obtuse-triangle-into-acute-triangles

Question: Is it possible to cut any n-gon into triangles that are both obtuse (largest angle strictly less than 90 degrees) and isosceles? If so how to do it into least number of obtuse isosceles triangles?

Further Question: It easy to see that we can partition any n-gon into triangles that are all obtuse - triangulate the n-gon and divide all resulting acute and right triangles into 3 by connecting each vertex to its Steiner point, resulting in at most 3n obtuse triangles.

Is the 3n a tight upper bound? If not, how can we cut any n-gon into the least number of obtuse triangles? Guess: for convex n-gons, one can do better.

Answer 1

A convex $n$-gon can be cut into $n-2$ triangles by just choosing a vertex and drawing all the diagonals from that vertex, so $3n$ obtuse triangles can be reduced to $3n-6$.

We can do a bit better. E.g., if a convex quadrangle is not a rectangle, then it has at least one obtuse angle, so we can cut off an obtuse triangle incorporating that angle, and just need three more triangles to finish the job, four triangles in all. Every convex $n$-gon, $n\ge5$, has one or more obtuse angles, which we can use to cut off triangles, to reduce the $3n-6$ further.

EDIT ––– Taking this observation to its logical conclusion, we can see that any convex $n$-gon, other than a rectangle, can be partitioned into $n$ obtuse triangles (a rectangle can be partitioned into six obtuse triangles).

This is certainly true for $n=3$. Any convex $n$-gon with $n\ge4$, except the rectangle, has at least one obtuse angle; cutting off the triangle containing this obtuse angle and the two adjacent vertices yields a convex $n-1$-gon, so induction yields the claimed result, provided that when going from a pentagon to a quadrilateral we can avoid forming a rectangle. But if a pentagon is an obtuse triangle glued to a rectangle, then the pentagon has two other obtuse angles in addition to the one we used, and using either one of these other obtuse angles, we get a quadrilateral that can't be a rectangle.

And, we're done.