Which regular polygons can be dissected into isosceles triangles by nonintersecting diagonals?
Okay, it seems if you lop off part of a regular polygon by cutting from point A to point E, you have a pentagon with 4 short sides AB=BC=CD=DE, with a long side AE. This can be dissected into isosceles triangles by cutting AC and CE.
If you lop off part of a regular polygon by cutting from point A to point I, you have a nonagon with 8 short sides AB=BC=CD=DE=EF=FG=GH=HI, with a long side AI. This can be dissected into isosceles triangles by cutting AC, CE, EG, GI, resulting in the above pentagon.
I made a crude diagram of the above shapes (which I shall name "halfgon"s) here.
While adjusting the angles between sides as necessary, you can create a regular polygon by either putting two identical halfgons together, or by putting two of them together with a slender additional isosceles triangle between them.
My earlier suspicion that if an ngon will work, a 2ngon will also work could be applied here as well. So if you create a halfgon with 2^(k1) short sides (where k is a positive integer >1), the polygon constructed will have 2^k sides, or 2^k+1 sides, or (2^k+1)*2^p sides (where p is a positive integer).
Or, more generally, since 1=2^0, 2^k sides or 2^k+2^p sides would cover all three cases.
EXCEPTIONS:
Obviously, 2^k only works for k>=2. k=1 creates a 2sided "polygon" (line).
2^k+2^p does not work of k=2 and p=1 (hexagon). I'm not sure what to do here, since k=3 and p=0 works (nonagon) and k=2 and p=2 works (octagon). I also can't find a way to make k=3 and p=2 work (dodecagon). Perhaps this option only works if kp<>1?
It's past 3am here, so I'm not going to try to plug this hole until the morning.
Even if I manage to fix my idea tomorrow, I still cannot prove these are the only such polygons that would fit the given criteria.

Posted by Dustin
on 20120121 05:48:20 