Which regular polygons can be dissected into isosceles triangles by non-intersecting 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 "half-gon"s) here.

While adjusting the angles between sides as necessary, you can create a regular polygon by either putting two identical half-gons together, or by putting two of them together with a slender additional isosceles triangle between them.

My earlier suspicion that if an n-gon will work, a 2n-gon will also work could be applied here as well.  So if you create a half-gon with 2^(k-1) 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 2-sided "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 |k-p|<>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 2012-01-21 05:48:20