Cutting Self-Similar Pentagons

Is it possible to cut a pentagon in two, along a straight line, in
such a way that one of the resulting pieces is geometrically
similar to the original pentagon?  Since the newly created (smaller)
pentagon has only one edge formed by the cut, its other four edges
must obviously be segments of the original pentagon.  

There are two general approaches that we can take.  The first and 
most obvious approach is to construct the original pentagon with 
consecutive edge lengths in geometric progression, from the smallest 
up to the second-largest.  Let A,B,C,D,E denote the five vertices 
of the original pentagon, and suppose that for some k less than 1 
the edge lengths are AB=1, BC=1/k, CD=1/k^2, and DE no greater than
1/k^3.  Also, let the interior angles at the vertices A,B,... be 
120, 120, 120, 120, and something near 60 degrees.  Rotating this 
figure 60 degrees and shrinking it by a factor of k makes four 
of the sides sit on top of the original sides, so this shows how
the original pentagon can be cut to give a self-similar pentagon.
Incidentally, if the reduction factor is 0.7548776... the figure 
we get is The Golden Pentagon, and the "other" piece that is cut 
off from the original pentagon is an equilateral triangle.  This 
enables us to tile the plane with a geometric spiral of unequal 
triangles.

The second approach is to consider whether there might be solutions
other than these simple "spiral" solutions, such that the edge 
lengths around the pentagon are arranged in an alternating sequence,
and such that the reduced pentagon must be rotated by TWO vertices
(instead of 1) to bring four of its edges into alignment with the
edges of the original pentagon.  This is indeed possible, and 
leads to a nice infinite family of solutions for any fixed scale 
factor k.

It's not hard to see that this can only work if the interior angles
of the pentagon have one of two values, either alpha or beta, and
they must be arranged in the sequence alpha, beta, alpha, beta, beta.
In addition, two of the edges must be a factor of k times the
respective edge in the unrotated orientation.  Thus, we require
a configuration as shown below

           

Notice that the two sizes of interior angles are

       alpha = 3w         beta = PI - 2w

and the sum of the five interior angles, 2alpha + 3beta, equals
3PI as required.  In order to make the edges meet, we must satisfy
the following two simultaneous conditions relating the edge lengths
s and t to the angle w:

   [cos(w) - k cos(2w)] s + [-cos(2w) + k cos(w)] t  =  1

   [sin(w) - k sin(2w)] s + [ sin(2w) - k sin(w)] t  =  0

Solving these for s and t gives

                       sin(2w) - k sin(w)
        s = -----------------------------------------
            sin(3w)k^2 - [sin(2w)+sin(4w)]k + sin(3w)


                      -sin(w) + k sin(2w)
        t = -----------------------------------------
            sin(3w)k^2 - [sin(2w)+sin(4w)]k + sin(3w)


So, for example, with k=3/4 we have an infinite family of pentagons
that can be cut into two parts along a straight line, with one part
being similar to the original pentagon.  This is illustrated below
for various values of w.



Likewise we can generate an infinite family for other values of k,
such that the family with k=2/3 illustrated below:



Notice that the "cut line" is always perpendicular to the base when
the angle w equals PI/6.