Hexahedron made with similar polygons The isohedra are made with congruent polygons. I wondered what could be built with similar polygons that are not congruent. Here is one where edgelengths are powers of $\sqrt{\psi}$, where $\psi$=supergolden ratio.  

Unfortunately, it's a flat 2D object.  From my List of substitution tilings, this can be seen as part of my psi-quad tiling. 
 
A similar 2D hexahedron can be made with a piece of my rho-quad tiling, where $\rho$=plastic constant

One goal is to find a polyhedron that sorta splits into 2 or more similar polyhedra where most of the vertices align, much like what is being done with polygons in the psi-quad and rho-quad tilings.  
But for this post, a simpler question: 

Are there 3D polyhedra where all polygons are similar but not congruent?  Is there a 3D hexahedron made with similar but not congruent polygons? Or a Herschel enneahedron?

EDIT:  That opening picture isn't a hexahedral graph. Silly me. I believe the only hexahedron that works has edges with powers $(a^{-1},0,0,a)\times4,(a^{-2},a^{-1},a^{-1},0)\times2$.
EDIT 2:  There are infinite solutions for the tetrahedron (rootbounds $1/\phi<a<\phi$) and octahedron (rootbounds $1/\sqrt2<a<\sqrt2$). Here are pictures with the power of the root labeling the edges. 
  
EDIT3: For the triangulated polyhedra (A000109), it should almost always be possible to 3-color the edges so that each face has three colors: 0,1,2. Change some 0 to 3, then let these edges be powers for a similar triangle representation. How many of these triangulated polyhedra are representable with similar triangles?
EDIT4:  I've posted Tetrahedral and Octahedral Similarohedrons for those that would like to explore this class.
 A: A hexahedron is possible with all faces similar, and some but not all congruent. But it is rather subtle.
A hexahedron with all faces similar must have either all faces quadrangular (a distorted parallelepiped) or all triangular (an irregular triangular bipyramid). I choose to work with the latter form because it is simpler and yields a solution by your own methodology.
We begin by assembling three similar triangles with sides in the ratio $1:r:r^3$ for some value of $r$ greater than $1$. The triangles are to share a common vertex $A$ and pairs will share edges $\overline{AB},\overline{AC},\overline{AD}$ so as to form the lateral surfaces of a pyramid with base $\triangle BCD$.
The edge lengths may then be rendered as follows:
Lateral: $AB=r^3, AC=r, AD=r^2$
Basal: $BC=1,CD=r^4, BD=r^5$
With these lengths, the pyramidal cap is formed provided that all four faces satisfy the strict triangle inequality and the Cayley-Menger determinant is positive to give a real volume for the tetrahedron. Intuitively we might expect the determinant to fail first as $r$ is increased, but in fact that does not happen for a reason we will shortly see.
The triangle inequalities require
Lateral: $r^3-r-1<0$
Basal: $r^5-r^4-1<0$
It happens that $r^5-r^4-1$ factors, equaling $(r^2-r+1)(r^3-r-1)$, so the two inequalities have a common bound and, geometrically, the cap collapses to a straight line rather than a flat triangular surface. This geometry prevents the Cayley-Menger determinant from changing sign, but it leaves the bound in place from the triangle inequalities. Carrying out the root calculation to three significant figures and rounding (the upper bound) down shows we can allow $1<r<1.32$.
We can also consider the case $r<1$ with the edge lengths as given above. In this case the triangle inequalities are rendered thusly:
Lateral: $r^3+r-1<0$
Basal: $r^5+r^4-1<0$
The Cayley-Menger determinant is more complicated, but as $r$ decreases from $1$ the determinant remains positive as long as a sixth-degree controlling factor does:
C-M determinant: $r^6+r^4-1>0$
Now, the Cayley-Menger determinant fails first as $r$ decreases; the tetrahedron goes flat (volume=0) with $0.86<r<0.87$ whereas the basal triangle would have degenerated with $0.85<r<0.86$! We accept the bound $0.87<r<1$ for this case.
Having formed our pyramidal cap, we simply combine it with its reflection through the base to form the bipyramid, eliminating the nonsimilar basal face by making it internal.
