hy.dodec wrote:Moving facet apart and outward, but also maintain the original vertices...

Can this operation possible in all higher-dimensions?

If possible, Could you give an example in 4D?

Here comes a detailed description of one of the 4D examples:

your quest quite generally comes down to this:

first bit = conway expansion, e.g. from xPoQoRo to xPoQoRx

second bit = but apiculating the new vertex figure facets

in such a way that the laterals of those apiculations become corealmic with those other cells

Just consider this one for example:

first bit: x3o3o3o = pen -> x3o3o3x = spid

second bit: plus apiculating one subset of the tets in such a pyramidal height, that the trips get symmetrical elongated triangular dipyramids here the height and thus the lacing edge size of those triangular pyramids has to be derived from the environment.

Here we consider the unit-edged spid first. Its 4D circumradius R then is 1 too. The 3D circumradius r of the trip cells is sqrt(7/12), accordingly the trip-center-radius of spid is sqrt(R^2-r^2)=sqrt(1-7/12)=sqrt(5/12)

Now consider some y-scaled pen y3o3o3o.

That one has a circumradius R of y*sqrt(2/5). Here the edges of those are the elements, which become the trips while expansion. Thus we have to consider the circumradius r of the edge, which is simply y/2. Accordingly the edge-center-radius of the y-pen is sqrt(R^2-r^2)=y*sqrt(2/5-1/4)=y*sqrt(3/20)

Now these 2 elements have to be corealmic, that is, we have to equate these radii, sqrt(5/12)=y*sqrt(3/20), therefrom deriving y=sqrt(100/36)=sqrt(25/9)=5/3

Therefore the searched for apiculated polychoron thus is the convex hull of the compound of x3o3o3x (x=unit edge) plus y3o3o3o (y=5/3 edge) or just

xy3oo3oo3xo&#zc, where the y surely is a pseudo-edge only, and the lacing edge c, i.e. the lacing edge of those apiculating pyramids are still to be calculated. But y=5/3 and the height of the unit trip is 1. Thus the pyramid itself has height h=(5/3-1)/2=1/3. The circumradius of the regular unit triangle is r=1/sqrt(3), therefore c=sqrt(r^2+h^2)=sqrt(1/3+1/9)=sqrt(4/9)=2/3

Further it can be observed, that xy3oo3oo3xo&#zc surely allows for a Stott contraction wrt. the first node, then resulting in oa3oo3oo3xo&#zc, where a=y-x=5/3-1=2/3 as well. But still those a's are pseude edges! - It appears that this contracted figure is not new, rather it is nothing but o3m3o3o or tibbid, i.e. the dual of o3x3o3o or rap.

Thus the above derived polychoron is nothing but a Stott expansion of tibbid. Its cells would be 10 elongated versions of the cells of tibbid, i.e. elongated triangular dipyramids xy2xo3oo&#zc, plus 5 tetrahedra x3o3o.

--- rk