Operations on polyhedra Is there an operation on polyhedra that add $1$ vertex and $1$ facet (thus, due to Euler's formula, add two edges)?
Here, a polyhedron is the convex hull of a finite number of non coplanar points in ${\mathbb R^3}$. If $P$ has a face $F$ that is a $n$-gon, then the blow-up of that face (i.e. gluing a pyramid over the face) add $1$ vertex, but also add $n$ edges and $n-1$ facets. Since $n$ is at least $3$, it adds too much edges and facets for my purpose. Is there an operation that would fill my need?
 A: Here is one idea:  Look for a vertex $v$ that has exactly three adjacent vertices $a,b,c$, if one exists.  Choose interior points points $a' \in \overline{va}$, $b' \in \overline{vb}$ and then cut your polytope $P$ with the plane that goes through $a',b,',c$.  Then you will have replaced the single vertex $v$ with two vertices $a',b'$, and there will be one new facet, the triangle with vertices $a',b',c$.  I'm not sure if this will do what you want, since I don't know what properties need to be preserved.  Also, not every polytope has a vertex with exactly three neighboring vertices - but perhaps there is a way around this.
A: In a dual way to the former answer by @JairTaylor you also could search for a triangular face of the polyhedron, if one exists. Then take some new vertex somewhere atop this face in such a way that it is incident to the span of one of the directly adjacent faces, but not to the other two. This is like attaching a shallow triangular pyramid ont that given face. 
Accordingly the bottom cancels out with the given face, one side is corealmic with the chosen adjacent face and thus itself does not change the face count, while the remaining 2 faces of the pyramid increase the face count by 2, while the now omitted base face is withdrawn. Obviously you have added a single vertex and no former is canceled. You also have added 3 lacing edges of the pyramid, but the former edge at the now extended adjacent face gets withdrawn.
--- rk
