I am looking for a polyhedron which consists only out of 15 quadrilateral faces? Does such a thing exist?
Let $ABCDE$ be a regular pentagon inscribed inside the unit circle on the x-y plane. Let $P = (0,0,1)$ and $Q = (0,0,-1)$ be two points on the $z$-axis.
The convex hull of $A,B,C,D,E$ and $P,Q$ is a pentagonal bipyramid.
Let $A'$ and $B'$ be the mid-point of $AB$ and $BC$ respectively. If one construct a vertical plane containing $A'$ and $B'$, this plane will intersect with the pentagonal bipyramid above in a small rhombus near vertex $B$. If one "chop off" the vertex $B$ along this rhombus and repeat the same thing for the remaining 4 vertices, one will obtain a convex polyhedron with 17 vertices, 30 edges and 15 quadrilateral faces as shown at end.
It is too bad I can't figure out what is its name.