Which geometric figure (polyhedron) has 15 quadrilateral faces? I am looking for a polyhedron which consists only out of 15 quadrilateral faces? Does such a thing exist?
 A: 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.

A: On a regular (having $D{_\infty h}$ symmetry) torus draw three "latitude" circles that loop around the symmetry axis. Then draw five "longitude" circles coplanar with this axis.
The torus is then divided into fifteen regions each of which has four coplanar vertices. Replace each region with the planar quadrilateral containing its vertices.
A similar method works for any product of two numbers each greater Tham or equal to three.
