I have just seen here the picture of a polyhedron with 15 quadrilateral faces. In some lists of polyhedra a big variety of quadrilateral sides can be found (12, 13, 15, 18, 20,...) but the number 16 is missing. I have some idea how it can be constructed but am not sure that all quadrilaterals would be plane. Can anybody help me with any information
Vladimir Sotirov, Bulgaria
www.math.bas.bg/~vlsot
Yes. start with a column with regular octagonal top $ABCDEFGH$ and bottom $A'B'C'D'E'F'G'H'$. Pick $P$ slightly above the center of the top face and let $B''$ be the intersection of plane $APC$ with $BB'$. Let $D''$ be the intersection of $CPE$ with $DD'$ and similarly find $F''$ and $H''$. Do the same symmetrically at the bottom.