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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

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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.

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  • $\begingroup$ Thank you, Hagen! That is exactly the construction in which I was not sure. How do you think, can we put P so to obtain rhombi on the top and bottom? I'm sorry, I am not a geometer and was surprised to read in the detailed list in the site polyhedra.org/poly/polygon/4/quadrilateral that only 5 polyhedra "are made exclusively out of quadrilaterals" (with 6, 12, 24, 30, and 60 sides). If 4-side vertices be cut, a beautiful polyhedron consisting of almost regular hexagons and squres will be obtained: en.wikipedia.org/wiki/File:Conway_polyhedron_t4daA4.png . An interesting topic! $\endgroup$ – Vladimir Sotirov Feb 18 '14 at 23:22
  • $\begingroup$ No, the variant with rhombi is not possible. Mea culpa, the polyhedron t4daA4 contains heptagons as well. Extremely interesting exemplar, indeed! $\endgroup$ – Vladimir Sotirov Feb 20 '14 at 21:08
  • $\begingroup$ The list at polyhedra.org seems to be based on netlib.sandia.gov/polyhedra whch also does not claim to list all polyhedra (or even just all convex polyhedra) upto a cretain vertex of face count, say. $\endgroup$ – Hagen von Eitzen Feb 21 '14 at 5:31
  • $\begingroup$ Having rhombi on top (and in fact everywhere) is possible if we do not require convexity. $\endgroup$ – Hagen von Eitzen Feb 21 '14 at 5:34
  • $\begingroup$ No, convexity/concavity of the roof does not change its general form. Moving the top point outside or inside, the result is symmetrical. The two "roof" edges of the quadrilaterals are always longer that their two peripheral edges laying on the prism. Hagen, I will send you a letter containing a ggb picture with moveable points together with some additional questions concerning polyhedra. $\endgroup$ – Vladimir Sotirov Feb 21 '14 at 20:33

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