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I am looking for a polyhedron which consists only out of 15 quadrilateral faces? Does such a thing exist?

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If you don't mind some of the faces lying in the same plane, you can take a cube and cut three of the sides into four smaller squares each. – Barry Cipra Aug 8 '13 at 16:52
up vote 4 down vote accepted

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.

truncated pentagonal bipyramid

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Very nice! Good to know that such a thing exists even though the name is still a mystery :) Does anyone know that? – A friendly helper Aug 8 '13 at 19:01
Nice picture, achille! +1 – Rick Decker Aug 8 '13 at 21:09

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