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 08 '13 at 16:52
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@ A friendly helper: Many possibilities. For example from a regular or scalene convex dodecahedron 3 small tetrahedrons around any 3 vertices are chopped off, it makes 12+3 =15 faces, .. right? – Narasimham Jan 27 '23 at 11:10
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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.
achille hui
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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 08 '13 at 19:01
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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.
Oscar Lanzi
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