Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

1 Answer 1

up vote 3 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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.