Math CentralQuandaries & Queries


Question from Jacquelin, a student:

I have 6 vertices and 10 edges. One of my faces is a regular polygon.


There are a great many answers -- the person raising the question should have imposed some further conditions. So let us say we want a convex polyhedron with 6 vertices and 10 edges. Euler's formula says that Faces + Vertices - Edges = 2. We see that we need ____ faces. Each triangle you use adds three edges, each quadrilateral adds four edges, and so on. Because each edge belongs to two faces we get the formula,
3*(Triangles) + 4*(Quads) + 5*(Pentagons) = 20.

You should check carefully, but I believe that there are two possible polyhedra that satisfy these conditions. One is a pyramid with a pentagon base. The other is a distorted triangular prism -- it starts with congruent triangles in parallel planes at the top and the bottom, whose corresponding vertices are joined by vertical edges. The initial polyhedron has two triangles and four squares as faces. Now rotate one of the vertical edges in the plane of one of the square faces and lengthen it a bit -- that face will no longer be a square, but it stays a quadrilateral, whereas the other face belonging to that edge turns into a pair of triangles. You can do this so that in such a way that you are left with a convex polyhedron having a square face, a quadrilateral face that is not a square, and four triangles.


About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS