# Polyhedra has more corners than facets [closed]

Let $$P$$ be a polyhedron. Is it true that $$P$$ has always more/as many corners than facets? I haven't found a counterexample in $$\mathbb R^2$$ and $$\mathbb R^3$$ and intuitively I think the statement is true. Is there a proof?

• Try using $V-E+F=2$?
– aras
Feb 20 at 17:12
• Nope. see this picture for an polyhedron with $12$ faces and $9$ vertices. Feb 20 at 17:23
• Is it also passible if $P$ has to be convex? @achillehui Feb 20 at 18:57
• Yes, just push the "fan" on top of the polyhedron in previous comment towards the face of pyramid it rest on as much as possible. Similar construction/procedure for other combination of $f \ge v$ will give you a convex polyhedron. Feb 20 at 19:05
• “I haven't found a counterexample” is a little bit puzzling, since you have missed both the regular octahedron and the triangular dipyramid (which is just two tetrahedra stuck together). I would have expected both of those to be among the first ten polyhedra you checked. Can I suggest that you assemble a more comprehensive list of examples? If you have a conjecture about polyhedra, you can check it with the 5 platonic solids, a couple of prisms and antiprisms, maybe the other five deltahedra, some things like that.
– MJD
Feb 20 at 23:48

In $$\Bbb R^2$$ the claim is trivial, since every polygon always has the same number of sides and vertices.
In $$\Bbb R^3$$, it's definitely false. For every polyhedron that has $$F$$ facets and $$C$$ corners, there is a so-called “dual” polyhedron that has $$C$$ facets and $$F$$ corners.