# Convex polyhedron - average of vertices inside polyhedron?

1. Jul 28, 2015

### pyroknife

I am interested in polyhedrons (mostly hexahedrons and pentahedrons). The shapes I am interested in are irregular, where none of the opposing faces are parallel to each other. However, the shapes I am dealing with all CONVEX.

I have the vertices of my polyhedrons and was wondering if I sum up all the vertices and divide by the number of vertices (averaging), would the resulting averaged (x,y,z) coordinate always be inside the convex polyhedron formed by the vertices that were averaged?

2. Jul 29, 2015

### HallsofIvy

Yes, that should be evident as result of the definition of "convex".

3. Jul 29, 2015

### aikismos

Seems to begs the question of what the relationship between the average vertex and the center of mass of a uniformly dense irregular convex polyhedron might be, and so naturally, pyroknife, you might be interested in general knowledge at https://en.wikipedia.org/wiki/Centroid

Last edited: Jul 29, 2015
4. Jul 31, 2015

### pyroknife

Is there a way to prove this?
I can't seem to come up with a proof for this for any generic n-faced CONVEX polyhedron.

5. Jul 31, 2015

### HallsofIvy

Well, what is the definition of "convex"?

6. Jul 31, 2015

### aikismos

Have you thought by analogy about this @pyroknife? Think about how you would prove that a polygon was convex on a plane, and then just generalize to 3D.

7. Aug 2, 2015

### pyroknife

"A polyhedron is said to be convex if its surface (comprising its faces, edges and vertices) does not intersect itself and the line segment joining any two points of thepolyhedron is contained in the interior or surface. A polyhedron is a 3-dimensional example of the more general polytope in any number of dimensions"
I mean I understand that the polyhedron is convex, but I am not understanding how the coordinate that is obtained by averaging the vertices of the polyhedron is defined to be within the 3D-volume.

8. Aug 3, 2015

### aikismos

I'm thinking the easiest way of doing that is to establish with certainty that there exists some some sphere with a center at the average in which the distance to all vertices exceeds the radius of the sphere. Only one of three conditions exist. The sphere exists with a nonzero radius indicating it is in the interior, it does not exist which mean the point is exterior to the polyhedron, or two vertices and the center of the sphere is linear, in which case the point lies in a plane or edge of the polyhedron.