A question about how to define the centroid of a plane convex body Let E(2) be the Euclidean plane with its standard metric. Let K be a compact convex subset of E(2) whose interior is non-empty. Is there a geometric or topological definition of the centroid of K, that does not require integration to be performed? Such a definition-by means of a weighted average of points of K-is well known, if K is a finite union of pairwise non-overlapping triangles. But K is not necessarily a polygon, and the only plane convex bodies which have this property are polygons. Furthermore, there is no non-constant mass density defined for the points of K. So the problem is not a problem of mechanics. It is a problem of geometry or topology. Finally, does K always have a unique centroid, and if so, is that centroid always an interior point of K?
 A: The centroid is certainly not a topological construct, since if $f$ is a nonlinear homeomorphism the centroid of $f(K)$ is not the image under $f$ of the centroid of $K$.  As for a "geometric" definition, that might depend on how you specify the set $K$ geometrically.  Essentially, you want to 
find a way to do the required integrations without explicitly integrating.
For example, in addition to polygons you could handle segments of circles.
For any bounded $K \subset R^n$ with positive $n$-dimensional Lebesgue measure, the centroid is defined in terms of integration, and that definition makes it unique.  It is 
also easy to show that the centroid of $K$ can't be separated from $K$ by a hyperplane, and therefore the centroid of a convex set with nonempty interior is in that interior.
EDIT: I should have been more careful here.  Suppose $\overline{x}$ is the centroid of a set $K$ with positive Lebesgue measure.  For any linear functional $f$ on $\mathbb R^n$, we have $f(\overline{x}) = m(K)^{-1} \int_K f(x)\; dm(x)$ where $m$ is $n$-dimensional Lebesgue measure.  Then the measures of $\{x \in K: f(x) > f(\overline{x})\}$ and $\{x \in K: f(x) < f(\overline{x})\}$ are either both $0$ or both $>0$.  If $K$ has nonempty interior and $f$ is nonzero, they can't both be $0$.  But if $\overline{x}$ was on the boundary of a convex set $K$, there would be a supporting hyperplane at $\overline{x}$, i.e. a nonzero linear functional $f$ with $f(x) \ge f(\overline{x})$ for all $x \in K$.
