What is the center of gravity of the perimeter of a triangle (as when a piece of wire is bent into triangular form)?

Let (xk, yk) be the coordinates of the three vertices (k=1,2,3). Then the centroids of the three sides are at:

( (xk1 + xk2)/2, (yk1 + yk2)/2 ) = Ck1,k2

weighing the three centroids by the respective fraction which the side contributes to the perimeter, then summing gives:

CM = Sum Lk1,k2 * Ck1,k2 / (L1,2 + L2,3 + L 3,1)

where Lk1,k2 ^2 = (xk1 - xk2)^2 + (yk1 - yk2)^2 and the sum is over 1 <= k1,k2 <= 3 with k1 <> k2.

I unaware of any special geometric significance that the point CM might have in relation to the triangle.

Edited on May 20, 2008, 9:22 am

Posted by FrankM on 2008-05-20 09:18:57