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

The centroid of a triangular area (i.e. the area enclosed within a plane triangle) is well-known to be at the intersection of the medians (=lines connecting each vertex with the mid-point of the opposite side).

We can use this fact to show that the centroid of a triangular perimeter is also at the intersection of the medians. To do this, we treat the triangular perimeter as a differential of triangular area.

ABC is a triangle whose medians intersect at O. Construct an infinite series of triangles within ABC, each having vertice along each of the ABC medians, and sides parallel to ABC. All such triangles are similar and share O as the common median intersection point.

Further, the centroid of the triangular area can be established by integrating the centroid positions of the perimeters of smaller triangles after having weighted these centroids by the perimeters. We see that the resultant centroid remains at O, whether or not we include the outer most perimeters of ABC. It therefore follows that O is the centroid of each of these perimeters.

Posted by FrankM on 2008-05-19 11:19:49