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

(In reply to Centroid invariance in the face of increasing triangle size by FrankM)

<<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.>>

But in the general scalene triangle case, the ratio of the thickness of the increment along any given side to that on a given other side is not 1.  The incremental perimeters will have different thicknesses on each side. Those on the shortest sides will have the "thickest" increments. I put "thickest" in quotes, as they all go to infinitesimal; but they go in proportion, and those that are twice as thick as others will be twice as thick all the way down to zero.

Since this frame, with differing thicknesses on the different sides, has the same CG as the overall triangle, if the thickest sides were shaved down so all sides would be uniform thickness, the CG would move away from the shortest sides and move towards the larger sides, of a uniform-thickness wire, away from the centroid of the triangle.

And in fact my Geometer's Sketchpad implementation of the formulas given in my first post does indeed show the calculated CG migrated from the centroid toward the longer sides.

Edited on May 19, 2008, 11:57 am

Posted by Charlie on 2008-05-19 11:52:37