What is the center of gravity of the perimeter of a triangle (as when a piece of wire is bent into triangular form)?
The center of gravity of each side would be that side's midpoint, and it would act as if its weight were concentrated there. Its weight can be considered to be its length.
Take the vertices of the triangle as being (x1,y1), (x2,y2), (x3,y3). The midpoints are then ((x1+x2)/2,(y1+y2)/2), ((x1+x3)/2,(y1+y3)/2), ((x2+x3)/2,(y2+y3)/2).
The weights of these points are sqrt((x2-x1)^2 + (y2-y1)^2), sqrt((x3-x1)^2 + (y3-y1)^2), sqrt((x3-x2)^2 + (y3-y2)^2), respectively.
Combining any pair of these midpoints to get their C.G., needs only taking the weighted average of their coordinates. Likewise, combining that resulting C.G. with the remaining midpoint results in merely the weighted average of the three midpoints.
Using Geometer's Sketchpad to plot such a point on a random triangle shows it not to be located on a given angle bisector and so the point is not at the center of an inscribed circle. It's not on a given perpendicular bisector of a side and is therefore not at the circumcenter. And its not along a median from a point to the opposite side. It's also not at the orthocenter of the triangle.
So it doesn't seem to be at any of the simply defined points within the triangle.
Posted by Charlie
on 2008-05-18 18:02:26