Points and slats
The center of mass of a triangle lies at the intersection of its medians. This proposition is true independently of the way we conceive the triangle:as a flat triangular slat of uniform density or as three vertices with equal weights attached to them. The fact that the choice makes no difference was proved by Archimedes in the 3rd century B.C. and it seems to be quite a remarkable fact.
Indeed, there are only few polygons with the property that the center of mass of the slat is the same as the center of mass of the vertices. In general the two points already differ in quadrilaterals. Which points are they then?
When the midpoints of the consecutive sides of a quadrilateral are joined, the resulting figure is a parallelogram. This was proved by Pierre Varignon in the 17th century and there is a good chance that most of our readers would be able to repeat the proof. In tribute to the original author of the proof the inner parallelogram used to be called Varignon's quadrilateral. Now, the center of mass of the four vertices of a quadrilateral lies exactly at the intersection of the diagonals of Varignon's quadrilateral.
This may not be the simplest way of describing the point in question (but it certainly is a good way; you might wish to check it yourself). We quote it here only because it is similar to the way the center of mass of a quadrilateral slat can be determined. In general, the problem of finding the center of mass of a figure-slat is a very difficult one.
Two hundred years after Varignon, Wittenbauer introduced his own parallelogram into geometry. He observed that if the sides of a quadrilateral are divided in three equal parts each and straight lines are drawn through partition points adjacent to one and the same vertex, then the straight lines will meet in the vertices of a parallelogram. (Certainly you will easily find a proof of this fact. It is quite similar to Varignon's proof). This new figure is sometimes called Wittenbauer's parallelogram.
The simplicity of the proof leads to believe that the fact that the figure was a parallelogram was not sufficient to attract interest. Indeed, Wittenbauer also proved that the center of mass of the quadrilateral slat lies at the intersection of the diagonals of Wittenbauer's parallelogram.
We invite you to verify this claim, though, to be honest, we must warn you that this exercise is by far more difficult than the previous ones.