Later in Geometry, we introduce midpoints all over the place, midpoints of a segment, as one end of the medians of a triangle, or as a point on the perpendicular bisector, but it is almost always in an abstract approach and they never actually have to ferret out the coordinates, so that seldom gets reinforced. I think that is a shame, because in the same course, we get another nice opportunity to introduce the 2 dimensional equivalent. One of the beautiful geometric ideas that still persists in a few geometry classes is that the three medians always intersect in a point. The point is called the centroid*, and sometimes the center of gravity. (Some teachers like to have students make triangles out of cardboard and after they find the centroid the students can cut it out and balance the cardboard triangle on a pencil tip.) If you use an interactive geometry software package, like sketchpad, you can actually construct the triangle and its medians, then grab a point and drag it around and see that...wow! no matter how you alter the shape of the triangle, the three medians still intersect in a single point. If you don't have the software to play with on your on, you can open a java applet I made here and click the mouse button on any of the triangle vertices and move them freely.

But almost no geometry class I know of goes on to show that the point is really sort of a "midpoint" of the triangle, and like the midpoint of a segment, it is just an "average" point. The x,y coordinates of the centroid are just the average of the x-coordinates of the three points followed by the average of the y-coordinates.

If, for example, we had triangle ABC with A=(1, 2); B= (5, 11) and C = (9, -1) then the medians will intersect at the point where x= (1+5+9)/3=5 and y= (2+11 - 1)/3 = 4; or more simply at the point (5,4)

About this time of year as I am introducing my pre-calc class to the beauty of three dimensional analytic geometry, I dust off their memory of this little jewel from when I taught it in Alg II, and then we extend it to show it works with triangles in three-space. This requires them to a) Find the midpoints of the sides; b) write the equation the medians in the space triangle; c) find the intersection of two vector lines in space; (d) and then confirm that that point is also on the third line; and finally e) insure that it is indeed the average of the relative coordinates. I think it is a pretty good illustration of the power of the linear algebra they are learning.

As a reward, I give them a discovery lesson on tetrahedra... we talk for a minute about the difference between a two dimensional object in three space (A triangle with (x,y,z) coordinates for the points) and a three dimensional object (like a cube) . I bring out a large cardboard model of a tetrahedron (triangle based pyramid) and hold it in front of me..... and I begin to wonder aloud..." This is sort of like an extension of a triangle in three dimensions.... so what would be like the medians of a triangle in this object?"

Frequently the first guess is a line from a vertex to a point which is the "face-center" or centroid. I usually channel them (or prompt them if needed) toward this idea. "Wouldn't it be wonderful", I add, " if there was something like the centroid for the tetrahedron; where all those lines would intersect." By now they suspect that there is, and perhaps with some fear as they realize that they are about to be challenged to find one, or worse, prove it is always true.

With much the same vector approach as before, I pass out some cards with a set of four points in space (carefully chosen to make the work a little less messy than it might be), and they find centroids of the four faces, hopefully using the "average point" method. Then they must write vector equations of the four lines which I call "medial" lines that run from a vertex of the tetrahedron to the centroid of the opposite face. Finally I suggest that they solve the intersections of the four medials in sets of two, and behold that the two solutions are the same point; all four lines intersect in a single point.

Then I ask, "What is the quickest way to find the midpoint of a line segment?" "Averages", they shout.

"What is the quickest way to find the centroid of a triangle?" "Averages", some reply, but others already have anticipated the next question, and have started to calculate the average of the four x values.... could it be?

As a finale I point out that there is a pattern in the division of the segments in each of the cases..

The midpoint divides the line segment (a one space object) into a ratio of 1:1

The centroid divides the segment of a triangle (a two space object) into a ratio of 2:1

The center of gravity of a tetrahedron ( a three space object) divides the medial segments in a ratio of 3:1

**Just another "average" day in mathematics class.**

If you want a little more about centroids, see my web page article.

*There are three common "centers of gravity" that are studied in math, science and engineering. The most common in math is the center of masses located at the vertices of a polygon. This is more common because the other two cases can be reduced to a variation of this approach. It is this case of point masses at the vertices that I mean when I use centroid or center of gravity in this note, unless otherwise stated. A second approach is to treat the area of the polygon as if it were a sheet of uniform density. The third, and least common, approach is to represent the sides of the polygon as wire rods of uniform density.

Most students are first introduced to the terms above in reference to a point in a triangle. Since the center of masses at the vertices in a triangle give the same point as a uniform sheet, they are often confused about the various distinctions. The three centers of gravity are usually different points in other non-symmetric polygons. It is this point, the center of balance for the uniform sheet and also of point masses at the vertices, that is almost universally referenced as the centroid of a triangle.