Centroids

In a beginning geometry course, students can find the centroids of triangles by constructing two or three medians. As an extension, I ask my students to think about finding the centroid of a non-special quadrilateral. Most quickly recognize that medians of a non-special quadrilateral are not concurrent, so that method is out. I offer the suggestion to treat a quadrilateral as two triangles:

A few stall out, but a couple students think to look at the other two triangles.

This method is very intuitive and lends itself immediately to the solution --

-- but this method is not scalable. Go ahead; try it for a pentagon. I dare you!

The key to a scalable method relies in understanding that the first pair of centroids is sufficient to find the center of gravity. The solution must lie on this line:

Furthermore, the center of gravity must be positioned on that line proportionally to the relative areas of the two triangles. If the two triangles are equal in area, it must be the midpoint. In my example, the left triangle has an area of 11 sq. units, and the right triangle has an area of 13 sq. units. Therefore, the center of gravity must be 13/24 of the way from the left triangle's centroid to the right triangle's centroid. In effect, this treats each triangle as a point-mass.

This is a method that will work for pentagons. Here is an example:

The centroids of each triangle are marked. Now all one has to do is use the three point-masses (location and area) and then find their collective center of gravity. Is there a way to do this simply with coordinates? Yes. And it is one of the more surprising results I think I have ever seen.

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Let's start with finding the centroid of a triangle. Is there any easy method, using the coordinates, to find it?

The easiest way to find the centroid is by vectors. The centroid is 2/3 of the way from A to D, the midpoint of side BC. That means we need to find 2/3 of vector w and add it to point A. One can show in several different ways that w = 0.5 u + 0.5 v. Using this, one can show that E = u/3 + v/3 + A.

Now it is possible to put everything in coordinates. If $A=(x_1,y_1), B=(x_2, y_2), C=(x_3, y_3)$, then . Thus, the coordinate of E is given by:

$\left(\frac{1}{3}\left(x_2-x_1\right)+\frac{1}{3}\left(x_3-x_1\right)+x_1,\frac{1}{3}\left(y_2-y_1\right)+\frac{1}{3}\left(y_3-y_1\right)+y_1\right)=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

It is quite easy to find the centroid of a triangle.

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How about area? Is there an easy way to find area using coordinates?

Once easy way to find the area using the coordinates is to think about vectors. Specifically, look at two vectors: u = a, b and v = c, d.

How do we find the shaded area between these two vectors? Copy the two vectors again to make a parallelogram:

Now box it in with a rectangle. The rectangle has an area of (a+c)(b+d). Now label all the other parts in terms of a, b, c, and/or d.

Therefore, the area of the parallelogram is (a+c)(b+d) - 2ad - 2 (0.5 ab) - 2 (0.5 cd), which simplifies down to bc - ad. Thus, the area of the triangle formed by vectors u and v is half the area of the parallelogram. Since we know that $a=x_2-x_1, b=y_2-y_1, c=x_3-x_1, d=y_3-y_1$, we can find the area of the triangle very simply in terms of the coordinates:

$\frac{\left|x_2{y}_1-x_1{y}_2+x_3{y}_2-x_2{y}_3+x_1{y}_3-x_3{y}_1\right|}{2}$

This pattern is remarkable, and it extends to any polygon. For example, try it on a quadrilateral.

Set up the areas for each triangle, and the amazing part is that two terms from the first triangle will cancel out two terms from the second: $x_2{y}_4, x_4{y}_2$. That leaves the area formula:

$\frac{\left|x_2{y}_1-x_1{y}_2+x_3{y}_2-x_2{y}_3+x_4{y}_3-x_3{y}_4+x_4{y}_1-x_1{y}_4\right|}{\mathrm{2 }}$

This formula extends easily to any polygon.

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Now we are ready to combine it all together. In overall form, the first step to find the center of gravity of a pentagon is to divide it into three triangles, find the centroid of each, and find the area of each. Given three point-masses, the center of gravity can be found by weighting the triangle centroid formula, like so:

$\left(\frac{a_1{x}_{c1}+a_2{x}_{c2}+a_3{x}_{c3}}{a_1+a_2+a_3}, \frac{a_1{y}_{c1}+a_2{y}_{c2}+a_3{y}_{c3}}{a_1+a_2+a_3}\right)$

Now all that remains is to substitute in.

It is an amazing monstrosity to try to simplify it. I only did the x-coordinate because it takes so much time. Here is the result:

Centroid

It is one of the most surprisingly simple results. This works for a polygon that does not intersect itself, whether it be convex or concave.

I also put together another post on centroids and area.