The Journal of Online Mathematics and Its Applications, Volume 8 (2008)

The Most Marvelous Theorem in Mathematics, Dan Kalman

Below you will find an animated graphic that shows how Marden's theorem works out in one example. The animation takes the form of short video clips, interspersed with pauses for brief explanatory comments. For each part of the animation you start the action by clicking the green arrow with the mouse.

I suggest you view the demonstration first, then read the explanation that follows. After reading it, you may wish to replay the demonstration. If you cannot view the entire demonstraton window in your browser, try pressing the F11 key for a full screen display. Don't forget to press F11 again when you are finished viewing the demo.

In the demonstration, use is made of an interesting fact: the average of the roots of a polynomial is equal to the average of the roots of the derivative (details here). This implies that the three roots of ` p`(`z`) have the same average as the two roots of `p′`(`z`). Now the average of `a`, `b`, and `c` is geometrically the center, (or more correctly, the `centroid`) of the triangle, while the average of the two roots of `p′`(`z`) is the midpoint of the line segment between them. Therefore, the two roots of `p`′(`z`) are equally spaced relative to the centroid of the triangle. In particular, if you pick one of the two roots, the other one must lie exactly opposite the first relative to the centroid of the triangle.

As the demonstration proceeds, the moving point that is attached to the mouse is being tested to see if it is one root of ` p`′(`z`). To see if it is, we construct a second point where the other root of the derivative is supposed to be -- opposite the first with respect to the centroid of the triangle. Then we use these points as focal points for three ellipses. If the original point was chosen correctly, these three ellipses will all coincide. But we don′t expect to see this initially, since the first point is initially selected at random.

Once we have the two focal points, there is a unique ellipse going through any other point `P` (except for points on the line segment between the foci). We draw three ellipses using this fact -- one for the midpoint of each side of the triangle. As you move the first focus around, the other focus and the three ellipses are also moved in response. The idea is to try to find a location for the moving point that makes all three ellipses combine into a single ellipse, tangent to the sides of the triangles at their midpoints. When such a point is found, it must be a root of `p`′(`z`), and the second focus is then the other root.