Where do the Marden ellipse foci go when the 3 vertices are collinear? Suppose A,B,C are the 3 triangle vertices, with their opposite line segments a,b,c, respectively, as usual. Suppose we go to the limit where a=b+c. As the triangle collapses, the foci move further and further into the corners B,C. In the limiting case, the foci are _at_ B,C. It should be possible to use Maxima (or equivalent) to make an animated .gif image that shows this process in action. Start with an equilateral triangle standing on its base, in which case the inscribed ellipse is actually a circle <=> the foci are at the same place in the center. Allow the top vertex to fall vertically until it hits the base and goes through to become an upside down equilateral triangle. Draw the inscribed ellipse and the 2 foci during this process. Q: What is the curve of the locus of points of the foci during this animation? At 05:03 PM 2/12/2013, James Cloos wrote:
"HB" == Henry Baker <hbaker1@pipeline.com> writes: HB> Actually, the _only_ connection between ellipses and the complex plane I can think of is Marden's Theorem: HB> http://en.wikipedia.org/wiki/Marden%27s_theorem
Silly question based on that:
The article doesn't mention what occurs when the three zeros of p(z) are collinear. Presumably if one allows for degenerate triangles, ellipses then the theorem holds? But what happens to the inellipse's foci?
Does it perhaps hold that if the zeros of p(z) are collinear, then they are also the union of the zeros of p'(z) and p''(z)?