Marden’s Theorem

Given a triangle and three vertices in x, y format, is there a systematic way to use Marden’s Theorem to get the vertices of the foci of the inscribed ellipse? It seems to involve the derivative but I cannot find any hardnumber examples online that clearly show how it works.

-

Just to be clear, there is no "the" inscribed ellipse. Marden's theorem tells you the foci of one particular inscribed ellipse (the one tangent to the midpoint of each side).

I think looking an example is probably clearer than trying to work it out in full generality. Say your triangle has vertices $(0,0)$, $(1,0)$, and $(0,1)$. Then the polynomial $p(z)=z(z-1)(z-i)$ has roots corresponding to the vertices of the triangle, so Marden's theorem tells you that the foci you're looking for are the roots of $p'(z)$.

Now, $$p(z) = z^3-(1+i)z^2+iz \, ,$$ so $$p'(z) = 3z^2-(2+2i)z+i \, .$$ Using the quadratic formula, this has roots $$\frac{2+2i \pm \sqrt{(2+2i)^2-12i}}{6}=\frac{1+i \pm \sqrt{-i}}{3}=\frac{1 \pm \sqrt{2}/2}{3} + \frac{1 \mp \sqrt{2}/2}{3}i \, ,$$ so your foci are at $\left(\frac{1+\sqrt{2}/2}{3}, \frac{1-\sqrt{2}/2}{3}\right)$ and $\left(\frac{1-\sqrt{2}/2}{3}, \frac{1+\sqrt{2}/2}{3}\right)$.

Something similar should work in general -- you'll always get some cubic, differentiate it to get a quadratic, and be able to apply the quadratic formula.

If you want to find the foci of some other inscribed ellipse, the wikipedia article on Marden's theorem mentions a generalization (by Linfield) that you might be able to use, as long as you know something about the locations of its points of tangency.

-
The type of ellipse i am after is a Steiner i believe – WhatsInAName May 26 '12 at 16:01
How do i set up p(z) in general based on the three points? I think that is part of my misunderstanding – WhatsInAName May 26 '12 at 16:06
You arrange for $p(z)$ to have roots at the points in $\mathbb{C}$ which correspond to vertices of your triangle. That is, if your triangle has vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, then you can take $p(z)=(z-(x_1+iy_1))(z-(x_2+iy_2))(z-(x_3+iy_3))$. – Micah May 26 '12 at 16:20