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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.

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up vote 2 down vote accepted

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.

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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

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