Steiner's Circumellipse in Barycentric Coordinates

Elsewhere we derived the equation in barycentric coordinates of Steiner's inellipse. Jacob Steiner's name is also associated with a unique ellipse that passes through the vertices of a given triangle and has its center at the centroid of the triangle. Note that the inellipse also has this property. This is because one is the image of the circumcircle, the other of the incircle, under an affine transformation that maps an equilateral triangles onto the given one.

Steiner's circumellipse of $\Delta ABC$ is described in the barycentric coordinates associated with the triangle by the following equation


The proof follows from the fact that Steiner's circumellipse of $\Delta ABC$ is Steiner's inellipse of the anticomplementary $\Delta A'B'C'.$ This is because the ellipses and the triangles share the same center (centroid, in the case of the triangles.) Moreover, the circumellipse is the image of the inellipse under the homothety with the center at the common centroid and the coefficient of $-2.$

29 November 2015, Created with GeoGebra

I'll be using capital letters to designate the affine coordinates associated with $\Delta A'B'C'.$ Thus Steiner's circumellipse is described by



We have to convert this into an equation in the barycentric coordinates $x:y:z$ associated with $\Delta ABC.$

In the barycentrics of $\Delta ABC,$ $A'=(-1:1:1),$ $B'=(1:-1:1),$ $C'=(1:1:-1).$ This means that, for a point $P=(X:Y:Z),$

$\begin{align} P &= XA'+YB'+ZC'\\ &= X(-A+B+C)+Y(A-B+C)+Z(A+B-C)\\ &= (-X+Y+Z)A+(X-Y+Z)B+(X+Y-Z)C, \end{align}$

so that we can take for the coordinates $P(x:y:z)$ the following expressions:

$x=-X+Y+Z\\ y=X-Y+Z\\ z=X+Y-Z.$

Thus, for example, adding the last two equations, we have $X=\frac{1}{2}(y+z).$ Similarly, $y=\frac{1}{2}(x+z),$ and $z=\frac{1}{2}(x+y).$ This can be substituted into (*):


It is straightforward to verify that this is equivalent to


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