The Siebeck/Marden theorem doesn't construct the ellipse itself, only the foci. Therefore, characteristics such as the _eccentricity_ of the ellipse aren't evident, since having only the foci don't tell us how long the string is. However, the triangle with its Steiner inellipse is an affine-transformed equilateral triangle with its incircle. Since we know that the centroid of the triangle is also the center of the Steiner inellipse, as well as the midpoint of the foci, we can translate the triangle to put the centroid at the origin, and then rotate the triangle to put the foci on the real axis. (If the foci coincide with one another, we are already done, as the triangle is already equilateral.) We can further standardize the triangle by stretching isotropically to place the foci at +-1. We can now easily compute the affine transformation, because the lengths of the vectors from the centroid to the 3 vertices should be equal for an equilateral triangle. We can compute a factor which stretches only in the Y dimension which will make all three of these lengths equal. Since the foci are at +-1, this stretch factor is just the eccentricity of the ellipse. (Note: since we have 2 equations in one unknown, we could actually use a least-squares method to get a numerically accurate solution for the stretch factor; this also handles the case where one of the equations may be degenerate.) We have thus also answered the question about the locus of points of all the triangles _having the same Steiner inellipse_: since these triangles are all affine-transformed from equilateral triangles _using the same affine transform_, this locus of points is an ellipse with the same eccentricity as the given Steiner inellipse. This affine transformation also shows that inellipse is tangent to the 3 triangle sides at their _midpoints_. Note that if the 3 triangle vertices are complex numbers that are the solution of a cubic polynomial, all of the steps up until the last one can be performed _without factoring the cubic polynomial_. I think it may be possible to compute the asymmetric stretch (=eccentricity) without factoring the cubic, but I don't see it right off hand. Note that the line of the foci is the _principle axis_ not just for the ellipse, but also for the triangle itself, as the line of the foci is the _least-squared error line through the triangle vertices_. Does this mean that if you throw up a triangle in a vacuum so that it spins, it will spin on the axis of its line of foci ?