I've been thinking again about minimum dilation stars (the subject of my paper arXiv:cs.CG/0412025 with my student Kevin Wortman).

Even for three points $$ABC$$, the problem of computing the “dilation center” $$O$$ minimizing $\max_{P,Q\in\{A,B,C\}} \frac{\mathrm{dist}(P,O)+\mathrm{dist}(Q,O)}{\mathrm{dist}(P,Q)}$ is interesting. From the three distances $$a=\mathrm{dist}(B,C)$$, $$b=\mathrm{dist}(A,C)$$, and $$c=\mathrm{dist}(A,B)$$ and the fact that the three pairwise dilations are equal at $$O$$ one can calculate that $d=\mathrm{dist}(C,O)=x(a+b-c),$ $e=\mathrm{dist}(B,O)=x(a+c-b),$ and $f=\mathrm{dist}(A,O)=x(b+c-a),$ where $$x$$ is some unknown scale factor; one can then plug these distances into Piero della Francesca's formula for the volume of a tetrahedron with side lengths $$a$$, $$b$$, $$c$$, $$d$$, $$e$$, $$f$$ and solve for the value of $$x$$ that gives volume zero (since $$O$$ is supposed to lie on the same plane as $$A$$, $$B$$, and $$C$$). This formula expresses the volume as a cubic polynomial in (the square of) $$x$$, so in principle $$x$$ itself, and the distances $$d$$, $$e$$, and $$f$$, can all be expressed using a closed form formula involving cube roots. Once the distances to $$O$$ are known, $$O$$ itself shouldn't be too hard to find.

There's a more visual and less computational way of describing O: given the three points $$A$$, $$B$$, $$C$$, and a number $$e\gt 0$$, one can draw three similar ellipses with eccentricity $$e$$ having each of three pairs of points as their foci. If $$e$$ is close to 1, the ellipses will be close to the edges of the triangle and will surround an empty region in the middle of the triangle. If $$e$$ is close to 0, the ellipses will surround some area of common intersection in the center of the circle. But for some critical value of $$e$$, the three similar ellipses will have only a single point of intersection. That point of intersection is $$O$$.

The dilation center $$O$$ of a triangle $$ABC$$ is a triangle center, a point defined from a triangle that is equivariant under similarity transformations of the plane (the center of a transformation of a triangle $$ABC$$ is the same as the point formed by applying the same transformation to the center of $$ABC$$). I thought to look it up in the Encyclopedia of Triangle Centers, but if my code and my interpretation of the ETC search page are correct, it doesn't match any of the known centers already listed there.

ETA 7/28: Now X(3513) in ETC, where Peter Moses has supplied trilinear coordinates for it. Moses observed that it's not the only point formed as an intersection of three similar ellipses with these foci: X(3514), the point corresponding to the dilation center under an inversion through the circumcircle, with conjugate trilinears, also has this property. The dilation center is always interior to the given triangle, though, while Moses' point is exterior to it. The two new points are collinear with the incenter and circumcenter of the triangle.