Given any two points x and y on a circle O, one can form four different lenses (regions between two circles, one of which is O) that have corners at x and y and make angles of 2π/3 at their corners. For three points x, y, and z, with O the circumcircle of triangle xyz, two of these lenses are outside O and two of them have z on the boundary, so there is a unique lens L(x,y) that is inside O and disjoint from z, and lenses L(y,z) and L(x,z) defined symmetrically.

These three lenses L(x,y), L(y,z) and L(x,z) intersect in a unique point c contained in O, defining c as a triangle center for the triangle xyz. There's also a closely related center c' that you can get by inversion of c through O (the intersection point of three lenses outside O). For an equilateral triangle xyz, c is (as for any triangle center) at the centroid of xyz, while c' is at the point at infinity in the one-point completion of the plane. {c,c'} is equivariant under Möbius transformations, an unusual property for a triangle center.

Another equivalent way of defining c, I think, is that it's the point for which angles xcy = π/3 + xzy, xcz = π/3 + xyz, and ycz = π/3 + yxz. This is similar to the fact that the circumcenter C is the point for which xCy = 2xzy etc.

Does anyone recognize these centers c and c'? Do they have entries in the Encyclopedia of Triangle Centers? The definitive way to answer this would be to compute trilinear coordinates and look them up using ETC's search feature, but I'm hoping that this looks familiar enough to someone here that I can skip that step. I tried searching for Möbius and inversion in the text of ETC but found nothing that way.


1 Answer 1


Unless I'm mistaken you seem to describe the isogonal conjugate of the Fermat point X(13). This is the first isodynamic point, or X(15) in ETC.

  • $\begingroup$ Thanks! The trilinear coordinates of that point match the angles I described, so I'm pretty sure it's the right answer. $\endgroup$ May 27, 2010 at 2:52

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