is dense in T yet excludes all trivial triangles (i.e. points) and degenerate triangles (i.e. lines) is the set of all scalene triangles. It is obtained by removing the planes b = c, c = a, a = b from T. Domain symmetry Not every subset D ⊆ T is a viable domain. In order to support the bisymmetry test D must be symmetric about the planes b = c, c = a, a = b. To support cyclicity it must also be invariant under 2π/3 rotations about the line a = b = c. The simplest domain of all is the line (t,t,t) which corresponds to the set of all equilateral triangles. Examples Circumcenter The point of concurrence of the perpendicular bisectors of the sides of triangle ABC is the circumcenter. The trilinear coordinates of the circumcenter are a(b2 + c2 − a2) : b(c2 + a2 − b2) : c(a2 + b2 − c2). Let f(a,b,c) = a(b2 + c2 − a2). Then f(ta,tb,tc) = (ta) ( (tb)2 + (tc)2 − (ta)2 ) = t3 ( a( b2 + c2 − a2) ) = t3 f(a,b,c) (homogeneity) f(a,c,b) = a(c2 + b2 − a2) = a(b2 + c2 − a2) = f(a,b,c) (bisymmetry) so f is a triangle center function. Since the corresponding triangle center has the same trilinears as the circumcenter it follows that the circumcenter is a triangle center. 1st isogonic center Let A'BC be the equilateral triangle having base BC and vertex A' on the negative side of BC and let AB'C and ABC' be similarly constructed equilateral triangles based on the other two sides of triangle ABC. Then the lines AA', BB' and CC' are concurrent and the point of concurrence is the 1st isogonal center. Its trilinear coordinates are csc(A + π/3) : csc(B + π/3) : csc(C + π/3). Expressing these coordinates in terms of a, b and c, one can verify that they indeed satisfy the defining properties of the coordinates of a triangle center. Hence the 1st isogonic center is also a triangle center. Fermat point Let Then f is bisymmetric and homogeneous so it is a triangle center function. Moreover, the corresponding triangle center coincides with the obtuse angled vertex whenever any vertex angle exceeds 2π/3, and with the 1st isogonic center otherwise. Therefore, this triangle center is none other than the Fermat point. Non-examples Brocard points The trilinear