Here are the 4 incenters/excenters, where a,b,c are the side lengths, and A,B,C are the coordinates of the vertices; A,B,C are complex numbers.
I0:(a*A+b*B+c*C)/(a+b+c); this is the incenter Ib:(a*A-b*B+c*C)/(a-b+c); this is the excenter on the b side Ic:(a*A+b*B-c*C)/(a+b-c); this is the excenter on the c side Ia:(-a*A+b*B+c*C)/(-a+b+c); this is the excenter on the a side
Yes, because the barycentric coordinates are (a, b, c), (a, -b, c), (a, b, -c) and (-a, b, c), respectively.
If you take the _centroid_ of these 4 in/excenters, you get the _circumcenter_ of the original triangle(!).
More specifically, the nine-point circle of any triangle formed by three in/excentres is the circumcircle of the original triangle. Your fact follows from mine, as those four in/excentres form an ortho- centric configuration.
If you form the polynomial (w-I0)(w-Ia)(w-Ib)(w-Ic) and clear fractions, a,b,c will only appear as a^2,b^2,c^2,a^4,b^4,c^4. Since the square of the side length for a can be computed as (C-B)(C-B)', we don't need any sqrt's to form the equation.
Yes. The polynomial is obviously a rational function in a, b, c, A, B, C. Moreover, it is unchanged by independently replacing a with -a, b with -b or c with -c, so can be expressed as a rational function in a^2, b^2, c^2, A, B, C.
I believe that a 4th degree polynomial in Q(i)[w] is the lowest degree that has any of these centers as roots. In other words, once you extend the field to contain one of them, you get the other 3 "for free".
This must be true. Given vertices alone as input, there is no (algebraic) method of distinguishing the sign of a side length, and therefore no way to distinguish between incentres and excentres. Also, this means that -s, s-a, s-b and s-c are indistinguishable, as are -r, r_A, r_B and r_C (the exradii and minus the inradius). You might also like the formula 4R = r_A + r_B + r_C - r.
Note that the coefficient of the cubic term of this (monic) quartic is minus *** 4x *** the centroid of the roots, and hence minus *** 4x *** the circumcenter of the original triangle.
This seems like a natural consequence of Vieta's formulas. Sincerely, Adam P. Goucher