I imagine that this is probably well-known by Euler/Gauss/..., but here goes anyway: If you take the incenter and the centers of the 3 excircles as four complex points, and then form a quartic polynomial with complex coefficients that has these 4 points as _roots_, you can compute the coefficients of this polynomial using only rational operations on the coordinates of the 3 vertices of the triangle. In other words, you never need to compute the perimeter itself, which seems to require 3 square roots, nor do you have to compute the lengths of any of the sides (by themselves). 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 If you take the _centroid_ of these 4 in/excenters, you get the _circumcenter_ of the original triangle(!). 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. Of course, to solve this quartic equation, we will then compute at least one square root. Of course, we could always utilize Newton with an initial guess of the original triangle's centroid/median. Since the roots are known to be well separated -- by the triangle sides themselves ! -- such a Newton iteration should be well-behaved. 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". Note that the coefficient of the cubic term of this (monic) quartic is minus the centroid of the roots, and hence minus the circumcenter of the original triangle.