Non-Isosceles Centers

This is done by number-crunching. I shall use the $uvw$ -representation of a triangle. Any triangle has sides of the form $a=v+w$ , $b=u+w$ , $c=u+v$ for $u,v > 0$ . The semiperimeter, area, inradius and circumradius of the triangle $ABC$ can be written in terms of $u,v,w$ as follows: $\begin{array}{rclcrcl} s & = & u+v+w & \hspace{2cm} & \Delta & = & \sqrt{uvw(u+v+w)} \\[2ex] r & = & \sqrt{\frac{uvw}{u+v+w}} & & R & = & \frac{abc}{4\Delta} \; =\; \frac{(u+v)(u+w)(v+w)}{4\sqrt{uvw(u+v+w)}} \end{array}$ and we know that the distances between the orthocentre $H$ , the incentre $I$ and the centroid $G$ are given by $\begin{aligned} IH & = \; \sqrt{2r^2 + 4R^2 - \tfrac12(a^2+b^2+c^2)} \; = \; \sqrt{2r^2 + 4R^2 - (u^2+v^2+w^2+uv+uw+vw)} \\ IG & = \; \tfrac13\sqrt{5r^2 - 16rR + s^2} \\ GH & = \; \tfrac23\sqrt{9R^2 - (a^2+b^2+c^2)} \; = \; \tfrac23\sqrt{9R^2 - 2(u^2+v^2+w^2+uv+uw+vw)} \end{aligned}$ We want to solve the equations $IG \; = \; IH \; = \; r \hspace{2cm} v + w = 6$ over the range $u > 0$ and $0 < v < 6$ . For definiteness, we assume that $v \le w$ , and so we solve over $u > 0$ and $0 < v \le 3$ . Solving these equations numerically yields three possible solutions: $\begin{aligned} (u,v,w) \; = \; & (1.0421399174810988,1.310353335593083,4.689646664406917)\\ & (1.3716700545694032,1.09090584856126,4.90909415143874)\\ & (11.960875955614997,2.6579627791559095,3.3420372208440905) \end{aligned}$ which give the following values of $GH$ : $1.7953426074814323 \hspace{0.5cm} 1.8793539310984508 \hspace{0.5cm} 4.578995340714356$ respectively. This makes the last solution of the three the one we want. The desired triangle has sides $a=6$ , $b=15.302913176459088$ and $c=14.618838734770906$ , and $d = GH = 4.578995340714356$ , so that $10^4d = 45789.95340714356$ , which rounds to $\boxed{45790}$ .