Well, since in an equilateral triangle angle bisectors, medians, altitudes, perpendicular bisectors are all the same, the circumcenter and incenter (and orthocenter) are the same point, what I guess you call "the triangle's center".
Now, by "triangle's radius" I'm guessing you mean the triangle's circumcircle's radius, and then: you need three point on the plane such that
i) They're equidistant from the circumcenter, and
ii) This common distance is 2/3 of each median's length.
Of course, you'll get infinite possibilities as any such solution can be rotated around the circumcenter in any angle $\,0<\theta<2\pi /3 $ and you'll get a new set of three vertices of an equilateral triangle with the same circumcenter.
If my guessing of your naming is incorrect then discard the above.