First, figure out the angle (~\theta~) and length (~L~) of the line from the origin to the given point ~P~ (use ~\tan^{-1}~ for the angle and the Pythagorean Theorem for the length). Points ~Q~ and ~R~ are also at a distance of ~L~ from the origin, but at angles ~\theta-60^\circ~ and ~\theta+120^\circ~ respectively. You can use ~\sin~ and ~\cos~ to get these points. (Another way to do this is to notice that the points are reflections of one another in the line ~y=x~, so once you have one you just flip the sign of the coordinates to get the other.)

The third point is trickier. Figure out where point ~S~ is (it's at angle ~\theta+60^\circ~ and at a distance of ~L~ from the origin). Now the final point you want is on the perpendicular bisector of ~PS~, at a distance of ~2h~ from the origin, where ~h~ is the height of the small triangles (use the Pythagorean Theorem to get ~h~). You can get the angle of the perpendicular bisector using the negative reciprocal of the slope of ~PS~ and ~\tan^{-1}~. And you know this line goes through the origin. Now use ~\sin~ and ~\cos~ to get the coordinates of ~T~.

Pitfalls

You have to remember that your computer probably thinks in radians, not degrees. So you have to do the proper conversions. Every time you compute a new angle for a point, you may have to correct it to make sure you are within ~0~ to ~360^\circ~, and you may also have to correct it to make sure it is in the same quadrant as the ~x~ and ~y~ you used to compute it. (e.g. if ~x=-10~ and ~y=-10~, ~\tan^{-1}(y/x) = 45^\circ~, but this puts you in Quadrant I. You need to add ~180^\circ~ in this case to get the real angle.) You also have to watch out for vertical and horizontal lines and you may need to treat them as special cases.