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I'm trying to get the exact coordinates of the resulting points from my transform:

Normal@GeometricTransformation[Point@nts, 
  RotationTransform[2 \[Pi]/3, pts[[3]]]]

but it doesn't evaluate to real points, it just gives this:

GeometricTransformation[ Point[{{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {Sqrt[ 5/8 + Sqrt[5]/8], 1/4 (-1 + Sqrt[5])}, {0, 1 - [Sqrt]((5/8 + Sqrt[5]/8 + (1 + 1/4 (1 - Sqrt[5]))^2) Sin[ 6 [Degree]]^2 + (5 - 2 Sqrt[5]) (5/8 + Sqrt[5]/ 8 + (1 + 1/4 (1 - Sqrt[5]))^2) Sin[ 6 [Degree]]^2)}, {-Sqrt[5/8 + Sqrt[5]/8], 1/4 (-1 + Sqrt[5])}, {-Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}}], {{{-(1/2), -(Sqrt[3]/2)}, {Sqrt[3]/ 2, -(1/2)}}, {-(1/2) Sqrt[ 3] (-1 + 2 Sqrt[5 - 2 Sqrt[5]] Sin[6 [Degree]]), -(3/ 2) (-1 + 2 Sqrt[5 - 2 Sqrt[5]] Sin[6 [Degree]])}}]

How to get the real points?

Here's my pts:

pts = {{Sqrt[5/8 - Sqrt[5]/8], 1/4 (-1 - Sqrt[5])}, {Sqrt[5/8 + Sqrt[5]/8], 
  1/4 (-1 + Sqrt[5])}, {0, 
  1 - \[Sqrt]((5/8 + Sqrt[5]/8 + (1 + 1/4 (1 - Sqrt[5]))^2) Sin[
        6 \[Degree]]^2 + (5 - 2 Sqrt[5]) (5/8 + Sqrt[5]/
         8 + (1 + 1/4 (1 - Sqrt[5]))^2) Sin[6 \[Degree]]^2)}, {-Sqrt[
   5/8 + Sqrt[5]/8], 1/4 (-1 + Sqrt[5])}, {-Sqrt[5/8 - Sqrt[5]/8], 
  1/4 (-1 - Sqrt[5])}}
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  • 1
    $\begingroup$ Why not transform the points directly instead of resorting to GeometricTransformation[]? $\endgroup$ – J. M. will be back soon May 30 '16 at 2:39
  • $\begingroup$ Yes, that works. $\endgroup$ – M.R. May 30 '16 at 2:43
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I forgot you have to remove the word Geometric!

b = pts;
r = RotationTransform[2 \[Pi]/3, pts[[3]]][pts[[{-1, 1, 2}]]];
l = RotationTransform[4 \[Pi]/3, pts[[3]]][pts[[{-1, 1}]]];

Graphics[{Green, Point@b, Red, Point@r, Yellow, Point@l}]

enter image description here

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