# How to get coordinates from GeometricTransform?

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])}}

• Why not transform the points directly instead of resorting to GeometricTransformation[]? – J. M. is away May 30 '16 at 2:39
• Yes, that works. – M.R. May 30 '16 at 2:43

b = pts;