>>12692946Take a vector {1,0,0}, duplicate 6 times, rotate by i*2*Pi/6 about the z axis to get the set of vectors {{1, 0, 0}, {1/2, Sqrt[3]/2, 0}, {-(1/2), Sqrt[3]/2, 0}, {-1, 0, 0}, {-(1/2), -(Sqrt[3]/2), 0}, {1/2, -(Sqrt[3]/2), 0}} See figure 1
Take each of those 6 vectors, and duplicate 6 times and rotate each about the y axis by j*2*Pi/6, but you only get 18 unique vectors:
{{1,0,0},{1/2,0,-(Sqrt[3]/2)},{-(1/2),0,-(Sqrt[3]/2)},{-1,0,0},{-(1/2),0,Sqrt[3]/2},{1/2,0,Sqrt[3]/2},{1/2,Sqrt[3]/2,0},{1/4,Sqrt[3]/2,-(Sqrt[3]/4)},{-(1/4),Sqrt[3]/2,-(Sqrt[3]/4)},{-(1/2),Sqrt[3]/2,0},{-(1/4),Sqrt[3]/2,Sqrt[3]/4},{1/4,Sqrt[3]/2,Sqrt[3]/4},{-(1/2),-(Sqrt[3]/2),0},{-(1/4),-(Sqrt[3]/2),Sqrt[3]/4},{1/4,-(Sqrt[3]/2),Sqrt[3]/4},{1/2,-(Sqrt[3]/2),0},{1/4,-(Sqrt[3]/2),-(Sqrt[3]/4)},{-(1/4),-(Sqrt[3]/2),-(Sqrt[3]/4)}} See figure 2 for the vectors and figure 3 for a triangulation of their convex hull.
Guess what? Take the angle between any vector and any other vector and you get the following possible angles(deduplicated for space)
{Pi/3,(2*Pi)/3,Pi,ArcCos[1/4],ArcCos[-(1/4)],ArcCos[7/8],ArcCos[5/8],ArcCos[-(7/8)],ArcCos[-(5/8)]}, the minimum of which is ArcCos[7/8] or 0.505 radians, 28.955 degrees.
Your proof is garbage. QED.
Figures 4 and 5 are the Thomson solution for 36 points and its convex hull, min angle 33.229 degrees
Figures 6 and 7 are the Thomson solution for 18 points and its convex hull, min angle 47.534 degrees
Figures 8 and 9 are the Thomson solution for 12 points and its convex hull, min angle 63.435 degrees
