TITLE: How to design/shape a polyhedron to be nearly spherically symmetrical, but not a platonic solid?
QUESTION [5 upvotes]: There are only 5 platonic solids, but take a look at these images:


How are these things designed? How are they shaped? It looks to me like those hexagons are all the same size and shape, and evenly distributed to approximate a sphere. Same thing with the triangles in the second picture.
So how is it possible? Are some of the hexagons actually slightly smaller or irregular? Can someone show me in the blueprint exactly where the properties of a platonic solid are NOT met?

REPLY [6 votes]: When you see a polyhedron like that, if it's not a platonic solid, then either the faces are not congruent or not exactly regular. There are many ways to do this. Usually they take a platonic solid, like the icosahedron or the dodecahedron and start truncating it or subdividing the faces into more polygons and then elevating the vertices to touch the original polyhedron's circumbscribed sphere (like a kleetope, for example).
The first image
Not all faces are hexagons:

This is a Goldberg polyhedron with Conway notation cctl:

The second image
Even ignoring the stellation (which makes the polyhedra non-convex), the triangles you're left with are not equilateral.
Constructing the Epcot dome is a bit complicated. You start with a dodecahedron and build its kleetope, which is the pentakis dodecahedron:

Alternatively, you can build this by starting with an icosahedron and building the dodecahedron as its dual polyhedra. The triangles are isosceles, but not equilateral—they differ in length by approximately 10%. This is excellent, because it looks almost regular to the naked eye.
Now you start a recursive construction of subdividing each triangle into 4 like this:

and raising the vertices to touch the original polyhedron's circumbscribed sphere.
The first iteration gives:

Notice the position of the pentagons. The second iteration gives:

And finally:

The image above corresponds to the Epcot dome without the pyramids that make it non-convex.
Here is the dome with the first iteration highlighted:

Click here for higher resolution.
In computer graphics, they call these polyhedra with triangular faces icospheres.
Here's the (badly written) Matlab code for plotting the Epcot dome:
phi = 0.5*(1+sqrt(5));
%%%%%%%% Icosahedron %%%%%%%%
Pc = [phi,1,0; -phi,1,0; phi,-1,0; -phi,-1,0;
      1,0,phi; 1,0,-phi; -1,0,phi; -1,0,-phi;
      0,phi,1; 0,-phi,1; 0,phi,-1; 0,-phi,-1]';
Pc = Pc/sqrt(1+phi^2);

%%%%%%%% Pentakis Dodecahedron %%%%%%%%
K = convhull(Pc(1,:),Pc(2,:),Pc(3,:));
Ksize = size(K);
for n=1:Ksize(1)
    x = Pc(1,K(n,:));
    y = Pc(2,K(n,:));
    z = Pc(3,K(n,:));
    [az,el,r] = cart2sph(mean(x),mean(y),mean(z));
    [x,y,z] = sph2cart(az,el,1);
    Pc = [Pc [x;y;z]];
end

MaxSubdiv = 3; %Change to 0, 1, 2 or 3 (3 is VERY slow)
%%%%%%%% Subdivisions %%%%%%%%
for subdiv = 1:MaxSubdiv
    K = convhull(Pc(1,:),Pc(2,:),Pc(3,:));
    Ksize = size(K);
    for n=1:Ksize(1)
        x = Pc(1,K(n,:));
        y = Pc(2,K(n,:));
        z = Pc(3,K(n,:));
        [az,el,r] = cart2sph(0.5*(x(1)+x(2)),0.5*(y(1)+y(2)),0.5*(z(1)+z(2)));
        [xnew,ynew,znew] = sph2cart(az,el,1);
        Pc = [Pc [xnew;ynew;znew]];
        [az,el,r] = cart2sph(0.5*(x(2)+x(3)),0.5*(y(2)+y(3)),0.5*(z(2)+z(3)));
        [xnew,ynew,znew] = sph2cart(az,el,1);
        Pc = [Pc [xnew;ynew;znew]];
        [az,el,r] = cart2sph(0.5*(x(3)+x(1)),0.5*(y(3)+y(1)),0.5*(z(3)+z(1)));
        [xnew,ynew,znew] = sph2cart(az,el,1);
        Pc = [Pc [xnew;ynew;znew]];
    end
end

%%%%%%%% PLOT %%%%%%%%
hold off
K = convhull(Pc(1,:),Pc(2,:),Pc(3,:));
Ksize = size(K);
for n=1:Ksize(1)
    x = Pc(1,K(n,:));
    y = Pc(2,K(n,:));
    z = Pc(3,K(n,:));
    h = fill3(x,y,z,[.3,.4,1]);
    set(h,'LineWidth',1);
    axis equal;
    hold on
end
alpha(0.85);