I have an Icosphere with 80 faces, 120 edges. Now i am looking to find out what the angle is of the bevel between all the faces. With the bevel i mean the following see the image below:
So i am looking on how to find the angle shown in pink.
Also would i be correct to assume that the bevel angle is the same on all sides of each individual triangle. For all the 80 triangles in the icosphere?
One more side question would i be correct to also state that the angles of the actual triangle would all be 120 degrees? See image below:
So that A1 = A2 = A3 = 60 degrees
after the comment of @Blue i thought maybe the shape is a combination of compound elements. A element consisting out of 5 triangles and then singles to fill the spaces between the compound elements that exist of 5 triangles. See Image below:
After doing some counting i came to the conclusion that it is possible to fit 12 compound shapes consisting of 5 triangles on the Icosphere. resulting in a leftover of 20 single triangles.
I am an art student so i have little knowledge when it comes to math. So if any clarification of my problem is needed i am happy to try and provide it!
This "icosphere" of yours should be a geodesic polyhedron, created from an icosahedron whose faces are divided into four equal triangles. New vertices (i.e. midpoints of icosahedron edges) are then projected onto the embedding sphere.
There are two kinds of dihedral angles, depending on the common edge. A quick construction with GeoGebra shows that (your bevel angle is half dihedral angle, I suppose):
angles having a common edge with an endpoint which is the center of a pentagon, measure about $157.54108°$;
angles having a common edge with endpoints which aren't the center of a pentagon, measure about $161.9709°$.
And there are two kinds of triangles:
triangles having a vertex which is the center of a pentagon, have angles of $55.56901°$, $55.56901°$, $68.86198°$.
triangles having no vertex which is the center of a pentagon, have three equal angles of $60°$.