A (not necessarily regular) solid has F faces, each one of which has A sides. It also has V vertices, each of which is the meeting place of B faces.
Show that
AF = BV
Call E the number of edges. We will count E in two seperate ways.
observations:
(1) each edge connects two vertices
(2) each vertex has B edges arriving in it
(3) each edge separates two faces
(4) each face has A edges
Combining (1) and (2):
E = V*B/2 (number of vertices times number of edges meeting there divided by two because otherwise all edges are counted twice)
Combining (3) and (4):
E = F*A/2 (number of faces times number of edges per face, divided by two because otherwise all edges are counted twice)
Combining these two results:
V*B = 2*E = A*F
q.e.d.
Solution
