Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on convex polyhedron.
A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?
Integers
Edges
Algebra
Answer: is 840.
AIME I, 1988, Question 10
Geometry Revisited by Coxeter
\({48 \choose 2}\)=1128
Every vertex lies on exactly one vertex of a square/hexagon/octagon
V=(12)(4)=(8)(6)=(6)(8)=48
each vertex is formed by the trisection of three edges and every edge is counted twice, once at each of its endpoints, the number of edges E=\(\frac{3V}{2}\)=72
each of the segment on face of polyhedron is diagonal of that face, so each square gives \(\frac{n(n-3)}{2}=2\) diagonals, each hexagon=9,each octagon=20. The number of diagonals is \((2)(12)+(9)(8)+(20)(6)\)=216
or, number of space diagonals =1128-72-216=840.
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