Definition D1
A regular polyhedron is a convex polyhedron all of whose faces are congruent regular p-gons, with q faces meeting at every vertex. Let V be the number of vertices, E the number of edges, and F the number of faces.
Proof.
(1)A polygon has at least 3 sides, so p ≥ 3convention
(2)With only 2 faces at a vertex no solid angle forms. Therefore q ≥ 3convention
(3)Around a vertex, faces and edges alternate. A vertex where q faces meet also has q edges meeting.convention
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(1)Count all pairs (f, e) of a face and one of its edges. There are F faces and each face has p edges, so there are pF pairs.D1
(2)Count the same pairs from the edge side. There are E edges and each edge touches exactly 2 faces, so there are 2E pairs.A2
(3)The same set was counted in two ways. pF = 2E(1), (2), E
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(1)Count all pairs (v, e) of a vertex and an edge meeting it. There are V vertices and q edges meet at each, so there are qV pairs.D1
(2)Count the same pairs from the edge side. There are E edges and each edge has exactly 2 end vertices, so there are 2E pairs.A3
(3)qV = 2E(1), (2), E
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Theorem T1
1/p + 1/q = 1/2 + 1/E, and therefore 1/p + 1/q > 1/2
Proof.
(1)p ≥ 3 > 0 and q ≥ 3 > 0, so V = 2E/q and F = 2E/pL1, L2, D1, E, N
(2)Substitute into Euler's formula. 2E/q − E + 2E/p = 2A1, (1), E
(3)E > 0, so divide both sides by 2E. 1/q − 1/2 + 1/p = 1/E(2), A4, E, N
(4)1/p + 1/q = 1/2 + 1/E(3), E, N
(5)E > 0, so 1/E > 0, and therefore 1/p + 1/q > 1/2(4), A4, A7, N
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Theorem T2
The integer solutions with p ≥ 3, q ≥ 3, 1/p + 1/q > 1/2 are exactly the five pairs (3,3), (3,4), (4,3), (3,5), (5,3).
Proof.
(1)Suppose p ≥ 6. Since q ≥ 3, 1/p ≤ 1/6, 1/q ≤ 1/3 and 1/p + 1/q ≤ 1/6 + 1/3 = 1/2D1, A5, A6, N
(2)This contradicts T1. Therefore p ≤ 5(1), T1
(3)Applying the same argument to q gives q ≤ 5D1, A5, A6, T1, N
(4)Therefore p and q are each one of 3, 4, 5, and there are nine candidate pairs.D1, (2), (3)
(5)(4,4) gives 1/4 + 1/4 = 1/2, which is not greater than 1/2. Excluded.(4), T1, N
(6)(4,5) and (5,4) give 1/4 + 1/5 = 9/20 < 1/2. Excluded.(4), T1, N
(7)(5,5) gives 1/5 + 1/5 = 2/5 < 1/2. Excluded.(4), T1, N
(8)Verify the remaining five pairs. (3,3): 2/3, (3,4) and (4,3): 7/12, (3,5) and (5,3): 8/15 — all greater than 1/2.(4), (5), (6), (7), N
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Corollary C
For each of the five pairs, E, V, and F are uniquely determined.
Proof.
(1)From the equation of T1, 1/E = 1/p + 1/q − 1/2, and the right side is positive, so E is determined uniquely as its reciprocal.T1, E, N
(2)Once E is determined, V = 2E/q and F = 2E/p are determined with it.(1), L1, L2, N
(3)(3,3): 1/E = 2/3 − 1/2 = 1/6. E = 6, V = 12/3 = 4, F = 12/3 = 4(1), (2), N
(4)(3,4): 1/E = 7/12 − 6/12 = 1/12. E = 12, V = 24/4 = 6, F = 24/3 = 8(1), (2), N
(5)(4,3): 1/E = 1/12. E = 12, V = 24/3 = 8, F = 24/4 = 6(1), (2), N
(6)(3,5): 1/E = 16/30 − 15/30 = 1/30. E = 30, V = 60/5 = 12, F = 60/3 = 20(1), (2), N
(7)(5,3): 1/E = 1/30. E = 30, V = 60/3 = 20, F = 60/5 = 12(1), (2), N
(8)These five are called, in order, the tetrahedron, octahedron, cube, icosahedron, and dodecahedron.(3), (4), (5), (6), (7), convention
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