Spherical f-Tilings by (Equilateral and Isosceles) Triangles and Isosceles Trapezoids

Avelino, Catarina P.; Santos, Altino F.

doi:10.1007/s00026-011-0110-9

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“…As for quadrilateral spherical tilings, Ueno and Agaoka [25] showed that the classification can be very complicated. For works on the spherical tilings by special types of quadrilaterals, see [1,2].…”

Section: Introductionmentioning

confidence: 99%

Spherical tiling by 12 congruent pentagons

Gao

Song

Yan

2013

Journal of Combinatorial Theory, Series A

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The tilings of the 2-dimensional sphere by congruent triangles have been extensively studied, and the edge-to-edge tilings have been completely classified. However, not much is known about the tilings by other congruent polygons. In this paper, we classify the simplest case, which is the edge-to-edge tilings of the 2-dimensional sphere by 12 congruent pentagons. We find one major class allowing two independent continuous parameters and four classes of isolated examples. The classification is done by first separately classifying the combinatorial, edge length, and angle aspects, and then combining the respective classifications together. * Research was supported by Hong Kong RGC General Research Fund 605610 and 606311. pears in the pentagon.Case 1: If the angles in the pentagon are α, α, α, α, β, then Lemma 14 tells us 4α + β = 5α, so that the angles are really α, α, α, α, α. This is the first case in the proposition.Case 2: The angles in the pentagon are α, α, α, β, γ, with β, γ = α. Note that β, γ are not yet assumed to be distinct.Lemma 14 becomes β +γ = 2α. If β 2 γ is a vertex, then 2β +γ = 2π = 3α, and we get β = γ = α, a contradiction. By the similar reason, βγ 2 is not a vertex. By Lemma 15, the only possible vertices are α 3 and αβγ, and β, γ must be distinct.Let {mα 3 , nαβγ} be the anglewise vertex combination. Since 12 tiles with angles α, α, α, β, γ in each tile have total of 36 angle α, 12 angle β and 12 angle γ, we get 3m + n = 36 and n = 12. Therefore m = 8 and n = 12. This is the second case of the proposition.Case 3: The angles in the pentagon are α, α, β, γ, δ, with β, γ, δ = α. Note that β, γ, δ are not yet assumed to be distinct.Lemma 14 becomes β+γ +δ = 3α. If αβγ is a vertex, then α+β+γ = 3α, and we get δ = α, a contradiction. The argument actually shows that there is no αβγ-type vertex.If β 2 γ is a vertex, then 2β + γ = 3α and β = 3α − β − γ = δ. Thus the β 2 γ-vertex is really a βγδ-vertex. The argument actually applies to any β 2 γ-type vertex. By Lemma 15, we conclude that any vertex is either an α 3 -vertex or a βγδ-vertex.Let {mα 3 , nβγδ} be the anglewise vertex combination. Without loss of generality, we may further assume that γ is different from α, β, δ. Since 12 tiles with angles α, α, β, γ, δ have total of 24 angle α and 12 angle γ, we get 3m = 24, n = 12. Therefore m = 8 and n = 12. Depending on whether β = δ, we get the third and the fourth cases of the proposition.Case 4: The angles in the pentagon are α, β, γ, δ, ǫ, with β, γ, δ, ǫ = α. Again β, γ, δ, ǫ are not yet assumed to be distinct.Lemma 14 becomes β + γ + δ + ǫ = 4α. We further divide the case by considering whether some among β, γ, δ, ǫ are equal. Case 4.1: β = γ = δ = ǫ. Lemma 14 becomes 4β = 4α, and we get β = α, a contradiction.Case 4.2: β = δ = ǫ = γ.

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Section: Introductionmentioning

confidence: 99%