By John Montroll

N this attention-grabbing advisor for paperfolders, origami specialist John Montroll offers uncomplicated instructions and obviously precise diagrams for growing remarkable polyhedra. step by step directions express find out how to create 34 diverse versions. Grouped in keeping with point of hassle, the types variety from the straightforward Triangular Diamond and the Pyramid, to the extra advanced Icosahedron and the hugely difficult Dimpled Snub dice and the very good Stella Octangula.

A problem to devotees of the traditional eastern artwork of paperfolding, those multifaceted marvels also will entice scholars and someone attracted to geometrical configurations.

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**Additional resources for A Constellation of Origami Polyhedra**

**Example text**

8). 3. Denote by A1 , . . 2). Let z be the intersection point. Consider what type of triangles must contain z. First, all triangles with one vertex in each of A1 , A3 and A4 , must contain z. The same is true for regions A2 , A4 and A6 . This gives 2k 3 triangles containing z. 2. Two types of triangles containing z. 2). This gives at least 2k 3 triangles containing z. Since the same argument holds for the other two pairs of opposite regions, we get at least 6k 3 triangles of this type, and 8k 3 in total.

Since X is simple, the only way the rhombi do not converge to a rhombus in the limit is when all four vertices converge to a point. Clearly, the latter is impossible for the limit of polygons. This completes the proof. 8. Every simple polygon in the plane has an inscribed rhombus with two sides parallel to a given line. Proof. Let X = [x1 . . xn ] ⊂ R2 be a simple polygon and let ℓ be a given line. We assume that ℓ is not parallel to any edge of X. Denote by ϕ : R2 → R a linear function constant on ℓ.

Centrally symmetric squares in polytopes. The following result is the first of the two main results in this section. 2 (Dyson). Let P ⊂ R3 be a convex polytope containing the origin: O ∈ P − ∂P . Then there exists a centrally symmetric square inscribed into P . Proof. Let us prove that the surface S = ∂P of the polytope P contains a centrally symmetric polygon Q. 1. Let P ′ be a reflection of P in O, and let S ′ = ∂P ′ . Since O ∈ P , then P ∩ P ′ = ∅, and thus either P ′ = P or S ∩ S ′ = ∅. Indeed, if S ∩ S ′ = ∅, then one of the polytopes would be contained in another and thus has a greater volume.