# Infinitesimally Flexible Octahedron

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This Demonstration illustrates the Blaschke–Liebmann theorem: Let be a polyhedron combinatorially isomorphic to the octahedron. Color the faces of red or white so that every pair of adjacent faces has a different color. Then is infinitesimally flexible if and only if the planes of four red (or four white) faces meet at a point.

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Contributed by: Izidor Hafner (July 2014)

Open content licensed under CC BY-NC-SA

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## Details

Definition: let be a polyhedron with triangular faces and vertex set . An infinitesimal isometric deformation of is a map such that at for all edges of [3, Definition 2.1]. This condition is equivalent to .

A polyhedron is infinitesimally flexible if it has an infinitesimal isometric deformation in which some of its dihedral angles change.

References

[1] P. R. Cromwell, *Polyhedra*, Cambridge: Cambridge University Press, 1997 pp. 222–223.

[2] M. Goldberg, "Unstable Polyhedral Structures," *Mathematics Magazine*, 51, 1978 pp. 165–170.

[3] I. Izmestiev, "Examples of Infinitesimally Flexible 3-Dimensional Hyperbolic Cone-Manifolds," *Journal of the Mathematical Society of Japan*, 63(2), 2011 pp. 363–713. projecteuclid.org/euclid.jmsj/1303737798.

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