A Bricard octahedron may be formed from three pairs of points, each symmetric around a common axis of 180° rotational symmetry, with no plane containing all six points. These points form the vertices of the octahedron. The triangular faces of the octahedron have one point from each of the three symmetric pairs. For each pair, there are two ways of choosing one point from the pair, so there are eight triangular faces altogether. The edges of the octahedron are the sides of these triangles, and include one point from each of two symmetric pairs. There are 12 edges, which form the
octahedral graph . As an example, the six points (0,0,±1), (0,±1,0), and (±1,0,0) form the vertices of a regular octahedron, with each point opposite in the octahedron to its negation, but this is not flexible. Instead, these same six points can be paired up differently to form a Bricard octahedron, with a diagonal axis of symmetry. If this axis is chosen as the line through the origin and the point (0,1,1), then the three symmetric pairs of points for this axis are (0,0,1)—(0,1,0), (0,0,−1)—(0,−1,0), and (1,0,0)–(−1,0,0). The resulting Bricard octahedron resembles one of the extreme configurations of the second animation, which has an equatorial
antiparallelogram.
As a linkage It is also possible to think of the Bricard octahedron as a
mechanical linkage consisting of the twelve edges, connected by flexible joints at the vertices, without the faces. Omitting the faces eliminates the self-crossings for many (but not all) positions of these octahedra. The resulting
kinematic chain has one
degree of freedom of motion, the same as the polyhedron from which it is derived. ==Explanation==