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The compound figure comprising two such dual tetrahedra form a stellated octahedron or stella octangula. Its interior is an octahedron, and correspondingly, a regular octahedron is the result of cutting off, from a regular tetrahedron, four regular tetrahedra of half the linear size (i.e., rectifying the tetrahedron).
Area#Area formulas – Size of a two-dimensional surface; Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities
Compound of eight octahedra with rotational freedom; Compound of eight triangular prisms; Compound of five cubes; Compound of five cuboctahedra; Compound of five cubohemioctahedra; Compound of five great cubicuboctahedra; Compound of five great dodecahedra; Compound of five great icosahedra; Compound of five great rhombihexahedra; Compound of ...
The apparent triangles formed from the figures are 13 units wide and 5 units tall, so it appears that the area should be S = 13×5 / 2 = 32.5 units. However, the blue triangle has a ratio of 5:2 (=2.5), while the red triangle has the ratio 8:3 (≈2.667), so the apparent combined hypotenuse in each figure is actually bent.
This polyhedron is topologically related as a part of a sequence of cantellated polyhedra with vertex figure (3.4.n.4), which continues as tilings of the hyperbolic plane. These vertex-transitive figures have (*n32) reflectional symmetry .
m also must be less than half of n; otherwise the lines will either be parallel or diverge, preventing the figure from ever closing. If n and m do have a common factor, then the figure is a regular compound. For example {6/2} is the regular compound of two triangles {3} or hexagram, while {10/4} is a compound of two pentagrams {5/2}.
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