# octahedron

octahedron Type of side surfaces equilateral triangles
Number of faces 8th
Number of corners 6th
Number of edges 12
Schläfli icon {3.4}
dual to Hexahedron (cube)
Body net Number of different networks 11
Number of edges in a corner 4th
Number of corners of a surface 3

The (also, especially Austrian : the) octahedron [ ɔktaˈeːdɐ ] (from ancient Greek ὀκτάεδρος oktáedros , German 'eight-sided' ) is one of the five Platonic solids , more precisely a regular polyhedron ( polyhedron , polyhedron ) with

The octahedron is both an equilateral four-sided double pyramid with a square base and an equilateral antiprism with an equilateral triangle as a base .

## symmetry Octahedron with examples of the axes of rotation and two planes of symmetry (red and green)${\ displaystyle C_ {4}, C_ {3}, C_ {2}}$ Because of its high symmetry - all corners , edges and surfaces are similar to each other - the octahedron is a regular polyhedron . It has:

• 3 fourfold axes of rotation (through opposite corners)${\ displaystyle C_ {4}}$ • 4 threefold axes of rotation (through the centers of opposing surfaces )${\ displaystyle C_ {3}}$ • 6 twofold axes of rotation (through the centers of opposite edges)${\ displaystyle C_ {2}}$ • 9 levels of symmetry (3 levels through four corners each (e.g. red), 6 levels through two corners each and two edge centers (e.g. green))
• 14 rotations (6 by 90 ° with planes through four corners each and 8 by 60 ° with planes through six edge centers each)

and is

In total, the symmetry group of the octahedron - the octahedron group or cube group - has 48 elements.

## Relations with other polyhedra

The octahedron is the dual polyhedron to the hexahedron ( cube ) and vice versa.

Two tetrahedra can be inscribed in a cube in such a way that the corners are at the same time cube corners and the edges are diagonals of the cube faces (see illustration). The three-dimensional intersection of the tetrahedron is an octahedron with half the side length. The union is a a star tetrahedron .

If, on the eight faces of the octahedron tetrahedron to also create a star tetrahedron .

With the help of octahedron and cube , numerous bodies can be constructed that also have the group of cube as a symmetry group . So you get for example

as the intersection of an octahedron with a cube (see Archimedean solids ) and

as a convex hull of a union of an octahedron with a cube . Three squares perpendicular to each other, each forming the base of a double pyramid.

## Formulas

Sizes of an octahedron with edge length a
volume ${\ displaystyle V = {\ frac {a ^ {3}} {3}} \ cdot {\ sqrt {2}} \ approx 0 {,} 471 \ cdot a ^ {3}}$ without solid angles in the corners${\ displaystyle \ Omega}$ Surface area ${\ displaystyle A_ {O} = 2 \ cdot a ^ {2} \ cdot {\ sqrt {3}} \ approx 3 {,} 464 \ cdot a ^ {2}}$ Umkugelradius ${\ displaystyle r_ {u} = h_ {p} = {\ frac {a} {\ sqrt {2}}} \ approx 0 {,} 707 \ cdot a}$ Edge ball radius ${\ displaystyle r_ {k} = {\ frac {a} {2}} = 0 {,} 5 \ cdot a}$ Inc sphere radius ${\ displaystyle r_ {i} = {\ frac {a} {6}} \ cdot {\ sqrt {6}} \ approx 0 {,} 408 \ cdot a}$ Ratio of volume
to spherical volume
${\ displaystyle {\ frac {V} {V_ {UK}}} = {\ frac {1} {\ pi}} \ approx 0 {,} 318}$ Interior angle of the
equilateral triangle
${\ displaystyle \ alpha = 60 ^ {\ circ}}$ Angle between
${\ displaystyle \ beta = \ arccos \ left (- {\ frac {1} {3}} \ right)}$ ${\ displaystyle \ approx 109 ^ {\ circ} \; 28 ^ {\ prime} \; 16 ^ {\ prime \ prime}}$ Angle between
edge and face
${\ displaystyle \ gamma = 45 ^ {\ circ}}$ 3D edge angles ${\ displaystyle \ delta = 90 ^ {\ circ}}$ Solid angles in the corners ${\ displaystyle \ Omega = \ arccos \ left ({\ frac {17} {81}} \ right) \ approx 1 {,} 3594 \; \ mathrm {sr}}$ ## Calculation of the regular octahedron

### volume

The octahedron basically consists of two assembled pyramids with a square base and edge length${\ displaystyle A_ {Gp}}$ ${\ displaystyle a.}$ For pyramids and thus for half the volume of the octahedron applies

${\ displaystyle V_ {p} = {\ frac {1} {3}} \ cdot A_ {G_ {p}} \ cdot h_ {p},}$ therein is the base area (square)

${\ displaystyle A_ {G} = a ^ {2},}$ and the height of the pyramid

${\ displaystyle h_ {p} = {\ frac {a} {\ sqrt {2}}},}$ with inserted variables and the factor 2

{\ displaystyle {\ begin {aligned} V & = 2 \ cdot \ left ({\ frac {1} {3}} \ cdot a ^ {2} \ cdot {\ frac {a} {\ sqrt {2}}} \ right) \; \ Rightarrow \\ & = {\ frac {a ^ {3}} {3}} \ cdot {\ sqrt {2}} \; \ approx 0 {,} 471 \ cdot a ^ {3} . \; \ end {aligned}}} ### Surface area

For the surface area of the octahedron (eight equilateral triangles) applies ${\ displaystyle A_ {O}}$ ${\ displaystyle A_ {O} = 8 \ cdot {\ frac {\ sqrt {3}} {4}} \ cdot a ^ {2} = 2 \ cdot a ^ {2} \ cdot {\ sqrt {3}} \ approx 3 {,} 464 \ cdot a ^ {2}.}$ ### Pyramid height

The height of the pyramid can be determined using the following right-angled triangle. ${\ displaystyle h_ {p}}$ The side lengths of this triangle are (see picture in formulas ): side height as a hypotenuse, pyramid height as a large side and half the edge length of the pyramid as a small side. ${\ displaystyle h_ {s}}$ ${\ displaystyle h_ {p}}$ ${\ displaystyle {\ tfrac {1} {2}} \ cdot a}$ The following applies to the height of the equilateral triangle ${\ displaystyle h_ {s}}$ ${\ displaystyle h_ {s} = {\ frac {\ sqrt {3}} {2}} \ cdot a}$ and according to the Pythagorean theorem applies

{\ displaystyle {\ begin {aligned} \; h_ {p} ^ {2} & = h_ {s} ^ {2} - \ left ({\ frac {1} {2}} \ cdot a \ right) ^ {2} = h_ {s} ^ {2} - {\ frac {1} {4}} \ cdot a ^ {2} = {\ frac {3} {4}} \ cdot a ^ {2} - { \ frac {1} {4}} \ cdot a ^ {2} \; \ Rightarrow \\ & = {\ frac {1} {2}} \ cdot a ^ {2} \ Rightarrow \\ h_ {p} & = {\ sqrt {{\ frac {1} {2}} \ cdot a ^ {2}}} = {\ sqrt {\ frac {1} {2}}} \ cdot a = {\ frac {a} { \ sqrt {2}}} \; \ approx 0 {,} 707 \ cdot a. \; \ end {aligned}}} This angle, marked with (see picture in formulas ), has its apex at one edge of the octahedron. It can be determined using the following right triangle. ${\ displaystyle \ beta}$ The side lengths of this triangle are: edge ball radius as a hypotenuse, incipple radius as a large leg and a third of the side height as a small leg. This value is determined by the position of the center of gravity of the triangular area, since the geometric center of gravity divides the height of the triangle in a ratio of 2: 1. ${\ displaystyle r_ {k}}$ ${\ displaystyle r_ {i}}$ ${\ textstyle {\ frac {1} {3}} \ cdot h_ {s}}$ The following applies to the angle${\ displaystyle \ beta}$ {\ displaystyle {\ begin {aligned} \; \ cos {\ frac {\ beta} {2}} & = {\ frac {{\ frac {1} {3}} \ cdot h_ {s}} {r_ { k}}} = {\ frac {{\ frac {1} {3}} \ cdot \ left ({\ frac {\ sqrt {3}} {2}} \ right) \ cdot a} {\ frac {a } {2}}} = {\ frac {{\ frac {1} {6}} \ cdot {\ sqrt {3}}} {\ frac {1} {2}}} = {\ frac {1} { \ sqrt {3}}} \; \ Rightarrow \\\ cos \ beta & = \ cos \ left (\ arccos \ left ({\ frac {1} {\ sqrt {3}}} \ right) \ cdot 2 \ right) = - {\ frac {1} {3}} \ Rightarrow \\\ beta & = \ arccos \ left (- {\ frac {1} {3}} \ right) \ approx 109 ^ {\ circ} \ ; 28 ^ {\ prime} \; 16 ^ {\ prime \ prime}. \; \ End {aligned}}} ### Angle between edge and face

This angle, marked with , has its apex at one corner of the octahedron. Angle can be determined using the following right triangle. ${\ displaystyle \ gamma}$ ${\ displaystyle \ gamma}$ The side lengths of this triangle are (see picture in formulas ): pyramid edge as hypotenuse, pyramid height as large cathetus and half the diagonal of a square with side length / edge as small cathetus. ${\ displaystyle a}$ ${\ displaystyle h_ {p}}$ ${\ displaystyle a}$ ${\ displaystyle {\ tfrac {d} {2}} = {\ tfrac {a} {\ sqrt {2}}}}$ The following applies to the angle${\ displaystyle \ gamma}$ {\ displaystyle {\ begin {aligned} \; \ cos \ gamma & = {\ frac {h_ {p}} {\ frac {d} {2}}} = {\ frac {\ frac {a} {\ sqrt {2}}} {\ frac {a} {\ sqrt {2}}}} = 1 \\\ gamma & = \ arccos \ left (1 \ right) = 45 ^ {\ circ}. \; \ End { aligned}}} ### 3D edge angle

This angle, marked with (see picture in formulas ), has its apex at one corner of the octahedron and corresponds to twice angle d. H. the interior angle of a square . ${\ displaystyle \ delta}$ ${\ displaystyle \ gamma,}$ Thus applies to the 3D edge angle of the octahedron

${\ displaystyle \ delta = 90 ^ {\ circ}.}$ ### Solid angles in the corners

The following formula, described in Platonic Solids, shows a solution for the solid angle${\ displaystyle \ Omega}$ ${\ displaystyle \ Omega = 2 \ pi -2n \ cdot \ arcsin \ left (\ cos \ left ({\ frac {\ pi} {n}} \ right) \ cdot {\ sqrt {\ tan ^ {2} \ left ({\ frac {\ pi} {n}} \ right) - \ tan ^ {2} \ left ({\ frac {\ alpha} {2}} \ right)}} \ right)}$ With the number of edges / faces at a corner and the interior angle of the equilateral triangle, the following applies ${\ displaystyle {n = 4}}$ ${\ displaystyle {a = 60 ^ {\ circ}}}$ {\ displaystyle {\ begin {aligned} {\ mathsf {(1)}} \; \ Omega & = 2 \ pi -2 \ cdot 4 \ cdot \ arcsin \ left (\ cos \ left ({\ frac {\ pi } {4}} \ right) \ cdot {\ sqrt {\ tan ^ {2} \ left ({\ frac {\ pi} {4}} \ right) - \ tan ^ {2} \ left ({\ frac {60 ^ {\ circ}} {2}} \ right)}} \ right) \ Rightarrow \\ & = 2 \ pi -8 \ cdot \ arcsin \ left ({\ frac {1} {\ sqrt {3} }} \ right) \ approx 1 {,} 359347638 \; \ mathrm {sr} \\\ end {aligned}}} because of it ${\ displaystyle \ cos (\ arcsin x) = {\ sqrt {1-x ^ {2}}}}$ {\ displaystyle {\ begin {aligned} {\ mathsf {(2)}} \; \ cos (\ arcsin x) & = {\ sqrt {1- \ left ({\ frac {1} {\ sqrt {3} }} \ right) ^ {2}}} = {\ frac {\ sqrt {6}} {3}} \ Rightarrow \\\ end {aligned}}} used in and formed ${\ displaystyle {\ mathsf {(2)}}}$ ${\ displaystyle {\ mathsf {(1)}}}$ {\ displaystyle {\ begin {aligned} {\ mathsf {(3)}} \; \ Omega & = 2 \ pi -8 \ cdot \ arccos \ left ({\ frac {\ sqrt {6}} {3}} \ right) \ approx 1 {,} 359347638 \; \ mathrm {sr} \; \ end {aligned}}} simplification

{\ displaystyle {\ begin {aligned} {\ mathsf {(4)}} \; \ cos \ Omega & = \ cos \ left (2 \ pi -8 \ cdot \ arccos \ left ({\ frac {\ sqrt { 6}} {3}} \ right) \ right) = {\ frac {17} {81}} \ Rightarrow \\\ end {aligned}}} {\ displaystyle {\ begin {aligned} {\ mathsf {(5)}} \; \ Omega & = \ arccos \ left ({\ frac {17} {81}} \ right) \ approx 1 {,} 359347638 \ ; \ mathrm {sr}. \; \ end {aligned}}} ## generalization

The analogs of the octahedron in any dimension n are called n-dimensional cross-polytopes and are also regular polytopes . The n -dimensional cross-polytope has corners and is bounded by (n − 1) -dimensional simplexes (as facets ). The four-dimensional cross polytope has 8 corners, 24 edges of equal length, 32 equilateral triangles as side surfaces and 16 tetrahedra as facets. The one-dimensional cross polytope is a segment , the two-dimensional cross polytope is the square . ${\ displaystyle 2 \ cdot n}$ ${\ displaystyle 2 ^ {n}}$ A model for the n -dimensional cross polytope is the unit sphere with respect to the sum norm

${\ displaystyle \ left \ | x \ right \ | _ {1} = \ left \ vert x_ {1} \ right \ vert + \ cdots + \ left \ vert x_ {n} \ right \ vert}$ For ${\ displaystyle x = (x_ {1}, \ dots, x_ {n}) \ in \ mathbb {R} ^ {n}}$ in vector space . The (closed) cross polytop is therefore ${\ displaystyle \ mathbb {R} ^ {n}}$ • the amount
${\ displaystyle \ left \ {x \ in \ mathbb {R} ^ {n} \ mid \ left \ | x \ right \ | _ {1} \ leq 1 \ right \} = \ left \ {(x_ {1 }, \ dots, x_ {n}) \ mid \ left \ vert x_ {1} \ right \ vert + \ cdots + \ left \ vert x_ {n} \ right \ vert \ leq 1 \ right \}}$ .
• the convex hull of the vertices , where are the unit vectors .${\ displaystyle 2 \ cdot n}$ ${\ displaystyle \ pm e_ {i}}$ ${\ displaystyle e_ {i}}$ • the intersection of the half-spaces divided by the hyperplanes of the shape${\ displaystyle 2 ^ {n}}$ ${\ displaystyle \ pm x_ {1} \ pm \ cdots \ pm x_ {n} = 1}$ can be determined and contain the origin .

The volume of the n-dimensional cross polytope is , where the radius of the sphere around the origin is with respect to the sum norm. The relationship can be proven using recursion and Fubini's theorem. ${\ displaystyle {\ frac {(2 \ cdot r) ^ {n}} {n!}}}$ ${\ displaystyle r> 0}$ ## Networks of the octahedron

The octahedron has eleven nets . That means, there are eleven ways to unfold a hollow octahedron by cutting open 5 edges and spreading it out in the plane . The other 7 edges connect the 8 equilateral triangles of the mesh. To color an octahedron so that no neighboring faces are the same color, you need at least 2 colors.

## Graphs, dual graphs, cycles, colors

The octahedron has an undirected planar graph with 6 nodes , 12 edges and 8 regions assigned to it, which is 4- regular , ie 4 edges start from each node, so that the degree is 4 for all nodes. In the case of planar graphs, the exact geometric arrangement of the nodes is not important. However, it is important that the edges do not have to intersect. The nodes of this octahedral graph correspond to the corners of the cube.

The nodes of the octahedral graph can be colored with 3 colors so that adjacent nodes are always colored differently. This means that the chromatic number of this graph is 3. In addition, the edges can be colored with 4 colors so that adjacent edges are always colored differently. This is not possible with 3 colors, so the chromatic index for the edge coloring is 4 (the picture on the right illustrates these coloring).

The dual graph (cube graph ) with 8 nodes , 12 edges and 6 areas is helpful to determine the required number of colors for the areas or areas . The nodes of this graph are assigned one-to-one (bijective) to the areas of the octahedral graph and vice versa (see bijective function and figure above). The nodes of the cube graph can be colored with 2 colors in such a way that neighboring nodes are always colored differently, so that the chromatic number of the cube graph is 2. From this one can indirectly conclude: Because the chromatic number is equal to 2, 2 colors are necessary for such a surface coloring of the octahedron or a coloring of the areas of the octahedron graph.

The 5 cut edges of each network (see above) together with the corners ( nodes ) form a spanning tree of the octahedral graph . Each net corresponds exactly to a spanning tree and vice versa, so that there is a one-to-one ( bijective ) assignment between nets and spanning trees. If you consider an octahedron network without the outer area as a graph, you get a dual graph with a tree with 8 nodes and 7 edges and the maximum node degree 3. Each area of ​​the octahedron is assigned to a node of the tree. Not every graph-theoretical constellation (see isomorphism of graphs ) of such trees occurs, but some occur several times.

The octahedron graph has 32 Hamilton circles and 1488 Euler circles .

## Room fillings with octahedra

The three-dimensional Euclidean space can be completely filled with Platonic solids or Archimedean solids of the same edge length. Such three-dimensional tiling is called room filling . The following space fills contain octahedra:

## Applications

In chemistry , predicting molecular geometries using the VSEPR model can result in octahedral molecules . The octahedron also appears in crystal structures , such as the face-centered cubic sodium chloride structure (coordination number 6), in the unit cell , as well as in complex chemistry if 6 ligands are located around a central atom .

Some naturally occurring minerals , e.g. B. alum , crystallize in octahedral form.

In role-playing games, octahedral game dice are used and are referred to as “D8”, ie a dice with 8 faces.