# Rhombohedron

(Redirected from Rhombohedra)
Rhombohedron
Type prism
Faces 6 rhombi
Edges 12
Vertices 8
Symmetry group Ci , [2+,2+], (×), order 2
Properties convex, equilateral, zonohedron, parallelohedron

In geometry, a rhombohedron (also called a rhombic hexahedron[1] or, inaccurately, a rhomboid) is a three-dimensional figure with six faces which are rhombi. It is a special case of a parallelepiped where all edges are the same length. It can be used to define the rhombohedral lattice system, a honeycomb with rhombohedral cells. A cube is a special case of a rhombohedron with all sides square.

In general a rhombohedron can have up to three types of rhombic faces in congruent opposite pairs, Ci symmetry, order 2.

Four points forming non-adjacent vertices of a rhombohedron necessarily form the four vertices of an orthocentric tetrahedron, and all orthocentric tetrahedra can be formed in this way.[2]

## Rhombohedral lattice system

The rhombohedral lattice system has rhombohedral cells, with 6 congruent rhombic faces forming a trigonal trapezohedron:

## Special cases by symmetry

Special cases of the rhombohedron
Form Cube Trigonal trapezohedron Right rhombic prism Oblique rhombic prism
Angle
constraints
${\displaystyle \alpha =\beta =\gamma =90^{\circ }}$ ${\displaystyle \alpha =\beta =\gamma }$ ${\displaystyle \alpha =\beta =90^{\circ }}$ ${\displaystyle \alpha =\beta }$
Symmetry Oh
order 48
D3d
order 12
D2h
order 8
C2h
order 4
Faces 6 squares 6 congruent rhombi 2 rhombi, 4 squares 6 rhombi
• Cube: with Oh symmetry, order 48. All faces are squares.
• Trigonal trapezohedron (also called isohedral rhombohedron):[3] with D3d symmetry, order 12. All non-obtuse internal angles of the faces are equal (all faces are congruent rhombi). This can be seen by stretching a cube on its body-diagonal axis. For example, a regular octahedron with two regular tetrahedra attached on opposite faces constructs a 60 degree trigonal trapezohedron.
• Right rhombic prism: with D2h symmetry, order 8. It is constructed by two rhombi and four squares. This can be seen by stretching a cube on its face-diagonal axis. For example, two right prisms with regular triangular bases attached together makes a 60 degree right rhombic prism.
• Oblique rhombic prism: with C2h symmetry, order 4. It has only one plane of symmetry, through four vertices, and six rhombic faces.

## Solid geometry

For a unit (i.e.: with side length 1) isohedral rhombohedron,[3] with rhombic acute angle ${\displaystyle \theta ~}$, with one vertex at the origin (0, 0, 0), and with one edge lying along the x-axis, the three generating vectors are

e1 : ${\displaystyle {\biggl (}1,0,0{\biggr )},}$
e2 : ${\displaystyle {\biggl (}\cos \theta ,\sin \theta ,0{\biggr )},}$
e3 : ${\displaystyle {\biggl (}\cos \theta ,{\cos \theta -\cos ^{2}\theta \over \sin \theta },{{\sqrt {1-3\cos ^{2}\theta +2\cos ^{3}\theta }} \over \sin \theta }{\biggr )}.}$

The other coordinates can be obtained from vector addition[4] of the 3 direction vectors: e1 + e2 , e1 + e3 , e2 + e3 , and e1 + e2 + e3 .

The volume ${\displaystyle V}$ of an isohedral rhombohedron, in terms of its side length ${\displaystyle a}$ and its rhombic acute angle ${\displaystyle \theta ~}$, is a simplification of the volume of a parallelepiped, and is given by

${\displaystyle V=a^{3}(1-\cos \theta ){\sqrt {1+2\cos \theta }}=a^{3}{\sqrt {(1-\cos \theta )^{2}(1+2\cos \theta )}}=a^{3}{\sqrt {1-3\cos ^{2}\theta +2\cos ^{3}\theta }}~.}$

We can express the volume ${\displaystyle V}$ another way :

${\displaystyle V=2{\sqrt {3}}~a^{3}\sin ^{2}\left({\frac {\theta }{2}}\right){\sqrt {1-{\frac {4}{3}}\sin ^{2}\left({\frac {\theta }{2}}\right)}}~.}$

As the area of the (rhombic) base is given by ${\displaystyle a^{2}\sin \theta ~}$, and as the height of a rhombohedron is given by its volume divided by the area of its base, the height ${\displaystyle h}$ of an isohedral rhombohedron in terms of its side length ${\displaystyle a}$ and its rhombic acute angle ${\displaystyle \theta }$ is given by

${\displaystyle h=a~{(1-\cos \theta ){\sqrt {1+2\cos \theta }} \over \sin \theta }=a~{{\sqrt {1-3\cos ^{2}\theta +2\cos ^{3}\theta }} \over \sin \theta }~.}$

Note:

${\displaystyle h=a~z}$3 , where ${\displaystyle z}$3 is the third coordinate of e3 .

The body diagonal between the acute-angled vertices is the longest. By rotational symmetry about that diagonal, the other three body diagonals, between the three pairs of opposite obtuse-angled vertices, are all the same length.