* No. of faces --                            20

* No. of edges --                           30

* No. of vertices/corner points -- 12

* At each vortex, 5 edges meet.

* Each face is an equilateral triangle.

* Dihedral Angle --                138°11'


The following Cartesian coordinates define the vertices of an icosahedron with edge-length 2, centered at the origin:

(0, 1, φ)
(1, φ, 0)
(φ, 0, 1)
When an icosahedron is inscribed in a sphere, it occupies less of the sphere's volume (60.54%) than a dodecahedron inscribed in the same sphere (66.49%). Herpes Simplex Virus has the shape of an icosahedron. These are identical protein subunits which are used again and again to build the overall unit by the genome. Icosahedron is also used as a 20 sided dice. Dihedral angle is the angle formed by the intersections of 2 planes. Buckminister fuller based his design of geodesic domes based on icosahedron.
conjugacy classes
I Ih
  • identity
  • 12 rotation by 72, order 5
  • 12 rotation by 144, order 5
  • 20 rotation by 120, order 3
  • 15 rotation by 180, order 2
  • 12 rotoreflection by 108, order 10
  • 12 rotoreflection by 36, order 10
  • 20 rotoreflection by 60, order 6
  • 15 reflection, order 2

Face-to-face angles on the icosahedron.

The angle between faces on an icosahedron can be found in the following way. Take the 5 sided pyramid formed by 5 of the triangular faces and remove it- the rest of the shape wont be necesary for finding the face-to-face angle. Label the five vertices of the pentagonal base of the pyramid A,B,C,D, and E. Label the tip of the pyramid X. The two sides between which the angle will be found are ABX and AXE. The midpoint of AX is F. So the triangle containing the face-to-face angle is BFE. Since the lines BF and FE are altitudes of the faces ABX and AXE, they have length √3/2. The line BE can be found with trigonometry- it is equal to 2sin(54) or 1.6180339875. 3 Now that we know the three sides of BFE, we can use trigonometry again to calculate the angle at point F. The angle is approximately 138.189685104.

Icosahedral Symmetry

A body with cubic symmetry possesses a number of axes about which it maybe rotated to give a number of identical appearances. The occurrence of icosahedral features in quite unrelated viruses is not a matter of chance, but a preference. An ICOSAHEDRON is composed of 20 facets, each an equilateral triangle, and 12 vertices, and because of the axes of rotational symmetry is said to have 5:3:2 symmetry. There are, in fact, six 5-fold axes of symmetry passing through the vertices, ten 3-fold axes extending through each face and fifteen 2-fold axes passing through the edges of an icosahedron.The simplest icosahedral capsids with 5:3:2 symmetry are built up by using 3 identical subunits to form each triangular face, thereby requiring 60 identical subunits to form a capsid. A few virus particles are constructed in this way, e.g. bacteriophage X174. Each unit would be related identically (equivalent) and asymmetrically with its neighbours, and none of the units would coincide with an axis of symmetry.

As soon as the first high resolution micrographs of negatively stained icosahedral viruses were obtained (Horne et al., 1959 - adenovirus; and Huxley and Zubay, 1960 - turnip yellow mosaic virus) it seemed that there was a structural paradox. The number of morphological units observed on the surface of known icosahedral viruses at that time was never 60 or multiples of 60, and was often more than 60 (The major reason is that the construction of a portein shell with a simpleicosahedral design (with "only" 60 units) severly restricts the size of the genome that can be packaged.); and more than 60 subunits cannot be arranged in an equivalent fashion around an icosahedron. Furthermore, the capsomers themselves appeared to be symmetrical and were located on symmetry axes, e.g. herpesvirus. There was direct evidence that capsomers of herpesvirus were hexagonal and pentagonal in section. It was therefore clear that the capsomers were not equivalent to the subunits of Crick and Watson (1956). An obvious solution to the problem was provided by supposing that the symmetrical capsomers are built from a number of ASYMMETRICAL SUBUNITS. In this way it is possible to build a variety of complicated bodies in which 5:3:2 symmetry is preserved and in which the number of subunits is a multiple of 60 as predicted by Crick and Watson.

In 1962 Donald Caspar and Aaron Klug developed two theoretical framework accounting for the structural properties of larger (> 60 units) particles with icosahedral symmetry.

1. Caspar and Klug proposed that when a capsid contains more than 60 subunits, each subunit occupies a quasi-equivalent position; that is, the bonding properties of subunits in different structural environments are similar (but not identical, as in the case of the simplest, 60-structure). 
2. The second important idea was that of triangulation, the description of the triangular face of a large icosahedral structure in terms of its subdivision into smaller triangels termed facets. This process is described by the TRIANGULATION NUMBER T, which gives the number of structural units per face. The icosahedron itself has 20 equilateral triangular facets and therefore 20 T structure units given by the rule: T=P x f(SQR) where P can be any number of the series 1,3,7,13,19,21,31 and f is any integer. Morphological units can be clustered as 20T trimers, 30T dimers or separated as 60T monomers. The number of morphological units that would be produced by a clustering into hexamers and pentamers can be calculated as follows: There are 10(T-1) hexamers plus 12. (and only 12) pentamers. Example


* Volume - L35(3+√5)/12 =  (√5/6)φ2L3         where φ is the Golden number whose value is  1.618034

* Area of each equlateral triangle - √3 L2/ 4

* Surface Area - 5√3 L3

* Each face has fcc-111 structure

* Icosahedron is one of the 5 Platonic solids consisting of 20 equilateral triangles.

* symmetries - ih(*532) . A regular icosahedron has 60 rotational or orientation preserving symmetries and a symmetry order of 120 including transformations that combine a reflection and a rotation.

Why? What is the basis for icosahedral symmetry being so strongly preferred by viruses? Following up on the implicit conclusions drawn by Crick and Watson and by Casper and Klug almost 50 years ago, we have recently argued (2003 PRL and 2004 PNAS) that there is indeed a simple physical basis for this special symmetry shown by viruses of so many different kinds, involving so many different capsid proteins. In particular we demonstrate that icosahedral symmetry allows for the lowest-energy configuration of particles interacting isotropically on the surface of a sphere. More explicitly, we find that the energy-per-particle is a minimum for configurations that involve 12 five-fold defects at the vertices of an icosahedron, and that these configurations are especially favored for "magic" numbers of particles corresponding to the "triangulation" ("T") numbers of Casper and Klug.

Numerical Characterstics of Regular Polyhedra


  n m f e v
Tetrahedron 3 3 4 6 4
Octahedron 3 4 8 12 6
Icosahedron 3 5 20 30 12
Hexahedron 4 3 6 12 8
Dodecahedron 5 3 12 30 20

m:- no. of polygons meeting at one vertex

n:- no. of vertices of each polygon

f:- no. of faces of each polyhedron

e:- no. of edges of a polyhedron

v:- no. of vertices of a polyhedron

e= nf/2 ;   v=nf/m;

f = 2+e-v ..... Euler's polyhedron theorem

* The number of dodecahedron faces is equal to the number of icosahedron vertices and no. of icosahedron faces is equal to the no. of dodecahedron vertices. The no. of planar angles on the surfaces of both are also same, i.e. 60 = 3x20=5x12

* Ri = radius of insphere that touches centroids of its faces.

  Rm = radius of mid sphere that touches centroids of its edges.

  Rc = radius of circumsphere that passes through the vertices.

(side length is 1)







φ2 / 23



φ2 / 2

φ2 / 2√[3-φ]

* Rc / Ri are equal for both icosahedron and dedecahedron.

* In Plato's cosmology, icosahedron symbolized water as the most fluid polyhedron whereas dedecahedron symbolized "The Real World"

Length of each side L
Find out  
Volume of icosahedron V
Radius of sphere with same vol. as above L'
Radius of circumscribed sphere rc =(L/2)*(φ√5)
Radius of inscribed sphere ri =φ2 L /(2√3)
Surface Area of icosahedron S
Area of each triangle a
Volume of a sphere of radius L V'
Surface area of sphere of radius L S'
Ratio S/S'