The DNA-surrounding capsid of the cold virus and the atomic arrangement of a certain allotrope of boron consists of 20 triangles arranged in a three dimensional shape known as a icosahedron.
If you form a three-dimensional structure with 8 triangles, you get an octahedron. An example is the molecular structure of the electrical insulator and potent greenhouse gas, sulfur hexafluoride, SF6. It can also appear in the mineral pyrite.
If you limit the number of triangles to four, you get a tetrahedron. The tetrahedral silicate unit, SiO42- , is the basic component of most silicates in the Earth’s crust. On our planet’s surface, any time carbon makes four single bonds with other carbons or other elements in a wide variety of life’s hydrocarbons, we also get a tetrahedral arrangement. This allows the four bonding pairs of electrons to get as far from each other as possible.
Changing polygon, we can use 6 squares to make a cube. The similarly-sized ions of cesium and chloride can form an arrangement where each whole ion centers 8 vertices occupied by an ion of the opposite charge. But viewed within the lattice, the boundaries of the cube are such that only 1/8 of each ion is at a vertex. Given that there are 8 vertices, this maintains the ratio of one cesium for every one chloride ion. In the mineral halite, which is composed of NaCl, the chloride ion is considerably bigger than the sodium. They don’t pack into a cube in the same manner, but overall they still form the same shape.
Then there is the dodecahedron, consisting of a dozen pentagons. Cubes of pyritohedron form the macro-illusion of a dodecahedron, but otherwise this “perfect” solid is rare in nature.
Are there more than these 5 possible perfect or Platonic solids? No. But why?
There is a relationship that holds for all solids of this type. If you let V represent the number of vertices, E = number of edges and F = the number polygonal faces,
you will observe that in a tetrahedron V= 4, E = 6 and F = 4.
For a cube,
Notice that for both solids, the following simple formula(called Euler’s Formula)holds true:
V + F – E = 2. Equation (1)
That doesn’t constitute a proof for why it should apply to all cases, but you can find one here. We can use this formula to prove that there are only a limited number of Platonic solids.
First let’s introduce two other variables, N = the number of sides in a polygonal face and R = number of edges that meet at a vertex.
Since each edge is shared by two vertices, if we multiply R by the number of vertices,V , and divide by two, we will get the total number of edges:
RV / 2 = E; Solving for V we get
V = 2E / R Equation (2)
The number of edges can also be obtained by the number of faces. Each face has one edge for each of the number of sides, N. But each edge is shared with another face, so again not to count things twice:
NF / 2 = E. Solving for F we get
F = 2E / N Equation (3)
Substituting equations (2) and (3) into equation 1:
2E / R + 2E / N – E = 2.
Now divide each term by 2E:
1/R + 1/N – 1/2 = 1/E.
We need at least three edges to get a 3D shape so R ≥ 3. Similarly to get a polygon, N ≥ 3. Interestingly N and R cannot simultaneously be greater than 3, because as they create progressively smaller fractions, 1/R and 1/N will add up to a maximum sum of 1/2 ( if R=N=4), which in the formula will yield 0 = 1/E.
Letting N = 3, if R = 3 then E = 6. Using this result and equation(3), F = 2E/N = 2(6)/3= 4: the tetrahedron.
Having no choice due to the restriction we mentioned, we have to keep either N or R at 3 while increasing the other, so
letting N = 3, if R = 4, E = 12 and F = 8, the octahedron.
Reversing the values and letting N = 4 and R = 3, E = 12 and F = 2(12)/4= 6, the cube.
We can then try the combinations of N= 3, R= 5 and N= 5, R= 3, which will solve for the icosahedron ( E = 30; F = 20) and dodecahedron ( E = 30; F = 12), respectively.
But 5 is the limit because if we try values of 3 and 6:
1/3 + 1/6 – 1/2 = 0, which means the impossibility of no edges. A value larger than 6 yields a negative value for E.
Given that there are only these 5 solutions to Euler’s formula , then only five Platonic solids can exist in three dimensional space.