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Platonic Solid's. (Wikipedia Part 2.)
created Oct 4th 2014, 01:05 by Nehemiah Thomas
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All other combinatorial information about these solids, such as total number of vertices (V), edges (E), and faces (F), can be determined from p and q. Since any edge joins two vertices and has two adjacent faces we must have:
pF = 2E = qV.\,
The other relationship between these values is given by Euler's formula:
V - E + F = 2.\,
This nontrivial fact can be proved in a great variety of ways (in algebraic topology it follows from the fact that the Euler characteristic of the sphere is two). Together these three relationships completely determine V, E, and F:
V = \frac{4p}{4 - (p-2)(q-2)},\quad E = \frac{2pq}{4 - (p-2)(q-2)},\quad F = \frac{4q}{4 - (p-2)(q-2)}.
Note that swapping p and q interchanges F and V while leaving E unchanged (for a geometric interpretation of this fact, see the section on dual polyhedra below). The classical result is that only five convex regular polyhedra exist. Two common arguments below demonstrate no more than five Platonic solids can exist, but positively demonstrating the existence of any given solid is a separate question – one that an explicit construction cannot easily answer. The following geometric argument is very similar to the one given by Euclid in the Elements:
Each vertex of the solid must coincide with one vertex each of at least three faces.
At each vertex of the solid, the total, among the adjacent faces, of the angles between their respective adjacent sides must be less than 360°.
The angles at all vertices of all faces of a Platonic solid are identical: each vertex of each face must contribute less than 360°/3 = 120°.
Regular polygons of six or more sides have only angles of 120° or more, so the common face must be the triangle, square, or pentagon. For these different shapes of faces the following holds:
Triangular faces: Each vertex of a regular triangle is 60°, so a shape may have 3, 4, or 5 triangles meeting at a vertex; these are the tetrahedron, octahedron, and icosahedron respectively.
Square faces: Each vertex of a square is 90°, so there is only one arrangement possible with three faces at a vertex, the cube.
Pentagonal faces: Each vertex is 108°; again, only one arrangement, of three faces at a vertex is possible, the dodecahedron.
Altogether this makes 5 possible Platonic solids. A purely topological proof can be made using only combinatorial information about the solids. The key is Euler's observation that V - E + F = 2, and the fact that pF = 2E = qV, where p stands for the number of edges of each face and q for the number of edges meeting at each vertex. Combining these equations one obtains the equation
\frac{2E}{q} - E + \frac{2E}{p} = 2.
Simple algebraic manipulation then gives
{1 \over q} + {1 \over p}= {1 \over 2} + {1 \over E}.
Since E is strictly positive we must have
\frac{1}{q} + \frac{1}{p} > \frac{1}{2}.
Using the fact that p and q must both be at least 3, one can easily see that there are only five possibilities for (p, q):
(3, 3),\quad (4, 3),\quad (3, 4),\quad (5, 3),\quad (3,5). There are a number of angles associated with each Platonic solid. The dihedral angle is the interior angle between any two face planes. The dihedral angle, θ, of the solid {p,q} is given by the formula
\sin{\theta\over 2} = \frac{\cos(\pi/q)}{\sin(\pi/p)}.
This is sometimes more conveniently expressed in terms of the tangent by
\tan{\theta\over 2} = \frac{\cos(\pi/q)}{\sin(\pi/h)}.
The quantity h (called the Coxeter number) is 4, 6, 6, 10, and 10 for the tetrahedron, cube, octahedron, dodecahedron, and icosahedron respectively.
The angular deficiency at the vertex of a polyhedron is the difference between the sum of the face-angles at that vertex and 2π. The defect, δ, at any vertex of the Platonic solids {p,q} is
\delta = 2\pi - q\pi\left(1-{2\over p}\right).
By a theorem of Descartes, this is equal to 4π divided by the number of vertices (i.e. the total defect at all vertices is 4π).
The 3-dimensional analog of a plane angle is a solid angle. The solid angle, Ω, at the vertex of a Platonic solid is given in terms of the dihedral angle by
\Omega = q\theta - (q-2)\pi.\,
This follows from the spherical excess formula for a spherical polygon and the fact that the vertex figure of the polyhedron {p,q} is a regular q-gon.
The solid angle of a face subtended from the center of a platonic solid is equal to the solid angle of a full sphere (4π steradians) divided by the number of faces. Note that this is equal to the angular deficiency of its dual.
The various angles associated with the Platonic solids are tabulated below. The numerical values of the solid angles are given in steradians. The constant φ = (1+√5)/2 is the golden ratio.
pF = 2E = qV.\,
The other relationship between these values is given by Euler's formula:
V - E + F = 2.\,
This nontrivial fact can be proved in a great variety of ways (in algebraic topology it follows from the fact that the Euler characteristic of the sphere is two). Together these three relationships completely determine V, E, and F:
V = \frac{4p}{4 - (p-2)(q-2)},\quad E = \frac{2pq}{4 - (p-2)(q-2)},\quad F = \frac{4q}{4 - (p-2)(q-2)}.
Note that swapping p and q interchanges F and V while leaving E unchanged (for a geometric interpretation of this fact, see the section on dual polyhedra below). The classical result is that only five convex regular polyhedra exist. Two common arguments below demonstrate no more than five Platonic solids can exist, but positively demonstrating the existence of any given solid is a separate question – one that an explicit construction cannot easily answer. The following geometric argument is very similar to the one given by Euclid in the Elements:
Each vertex of the solid must coincide with one vertex each of at least three faces.
At each vertex of the solid, the total, among the adjacent faces, of the angles between their respective adjacent sides must be less than 360°.
The angles at all vertices of all faces of a Platonic solid are identical: each vertex of each face must contribute less than 360°/3 = 120°.
Regular polygons of six or more sides have only angles of 120° or more, so the common face must be the triangle, square, or pentagon. For these different shapes of faces the following holds:
Triangular faces: Each vertex of a regular triangle is 60°, so a shape may have 3, 4, or 5 triangles meeting at a vertex; these are the tetrahedron, octahedron, and icosahedron respectively.
Square faces: Each vertex of a square is 90°, so there is only one arrangement possible with three faces at a vertex, the cube.
Pentagonal faces: Each vertex is 108°; again, only one arrangement, of three faces at a vertex is possible, the dodecahedron.
Altogether this makes 5 possible Platonic solids. A purely topological proof can be made using only combinatorial information about the solids. The key is Euler's observation that V - E + F = 2, and the fact that pF = 2E = qV, where p stands for the number of edges of each face and q for the number of edges meeting at each vertex. Combining these equations one obtains the equation
\frac{2E}{q} - E + \frac{2E}{p} = 2.
Simple algebraic manipulation then gives
{1 \over q} + {1 \over p}= {1 \over 2} + {1 \over E}.
Since E is strictly positive we must have
\frac{1}{q} + \frac{1}{p} > \frac{1}{2}.
Using the fact that p and q must both be at least 3, one can easily see that there are only five possibilities for (p, q):
(3, 3),\quad (4, 3),\quad (3, 4),\quad (5, 3),\quad (3,5). There are a number of angles associated with each Platonic solid. The dihedral angle is the interior angle between any two face planes. The dihedral angle, θ, of the solid {p,q} is given by the formula
\sin{\theta\over 2} = \frac{\cos(\pi/q)}{\sin(\pi/p)}.
This is sometimes more conveniently expressed in terms of the tangent by
\tan{\theta\over 2} = \frac{\cos(\pi/q)}{\sin(\pi/h)}.
The quantity h (called the Coxeter number) is 4, 6, 6, 10, and 10 for the tetrahedron, cube, octahedron, dodecahedron, and icosahedron respectively.
The angular deficiency at the vertex of a polyhedron is the difference between the sum of the face-angles at that vertex and 2π. The defect, δ, at any vertex of the Platonic solids {p,q} is
\delta = 2\pi - q\pi\left(1-{2\over p}\right).
By a theorem of Descartes, this is equal to 4π divided by the number of vertices (i.e. the total defect at all vertices is 4π).
The 3-dimensional analog of a plane angle is a solid angle. The solid angle, Ω, at the vertex of a Platonic solid is given in terms of the dihedral angle by
\Omega = q\theta - (q-2)\pi.\,
This follows from the spherical excess formula for a spherical polygon and the fact that the vertex figure of the polyhedron {p,q} is a regular q-gon.
The solid angle of a face subtended from the center of a platonic solid is equal to the solid angle of a full sphere (4π steradians) divided by the number of faces. Note that this is equal to the angular deficiency of its dual.
The various angles associated with the Platonic solids are tabulated below. The numerical values of the solid angles are given in steradians. The constant φ = (1+√5)/2 is the golden ratio.
