Euclid: Picture courtesy of Lexington High
School
Leanna Ho and Laura Keating
Student # (82708017) and (84323013)
MATH 308, Bill Casselman
Final Project
Table of
Contents:
The Construction of a Regular Icosahedron
The Proof of Regularity
Symmetry
Related Proofs
References
The Construction of a Regular Icosahedron - Euclid Book
XIII Proposition 16
Proof that an Icosohedron is a Regular Polyhedron
To prove that the icosahedron is a regular polyhedron, we need to show that the figure satisfies four conditions:
a) All of its faces are regular polygons.
b) All the faces are congruent
c) The figure is convex.
d) All of its vertices are congruent.
By construction by Euclid’s proof, we know that the icosahedron is constructed of 20 equalateral triangles. Therefore all of its faces are all the same equilateral triangle, and condition (a) and (b) are satisfied.
We recognize a convex polyhedron by the fact that any segments which join two vertices not joined by an edge, the diagonals of the polyhedron, are all inside the figure.
By definition, a set in Euclidean space is convex if it contains all the line segments connecting any pair of its points. From a given point, all the diagonals from that point of the icosahedron are drawn, and shown to be inside the solid, and therefore (c) is satisfied.
We show that all the vertices are congruent by showing that the same number of faces around each vertex is the same for all vertices.
Let A be the interior angle of a face at a vertex. All of these angles are equal since we have already shown that each face is constructed with a regular polygon, and by definition all its angles are equal.
Let p be the number of edges around each face. Since each face is the same regular polygon, the number of edges is the same for each face.
Let q be the number of faces around each vertex.
We know that
A = 180 – 360/p
For example, with the icosahedron, each face is a triangle, and therefore p=3 and A = 60^{o}.
We also know that qA < 360.
For example, we look at equilateral triangles. We must have a minimum of three triangles to form a solid, since two will only form a flat plane. The total angle with three triangles is 180^{o } (3 x 60^{o}), and this is the vertex of a tetrahedron. Four triangles gives 240^{o}, which is the vertex of an octahedron, five triangles gives 300^{ o }, the vertex of an icosahedron. However, since an equilateral triangle ia two-thirds of a right angle, six triangles gives 360^{ o} and is flat. Therefore six triangles cannot form an angle of a solid because they must all be planar.
In a similar manner, we can show that for squares and pentagons that qA is less than 360^{o}.
We also know that p>2 and q>2, since each face must have at least 3 sides, and at least 3 faces must meet at each vertex.
Then using these facts we can construct a table
p = 3 |
A = 60 |
60q < 360, q < 6 |
q = 3,4,5 |
p = 3 |
p = 4 |
A = 90 |
90q < 360, q < 4 |
q = 3 |
p = 4 |
p = 5 |
A = 108 |
108q < 360, q < 10/3 |
q = 3 |
p = 5 |
p = 6 |
A = 120 |
120q < 360, q < 3 |
no q |
p = 6 |
Therefore there are only five possibilities
For:
p = 3, q = 3, the polyhedron is a tetrahedron, a four faced figure, for which 3 equilateral triangles meet at each vertex
p = 3 q = 4, the polyhedron is an octahedron, an eight faced figure for which 4 equilateral triangles meet at each vertex.
p = 3, q = 5, the polyhedron is a icosahedron, a twenty faced figure for which 5 equilateral triangles meet at each vertex.
p =4, q = 3 the polyhedron is a cube, an eight faced figure for which 3 squares meet at each vertex.
p = 5, q = 3, the polygon is a dodecagon, a twelve faced figure for which 3 pentagons meet at each vertex.
Therefore there are only five such combinations, and therefore there are only 5 regular polyhedra.
Since {3, 5} is the description of the icosahedron, we know that the solid consists of faces which have 3 sides, and therefore are triangles, and that 5 triangles meet each other at the vertex. Therefore at every vertex, 5 equilateral triangles meet, and therefore every vertice is the same.
Proof by Euler’s
Theorem
Euler’s formula gives a prescribed ratio of face, edges and vertices for a polyhedron. It is:
F – E + V = 2
For F = number of faces
E = number of edges
V = number of vertices
Proof that there are
only five regular convex polyhedra
Let a convex polyhedron have p edges on each face. Then there are a total number of pF edges. But since edge is shared by 2 faces, then there is only half of this number, or pF = 2E. Let q be the number of edges that meet at every vertex. Since two vertices are connected by an edge, qV = 2E.
We substitute F = 2E/p and V = 2E/q into Euler’s formula:
F – E + V = 2
2E/p – E + 2E/q = 2
1/p + 1/q = ½ + 1/E
We know that p3 and
q3,
because a polygon cannot have less than 3 vertices and 3 sides. Both p and q
cannot be simultaneously larger than 3,
1/p + 1/q > ¼ + ¼ = ½ < ½
+1/E
since the number of edges cannot be 0.
Therefore p = 3 or q = 3.
If we let p = 3
1/3 + 1/q = ½ + 1/E
1/q – 1/6 = 1/E
Letting q = 3
1/3 + 1/3 = ½ + 1/E
gives E = 6
Letting q = 4
1/3 +1/4 = ½ + 1/E
give E = 12
Letting q = 5
1/3 +1/5 = ½ + 1/E
gives E= 30.
We show that q cannot equal 6 by
1/3 +1/6 = ½ + 1/E
and E = 0, which is not possible.
Similarly, we can let q = 3, then we find that p can be 3, 4 or 5.
These 5 combinations represent the 5 regular polyhedra:
For {p,q}
{3, 3} is a tetrahedron. Four faced figure, for which 3 equilateral triangles meet at each vertex
{3, 4} is an octahedron. Eight faced figure for which 4 equilateral triangles meet at each vertex.
{3, 5} is a icosahedron. Twenty faced figure for which 5 equilateral triangles meet at each vertex.
{4, 3}is a cube. Eight faced figure for which 3 squares meet at each vertex.
(5, 3) is a dodecagon. Twelve faced figure for which 3 pentagons meet at each vertex.
Therefore there are only five such combinations, and therefore there are only 5 regular polyhedra.
Since {3, 5} is the description of the icosahedron, we know that the solid consists of faces which have 3 sides, and therefore are triangles, and that 5 triangles meet each other at the vertex. Therefore at every vertex, 5 equilateral triangles meet, and therefore every vertice is the same.
Take the 5 sided pyramid shape constructed by five equilateral triangular faces of an icoshedron which join together at one vertex. Each side is an equilateral triangle and the base is a pentagon. The five vertices of the pentagon are ABCDE, with the top of the pyramid, X. We are looking for the angle between the face ABX and the face AXE.
Let F be the midpoint between A and X. Then the angle BFE is the angle between the two faces ABX and AXE.
BF and FE are the altitudes of the faces ABX and AZE, and they have the length. Using trigonometry, the line BE is equal to 2·sin(54°).
References:
Avnet, Jeremy and Chisholm, Matt. Finding Face-to-Face Angles.
[Theory.org]
Bogomolny, Alexander. Interactive Mathematics Miscellany and Puzzles.
[Cut The Knot]
Bourke, Paul. Platonic Solids (Regular polytopes in 3D).
[Platonic Solids]
Casselman, Dr. W.. A Manual of
Mathematical Illustration. [MATH 308
text]
Casselman, Dr. W..
Geometrical Symmetry and the Fine Structure of Regular Polyhedra.
[Notes on
Symmetry]
Joyce, David E..
Euclids Elements.
[Euclid]
MacLean, Kenneth J. M.. The Icosahedron.
[Geometry Web Page]
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