Euler’s Formula

V + F – E = 2

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This project contains:

I.                    Introduction

II.                  The Definition of a Polyhedron

III.                Euler’s Own Proof

i.                    Explanation

ii.

IV.               Legendre’s Proof Using Radial Projections

i.                    Explanation

ii.

iii.                Difficulties

V.                 A Proof Based Principally on Illustrations

i.                    Explanation

ii.                  Proof in Pictures

a.

b.

VI.               Conclusion

VII.             Bibliography

I.                   Introduction

It is said that in 1750, Euler derived the well known formula V + F – E = 2 to describe polyhedrons.[1] At first glance, Euler’s formula seems fairly trivial.  Edges, faces and vertices are considered by most people to be the characteristic elements of polyhedron.  Surprisingly however, concise labelling of such characteristics was not presented until the 1700’s.  Leonhard Euler, recognizing the deficiency, began his investigation of general polyhedra, and the relationship between their elements.

Euler emphasized five major components of a polyhedron in an attempt to find the relationship between them.  These five components were vertices, (a location where 2 or more edges meet), faces (contained and defined by 3 or more edges), edges (defined as the “ridge or sharp edge”[2] of a polyhedron), sides (used to refer to the sides of each face), and plane angles (the angle found at a vertex, contained by 2 sides).  These definitions, contrasted with the features that Euclid had relied on previously, right angles, and bases, led to far more possible relationships between characteristics.

From above, green denotes the edge of the cube; red denotes the plane angle;

the yellow sphere occurs at a vertex; and all faces are shaded purple.

Sides are in black, they surround the purple faces.

In the remainder, let:               -    V be the number of vertices,

-         F be the number of faces,

-         E be the number of edges,

-         S be the number of sides, and

-         P be the number of plane angles.

By naming each component, Euler observed some general relationships that occur for all polyhedron.  For example, the definition of a side leads to the equation 2S = E, where S is the number of sides and E is the number of edges.  It can then be stated that the number of edges must be an even number, for if this were not the case, it would be possible to have a half of a side, which contradicts the definition of S.

By looking at the condition to form a face, we can derive another formula.  Since a face is a surface enclosed by at least 3 sides, we can state that S ≥ 3F.  Furthermore, there must be at least 3 faces surrounding every vertex ( F ≥ 3V ).  Additionally, it can be stated that the number of plane angles equals the number of sides, P = S, that 2E ≥ 3F, and that 2E ≥ 3V

In summary:            2S = E

S ≥ 3F

F ≥ 3V

P = S

2E ≥ 3F

2E ≥ 3V

II.                 The Definition of a Polyhedron

There are many definitions of a polyhedron.  The definition used in this project given by P. Cromwell in his book entitled “Polyhedra” is the following:

“A polyhedron is the union of a finite set of polygons such that

i.                    Any pair of polygons meet only at their sides or corners.

ii.                  Each side of each polygon meets exactly one other polygon along an edge.

iii.                It is possible to travel from the interior of any polygon to the interior of another.

iv.                Let V be any vertex and lef F1, F2, …, Fn be the n polygons which meet at V.  It is possible to travel over the polygons Fi from one to any other without passing through V.”

III.              Euler’s Own Proof

i.                    Explanation

Although Euler presented the formula, he was unable to prove his result absolutely.  His proof is based on the principle that polyhedrons can be truncated.  Euler proceeds by starting with a polyhedron consisting of a large number of vertices, faces, and edges.  By removing a vertex, you remove at least 3 faces (while exposing a new face), and at least 3 edges.  As you continue, more vertices are removed, until eventually you will find that Euler’s proof degenerates into an object that is not a polyhedron.  A polyhedron must consist of at least 4 vertices.  If there are less than 4 vertices present, a degenerate result will occur, and Euler’s formula fails.  While the proof fails to prove his formula, it does show that truncated and augmented platonic polyhedra satisfy the equation, (thereby including classes such as the Archimedean solids, and pyramids).

ii.                  Proof in Pictures

Click here to see an example of Euler’s proof using an octahedron.

IV.             Legendre’s Proof Using Radial Projections

i.                    Explanation

Another interesting proof [3] that has wider applications, is a proof credited to Adrien Marie Legendre.  This proof inscribes a convex polyhedron inside a sphere, where the centre is found in the region enclosed by the polyhedron, and the polyhedron is fully inscribed.  We then proceed to project light rays from the origin through each of the vertices on the polyhedron.  By finding where these rays intersect the sphere, and connecting the points of intersection by the arcs that characterize the shortest distance between two points along the sphere, we produce a radial projection of the polyhedron.  Vertices are now the points where the arcs meet, edges are now arcs, and faces are now spherical polygons.  By combining the characteristics of a sphere, spherical polygons, and the summarized characteristics of vertices, edges, and faces (from above), Euler’s formula can be derived.

Main facts:

-  If you sum all of the angles surrounding a vertex, the resulting angle will be 2π.  It follows that the total angle sum around all the vertices will be 2πV.

-  From above, 2S = E.

-  Each side has an angle of π.

-  Each face has an angle of 2π.

-  The surface area of a sphere of R=1 is 4 π.

Let’s now establish the equation for the surface area of a sphere.  Let us assume the radius of the sphere is 1.

# Surface Area = V( angle sum) - Sπ +  2πF

4 π = 2 πV – 2E π + 2πF

If we divide both sides of the equation by 2π, we get the formula:

2 = V – E + F

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**This formula is from Girard’s Theorem for spherical polygons, although no further discussion of this theorem is provided in this project.

ii.                  Proof in Pictures

The following demonstrates elements of Legendre’s proof of Euler’s formula through pictures, by inscribing a tetrahedron inside a unit sphere.  Click here to view it.

iii.                Difficulties

At first glance, it seemed like an easy task to solve for the path along the surface of a sphere that connected two points.  However, upon attempting to code such a procedure, it turns out there are some interesting questions that need to be answered.

Let us first look at how to parameterize a unit sphere of form x2 + y2 + z2 = 1.  Let u, v be the parameters we introduce to describe this curve.  This being said, any point on the sphere satisfies the equation

( cos(u)cos(v) , cos(v)sin(u), sin(v) )  where 0o ≤ u, v ≤ 360o.

However, knowing how to parameterize the unit sphere is unnecessary when trying to derive the direct path from any two points, A and B, because there is an additional condition that must be satisfied.  To find the appropriate arc, we must find the plane that contains A, B, and the origin of the sphere.  This plane will also contain all of the points that describe the path between the two points.

After looking at different cases, I found the easiest way to find the arc along the great circle was to calculate the midpoint between the two points we wish to connect.  Let C be the centre of the sphere.

Let A = (xA, yA, zA), B = (xB, yB, zB), and C = (0, 0, 0).

### The direction of the midpoint is therefore M, where

M = [ ((xA+ xB)/2), ((yA+ yB)/2),  ((zA+ zB)/2) ].

To find the midpoint along the arc of the great circle, we must therefore find where the vector of the midpoint intersects x2 + y2 + z2 = 1.

To do accomplish this, you can find the unit length M, and divide M by it.  This will result in the midpoint.  Therefore, the midpoint along a great arc, m, is:

m = M / ||M||

V.               A Proof Based Principally on Illustrations

i.                    Explanation

I found Von Staudt’s proof to be one of the nicer proofs for Euler’s formula.  Von Staudt’s proof begins by picking any vertex on a convex polyhedron.  From this vertex, we look for a vertex that has never been coloured, and connect the vertices by colouring along the edge that connects them.  Now at the new vertex, we look for another vertex that has never been coloured.  If we find one, we connect the vertices and colour along the edge that connects them again.  We proceed in this fashion until there are no more vertices to be visited.

If there are no more vertices to colour, then all vertices have been coloured.   When traversing a polyhedron in this manner, we can see that this proof requires that there is always a path that touches all vertices only once.

Around the same time that this proof was being published, an Irish mathematician Sir William Rowan Hamilton (1805 – 1865) was examining the problem of finding a path in a graph, or along a solid, where each vertex is traversed only once.  This path is called a Hamiltonian circuit, and finding whether or not a circuit exists in a figure is quite a challenge.  To solve the problem on an arbitrary graph is known to be intractable; no efficient algorithm is known.  Hamilton actually created this idea as a sort of puzzle, find the path to connect every vertex on a dodecahedron, visiting every vertex only once, and sold his dodecahedron puzzle to an Irish merchant who began to sell it.  It turns out that it is quite a difficult task to identify whether or not a solid has a Hamiltonian circuit or not, and it may require a proof by exhaustion to determine whether or not a path is possible.  However, it has been proven that all Platonic solids, Archimedean solids, and planar-4-connected graphs have Hamiltonian circuits.[4]

As we begin our proof [5], we initially connect 2 vertices, and for every other edge we shade (let’s use red), we find another vertex that we previously had not visited.  Therefore, by counting the number of edges that we shade red we can determine the number of vertices.  The formula is:

## Red Edges = V – 1

Furthermore, Von Staudt states that all the vertices have been coloured.  To prove this, assume this is not the case.  Then there would be a vertex that is not coloured, which means we were not done colouring our edges red.

Next we will begin by examining faces.  Pick a face and shade it green, as well as any edges that have not yet been coloured red.  Proceed by finding another face who has only one green side, and shade green as before.  Continue until there are either no more faces to colour that satisfy the condition.  Since it is impossible that a face exists with all four sides coloured, all faces must be shaded green.  The relationship between green edges and faces can be described as:

## Green Edges = F – 1

The total number of edges, E, will equal the number of green edges and red edges.

## E = Red Edges + Green Edges

E = V – 1 + F – 1

Therefore,

# E = V + F – 2

i.                    Proof in Pictures

I have illustrated Von Staudt’s proof for a cube initially to clearly demonstrate how the proof is to proceed.  Next, the same proof is illustrated on a dodecahedron.

VI.             Conclusion

For over two centuries people have discussed, proven and disproved Euler’s formula.  The sample proofs I have provided by no means represent the sole representations of Euler’s formula.  It is possible to find proofs based on electrical charge that will prove the formula.  There are others that use Jordan curves and planar graphs, and there are other proofs that rely on topology.  Furthermore, by carefully declaring the case and family of solids you wish to explain, the formula  V + F – 2 = E can be generalized (by changing the 2) to explain more topologically complicated solids.  (L’Huilier, a Swiss mathematician, investigated such figures and discovered that this was possible.)  It is this variety of proofs and applications of Euler’s formula that makes it such an interesting topic to study.

VII.          Bibliography

1.      Casselman, Bill, A Manual of Mathematical Illustration, http://www.math.ubc.ca/~cass/graphics/text/www/

2.      Cromwell, Peter, Polyhedra, Cambridge University Press, Cambridge, UK, © 1997.

3.      Dr. Math, Regular Polyhedra, http://mathforum.org/dr.math/faq/formulas/faq.polyhedron.html#icosahedron

4.      Geometry Junkyard, 17 Proofs of Euler’s Formula, http://www.ics.uci.edu/~eppstein/junkyard/euler/

5.       Polking, John C., The Geometry of a sphere, http://math.rice.edu/~pcmi/sphere/gos6.html

6.      School of Mathematics and Statistics: University of St. Andrews, Scotland, Leonhard Euler, http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Euler.html

7.      Wolfram Research, Hamiltionian Circuit, http://mathworld.wolfram.com/HamiltonianCircuit.html

8.      Wolfram Research, Polyhedron, http://mathworld.wolfram.com/Polyhedron.html

[1] Cromwell, Peter, Polyhedra, Cambridge University Press, Cambridge, UK, © 1997. pg 189

[2] Cromwell, Peter, Polyhedra, , Cambridge University Press, Cambridge, UK, © 1997. pg 189

page 189

[3] Cromwell, Peter, Polyhedra, Cambridge University Press, Cambridge, UK, © 1997. pg 198-200

[4] Wolfram Research, Hamiltionian Circuit, http://mathworld.wolfram.com/HamiltonianCircuit.html

[5] Cromwell, Peter, Polyhedra, Cambridge University Press, Cambridge, UK, © 1997. pg 210-213