Euler's Theorem

F + V - E = 2

An algebraic argument

 

Objective::

Students will be introduced to the first of several proofs for Euler's theorem. Students will use counting methods for the sum of the angles of a polyhedron, equate them and then use algebraic manipulation to demonstrate Euler's theorem.

Materials :

Optional:

Java, director or powerpoint script can be used to illustrate the algorithm through a computer animation.

Lesson:

One of the many concepts lumped into the category of discrete mathematics topics at the secondary school level is counting methods. Using the interior angle sum theorem, the exterior angle sum theorem and other properties of polyhedra, students are to count for the general case the sum of the angles in a polyhedron. Listen for students who get lost in the algebra and summation terminology. Students need to focus on the best method for illustrating their sums. Listen for students who are struggling with variable representation. Moving back and forth from three dimensions to two is difficult and certain properties that are straight forward for a simple polygon become more cumbersome summations for a polyhedron.

 

The first day spend time reviewing the theorems being used in the counting methods. Focus on only one method through to completion. Draw the students attention to the conclusion and set up the second counting method thoroughly. Emphasize for the students the need to have a vision so that the two counting methods include the necessary pieces to demonstrate Euler's theorem. Be sure to point out that this is a deductive argument and should be an algorithmic sequence, there should be no surprises from step to step.

 

Day two, through an open discussing work through the second counting method. Once the class has reached the desired result from their method, equate the two sums. Walk through the algebra to manipulate Euler's theorem. Place emphasis on definitions and the switch from variables in a general summation to the traditional representations: F for the number of faces, V the number of vertices and E the number of edges. Demonstrate an Inductive proof on edges for the same formula for the class.

Assignment:

(Day 1) Students should finish the second counting method through to completion for homework

(Day 2) Students should rewrite in sequence both counting methods with explanations for each step to hand in to the instructor for practice and closure.

 


Algebraic/Deductive
Argument
Euler Theorem

F + V - E = 2

Sum up the angles in each face of a straight line drawing of the graph {including the outer face}

Note: the planar graph formed by the polyhedron can be embedded so all edges form straight line segments.


Ex: From regular tetrahedron to planar graph

........................


We will sum the angles in two ways and then equate them in order to demonstrate Euler's Theorem is true for the general case algebraically.

In the first counting method, begin with the following:

The sum of the interior angles in a k-gon is (k - 2) {Interior Angle Sum Theorem}, and each edge contributes to two faces of the polyhedron, so the total sum is

and can be rewritten as by distributing and

using properties of sums. Since each edge contributes to two faces of the polyhedron , and is the counter for each polygon that is a

face of the polyhedron. By substitution we have 2E - 2F which can be rewritten as

(2E - 2F)

* the ' outer face' of the polyhedron


Another method for counting the same angles.

Each interior vertex is surrounded by triangles and contributes a total angle of 2 to the sum. The vertices v on the outside face contribute
2( -(v)), where is the exterior angle of the polygon at that vertex.

Each interior vertex has an exterior angle or equal to zero.

Therefore the sum of all of the angles of a polyhedron can be written as:

and rewritten as and again as

since , and the total exterior angle of any polygon is 2.

{Exterior Angle Sum Theorem}.

Then 2V - 2*2 or 2V - 4


By equating the separate counting methods one can find:

2V - 4 = (2E - 2F)

2V - 4 = 2E - 2F distribute

V - 2 = E - F divide through by 2

V + F - E = 2 algebraic manipulation

 


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