Counting faces and edges of 3D shapes (video) | Khan Academy
A cube has 6 Faces, 8 Vertices, and 12 Edges,. so: Let's try with the 5 Platonic Solids (Note: Euler's Formula can be used to prove that there are only 5 Platonic . So much power from just the simple relationship: Vertices + Faces - Edges = 2. graph theory, computational geometry, and other parts of mathematics. . Grinberg's Theorem states that the following relationship must hold between the faces. A Polyhedron is a closed solid shape which has flat faces and straight edges. Where F= Number of faces, V= Number of vertices (Corners), E=Number of edges part of most undergraduate mathematics degrees as part of a topology course. A topologist can't tell the difference between a coffee cup and a donut, or so.
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 planarconnected graphs have Hamiltonian circuits.
Therefore, by counting the number of edges that we shade red we can determine the number of vertices. 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.Learning about Faces, Edges, and Vertices - Three Dimensional Figures
The graph below is 3-valent and has faces which are at least three-sided. Can you convince yourself that it does not correspond to any convex 3-dimensional polyhedron, but that there is a non-convex 3-dimensional polyhedron which does have this as its edge-vertex graph?
The polyhedron, when you find it, has four triangular faces and two hexagonal faces! In a rather curious situation, the solution to this problem, finding conditions on a graph that it be isomorphic to have the same structure as the vertex-edge graph of a convex 3-dimensional polyhedron, was completely answered in the early part of the 20th century. This work was done by the great German geometer and algebraist Ernst Steinitz Steinitz's work was published in German and, unfortunately, did not become widely known for quite some time.
The catalyst for the reformulation of what Steinitz had done was the "translation" of his work into modern graph theory terminology. A graph G is isomorphic to the vertex-edge graph of a 3-dimensional polyhedron i. G is 3-polytopal if and only if G is planar and 3-connected. The property of being 3-connected requires that for any pair of vertices u and v of the graph, there are at least three paths between u and v whose only vertices in common are u and v.
The diagram below offers a "schematic" view of what such paths might look like for two typical vertices in a graph. The diagram omits other edges that might be present at the dots shown.
Amazingly, Steinitz's Theorem enables one to study the combinatorial theory of 3-dimensional convex polyhedra by drawing diagrams in 2 dimensions! These appeared in areas involving Hamiltonian circuits a tour of the vertices, starting and ending at the same vertex, visiting each vertex once and only oncecoloring problems, and matchings disjoint sets of edges. Perhaps the greatest progress concerned existence theorems for 3-dimensional convex polyhedra. Such questions now were reduced to constructions of planar graphs.
The study of Hamiltonian circuits was spurred by the graph theory version of Steinitz's Theorem. Thus, David Barnette and others found a vertex, 3-valent, 3-polytopal non-Hamiltonian graph, and was led by his work to the still open conjecture that planar 3-valent 3-connected bipartite graphs have a Hamiltonian circuit.
Another milestone in the theory of Hamiltonian circuits was Grinberg's amazing result, described below. Let G be a plane graph with Hamiltonian circuit H, let pk' denote the number of k-gonal faces of G which are interior to the circuit H, and let pk'' denote the number of k-gonal faces of G which are exterior to the circuit H. Figure 2 For the diagram above and the Hamiltonian circuit shown in blue we have four interior faces labeled in red which are a 3- 4- 6- and 7-gon. The faces which are exterior to the blue Hamiltonian circuit labeled in green are a 3-gon, two 4-gons, a 5-gon, and a 6-gon.
Grinberg's Theorem states that the following relationship must hold between the faces lying in the interior and exterior of such a Hamiltonian circuit: You should verify that the numbers associated with the Hamiltonian circuit in Figure 2 satisfy the equation above. As an example of the power of Grinberg's Theorem, one can use it to show that the 3-valent polyhedron whose graph is shown below has no Hamiltonian circuit. The lack of a Hamiltonian circuit follows from the fact that all of the faces in the graph except one have a number of sides which leaves a remainder of 2 when divided by 3.
Put differently, all the faces have a number of sides congruent to two mod 3, except for one face, the 9-gon. Although Tutte's vertex graph contains sets of 3 edges which will disconnect the graph into two pieces, each of which contains a circuit, for the graph below, one has to cut 5 edges to disconnect the graph into two pieces, each of which contains a circuit.
Eberhard Type Theorems Look again at the equation that a 3-valent plane graph must satisfy: Given a solution of this equation in non-negative integers, does there exist a convex 3-dimensional polyhedron P such that the number of sides of the faces of P are the given ones?
However, because there is no restriction on the value of p6 for a plane graph, the following question arises. It was this problem that a blind 19th century geometer, Victor Eberhard, raised and thought he had solved in his book Zur Morphologie der Polyeder. Although Eberhard's "Theorem" was given by Eberhard, his proof does not meet modern standards of rigor. Now you have this triangular face on top.
So let me color that in. So you have this triangular face on top. So that's going to be our third face, third face. And then you have this triangular face on the bottom.
That's gonna be our fourth face.
That's going to be our fourth face. And then the key question is are we done? Looks like I've colored all the ones that I can see, but there's one a little bit tricky here.
There's the one that we are actually seeing through. There is the face, there's, let me pick a color, there's the face out front that we can see through so that we can see faces one, two, and four. So that's actually going to be our fifth face. The way they've drawn it, it's like it's made out of glass, so we can see faces one, two, and four. But that is our fifth face.
And so this thing has five faces.
Vertices, Edges and Faces
All right, let's do another example, but instead of faces, we're gonna think about edges. So how many edges does the following shape have?
Pause the video, and see if you can work through this.