Deriving Euler's Polyhedron Equation for middle school/high school students

I've been perusing the 20 proofs, but they all seem to involve graph theory and complexities... How would you derive this simple formula for high school or middle school students?

Specifically: $V-E+F = 2$ (Vertices, Edges, Faces)

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I had thought that a simple system of linear equation such as $aV + bE + cF = Q$ could be used to derive the linear combination from a table of Platonic solids $V,E,F$'s - but I guess even my linear algebra is rusty...

• Which formula are you asking about? Is it $V+F=E+2$? Commented Nov 16, 2014 at 21:43
• updated q again with V-E+F=2 and how other than "ansatz" there does not seem to be a methodical way to deduce this from tabulating in a table?
– ina
Commented Nov 16, 2014 at 21:47
• Are you sure that the proofs with graph theory wouldn't go over well? The inductive proof where you repeatedly remove vertices from a graph, showing that $V-E+F$ is invariant is very visual. (Someone taught it to me in high school, for what it's worth - but then again, I was familiar with graph theory at that point) Commented Nov 16, 2014 at 21:51
• Is there a way to arrive at that formula from a linear system of equations from counting V, E, F?
– ina
Commented Nov 16, 2014 at 21:55
• I'm sure you can derive a formula if you know there is one - noting that the tetrahedron is self dual and the others are in dual pairs there will be something. The powerful result is that this does not apply just to figures which are regular, but to any figures. One application within reach is the six or five colour theorem (or why a 'spherical' football has pentagonal sections, why you can find pentagons in the Eden Project domes - Cornwall, England) Commented Nov 16, 2014 at 22:19

Not a bad idea.\begin{align} &&Va+ Eb+ Fc=d\\ &Tetrahedron &4a+ 6b+ 4c=d \\ &Hexahedron &8a+ 12b+ 6c=d \\ &Octahedron &6a+ 12b+ 8c=d \\ &Dodecahedron &20a+ 30b+ 12c=d \\ &Icosahedron &12a+ 30b+ 20c=d\end{align} As the relation is homogenous in the unknowns, you can arbitrarily choose one of them, let $d$.

Solving the first three equations by Cramer, $$a=\frac d2, b=-\frac d2,c=\frac d2,\\ \frac{V-E+F}2d=d.$$ As one can check, the last two solids are also verify the relation.

For convenience, use $d=2$ and $$V-E+F=2.$$

The hard part is to prove that this relation applies to all polyhedra.

• ah, i was trying to solve it as a 4x4 - can you explain more about why you can set d to 1?
– ina
Commented Nov 16, 2014 at 22:29
• I have reformulated to get rid of $d$.
– user65203
Commented Nov 16, 2014 at 23:01
• Thanks - but how do you explain the homogenous constant -> 1 "fudge" to middle school students?
– ina
Commented Nov 17, 2014 at 0:10
• If you don't want to, just leave $d$ in the RHS and solve like this, assuming you know $d$. Then you'll obtain $a,b,c$ as a function of $d$.
– user65203
Commented Nov 17, 2014 at 7:34

"Proofs and Refutations: The Logic of Mathematical Discovery" by Imre Lakatos has an excellent discussion of this theorem and its history.

It's even available for Kindle.

You can allow the following moves on any connected configuration of vertices and edges on a plane or on the surface of a sphere:

1) If a vertex is on only two distinct edges, remove the vertex, and merge the edges into one. This reduces $V$ and $E$ by one, and keeps $V-E+F$ the same.

2) If an edge separates two faces, delete the edge and merge the faces into one. This reduces $F$ and $E$ by one and keeps $V-F+E$ the same. Note that the configuration of edges and vertices remains connected because the union of the boundaries of the two faces is connected even when the separating edge is removed.

3) If there is a vertex connected to just one edge at one end, remove edge and vertex. This reduces $V$ and $E$ by one, and keeps $V-E+F$ the same.

Applying these reductions leaves us with only one face (otherwise there would be an edge separating two faces). This means there are no cycles of edges, so if there are edges at all there must be an edge with a lone vertex at one end (otherwise you can go along edges until you have made a cycle - either the cycle is complete because you have visited the vertex before, or there is a way to continue the path because you haven't visited the vertex before). But such edges don't exist by $3$.

The last move was to remove a vertex and an edge leaving a lone vertex behind (by rule 3) or to remove a loop edge joined to one vertex and separating two faces (rule 2). Since rule 1 leaves the graph with an edge, and the final graph has no edges, it can't provide the last move. With either possibility of last move you are left with one vertex and one face and $1-0+1=2$.

• I wonder if this geometric proof can be extended to help students understand systems of linear equations in a more intuitive way
– ina
Commented Nov 16, 2014 at 22:31

We can prove this by imagining reassembling the polyhedron from its faces. In particular, consider that an $n$-gon face consists of $n$ vertices, $n$ edges and one face and hence has an "Euler characteristic" of $n-n+1=1$. So, when we set in place our first face, we have Euler characteristic one.

Next, we line up another face of the polygon such that it shares an edge with the original one, and we glue it up real nice and such - now, naively, since we just added another polygon, which has Eulcer characteristic one, we would expect the characteristic of this figure would be $1+1=2$ - however, we need to correct for the fact that the two faces share $2$ vertices and $1$ edge, implying that our characteristic is $(2-1+0)=1$ higher than it ought to be - so it still has characteristic one.

Next, we can proceed to choose one of the vertices shared between these first two shapes, and one by one add all the faces around this vertex. In general, these new faces will share $2$ vertices and $1$ edge and with the previously existing shape (and will add a face), leaving the Euler characteristic unaltered. However, when we put in the last face required to enclose this vertex, it will adjoin to two edges (completing a little cycle around the vertex), and thus share $3$ vertices and $2$ edges with the previous shape - which still leaves the characteristic unchanged, mind you.

More generally, if a new face joins with $n$ pre-existing edges, we want it to be true that it will share $n+1$ vertices with the pre-existing shape. To do this, we need that those $n$ pre-existing edges form a path between $n+1$ of vertices of the added shape. If this didn't happen, it would mean that there were two, disconnected, places where the new face intersected the previous shape - thus creating a "hole" in the figure (read: the shape wouldn't be simply connected). Rather than try to deal with this, we just notice that we can always avoid creating a hole, and thus decide that we will never create a hole while making the figure, ensuring that our invariant is actually invariant.

Then, using this method, we can build up the shape minus a single face, and it will have characteristic $1$. However, when we join the new face on, it will line up with all its vertices and edges - thus will leave $V$ and $E$ unchanged - but it constitutes a new face, increasing the characteristic to $2$.

This may be difficult to visualize on paper - but I have a solution to that: You can carry out this construction physically and watch it work.

My geometry professor in college gave an argument involving spherical geometry. I don't remember the details, but you basically "inflate" the polyhedron so that it is a sphere and each face is a spherical polygon. Then you use properties of spherical triangles (angle surplus, etc.) to derive the invariant. I think it would be reasonably accessible to high schoolers.

• An argument would be that an $n$-gon has interior angles summing to $(n-2)\pi$, and each edge being used twice as an edge of a face, you get that the interior angles of the faces should sum to $(2E-2F)\pi$. However, the sum of the angles on in spherical geometry would be $2V\pi$, the difference being $2\pi(V-E+F)$ which must be $4\pi$, since that is the total spherical excess. (Of course, short of using Gaussian curvature, I'm not sure how you'd prove that spherical excess ought to be $4\pi$ without using the Euler characteristic, which would be circular) Commented Nov 17, 2014 at 3:14