2013 Putnam A1 Proof understanding (geometry) Problem A1:

Recall that a regular icosahedron is a convex polyhedron having 12 vertices and 20 faces; the faces are congruent equilateral triangles. On each face of a regular icosahedron is written a non-negative integer such that the sum of all $20$ integers is $39.$ Show that there are two faces that share a vertex and have the same integer written on them.

Beginning with a contradiction. Suppose there aren't two faces that touch one vertex. Each vertex has $5$ faces attached to it. Hence the $5$ take on different values (since they are touching each other). The least possible is: $$0, 1, 2, 3, 4$$ For corresponding sides such that another side touching cannot have the same value. So for example

If I have $5$ faces abcde, on the side of $e$ there is a different face (not part of the original 5) that cannot take on the value of $e$.
Hence,
$$a + b + c + d + e = 0 + 1 + 2+ 3 + 4 = 10$$
Same for the other 15 faces, the least becomes $40 > 39$.
Here is my problem.

I don't see this properly. If two faces did have the same number with the same vertex, how would that have changed the sum from $40$? Can somebody give me an example of the original statement of the problem for an intuitive understanding?

Thanks!
 A: This is an alternative answer. I don't think it answers your specific question about the existing proof.
Consider the set $P$ of pairs $(f,v)$ of faces and vertices on the face. Let $n(f)$ be the number written on the face.
Let $$S=\sum_{(f,v)\in P} n(f)$$
Counting by each face we get:
$$S=\sum_{f} 3n(f) = 3\cdot 39$$
Counting by each vertex, we get:
$$S=\sum_{v} \sum_{f\mid(f,v)\in P} n(f)$$
But you've shown that $\sum_{f} n(f)\geq 10$.
Since there are 12 vertices, this means that $S\geq 120$. 
This is a contradiction, since $3\cdot 39<3\cdot 40=120$.
More generally, if $G$ is any trianglulation of the sphere, then we can define $P$ again ad we get:
$$\sum_f n(f) \geq \frac16\sum_{v} d(v)(d(v)-1)$$
where $d(v)$ is the degree of the vertex - the number of edges incident with the vertex.
For example, if you triangluate each of the faces of the icosahedron by adding one node internally, you end up with 12 nodes or degree $6$ and 20 nodes of degree $3$. So you have as lower bound a sum:
$$\frac 16\left(12\cdot 30 +20\cdot 6\right)=80$$
You can definitely do better than that - $90$ is a better lower bound. easily proven.
That doesn't mean you can achieve this value - the value for the tetrahedron by this formula, with $4$ nodes each of degree $3$, would be $\frac{1}{6}\cdot 4\cdot (3\cdot 2)=4$, while the actual minimal value is $6$.
It does come up with the exact right answer for the octahedron of $12$.
A: What the proof essentially shows is that any valid arrangement that satisfies the incidence requirement must have a sum at least 40; therefore, no arrangement with a sum of 39 is possible without violating the incidence requirement.  The proof does not explicitly construct an arrangement with a sum of 39, because there are in general a very large number of such arrangements, and in order to show that all of them violate the incidence requirement (i.e., there exist two faces of equal value that share a vertex), you would have to enumerate them exhaustively.  That's inefficient.
Your question, therefore, isn't really the most pertinent one to ask:  it merely suffices to show that any valid arrangement of numbers must have a sum strictly greater than 39.  There are arrangements with sum greater than 40, for example.  If two faces of such an arrangement have the same number and share a vertex, that does not guarantee the sum will be less than 40.  For example, choose all the faces to have the value 1000.
For those interested in the geometry of the problem, it should be noted that the twenty faces of the icosahedron can be matched into five groups of four faces, such that each group does not share a vertex among any of its faces (the faces in a group are disjoint), and that each vertex is incident to exactly one face from each group.  For each group, draw an edge from the center of each face to the center of the other faces.  These edges then form five regular tetrahedra and the resulting compound of five tetrahedra comes in two enantiomorphs.


