Can you place 20 distinct numbers on an icosahedron so that the faces touching each vertex add up to the same amount? This is a problem about creating balanced dice.
It's easy to prove that you can't place the numbers 1 to 20 on an icosahedron so that each vertex has five faces incident with it that add up to the same number - their sum is 210, which is 630 when tripled to account for counting each face three times, and 630 is not divisible by 12, the number of vertices.
But could you do it with, say, the odd numbers from 1 to 39? The square numbers? Some other set of numbers?
For that matter, is it possible for other Platonic solids?
 A: Impossible for cube, dodecahedron, tetrahedron, simply because each vertex has valence three; a small diagram shows that the faces at two ends of an edge must have equal values. 
Meanwhile, octahedron works, here it is spread out with the open part of the page corresponding to the triangle in back. I will think about the icosahedron tomorrow.

A: Since the average sum per vertex is 52.5 (10.5 per face), it's not possible to place the numbers 1-20 such that each vertex has the exact same same. However, a numbering has recently been found by Bob Bosch that has ideally-balanced vertex sums of 52 for six vertices and 53 for the other six. In addition, his solution retains the tradition in dice numbering of having the largest number opposite the smallest (1/20), next largest opposite next smallest (2/19), etc. Physical numerically-balanced d20 dice of this sort are in the works. 
A: With a regular icosahedron we can make all the sums the same using 20 of the numbers from 1 through 25. And the result is connected with, of all things, magic squares.
In this answer a method is described to partition the faces of a regular icosahedron into five groups of tetrahedral faces. It turns out that there are two solutions which are mirror images of each other.
Take one solution and number the distinct tetrahedral face groups 1, 2, 3, 4, 5 so that there are four faces on the icosahedron with each of these numbers (each number corresponding to one of the tetrahedra). Then in the mirror image solution number those tetrahedral face groups 0, 5, 10, 15, 20.
Then simply add up the two numbers assigned to each face. After this addition you get all but five of the numbers from 1 to 25, the missing numbers depending on how the numbers in each group of tetrahedra were permuted before the addition stage. Since each face touching any vertex comes from a different tetrahedron in both groups of five, thecsum surrounding each vertex must be
$1+2+3+4+5+0+5+10+15+20=65.$
The sum of 65 is also found in a pandiagonal 5×5 magic square using all of the numbers 1 through 25. This is not a coincidence; rather, it results from the fact that the magic square is constructed from a similar superposition of mirror-image patterns.
Let's look at a concrete case of this. In the figure below, the faces of a regular icosahedron are split into five tetrahedrally symmetric groups, in each of the two mirror-image configurations. The groups on the left are numbered 1, 2, 3, 4, 5; their mirror images on the right are numbered 0, 5, 10, 15, 20. The exterior numbers 3 and 10 represent the back face of the icosahedron, which maps into the exterior of the triangles shown in the graph.

When the two sets of numbers are added together the sums are unique to each face and include all the numbers from 1 to 25 except 3, 9, 11, 17, 25. These may be fitted into a pandiagonal magic square as shown on the right in the figure below; the missing numbers have been shaded. When we select any five faces surrounding one vertex, such as the blue and gold sets shown on the left, they correspond to one of the 120 magic sum patterns in the 5×5 pandiginal magic square, including quincunx sums. The vertex sums are magic sums that do not involve the missing numbers.

