Icosahedronimoes?

[bonanova]

A game of my grandson's uses a plastic icosahedron. To play, pegs numbered 1-12 are first inserted at its 12 vertices. Then twenty (removable) triangles, having 0, 1, 2 or 3 dots on each corner, are fastened to its faces. You win if the peg's number at each vertex equals the sum of the dots on the 5 triangle corners that meet there. The game maker asserts that there is a solution (appropriate placement of triangles) for any arrangement of the pegs, which is surprising. I wonder whether it is provable.

The image shows a position for which the vertex 10 is solved, but not the others.

The sets of numbers on the triangle corners are (symmetrical, but reading CW if you like):

000 111 222 333

001 002 003

011 022 033

012 012 012

021 021 021

123 123

132 132

Note the pegs and dots both sum to 45, the 12th triangular number, as they must.

[gavinksong]

To start off the conversation...

 

It's interesting that there are only 20 triangles, and they are not distinct.

 

If my reasoning is correct there should be 24 distinct possible triangles:

 

(4 * 3 * 2) / 3 = 8 triangles with three different numbers (e.g. 012)

4 * 3 = 12 triangles with two different numbers (e.g. 011)

4 triangles with all same numbers (e.g. 000)

 

Those 20 triangles must have been chosen specifically for a reason.

I guess two obvious constraints, as bonanova mentioned, is that there must be 20 triangles and 45 dots.

However, I don't think they are the only constraints. Either way, the guarantee of a solution must not arise naturally but by design.

So I say we should look carefully at the triangles.

 

012 and 021 are duplicated thrice each.
123 and 132 are duplicated twice each.

112, 113, 221, 223, 331, 332, 013, 031, 023, and 032 are missing.

[Logophobic]

It certainly is provable, though a brute force approach would be time-consuming.

There are 11 * 2 * 9! = 7,983,360 unique arrangements of pegs. And, if I counted correctly, there are 68,600 unique solutions for peg #1, with five rotations of each solution for each arrangement of pegs. That's 343,000 positions to check for each arrangement of pegs. More than 2.7 * 10¹² positions to check, and we still have 15 more triangles to place.

Should we go ahead and reserve time on some super-computer?

BTW, the 12^th triangular number is 78, not 45.