How many distinct figures are there with 3 light bulbs on the vertices of an icosahedron? Place only 1 light bulb per vertex.

How many distinct figures are there with 3 light bulbs on the vertices of a cube? 2 lightbulbs?

Source: POTW reader "Tulsa Mclain". Original cube extensions.

Solutions were received from Alan O'Donnell, Joseph DeVincentis, and Kirk Bresniker. Joseph's solutions and extension follow.

How many distinct figures are there with 3 light bulbs on the vertices of an icosahedron? Place only 1 light bulb per vertex.

If two bulbs are on opposite vertices, then any placement for the third
bulb can be rotated to any other one, so there is one such placement.

If the three bulbs are one of the regular pentagons formed by the five
faces around one vertex, there are two configurations: either the three
bulbs are adjacent, or only two of them are and the other one is opposite
those two on the pentagon.

There are twelve such pentagons, one around each vertex. Call the
vertices which these pentagons surround their centers for the purposes of
this proof. No pair of opposite vertices is on a pentagon (the opposite
vertices to a pentagon form a second pentagon around the vertex opposite the
first) and each vertex is on five pentagons (one for each of the five
adjacent vertices as centers). There are two such pentagons containing any
given pair of non-opposite points (for adjacent points, the relevant centers
each form a face with those two points; for non-adjacent ones, the centers
as a pair form two faces, one with each of the two given points). These two
rings contain 6 of the other 8 points which are not opposite either of the
two given points. This reveals two configurations of vertices not on any
pentagon and with no two vertices opposite. One is the three vertices
forming a face, and the other is three vertices on a plane parallel to a
face but not themselves forming a face. (The latter configuration can also
be seen as the vertices of a large, bent triangle comprising four faces of
the icosahedron.)

So there are five configurations.

How many distinct figures are there with
3 light bulbs on the vertices of a cube? 2 lightbulbs?

The easier one first. Two light bulbs on a cube form three
configurations. They can define either an edge, a face diagonal, or a body
diagonal.

With three light bulbs on a cube, much as with the icosahedron, if two
are opposite then the third bulb can be rotated to any position, making one
configuration of this type.

Otherwise no two points are on a body diagonal, and since you cannot have
a triangle made of three cube edges, at least one pair must be on a face
diagonal. Starting with a face diagonal, two of the other six vertices are
opposite the first two, two of them form a triangle within a face of the
cube, and two of them form equilateral triangles where every edge is a face
diagonal.

This is probably an easier way of solving the first problem, since there also there are only three types of pairs of points: edges, short diagonals (not face diagonals in this case, but points that are two edges apart), and long diagonals made of two opposite points. The five configurations can be seen as (1) any long diagonal, with the third poitn necessarily forming an edge and a short diagonal, (2) two edges and a short diagonal, (3) one edge and two short diagonals, (4) three edges, and (5) three short diagonals.

We might as well extend this to the other platonic solids.

There is only one configuration of three light bulbs on a tetrahedron; they
always form a face.

There are only two configurations of three light bulbs on an octahedron.
Since any two non-opposite points are connected by an edge, either there are
two opposite points or the points form a face.

The dodecahedron is harder.

There are five configurations of two light bulbs on a dodecahedron. The
vertices can be 1, 2, 3, 4, or 5 edges apart, the first being edges, the
second being face diagonals, and the last being points directly opposite one
another.

Among any three vertices, if two are opposite, the third is either adjacent
to the one of the first two, or it is a face-diagonal away from one of the
first two. This gives two configurations.

If two vertices are distance 4 apart and none are opposite one another, the
third vertex can be, relative to the other two: (a) adjacent to one and 3
edges away from the other (4 vertices), (b) two edges away from each (2
vertices), (c) two edges away from one and three from the other (4
vertices), (d) two edges away from one and four from the other (4 vertices),
or (e) three edges away from each (2 vertices). This gives five more
configurations.

If two vertices are distance 3 apart and nothing is farther, then the third
vertex can be (a) adjacent to one and 2 edges away from the other (2
vertices), (b) adjacent to one and 3 away from the other (2 vertices), (c)
two edges away from each (2 vertices), (d) two edges away from one and three
from the other (2 vertices), or (e) three edges away from each (2 vertices)
for another five configurations.

If no vertices are more than a distance 2 apart, there are three
configurations: (a) the three vertices adjacent to one vertex, all 2 apart,
(b) three adjacent vertices in one face, or (c) two adjacent and one
opposite vertex all in one face.

So there are a total of 15 configurations of three light bulbs on a
dodecahedron.

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