How to find the number of intersections of diagonals in icosahedron?

Question

How to find the number of points of intersection of the diagonals in icosahedron?

Answer 1

(I’m assuming that by “icosahedron” you mean the regular icosahedron; that by “diagonals” you mean line segments whose endpoints are vertices but which are not themselves an edge; and that by “all intersections” you mean all points which lie at the intersections of two or more diagonals, except for the vertices themselves.)

There are 21 such intersections. One is the centre of the icosahedron; the rest are each an intersection of diagonals in one of the “obvious pentagons”.

To see that these are all, stare at this picture of the icosahedron, and note that:

1. Each intersection of diagonals must come from a set of 4 coplanar vertices.
2. Any set of 4 vertices must contain a pair that are adjacent. (Let $v_1$, $v_2$ be non-adjacent. If they are antipodal, then every other vertex is adjacent to one of them. If they are not antipodal, i.e. are at distance 2, then there are two vertices not adjacent to either of them; but those two are adjacent to each other.)
3. So now, we can enumerate all the sets of ≥4 coplanar vertices (up to symmetry) by picking an adjacent pair $v_1$, $v_2$, and looking at the planes containing this pair together with at least one other vertex. There are ten other vertices, which turn out to give five planes: two pentagons, each containing three other vertices besides $v_1$, $v_2$; a rectangle, containing their two antipodal points; and two triangles, the faces of which $v_2$ is an edge.
4. So, in sum: every intersection of diagonals lies either in one of the pentagons, or in a rectangle whose opposite vertices are antipodal.
5. The diagonals of these “antipodal rectangles” always intersect in the centre of the icosahedron — so there is only 1 intersection point of this form.
6. There are 12 pentagons (one around each vertex); drawing the diagonals of a pentagon, each one gives 5 diagonal intersections. However, we’ve overcounted: each such intersection lies in 3 pentagons(*). So the number of such intersections is $(12 \cdot 5)/3 = 20$.
7. Point (*) is a bit tricky to show. Consider a pentagon with vertices $(v_1,…,v_5)$ (ordered cyclically), and centered around $v_0$. A typical intersection point is now given by $x = (v_1v_3) \cap (v_5v_2)$. We need to find: what other pentagons does $x$ occur as an intersection in? The diagonal $(v_1v_3)$ lies in one other pentagon: the one centred around $v_2$. The diagonals of this pentagon will clearly divide $(v_1v_3)$ up in the same proportion as the diagonals of the first pentagon — so $(v_1v_3) \cap (v_5v_2)$ will indeed occur as an intersection of this pentagon again, specifically as $(v_1 v_3) \cap (v_0 w)$, where $w$ is some vertex of the new pentagon. So it will also occur in a third pentagon (the one centred on $v_1$) as $(v_5 v_2) \cap (v_0 w)$. But note that (by looking at the three pentagons it occurred in) $x$ is equidistant from $v_0$, $v_1$, and $v_2$, and closer to these than to any other vertices; so it cannot appear in any other pentagons than the ones centred on these three points.

(Can this last step be simplified somehow, perhaps?)