Italian Mathematical Olympiad, 2003

Solution:

The answer is 100. We need to count the (unordered) pairs of vertices that do not belong to the same face. At each vertex, 3 faces meet, and in each of them there are 2 vertices that are not on a face that also contains the initial vertex (total 6), plus 2 shared with another face (which are counted twice and are therefore actually 3 in total), plus the vertex itself: so we must exclude from the count 10 vertices that do not give rise to diagonals. But the vertices of the dodecahedron are $12 \cdot 5 / 3 = 20$ (12 pentagonal faces, but meeting 3 at a time), so for each of these we must consider $20 - 10 = 10$ possible vertices as the other endpoint of a diagonal. In total the diagonals are $20 \cdot 10 / 2 = 100$, where we divided by 2 since, by multiplying, each diagonal is counted twice.