Shortest distance on the surface of a dodecahedron

The following animation illustrates unfolding of the faces of the dodecahedron.

The last frame from above is repeated below with the distance $d$ as one side of the triangle, the other two sides are of lengths $2 h$ and $1$ where $h$ is the altitude of the pentagonal face, and the angle between these two sides is $54^{\circ} + 108^{\circ}$ .

The altitude of a unit edge length pentagon is $h = \sin(108^{\circ} ) + \sin(108^{\circ} - 72^{\circ} ) = 1.538841769$

Applying the Law of Cosines, the required distance is

$d = \sqrt{ (2 h)^2 + 1 - 4 h \cos(54^{\circ} + 108^{\circ} )} = 4.040573959$ , which makes the answer $\boxed{4040}$ .