Brendan Owens has made a 3D image of these.
You will need to be able to display vitual reality files there is some info here.
I found four by hand, using a procedure
that is designed to be
exhaustive. But my track record on such hand work is not terribly
good, so I disparage the results with a grain of salt. Here are
nets of the ones I found:
/\ /\ \ \
/ \/ \ \ \
\ / / / /\
_______\____/_______ / /___________/ \
/ / \ / \ / \ \
/ /__________\ / \/__________\_______\
\ \ / / / \
/\ \ \ /___/ / \
____/ \ \ / ________ ____ \ /\ /
\ / /\ \/ \ / \ \ \/ \/
\ /_______/ \____ \ /______\ \____
\ / \ \ / / \ /
/ \ / \
/ \ / \
\ /\ / \ /\ /
\/ \/ \/ \/
The pentiamond that closes on itself can of course be rotated to any of five positions, so we might count these as twenty solutions. Dan Hoey
Ugo Di Girolamo has found 20 solutions.
I'm new to pentiamonds, so I will refer to this figure: I think that you can make 2400 different icosahedrons using pentiamonds.
However, some of them are obtained one from the other with rotations and mirrorings, so that only 20 can be considered the primitive ones.
Consider the pentiamonds A: there is only one way in which it can be arranged on an icosahedron, which is closing it to form the upper part of a pentagonal pyramid. So we can put A on an icosahedron in 12 (number of vertices) times 5 (number of sides incoming in the vertex, one of which is to be cut).
The remaining part of the icosahedron is this:
Then we can see that D cannot have as a vertex the one opposed to the one chosen for A. It is easy to show that if D has as vertex the one opposed to the one chosen for A, this vertex belongs to 3 of its triangles. Then the figures below show that it is impossible. Impossible! needs two D! Impossible! red part is not a pentiamond!
Then D lies in the part that separates A from the opposite pentagonal pyramid. It is a close stripe of ten triangles, five with a side in
common with A and five without. So D can have three sides in common with A (in five different ways) or two (yet in in five different ways). In the first case, we can arrange the remaining two pentiamonds in six different ways: the remaining region has this shape, and the big triangle in B cannot be in the center: it would leave two unconnected regions. For each of the three possible positions and the big triangle in B cannot be in the center: it would leave two unconnected regions. For each of the three possible positions of that
triangle, both positions of A lead to a valid solution. So we have 6*5*12*5=1800 solutions of this type. In the second case for D, the remaining region is this: Observing that the numbered vertices are the only ones in which are coming at least 4 triangles, we can easily show that B & C can be located in only two different ways (the one evident from the figure, and the symmetric (respect the plane per 2-4 and the center of the icosahedron) one! So we have other 2*5*12*5=600 solutions, which yield the total of 2400
solutions. If we don't want solutions that can be obtained one for the other with a rigid movement of the icosahedron, it's enough to fix the vertex and the side for A, obtaining 40 different solutions. Moreover, to cut away the symmetric ones, we can notice that
any solution has a different one which is its mirroring along the plane for the cut-side of A and the center of the icosahedron. So the number is reduced to 20.