**Wei-Hwa's Pentiamonds**

Andrew Clarke

Brendan Owens has made a 3D image of these.

http://www.egroups.com/files/polyforms/Brendan/icosahedron1.wrl

You will need to be able to display vitual reality files there is some info
here.

http://www.georgehart.com/virtual-polyhedra/vrml-info.html

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.