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Topic: Deformable platonic "solids"
Replies: 23   Last Post: Mar 12, 2013 8:11 PM

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 David Bernier Posts: 3,683 Registered: 12/13/04
Re: Deformable platonic "solids"
Posted: Feb 27, 2013 9:44 PM

On 02/27/2013 04:02 PM, Frederick Williams wrote:
> Ken Pledger wrote:
>>
>> In article <512E5CBC.9738E289@btinternet.com>,
>> Frederick Williams <freddywilliams@btinternet.com> wrote:
>>

>>> Suppose the platonic solids aren't solid at all but are made of rigid
>>> line segments with completely flexible hinges at the vertices. The cube
>>> can be flattened into a... um... non cube. The tetrahedron, octahedron
>>> and icosahedron cannot be deformed at all. But what about the
>>> dodecahedron, can it be deformed?

>>
>> Here's an intuitive line of thought, not a complete proof.
>>
>> Starting from a face of the cube, the four adjacent edges are
>> parallel. That seems to be what permits the deformation. None of the
>> other regular polyhedra has parallel edges adjacent to a face, so I
>> suspect none can be deformed.
>>
>> Ken Pledger.

>
> Hmm... Suppose, instead of regular polyhedra we consider others some of
> the faces of which may be quadrilaterals which have no edges parallel.
> Such polyhedra may be deformable. So it seems to me that some property
> other than having parallel edges adjacent to a face is relevant. But
> thank you for considering the matter.
>
> I can imagine twisting a dodecahedron so that of two parallel faces one
> remains fixed while the other is turned about the axis that runs through
> the center of them both. If I was good with my hands I'd make a model.
>

I've been looking at the image model at Wikipedia here:
http://en.wikipedia.org/wiki/Dodecahedron

Suppose looking down we see the top pentagonal face T.
Opposite, at the lowest height, we see the botton pentagonal
face B.

Then the T and B faces lie in planes that are parallel to each
other.

Emanating from T downwards, we have five edges
e1, e2, e3, e4 and e5 that are incident on a vertex of
T but are not in the plane of T (they're all "lower").

The five edges e1, e2, e3, e4 and e5 terminate going down
at vertices V1, V2, V3, V4 and V5: these vertices are
vertices of the dodecahedron, but are not vertices of
the top face T.

Then V1, V2, V3, V4 and V5 lie in a plane and in the
rigid form in the beginning, before any deformation(s),
can be joined in a natural way by five dotted line segments
l1, l2, l3, l4 and l5 to produce a big pentagon P_t , t
for 'top'.

Ignoring everything below the plane through the big pentagon
P_t, we can imagine a bowl-like structure with a pentagonal
base as the face T, and the (rotated) edges
e1, e2, e3, e4 and e5 going upwards and joining with the
(rotated) big pentagon P_t:

\_____/ represents the bowl,

with the bottom part face T seen edge-on. The \ and / represent
just two of the five edges e1, e2, e3, e4 and e5.

The bowl-shape, with one pentagonal "plate" at bottom and
five non-regular quadrilateral "plates" and an "open" top,
really does look like a bowl, a soup-bowl.

The bowl accounts for 5+5 = 10 vertices of the dodecahedron.

Starting with the bottom face B, we can do a similar
procedure with the five lowest vertices (looking down on the figure)
http://en.wikipedia.org/wiki/Dodecahedron
higher up than the plane containing B and obtain
another bowl-shape. That bowl-shape contains 10 vertices
also.

The two bowls together account for all 20 vertices of the dodecahedron,
at 10 vertices each.

Each bowl accounts for 10 edges of the dodecahedron.
The dodecahedron has 30 edges, so with two bowls at
10 edges each, we're missing 30-(10+10) = 10 edges.
Those will be edges/wire joining the two bowls.

Initially, the two bowls face each other and perpendicular
to the bowl bases, are rotated 360/10 = 36 degrees one
bowl from the opposite bowl.

It occurs to me that the extra 10 edges form a zig-zagging
formation like this:

\/\/\/\/\/

where the left extremity of the leftmost '\' is glued to
the right extremity of the rightmost '/' .

So those 10 wire-pieces represent the 10 edges missing from
the two bowl-shapes. That way, we've accounted for all
30 edges.

Let's say your twisting idea can be done. Then how would the
10 edges that aren't part of the top and bottom faces T and B
move?

In my discussion, they are the 10 zig-zagging edges:
\/\/\/\/\/

David Bernier

--
dracut:/# lvm vgcfgrestore
File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID
993: sh
Please specify a *single* volume group to restore.