
Platonic Solids - Why Five? - ColinWright
http://www.mathsisfun.com/geometry/platonic-solids-why-five.html
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gregfjohnson
As the article points out, it is easy to see that there are at most five
platonic solids in 3 dimensions. One thing I've wondered about is why all five
possibilities actually work.

As an example, consider the icosahedron. You could imagine trying to build one
as follows. Take five equilateral triangles, and attach them together so that
they meet at one vertex and the respective sides adjacent to the vertices are
shared. This gives you a sort of cap-like shape consisting of the five
triangles.

How, attach more triangles to the cap as above. This will create a strip of 10
triangles around the "equator".

Then, keep adding more triangles. This step will end up creating another cap
at the bottom of the icosahedron.

The question is, why do things inevitably match up on the bottom if you fill
things in this way? You can do the same construction with 3, 4, or 5
triangles, 3 squares, or 3 pentagons, and in all cases things line up exactly
to give you a complete polyhedron.

~~~
T-hawk
One answer: Things don't inevitably match up on the bottom. The tetrahedron is
the counterexample. Constructing it based around one vertex as you describe
(call it the north pole) does not result in a matching cap around the south
pole. Of course, the gap around the south pole is still congruent to the three
original triangles. It must be by the rules of Platonic solids, because the
other three vertices must each behave identically to the original north pole,
meaning the south-pole-centered triangle must be identical to the other
triangles.

Here's a more general answer. The solid always has N-way radial symmetry when
viewed from above any vertex, such as our north pole. All longitude lines at
360°/N intervals are identical from this method of construction. There is
nothing to differentiate the triangle of your icosahedron on the 0° meridian
from its counterpart at 72°, so they will all fall into the same relationship
with the south pole. The polygons can't overlap the south pole from one
direction while falling short from another. In the tetrahedron, the south pole
gets overlapped equally from all three radial directions. In the other four
solids, the edges and faces meet exactly at the south pole.

I can't answer exactly why that last sentence is true. It reduces to, why does
this construction method always produce symmetry across the equator for the
four non-tetrahedron solids? Your strip of 10 triangles on the icosahedron: if
that is indeed equatorial, then we have both latitudinal and longitudinal
symmetry which renders your north and south hemispheres identical so things
will match up. But why does that strip of 10 triangles end up centered on the
equator? I do not have the answer to that, but maybe someone does.

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gus_massa
In 4D there are 6, the five expected ones and an additional strange version of
the octahedron.

In 5D and more, there are only 3, the equivalents of the tetrahedron, cube and
octahedron.

~~~
stephencanon
I wouldn't really call the 24-cell "a strange version" of the octahedron. It
really isn't anything like any of the 3d Platonic solids. (It does have
octahedral faces, though.)

~~~
oscilloscope
Here's a 24-cell you can explore:

[http://fleetinbeing.net/hypersolid/examples/24cell.html](http://fleetinbeing.net/hypersolid/examples/24cell.html)

It doesn't bring out the cells or faces, but it does show edges in parallel
coordinates. Edges show up as multi-points in the plot below. Above is a 3-d
projection of the 24-cell which you can click+drag to rotate.

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ravich2_7183
Thanks for posting.

Now I get why Kepler was hell bent on trying to fit the orbits of the 5 known
planets (at his time) with the 5 platonic solids [1]. This part of Kepler's
life is very nicely depicted in Carl Sagan's Cosmos [2]. It seemed interesting
when I watched it, but I didn't give it much thought back then.

[1]
[http://en.wikipedia.org/wiki/Johannes_Kepler#Mysterium_Cosmo...](http://en.wikipedia.org/wiki/Johannes_Kepler#Mysterium_Cosmographicum)

[2]
[https://www.youtube.com/watch?list=PLBA8DC67D52968201&featur...](https://www.youtube.com/watch?list=PLBA8DC67D52968201&feature=player_detailpage&v=ICHDl0olsS0#t=2007)

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jonsen
You could say the sixth is an infinitely large one made of regular hexagons.

~~~
T-hawk
By that method, there's eight. Also included would be the constructions of
regular triangles meeting six at a vertex, and regular squares meeting four at
a vertex.

~~~
nickzoic
It occurs to me that (per the second argument) these have Schläfli symbol
{s,m} = {6,3}, {3,6} and {4,4} ...

for each of these, working out

    
    
        1/s + 1/m - 1/2 = 1/E
    

gives

    
    
        1/E = 0
    

... which makes sense (kind of) since those three cases are infinite regular
tilings instead of regular polyhedra ...

(EDIT: Just noticed jonsen's comment here too, "infinihedron" suddenly makes
sense to me :-) )

