Platonic Solids--circumspheres

[Jenn]

I recently purchased little clear sphere ornaments in hopes of
figuring out how to construct the Platonic Solids inside to represent
their circumspheres....

My 9 year old and I did this when he was 5 with loads of help....I
have lost the directions! However, my son would love to do this
again. Daud Sutton's book on Platonic and Archimedean solids is his
all time favorite book...ever.

Is there anyone here who would be willing/able to help walk me through
how to get the edge value of the platonic solids to fit inside the
sphere.

I hope this all makes sense....

OR is there a better place to ask this question?

Thanks in advance~
Peace~ Jenn

[John Sharp]

Hi Jenn

This is a way to solve such problems and then this one in particular

1) go to Google and search for Platonic solid

2) Find something like the Wikipedia entry
http://en.wikipedia.org/wiki/Platonic_solid#Radii.2C_area.2C_and_volume

3) Scroll down until you see a table that has the properties you require, or click on the appropriate heading

4) you need     Radii, area, and volume

5) get to the table which says

The following table lists the various radii of the Platonic solids together with their surface area and volume. The overall size is fixed by taking the edge length, a, to be equal to 2.

6) you want the Circumradius (R)  so with a side = 2 for a cube, then the radius is

7) You have the info which you are now going to have to adjust to you spheres

I hope you can do the calculation from here. If not please ask for more help and provide the size of the sphere (you will probably measure the diameter - let us know how you are going to do that if you don't know it)

John S

PS don't forget to post a picture

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[Edward Cherlin]

On Wed, Jan 19, 2011 at 13:23, Jenn <jenniyo...@yahoo.com> wrote:
> I recently purchased little clear sphere ornaments in hopes of
> figuring out how to construct the Platonic Solids inside to represent
> their circumspheres....
>
> My 9 year old and I did this when he was 5 with loads of help....I
> have lost the directions!  However, my son would love to do this
> again.  Daud Sutton's book on Platonic and Archimedean solids is his
> all time favorite book...ever.

An excellent choice. Your son can enjoy the Renaissance artists'
renderings of regular solids in perspective, as explained by Luca
Pacioli, and can look forward to reading Kepler (who tried to fit the
Platonic solids to the geometry of the solar system), Coxeter on
higher-dimensional regular polytopes, and at a higher level Felix
Klein (who tied the icosahedron and solving fifth degree equations
together), among others.

> Is there anyone here who would be willing/able to help walk me through
> how to get the edge value of the platonic solids to fit inside the
> sphere.

Where is the center of a Platonic solid? How far from the vertices?
All of this is laid out in the 13th book of Euclid's elements, but
from the opposite direction. Given the radius of the sphere, he
constructs the edges of the solids. The first three are easy, the
other two a little more difficult.

Tetrahedron: 3/4 of the altitude, in much the same way that the
circumradius of an equilateral triangle is 2/3 of the altitude. That
is, we divide the tetrahedron into four equal pyramids, each with one
fourth of the volume, and therefore one fourth of the height.

Cube: If the side is 2, then by the Pythagorean theorem half of the
diagonal is sqrt(3)

Octahedron: Half of an octahedron is a square pyramid with equilateral
triangle faces. Apply Pythagoras to find the diagonal of the base,
which is the diameter of the circumsphere, and also twice the altitude
of the pyramid.

Dodecahedron: http://en.wikipedia.org/wiki/Dodecahedron
If the edge length of a regular dodecahedron is a, the radius of a
circumscribed sphere (one that touches the dodecahedron at all
vertices) is

r_u = \frac{a}{4} \left(\sqrt{15} +\sqrt{3}\right) \approx
1.401258538 \cdot a

The following Cartesian coordinates define the vertices of a
dodecahedron centered at the origin:

(±1, ±1, ±1)
(0, ±1/φ, ±φ)
(±1/φ, ±φ, 0)
(±φ, 0, ±1/φ)

where φ = (1+√5)/2 is the golden ratio (also written τ) = ~1.618. The
edge length is 2/φ = √5–1. The containing sphere has a radius of √3.

Icosahedron: http://en.wikipedia.org/wiki/Icosahedron
If the edge length of a regular icosahedron is a, the radius of a
circumscribed sphere (one that touches the icosahedron at all
vertices) is

r_u = \frac{a}{2} \sqrt{\varphi \sqrt{5}} = \frac{a}{4} \sqrt{10
+2\sqrt{5}} \approx 0.9510565163 \cdot a

The following Cartesian coordinates define the vertices of an
icosahedron with edge-length 2, centered at the origin:

(0, ±1, ±φ)
(±1, ±φ, 0)
(±φ, 0, ±1)

where φ = (1+√5)/2 is the golden ratio (also written τ). Note that
these vertices form five sets of three concentric, mutually orthogonal
golden rectangles, whose edges form Borromean rings.

> I hope this all makes sense....
>
> OR is there a better place to ask this question?
>
> Thanks in advance~
> Peace~ Jenn
>

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--
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Silent Thunder is my name, and Children are my nation.
The Cosmos is my dwelling place, the Truth my destination.
http://www.earthtreasury.org/

[kirby urner]

Off topic but another thing you might do with plastic spheres 

all the same size, like ping pong balls, is glue them together

into Platonics, like four make the self-dual tetrahedron.  Making

the two other dual pairs (cube-octa; icosa-dodeca) takes more

doing (like your icosa probably needs to be hollow).  The octa-

hedron ain't hard.  Twelve-around-1, all intertangent, in two 

different ways, gets you off on a lot of topics, is a grand central 

station of math threads (a STEM bonanza).

Kirby

[Maria Droujkova]

While you are at it, draw some lines on a sphere with a permanent marker, find a strong flashlight and see if you can't reproduce this thing of beauty in physical space (Mobius Transformation):
http://www.youtube.com/watch?v=JX3VmDgiFnY

Cheers,
Maria Droujkova

Make math your own, to make your own math.

[jennifer kurtz]

Thanks for all the wonderful suggestions~   We will start constructing as soon as we are done with the flu.   Peace~ Jenn   I will post pics when the circumspheres are finished.   PS Maria--the Mobius strip youtube link has been a favorite of my son's for quite some time...that one and there is one about turning a sphere inside out that fascinates him as well.

[Alexander Bogomolny]

Jenn,

 

if you boy likes playing with the Mobius strip, he may like the applet at

 

http://www.cut-the-knot.org/Curriculum/Geometry/Moebius.shtml

 

There are several parameters available to play with, the number of half turns is just one. You may create the strip from several layers of rectangles and then remove rectangles one at a time, simulating cutting the strip any other way.

 

Alexander Bogomolny

[John Sharp]

Alex

It looks like you are generating the strip by rotating a line about a circle.

You may not realise this, but this method gives a surface which cannot be developed (ie cut and opened into a flat strip).
It is a ruled surface but the lines do not meet on a curve.

The only reference I know to this (even in books)  is in French at http://www.mathcurve.com/surfaces/mobius/mobius.shtml

John S

[Alexander Bogomolny]

John,

 

Yes, of course, this is how it comes around. I understand the first part of your remark (that it can't be deveoped), but not the second part. What lines do not meet on what curve?

 

I have just finished an applet for the auxetic tessellation - thanks for pointing this out to me. Now I have to put it on a page. I'll check the French site when done.

 

Alex

[Alexander Bogomolny]

Ah, I understand what you are saying.

 

It would have been possible to compose the surface of triangles and not of rectangles. I agree. The effect would be different, though, and handling/playing with the surface would be less intuitive.

 

Alex

On Fri, Jan 21, 2011 at 12:48 PM, John Sharp <js.sli...@gmail.com> wrote:

[John Sharp]

I wasn't suggesting that you should make it triangles because your applet is designed to show how a Mobius strip looks.

But if you do, then it is possible to unfold. It is a slightly different surface.

There is a wonderful program that allows you to unfold such objects, called Pepakura

http://www.tamasoft.co.jp/pepakura-en/

I have attached an image of the result, which is a curve not a straight strip.

It would make a different cut-the-knot entry though

John S

PS someone mentioned a YouTube Video which shows a Mobius Transformation. This is mathematics from Mobius but not related to the Mobius strip

[Alexander Bogomolny]

John,

 

I only mentioned a triangulation just to show that I understood what you were saying. The surface would be different, subject to flattening, but would create a visual complication. 

 

http://www.tamasoft.co.jp/pepakura-en/

 

This is another great link. Thank you.

 

Alex

 

P.S. Moebius' name is attached to quite a few mathematical objects, not all related. A Moebius transform is nicely explained with the Riemann sphere. And so is the inversion, albeit in a less spectacular manner:

 

http://www.cut-the-knot.org/pythagoras/StereoProAndInversion.shtml

[Sue VanHattum]

Hi Jenn,

Here's a great coincidence! Pat Bellew just wrote about this topic. The thing that puzzled him is that a dodecahedron hugs the sphere most closely when the Platonic solid is inside the sphere (like you're doing). But when you have the sphere inside the Platonic solid, then the icosahedron hugs it most closely.

Warmly,
Sue

[kirby urner]

The Platonics are a wonderful starting point.  You've got the three dual pairs, with the tetrahedron self-dual.  When you combine a shape with its dual, by crossing edges, you get more shapes.

The tetrahedrons combine to keep it in the family of Platonics:  tetra + tetra = cube.

Then it's cube + octa = rhombic dodeca

And icosa + pentagonal dodeca = rhombic triacontahedron.

The latter is more spherical yet.

If you dare to make the tetrahedron your unit of volume then you have the following Verboten Math table:

Tetrahedron: 1
Cube: 3
Octahedron: 4

Rhombic dodecahedron: 6

Icosahedron:  5  phi^2  2^(1/2)

Rhombic triacontahedron: 15 2^(1/2)

Cuboctahedron: 20

The last shape is the dual of the rhombic dodecahedron and is here sized

to have its 12 vertexes coincide with the centers of closest packed spheres

jammed around a nuclear sphere, ala the cubic closest packing.  Brad 

knows about this.

Anchoring the polyhedrons to a sphere, perhaps unit radius, has been a 

way to size them relative to one another, with in-sphere and circum-sphere

two obvious options.

The above volumes table, with the unit volume tetrahedron, is likewise 

anchored to spheres in that the rhombic dodecahedron is circum-sphere

whereas the tetrahedron is made from four sphere packed together.

Spheres definitely stay in this picture.

Interesting that the tetravolume 5 rhombic triacontahedron (not listed

above) is extremely close to being cirum-spherical.  Its radius is ~0.9995 

versus 1.0.

Using "tetravolumes" is of course non-standard.  Our ethnicity, a 

small minority, has to do homeschooling to supplement.  We were 

happy when the NCTM touched on our curriculum, at least a little **,

but for the most part we're strangers in a strange land.

Thanks to the diversity protections in our system of government, 

we're allowed to keep promulgating these radical alternative 

curricula, including with Python for an executable math 

notation (MN).

Kirby

** http://bfi.org/news-events/community-content/nctm-embraces-synergetics