Is this a new projection of a 4D cube ??

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Is this a new projection of a 4D cube ??

Postby Firebird » Mon Sep 05, 2005 6:13 am

Hi Everyone
I am a total amatuer in terms of 'real' mathematics, and totally in awe of some of the people who post on this board :shock:
As a teenager though (nearly 30 years ago) I came up with what I think is a 3d projection of the hypercube that I have yet to see in any books or discussions on the subject. Could I have discovered something?
I would love to have your considered opinions on it.
I have recently had it made into a piece of jewelry that is for sale.
You can check it out here http://www.sacredpatterns.com/hypercube
and please dont miss the stereo projections of it here http://www.sacredpatterns.com/hypercube ... rStuff.htm

Thanks
Firebird
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Postby wendy » Mon Sep 05, 2005 11:23 am

It's a well known projection of the tesseract, using isometric projections. If you look around the list, you will see that tetraspace people (ie people wiþ more than 256 posts, have it as an icon.
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Postby Firebird » Mon Sep 05, 2005 10:44 pm

Uhh... Hi Wendy... You are one of the people I am most :shock: about - so Im almost scared to ask but...
Yes the icon is the same 2D projection but my 3D projection has features I have never seen anywhere...
First, all 8 cubes are identically distorted. At every vertex all 4 lines meet at the same angles to every other vertex.
The cost of this is that some of the 'squares' are bent into 3 dimensions.

:cry: Is this really a common well known 3D projection? :oops: I have read every bit of info I can find and never stumbled across any mention of it. They usually have 2 proper cubes connected by lines between their points.

Also, the animated online ones never 3 dimensionalise in this way. Please refer me to another description or perspective drawing of it if you know of one, on this board or elsewhere. :arrow:
Thanks
Firebird

PS where does one go to learn all the terminology that you people use for higher space figures? Is there a hyper-glossary?
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Postby wendy » Mon Sep 05, 2005 11:02 pm

Don't be too afraid. I'm just learning too.

There are a couple of glossaries etc around. Most of these deal with good old 4d euclidean geometry.

George Olshevsky maintains a multi-dimensional glossary at his web site. It's pretty much what the convention is, and it has a few pictures. Much of his site is pretty good too.

http://members.aol.com/Polycell/glossary.html

Jonathan Bowers has discovered all of the stary uniforms in 4d, he has lots of pretty pictures as well.

http://hometown.aol.com/hedrondude/polychora.html

You could always forrage around the geometry-junkyard. It is a facinating site, where people go to get to other sites. It's pretty much a link-farm with descriptions.

http://www.ics.uci.edu/~eppstein/junkyard/

Other folk play around with stranger things, too.

I tend to wander around the more bisarre things, like hyperspace. ye find my page at

http://www.geocities.com/os2fan2/index.html

There's a glossary there and some pretty old stuff i have not rewrote as yet, but it's worth the look.
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Postby wendy » Mon Sep 05, 2005 11:10 pm

Regarding the three-dimensional projection.

I popped over and had a look at your site.

The projection that you use is described in Coxeter's "regular polytopes" from way back when ('37 or '48 or something). He talks figures line zonohedra, which are isometric projections of 4d and higher into polyhedra. But he is more interested in the ones that have highly regular symmetry (eg the rhombic tricontahedron is a projection of a 15-dimensional cube).

The book is widely held to be a classic, and it's got some heavy parts if you don't either have mathematics or some models to maul. But many people would get benefit out of at least the early chapters.

Author: HSM Coxeter
Title: Regular Polytopes
Publisher: Dover

Dover does reprints of older classics for relatively cheap prices. I picked this one when i was out on the moon one time, so it should be fairly easy to get new or used.

My own notation is more a reaction to the pibald notation Coxeter uses, but then i went a tad further in some areas....

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Postby Marek14 » Tue Sep 06, 2005 5:36 am

Ah, this projection :) I remember I stumbled upon it when I was at high school.

Sadly, I never managed to put a 5D cube into a decagon :(
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Postby Firebird » Tue Sep 06, 2005 5:53 am

Since everyone seems to already know about my great discovery :( maybe someone will know about another similar shape.

I suspect there must be a 5D cube that has a projection into 2D that is bounded by a decagon (I hope thats the right word for a 10 sided polygon).

Is there a neat symetrical version of this somewhere on the web?

When I've done enough posts will I get one as my icon? :)

Cheers
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Postby wendy » Tue Sep 06, 2005 10:39 am

Seems you go up a dimension every power of 4 completed. So your next post ought make you a line citizen,

None the same, i am still impressed you and Marek (and a few others found it whike ye were so young.

I actually described the 24-cell when i was 20, but totally missed it. :(

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Postby jinydu » Tue Sep 06, 2005 1:44 pm

wendy wrote:Seems you go up a dimension every power of 4 completed. So your next post ought make you a line citizen,


Indeed (well, almost)

1-3 pointspace citizen (0D)

4-15 linespace citizen (1D)

16-63 planespace citizen (2D)

64-255 realmspace citizen (3D)

256- 1023 tetraspace visitor (4D)

>= 1024 tetraspace citizen (4D)

(a long road ahead for me...)
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5D cube as decagon

Postby Firebird » Tue Sep 06, 2005 11:40 pm

Hi
Thanks Wendy for the above info.
I will order Coxeter's "regular polytopes" from Amazon second hand for about $6 :-)

I would love to find the 5D cube decagon.

Another figure I played around with long ago was doing stereo drawings of progressively truncating the hypercube at the centre point of its edges.
After about 4 truncations you end up with 8 visually spherical figures and the whole thing looks very toroidal.
Is there anywhere here to post graphics? I guess I could put up a page on my own site and just link to it.

Do you work in the field of computers or maths?

Cheers
Firebird
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Postby wendy » Wed Sep 07, 2005 3:41 am

I don't work in computers or maths. It's just a hobby, like knitting or astrology.

On the other hand it is let loose, because i wanted to get a bit of þ on the net, and learn some html, so i needed filler-text.

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Postby Marek14 » Wed Sep 07, 2005 5:48 am

I see the problem in the fact that 4D cube has 16 vertices - that is 8 vertices of octagon and another 8 of the inner octagon. But if one would try to inscribe 5D cube into decagon, he would have to fit 22 points into it - and that's just not the same symmetry :)

Maybe a 16-gon would work, though...

In general, I tried to find out the "smallest 5D cube that can be well-drawn", but I didn't manage to do so yet. My criteria for "well-drawing" are:

1. Each of the five directions (x,y,z,w,v) must be translated in a 2D vector with integer components. (in case of 4D, the symmetrical picture is produced with vectors (2,1),(2,-1),(1,2), and (1,-2). The version which uses regular octagon looks better, but is hard to draw on checkered paper)
2. Every possible sum of these vectors (with 0 or 1 at each) must yield a distinct point.
3. Parallel lines connecting the points shouldn't merge.
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Postby wendy » Wed Sep 07, 2005 6:26 am

16-gon won't work. any one can see that. (it has 8 sets of parallels, while the penteract has only five.

the thing quite happily exists in the pentagon, but i suppose that you are not interested in lots of shared incidences.

next one to turn up discrete is octacube (where 256 vertices squeeze into þe sixteen-gon) So you could pick out one of the inscribed {4,3,3,3}.

I mean, you could take an ordinary {4,3,3}, and add an 16-gon segment to it: that's pretty much what it does anyway.

Ie {4,3,3} is made out of the 4 vectors corresponding to 0, 45, 90, 135 degree rays from the centre. All you need is any binary angle, and you will _never_ get co-incident vertices, eg

0 45 67.5 90, 135

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Postby Firebird » Wed Sep 07, 2005 8:15 am

Hi Marek14
You are inspiring me. I think Im on to something. I have a first sketch that is getting scarrilly close. Give me a couple of days and I'll do a proper computer sketch of it.

Where do I go to learn how this numbering convention works? {1,2,3,4,etc} :oops:

I cant find a frequently asked questions for this board except the one about how to use the board itself so I'll have to keep asking these embarrassingly dumb questions :(

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Postby Firebird » Wed Sep 07, 2005 10:13 am

Okay I got carried away and couldnt stop...
http://www.sacredpatterns.com/hypercube/5D.htm

Im sure this has been done before as its pretty simple and obvious.

Anyone like to say if its really a projection of the 5D cube?

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Postby wendy » Thu Sep 08, 2005 1:24 am

It's missing a whole mob of lines: there should be 32 vertices, 80 lines, 5 lines at a vertex.

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Postby Marek14 » Thu Sep 08, 2005 6:06 am

While Wendy is right and one wouldn't get duplicate points from the sets she is proposing, my approach works only with vectors you can draw in square lattice - which ensures that any two vertices will be always at least 1 unit apart.

Your picture on your page is definitely wrong, though, as some of the points on the circumference have 4 edges, and some have 6. Not only 5D cube (penteract) has five edges at every vertex, but NO 5D polytope can have 4 edges at a vertex.

As for the counting, it goes like this:

We start with [0,0] and use vectors x=[1,2], y=[2,1], z=[2,-1], and w=[1,-2].

So, the points will be:

1 origin [0,0]

4 vertices of first order: [1,2], [2,1], [2,-1], [1,-2]

6 vertices of second order that combine two vectors: [3,3], [3,1] [2,0] [4,0], [3,-1], [3,-3]

4 vertices of third order that combine three vectors: [5,2], [4,1], [4,-1], and [5,-2]

1 vertex which combines all four and closes the thing up: [6,0]
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Postby Marek14 » Thu Sep 08, 2005 6:50 am

Update: I have produced an image of penteract from regular decagon. It was telling. This thing has mostly distinct vertices (except for two coincident ones in the middle of the picture), but many edges lie in the same line.

How can I send it here? I don't have webspace...[/img]
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Postby wendy » Thu Sep 08, 2005 7:34 am

Marek, i believe, drew the correct thing. It basically has an extra decagon of vertices to the one on firebird's page.

still.

don't be too afraid of dumb questions, they're really intellegent issues poorly phrased. Most people don't see past the phrasing.

Basically, the penteract has five sets of parallels, and you can only use each line once. For all polygons except the powers of 2, the set of parallels for the edges lead to the centre-point: it can't be helped.

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Postby wendy » Thu Sep 08, 2005 7:41 am

The numbering that Marek used, eg {4,3,3,3} is the Schläfli symbol. It works like this:

{p} is a polygon with p sides

{p,q} has faces {p}, q at a corner. A cube is {4,3}

{p,q,r} is a polychoron, with faces {p,q}, r at an edge. {4,3,3} is a tesseract.

{p,q,r,s} is a polyteron, (5d) with faces {p,q,r}, s at a margin.

and so on.

The platonics are {3,3}, {3,4}, {4,3}, {3,5}, {5,3}. The order is increasing vertex count and increasing volume.

In 4d, we have {3,3,3}, {3,3,4}, {4,3,3}, {3,4,3}, {3,3,5}, {5,3,3}.

In 5d, {3,3,3,3}, {3,3,3,4}, {4,3,3,3}

and thereafter, add a {3} to the string of threes.

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Postby Keiji » Thu Sep 08, 2005 10:05 am

wendy wrote:{p,q} has faces {p}, q at a corner. A cube is {4,3}


A cube has 6 faces... :?
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Postby Marek14 » Thu Sep 08, 2005 5:09 pm

Although cube has six faces, they are squares (therefore the 4), and there are three squares meeting at every vertex of the cube (therefore the 3).

One way to look at Schlafli symbol is to understand it in term of angles. {4,3} would mean (square with vertex angle 360/3). This is 120-degree square which exists in spherical geometry. You can fit six of them on the surface of sphere. What we call "cube" is the same thing but with circle arcs replaced by straight lines.

Let's look at the possibilities: we take the triangles first.

{3,3} is 120-degree triangle, and you can cover the whole sphere with just 4 of these - thus, tetrahedron. As you decrease the angle down to 90, you can cover it with eight triangles, leading to octahedron. Shrinking further, we get to 72 degrees, and {3,5} covering - or icosahedron. If we go even further, then {3,6} are common 60-degree Euclidean triangles. {3,6} is, therefore, covering of the plane.
If we reduce the angle even more, we get into hyperbolic realm. The edge of triangle grows as the angle shrinks. The limit is {3,oo} covering with infinite triangles that have 0-degree angles.

Analogically, we find that 120-degree squares cover sphere, 90-degree ones cover plane and anything more covers H-plane. {5,3} is spheric (dodecahedron), while {5,4} already passes over 108-degree mark of Euclidean pentagons and is well into hyperbolic realm. {6,3} is Euclidean, while {6,4} and anything {m,n} with m>6 is hyperbolic.

Now, we can apply it in four dimensions too, if we treat the first two numbers as a polyhedron, and the last one as its "dihedral angle" - the angle between neighbouring faces.

For example, the Euclidean regular tetrahedron has dihedral angle around 70 degrees - the exact value is not important here, just that it's more than 60 and less than 72. This means that {3,3,3}, {3,3,4}, and {3,3,5} are all spherical tetrahedra which tile the hypersphere, and they can be made into polytopes.
If we start to shrink the dihedral angle of tetrahedron, we find that it can't be shrunk down to zero - not even in hyperbolic space. The limit is 60 degrees here - corresponding to {3,3,6} which is made up of infinitely large tetrahedra (but still with finite volume).

The same way, Euclidean cube has dihedral angle 90 degrees - 120-degree cube is part of {4,3,3} - tesseract, while 72-degree cube {4,3,5} tiles H3 space and {4,3,6} is infinite. 120-degree dodecahedron is spherical dodecahedron which is the cell of 120-cell {5,3,3}. {5,3,4} and {5,3,5}, made up of 90-degree and 72-degree dodecahedra, are hyperbolic tilings, {5,3,6} is once again infinite.

{3,4,3} is icositetrachoron, or a spherical tiling made up of 120-degree octahedra. 90-degree ones from {3,4,4} are hopelessly infinite, though.

{3,5,3} is already hyperbolic, as the dihedral angle of Euclidean icosahedron is greater than 120 degrees. You won't even get to 90 here by the time you hit infinity.

Shrinking dihedral angles of infinite constructions like {6,3} is also interesting, but beyond the scope of this topic.

The only interesting dimension remaining is 5 - it has three spherical tilings (and therefore three regular polytopes) {3,3,3,3}, {3,3,3,4} and {4,3,3,3}, but it also has three tilings of E4 - {3,3,4,3}, {3,4,3,3}, and {4,3,3,4}, five tilings of H4 - {3,3,3,5}, {4,3,3,5}, {5,3,3,3}, {5,3,3,4}, and {5,3,3,5} and a single tiling of H4 with infinite tiles - {3,4,3,4}.
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Postby Keiji » Thu Sep 08, 2005 5:34 pm

Marek14 wrote:Although cube has six faces, they are squares (therefore the 4), and there are three squares meeting at every vertex of the cube (therefore the 3).


Gotcha.

If we reduce the angle even more, we get into hyperbolic realm. The edge of triangle grows as the angle shrinks. The limit is {3,oo} covering with infinite triangles that have 0-degree angles.


What exactly is a hyperbolic realm? It's impossible to have a {3,7} thing... you just can't fit seven triangles round a point.
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Postby Firebird » Fri Sep 09, 2005 2:17 am

Hi Again
:D Okay this time Im pretty sure Ive got a proper 5D cube with no lines obscured, though after that embarassing last attempt I dont suppose anyone will believe me . I must have been tired and stressed out and very :?
Anyway heres the new one.
http://www.sacredpatterns.com/hypercube/5D.htm

Thanks for all the above explaination on the vertex numbering system.

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Postby wendy » Fri Sep 09, 2005 3:49 am

I'll bite.

The connectivity is correct, but the edges ought be parallel in 5 sets of 16. Some of the squares are a tad hard to follow.

I suppose the way around this is to distort the pentagon by shortning one of the sides or something. This will move things off-centre.

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Postby Marek14 » Fri Sep 09, 2005 5:18 am

iNVERTED wrote:
Marek14 wrote:Although cube has six faces, they are squares (therefore the 4), and there are three squares meeting at every vertex of the cube (therefore the 3).


Gotcha.

If we reduce the angle even more, we get into hyperbolic realm. The edge of triangle grows as the angle shrinks. The limit is {3,oo} covering with infinite triangles that have 0-degree angles.


What exactly is a hyperbolic realm? It's impossible to have a {3,7} thing... you just can't fit seven triangles round a point.


The important thing to realize here is that Euclid's fifth postulate is just a convention :)

The Euclidean world we live in (or if it's not Euclidean, then it's so close that there is no obvious difference) demands that sum of angles in the triangle is 180 degrees. We say that Euclidean realm has zero curvature. Other worlds, though, don't have this limitation.

One of them is a spheric world. It's a well-known fact that triangle drawn on the surface of a sphere has sum of angles greater than 180 degrees. Moreover, there is a relation between this sum and the area of triangle. If two triangles on sphere have the same angles, they must have the same edges, too.

Now, realize this: While it's easy (and done frequently) to picture a world with positive curvature in our realm, it's impossible to show Euclidean plane correctly in spheric 3D space. That is because you can only immerse surfaces with greater curvature than your own into your space.

Which is a problem, because there are also hyperbolic spaces whose curvature is NEGATIVE. It's not possible to immerse these realms in our space without distortion.

The basic rules of hyperbolic space are wretchedly inverted rules of spheric geometry, and they are as follows:

1. Sum of angles in a triangle is always smaller than 180 degrees. This is why we can have fit seven triangles around a vertex - the triangle's angles are 51 3/7 degree instead of sixty, and so it fits.

2. There is a relation between angle sum and area. However, in hyperbolic realm, the area grows as the angle sum SHRINKS. One weird result is that even when you shrink the angle sum all the way to zero, putting the vertices at infinity, the area will stay finite - which means that there is an UPPER LIMIT of area of hyperbolic triangle which can be never surpassed.

If you have a line and two lines perpendicular to it, then those lines are parallel in Euclidean world - but they eventually cross in spheric world, and in hyperbolic world, they diverge, as if more and more space was "born" between them as you follow them.

In hyperbolic 3D space, or H3, you can have spheres, but you can also have HOROSPHERES - spheres of infinite size, in effect, where the surface geometry becomes Euclidean. That's because zero curvature is larger than a negative curvature, and so curvature-0 surface can be immersed in hyperbolic realm. Then you have a scale of "pseudospheres" which have hyperbolic surface geometry. A certain triangle's edges will be larger when drawn on a pseudosphere than if you drew it in plane. Compare it to spheric world, where the triangle drawn in plane is LARGER then any triangle with same angles drawn on a surface of smaller sphere. (In other words, a hyperbolic architect couldn't draw smaller versions of his plans.)

There also horohedra and pseudohedra, which you may imagine as follows:

What happens when you put six triangles to a vertex when each triangle's inner angle is LESS than 60 degrees? You get a thing that is non-planar, i.e. curved, yet has the same topology as {3,6} planar tiling. This is a horohedron. Pseudohedron results when you do thing like {3,7} with triangles of angle 45, making it once again curved. But because 45-degree triangle is larger than 51 3/7-degree one, the edge of the pseudogon is always larger than if you just drew the tiling in plane.

How to picture hyperbolic plane? There are several standard representations, for example the Poincare disc.
This representation takes a unit circle and claims that hyperbolic plane H2 are all points within the circle, not including the boundary. The hyperbolic straight lines are straight line segments and circle arcs that cross the boundary in two points and are right-angled.

There are several programs for visualizing hyperbolic geometry that use this representation, for example Tyler (accessible from http://www.superliminal.com/geometry/geometry.htm)
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Postby Keiji » Fri Sep 09, 2005 10:16 am

Ok, I understand all that now, but...

Marek14 wrote:2. There is a relation between angle sum and area. However, in hyperbolic realm, the area grows as the angle sum SHRINKS. One weird result is that even when you shrink the angle sum all the way to zero, putting the vertices at infinity, the area will stay finite - which means that there is an UPPER LIMIT of area of hyperbolic triangle which can be never surpassed.


The upper limit is a constant? Or does it depend on the curvature of the realm?

http://www.superliminal.com/geometry/geometry.htm


hmm, interesting! I like the hyperbolic planar tesselations ;)
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Postby wendy » Fri Sep 09, 2005 12:54 pm

In the spheric plane, the excess of angle in a polygon is equal to its area. For example, consider the pentagon of the dodecahedron.

We have vertex angle = 1/3 circle = 120 deg

this exceeds the pentagon (108) by 12 degrees.

Pentagon has 5 vertices, so 5*12 = 60 degrees excess.

Now, there are 720 degrees excess on the sphere, so we have 720/60 = 12 faces.

It works in all geometries, but for the euclidean, there is no defect/excess, and so the face-areas (relative to all-space) is zero (ie 1/oo).

There is indeed an upper limit on the area of every polygon, equal to (n-2)*180 degrees defect.

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Postby Marek14 » Fri Sep 09, 2005 4:25 pm

iNVERTED wrote:Ok, I understand all that now, but...

Marek14 wrote:2. There is a relation between angle sum and area. However, in hyperbolic realm, the area grows as the angle sum SHRINKS. One weird result is that even when you shrink the angle sum all the way to zero, putting the vertices at infinity, the area will stay finite - which means that there is an UPPER LIMIT of area of hyperbolic triangle which can be never surpassed.


The upper limit is a constant? Or does it depend on the curvature of the realm?


The upper limit depends on chosen area units. The curvature is a matter of scale (there is no difference between small triangle on small sphere and big triangle on a big sphere). But the area of hyperbolic triangle grows linearly with its defect - i.e. (180-angles sum). You could say, for example, that triangle with defect 1 degree has area 1. Then, the maximum possible area would be 180. Triangle with defect 90 (for example equilateral triangle with inner angle 30 degrees, that is a triangle of {3,12} tiling) will be exactly half as large as the maximum triangle.

Note, however, that many laws familiar for Euclidean beings doesn't hold. Like the simple notion that doubling the size of figure quadruples its area. In fact, exact doubling of most figures is impossible because changing the size in hyperbolic space also changes the shape. As the angle sum of triangle goes to zero, its sides grow to infinity, but the area stays within the limit.

Another thing is that circumference of hyperbolic circle is not equal to 2 pi r - it's larger: the bigger r, the larger is the factor we must multiply it with.
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Postby Keiji » Fri Sep 09, 2005 9:47 pm

As the angle sum of triangle goes to zero, its sides grow to infinity, but the area stays within the limit.


If the sides are infinitely long, and the area stays finite, that means that there must be an infinate length of each side where the distance to the adjacent side is 1/inf (I use 1/inf instead of 0 to mean an infinately small number greater than zero), which makes the area undefined...
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