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aero_guy
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 Re: Intersecting Spheres   « on: Apr 10th, 2003, 11:51am »

Are you saying that the radii of any sphere is equal to the length of one edge of the tetrahedron?
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aero_guy
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 Re: Intersecting Spheres   « Reply #1 on: Apr 11th, 2003, 9:18am »

OK, I had a hard time visuallizing what you were talkiing about.  So, each side is PART of a theoretical sphere with center at the OPPOSITE vertex.  The rest of the sphere, aside from the part that makes that one face, does not exist, no?
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ThudnBlunder
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 Re: Intersecting Spheres   « Reply #2 on: Apr 11th, 2003, 9:26am »

Quote:
 So, each side is PART of a theoretical sphere with center at the OPPOSITE vertex.

Yes

Quote:
 The rest of the sphere, aside from the part that makes that one face, does not exist, no?

As far as the spherical tetrahedron is concerned, the rest of the spheres don't exist. But we need to consider the whole of the spheres in order to find out where any two of them intersect.
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Icarus
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 Re: Intersecting Spheres   « Reply #3 on: Apr 11th, 2003, 3:56pm »

Another way of describing it is to take a regular tetrahedron of edgelength R. Place spheres of radius R centered at each of the vertexes. Throw away the original tetrahedron, and intersect the 4 spheres. What's left is the Spherical Tetrahedron.
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cho
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 Re: Intersecting Spheres   « Reply #4 on: Apr 11th, 2003, 8:09pm »

Speaking as one who has none of the math background many of you have, I'm going to take a wild guess and say 30 degrees?
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ThudnBlunder
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 Re: Intersecting Spheres   « Reply #5 on: Apr 12th, 2003, 10:35am »

cho, why is your guess any better than a guess of (say) 12.3456789...0?
 « Last Edit: Apr 12th, 2003, 11:56am by ThudnBlunder » IP Logged

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cho
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 Re: Intersecting Spheres   « Reply #6 on: Apr 12th, 2003, 11:30am »

My guess is based on what seems like common sense, but could be completely off the mark (hence the disclaimer).
With tetrahedron ABCD, imagine the sphere with point A at its center. It's just a matter of orienting it so that point D is at the North Pole. It would seem that the line connecting B and C is all the points equidistant from the North Pole, ie, at the same latitude as B and C. The tetrahedron is made of equilateral triangles, so that's 60 degrees south of the pole, or 30 degrees north latitude.
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Chronos
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 Re: Intersecting Spheres   « Reply #7 on: Apr 13th, 2003, 1:39pm »

Quoth THUDandBLUNDER: Quote:
 I'm saying that each sphere, with its centre at an apex of the pyramid, also passes through all 3 points of the opposite triangular face.
This is equivalent to what aero_guy said.  You have the center at one vertex, and another vertex on the surface.  The distance from the center of a sphere to a point on the surface (any point) is the radius.  The distance from one vertex to another is the length of the edge.  So the radius is equal to the length of the edge.

I don't know if 30 degrees is right, but I know that it can't be something like 12.3456789.  Whatever the answer is, it's a constuctable angle (since we just constructed it), and most angles are not constructable.  In particular, *most constructable angles are of the form (m/2n)*24o (m and n integers).  So I can accept 30o[/sup, but not 12.3456789[sup]o.

*(Stictly, you can get (many) other angles, since the regular 17-gon is also constructable, and I think I recall seeing that there's a 1-1 correspondence between perfect numbers and constructable prime-gons, so there's probably an infinite number of constructable prime-gons.  But those hardly ever show up in real problems.  In practice, most constructable angles you'll ever encounter are multiples of 1.5o, and this problem is just interesting enough that I suspect it might be a multiple of .75o.)
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ThudnBlunder
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 Re: Intersecting Spheres   « Reply #8 on: Apr 13th, 2003, 2:16pm »

Yes, silly me. You are right (and aero_guy, too) about the radius.

When he was just 19(!), Gauss proved that all regular n-gons are constructible (using an unmarked straightedge and compass only) iff n is a Fermat prime (whereas the pefect numbers that you mentioned are related to Mersenne primes). According to Coxeter (recently deceased), a guy called Hermes spent more than 10 years constructing a 65537-gon!!
 « Last Edit: Apr 14th, 2003, 2:59am by ThudnBlunder » IP Logged

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SWF
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 Re: Intersecting Spheres   « Reply #9 on: Apr 13th, 2003, 4:49pm »

Why would the angle be any different from interesecting circles formed by drawing circles on a equilaterial triangle with center of each circle at a vertex and radius equal to a side of the triangle? Unless there is some trick to this question, the answer is simply 30 degrees as already suggested by cho.
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Chronos
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 Re: Intersecting Spheres   « Reply #10 on: Apr 14th, 2003, 10:57pm »

Because the arcs of interest are not in the same plane as an edge.  The first thing to do would be to remember, re-derive, or look up the angle between two edges of a regular tetrahedron, I think.  After that, I'm getting memory overflows when I try to solve this mentally...  I would try to draw a diagram, but all of the whiteboards here are stubbornly two-dimensional .
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SWF
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 Re: Intersecting Spheres   « Reply #11 on: Apr 15th, 2003, 7:56pm »

on Apr 14th, 2003, 10:57pm, Chronos wrote:
 Because the arcs of interest are not in the same plane as an edge.  The first thing to do would be to remember, re-derive, or look up the angle between two edges of a regular tetrahedron, I think.  After that, I'm getting memory overflows when I try to solve this mentally...

The faces of a regular tetrahedron are equilateral triangles, so the angle between two intersecting edges is 60 degrees. Did you mean the angle between radial lines from tetrahedron centroid to vertices?  That is around 109.5 degrees, but I don't see how that is relevant.

Since you were unable to find that angle mentally, there is a way to visualize this so that a simple inverse trig expression for the ~109.5 degree angle can be found mentally. I'll leave that as a riddle, since most people would probably find that angle in a more complicated way.
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Chronos
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 Re: Intersecting Spheres   « Reply #12 on: Apr 17th, 2003, 4:52pm »

D'oh!  I meant the angle between two faces, not edges!  That should be a little less than 60 degrees.
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SWF
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 Re: Intersecting Spheres   « Reply #13 on: Apr 17th, 2003, 7:13pm »

Don't you mean a little more than 60 degrees?  More precisely it is 180 minus the ~109.47 angle I described earlier, or ~70.53 degrees.
 « Last Edit: Apr 17th, 2003, 7:14pm by SWF » IP Logged
cho
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 Re: Intersecting Spheres   « Reply #14 on: Apr 18th, 2003, 4:54am »

I still don't see why my original answer isn't the self-evident solution. The angle between 2 faces (measured in the middle, I presume) is irrelevant because the middle of the edge is where the new curved edge will differ the most from the standard tetrahedron.
Look at it this way, the spherical surface that makes up side BCD is equidistant from A, so imagine I have a small metal bar of length r, anchor one end at A. I can pivot and turn that bar anywhere within side BCD and it will always touch the surface (including along the edges). I could do the same thing with a radius bar from point D touching all points on face ABC. So the edge BC is found by hooking together the ends of those two bars at angle ABD and pivoting it to ACD.
How are lines of latitude drawn? By drawing concentric rings around the North Pole. What is the latitude that is one radius away from the North Pole? 30o north latitude.
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Eigenray
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 Re: Intersecting Spheres   « Reply #15 on: Apr 28th, 2003, 7:58pm »

on Apr 13th, 2003, 2:16pm, THUDandBLUNDER wrote:
 Gauss proved that all regular n-gons are constructible (using an unmarked straightedge and compass only) iff n is a Fermat prime

This is partially true, but more needs to be said.
It's clear(?) that if an m-gon and an n-gon can be constructed, with m, n relatively prime, then so can an mn-gon.
Thus it suffices to determine whether a pk-gon can be constructed; this is possible iff p is a Fermat prime and k=1, or p=2.
Thus a regular n-gon is constructible iff n is a product (possibly empty) of distinct Fermat primes, possibly times a power of 2.
This is equivalent to the statement that phi(n) is a power of 2, which is actually where the result comes from (using Galois theory).
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rmsgrey
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 Re: Intersecting Spheres   « Reply #16 on: May 16th, 2003, 1:24pm »

on Apr 13th, 2003, 1:39pm, Chronos wrote:
 I don't know if 30 degrees is right, but I know that it can't be something like 12.3456789.  Whatever the answer is, it's a constuctable angle (since we just constructed it), and most angles are not constructable.  In particular, *most constructable angles are of the form (m/2n)*24o (m and n integers).  So I can accept 30o, but not 12.3456789o.

I'm not clear on whether the spherical tetrahedron's construction fits the limited definition of "constructible" (ie constructible using compass and (unmarked) straight edge in the plane) generally used. It is possible, for example, using a tomahawk-like tool to trisect any general angle (I forget the specific design and technique). I'm not sure how valid the limits on constructablity are when you allow 3D construction (quite apart from the problem of using a pair of compasses in 3D!). It's possible that they continue to apply, but if so, I've not heard.
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Icarus
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 Re: Intersecting Spheres   « Reply #17 on: May 17th, 2003, 11:15am »

The rules of construction can be stated set theoretically:

A construction is a sequence of sets of geometric objects satisfying the following conditions:
1) The first set consists of whatever was "given".
2) For each set Sn other than the first, either
• Sn C Sn-1, or
• There are points A, B in Sn-1 such that Sn is Sn-1 U {L}, where L is the line through A and B, or
• There are points A, B, C in Sn-1 such that Sn is Sn-1 U {C}, where C is the circle centered at A whose radius is the distance from B to C, or
• Sn is Sn-1 U {I}, where I is one of the discrete components of the intersection of a finite number of objects in Sn-1.
• Sn is Sn-1 U {I}, where I is one of the discrete components of A - B, for some A, B C Sn-1.

To generalize this to three dimensions, you just add a couple more possibilities:
• There are points A, B, C in Sn-1 such that Sn is Sn-1 U {P}, where P is the plane through A, B, and C.
• There are points A, B, C in Sn-1 such that Sn is Sn-1 U {Sp}, where Sp is the sphere centered at A whose radius is the distance from B to C.

It no longer has the direct tie-in to actual physical tasks, but the gist of the situation remains the same - you are able to construct objects of constant curvature.
 « Last Edit: Jul 30th, 2003, 4:18pm by Icarus » IP Logged

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Icarus
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 Re: Intersecting Spheres   « Reply #18 on: Jul 29th, 2003, 4:16pm »

Cho is right. The latitude is exactly 30o.

Proof: By definition the latitude is the angle at the center of the sphere from the circle to the equator, where the equator is the unique great circle parallel to the given one.

Choose any point on the intersection of two of the spheres, and look at the triangle formed by the point and the two centers. All three sides of the triangle are radii of one or both spheres, and thus they are all the same length - so the triangle is equilateral. The great circle in the definition of latitude is on a plane perpendicular to the radius connecting the two sphere centers. Therefore the latitude angle is complementary to the equilateral triangle angle at the sphere center. So it must be 30o.

This is effectively the same argument Cho made in his last post, but stated in a more mathematical fashion.
 « Last Edit: Jul 29th, 2003, 4:28pm by Icarus » IP Logged

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James Fingas
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 Re: Intersecting Spheres   « Reply #19 on: Jul 30th, 2003, 1:38pm »

What happens when you roll one of these "spherical tetrahedra" down a ramp?
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