

Title: Re: Intersecting Spheres Post by aero_guy on Apr 10^{th}, 2003, 11:51am Are you saying that the radii of any sphere is equal to the length of one edge of the tetrahedron? 

Title: Re: Intersecting Spheres Post by aero_guy on Apr 11^{th}, 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? 

Title: Re: Intersecting Spheres Post by THUDandBLUNDER on Apr 11^{th}, 2003, 9:26am Quote:
Yes Quote:
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. 

Title: Re: Intersecting Spheres Post by Icarus on Apr 11^{th}, 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. 

Title: Re: Intersecting Spheres Post by cho on Apr 11^{th}, 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 [hide]30 degrees?[/hide] 

Title: Re: Intersecting Spheres Post by THUDandBLUNDER on Apr 12^{th}, 2003, 10:35am cho, why is your guess any better than a guess of (say) 12.3456789...^{0}? 

Title: Re: Intersecting Spheres Post by cho on Apr 12^{th}, 2003, 11:30am My guess is based on what seems like common sense, but could be completely off the mark (hence the disclaimer). [hide]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.[/hide] 

Title: Re: Intersecting Spheres Post by Chronos on Apr 13^{th}, 2003, 1:39pm Quoth THUDandBLUNDER: Quote:
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/2^{n})*24^{o} (m and n integers). So I can accept 30^{o[/sup, but not 12.3456789[sup]o}. *(Stictly, you can get (many) other angles, since the regular 17gon is also constructable, and I think I recall seeing that there's a 11 correspondence between perfect numbers and constructable primegons, so there's probably an infinite number of constructable primegons. But those hardly ever show up in real problems. In practice, most constructable angles you'll ever encounter are multiples of 1.5^{o}, and this problem is just interesting enough that I suspect it might be a multiple of .75^{o}.) 

Title: Re: Intersecting Spheres Post by THUDandBLUNDER on Apr 13^{th}, 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 ngons 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 65537gon!! 

Title: Re: Intersecting Spheres Post by SWF on Apr 13^{th}, 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. 

Title: Re: Intersecting Spheres Post by Chronos on Apr 14^{th}, 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, rederive, 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 twodimensional :(. 

Title: Re: Intersecting Spheres Post by SWF on Apr 15^{th}, 2003, 7:56pm on 04/14/03 at 22:57:17, Chronos wrote:
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. 

Title: Re: Intersecting Spheres Post by Chronos on Apr 17^{th}, 2003, 4:52pm D'oh! I meant the angle between two faces, not edges! That should be a little less than 60 degrees. 

Title: Re: Intersecting Spheres Post by SWF on Apr 17^{th}, 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. 

Title: Re: Intersecting Spheres Post by cho on Apr 18^{th}, 2003, 4:54am I still don't see why my original answer isn't the selfevident 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? 30^{o} north latitude. 

Title: Re: Intersecting Spheres Post by Eigenray on Apr 28^{th}, 2003, 7:58pm on 04/13/03 at 14:16:34, THUDandBLUNDER wrote:
This is partially true, but more needs to be said. It's clear(?) that if an mgon and an ngon can be constructed, with m, n relatively prime, then so can an mngon. Thus it suffices to determine whether a p^{k}gon can be constructed; this is possible iff p is a Fermat prime and k=1, or p=2. Thus a regular ngon 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). 

Title: Re: Intersecting Spheres Post by rmsgrey on May 16^{th}, 2003, 1:24pm on 04/13/03 at 13:39:19, Chronos wrote:
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 tomahawklike 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. 

Title: Re: Intersecting Spheres Post by Icarus on May 17^{th}, 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:
2) For each set S_{n} other than the first, either To generalize this to three dimensions, you just add a couple more possibilities:
It no longer has the direct tiein to actual physical tasks, but the gist of the situation remains the same  you are able to construct objects of constant curvature. 

Title: Re: Intersecting Spheres Post by Icarus on Jul 29^{th}, 2003, 4:16pm Cho is right. The latitude is exactly 30^{o}. 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 30^{o}. This is effectively the same argument Cho made in his last post, but stated in a more mathematical fashion. 

Title: Re: Intersecting Spheres Post by James Fingas on Jul 30^{th}, 2003, 1:38pm What happens when you roll one of these "spherical tetrahedra" down a ramp? 

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