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Thread: Not really sure where to put this, its pretty much just a spherical geometry question

  1. #1

    Not really sure where to put this, its pretty much just a spherical geometry question

    So I am currently in 11th grade, and taking precalculus, college algebra/trigonometry, and physics. I have asked all of my teachers this just random question that popped into my head one day, and no one has been able to even come close.

    (keep in mind this is kinda hard to explain) So say you have a sphere with arbitrary radius r, and say I were to "draw" a one dimensional line on the surface of that sphere, with a length equal to the radius, r. and then I draw two more of these lines on this sphere, with the points at the end of each line ending on the end of another. Now I would have an equilateral triangle graphed on the surface of a sphere, with the triangle having sides equal to the radius of the sphere. My question is, how many of those triangles would i be able to fit on the surface of the sphere? Would it be a 'nice' number like 6pi or 8pi? Proof / Work?
    Last edited by haffofhayes; 02-05-2018 at 10:57 PM.

  2. #2
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    Quote Originally Posted by haffofhayes View Post
    So I am currently in 11th grade, and taking precalculus, college algebra/trigonometry, and physics. I have asked all of my teachers this just random question that popped into my head one day, and no one has been able to even come close.

    (keep in mind this is kinda hard to explain) So say you have a sphere with arbitrary radius r, and say I were to "draw" a one dimensional line on the surface of that sphere, with a length equal to the radius, r. and then I draw two more of these lines on this sphere, with the points at the end of each line ending on the end of another. Now I would have an equilateral triangle graphed on the surface of a sphere, with the triangle having sides equal to the radius of the sphere. My question is, how many of those triangles would i be able to fit on the surface of the sphere? Would it be a 'nice' number like 6pi or 8pi? Proof / Work?
    The problem is that only certain particular sizes of triangles would fit at all; most sizes would leave gaps that can't be filled with these triangles. To find out about this, look up "regular polyhedra". One example would be if the "side" of the "triangle" is 1/4 of the circumference of the sphere, forming a regular octahedron. Yours will not fit.

    If you are not really requiring that the triangles fit together, and just want the ratio of the surface area of the triangle to that of the whole sphere (in terms of the ratio of the "side" of the "triangle" to the radius of the sphere), then your question is about the area of a "spherical triangle"; you can look that up.

  3. #3
    Quote Originally Posted by Dr.Peterson View Post
    If you are not really requiring that the triangles fit together, and just want the ratio of the surface area of the triangle to that of the whole sphere (in terms of the ratio of the "side" of the "triangle" to the radius of the sphere), then your question is about the area of a "spherical triangle"; you can look that up.
    That is what I was going for, I've tried a couple of methods for solving this. I've found the area of a spherical triangle to be r^2*E. since I don't know spherical excess, just the side lengths, I tried to use L'Huilier's theorem. I rearranged the formula to isolate E and plugged that into (4pir^2) / r^2* (lhuiliers) this yielded a ungodly fraction. After a fair amount of careful simplification I arrived at my (wrong) 'answer'. I tried graphing it on desmos, and got a ridiculous graph, proving my wrongness. Where did I go wrong?

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