Re: [math-fun] geometry puzzle
What's the story for the sphere instead of the circle? That is, what ways are there, if any, to dissect a sphere into a finite number of congruent pieces, other than the pieces just being fundamental domains of some finite subgroup of O(3)?
I suppose it is possible that you intended "unit ball", i.e. solid sphere, rather than sphere. However, it doesn't really matter for now, as all congruent dissections I know of the ball are formed from congruent dissections of the sphere, and then "conifying". I will work with the (hollow) sphere below, but dissections of the solid sphere follow. Independently of that, I think you probably want to phrase it differently. As asked, the dissection into twenty spherical triangles corresponding to a "spherical icosahedron" would be such a dissection. That is, there's no group of order 20 for which the spherical triangles are fundamental domains. Instead, let me suggest an alternative: Dissections whose symmetry groups act transitively on the pieces are "uninteresting" for this discussion. If I have missed your intent, or overlooked some other issue, feel free modify it further. Even with this modification, there are still plenty of other dissections. For example, first dissect into twelve spherical pentagons, corresponding to a "spherical dodecahedron". Each pentagon has bilateral symmetry, so can be bisected, and each has five ways to do it. There are then many ways to bisect all twelve pentagons, and in most, the symmetry group of the dissection does not act transitively on the pieces. Likewise, there are many such dissections starting from spherical tetrahedra, hexahedra, octahedra, and further subdividing. There are also dissections formed by joining pieces. Indeed, the spherical icosahedron dissection is formed by joining sets of three fundamental regions for the action of the icosahedral group. Here's a different type of example, perhaps closer in spirit to what was requested. First split the sphere into two hemispheres, and then dissect each into n congruent spherical triangles using great circles emanating from the poles. If n = 4m , then m of these spherical triangles form an equiangular spherical triangle, which has three-fold rotational symmetry, so that group can be rotated in place to give different dissections. If n = 2k is even, then k of them form a quarter of a sphere, which can also be rotated in place. (Hmmm, this gives me some ideas for rotational puzzles!) Michael Reid
Hmm -- you're quite right that I meant the sphere, and also that the space of uninteresting dissections is larger than I was thinking. Perhaps the good question again is to ask for one with not all pieces touching the center? --Michael On May 18, 2012 7:57 PM, "Michael Reid" <reid@gauss.math.ucf.edu> wrote:
What's the story for the sphere instead of the circle? That is, what ways are there, if any, to dissect a sphere into a finite number of congruent pieces, other than the pieces just being fundamental domains of some finite subgroup of O(3)?
I suppose it is possible that you intended "unit ball", i.e. solid sphere, rather than sphere. However, it doesn't really matter for now, as all congruent dissections I know of the ball are formed from congruent dissections of the sphere, and then "conifying". I will work with the (hollow) sphere below, but dissections of the solid sphere follow.
Independently of that, I think you probably want to phrase it differently. As asked, the dissection into twenty spherical triangles corresponding to a "spherical icosahedron" would be such a dissection. That is, there's no group of order 20 for which the spherical triangles are fundamental domains.
Instead, let me suggest an alternative:
Dissections whose symmetry groups act transitively on the pieces are "uninteresting" for this discussion.
If I have missed your intent, or overlooked some other issue, feel free modify it further.
Even with this modification, there are still plenty of other dissections. For example, first dissect into twelve spherical pentagons, corresponding to a "spherical dodecahedron". Each pentagon has bilateral symmetry, so can be bisected, and each has five ways to do it. There are then many ways to bisect all twelve pentagons, and in most, the symmetry group of the dissection does not act transitively on the pieces.
Likewise, there are many such dissections starting from spherical tetrahedra, hexahedra, octahedra, and further subdividing. There are also dissections formed by joining pieces. Indeed, the spherical icosahedron dissection is formed by joining sets of three fundamental regions for the action of the icosahedral group.
Here's a different type of example, perhaps closer in spirit to what was requested. First split the sphere into two hemispheres, and then dissect each into n congruent spherical triangles using great circles emanating from the poles. If n = 4m , then m of these spherical triangles form an equiangular spherical triangle, which has three-fold rotational symmetry, so that group can be rotated in place to give different dissections. If n = 2k is even, then k of them form a quarter of a sphere, which can also be rotated in place. (Hmmm, this gives me some ideas for rotational puzzles!)
Michael Reid
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