Re: [math-fun] geometry puzzle
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?
Nah, ... the right question is to ask for a classification of congruent dissections. I'll do this for the 2d case of a unit disc, but of course it only encompasses *known* congruent dissections. And, to be explicit, I'm only considering pieces that have connected interior. First, there are the obvious, uninteresting dissections: use n equally spaced segments (possibly curved) from the center to the boundary. In this case, I think it's OK to describe the "uninteresting" cases as fundamental regions of a finite subgroup of SO(2) ; we don't even need all of O(2) . Second, start by dissecting into six congruent pieces with arcs of unit circles. These pieces have mirror symmetry, so can be bisected with straight line segments. This is the figure in UPIG, and the first that Michael sent last month. Third, start with the same six-piece dissection as before. Now further dissect each into n congruent pieces using arcs of unit circles from the center. Since n of these make a shape with reflective symmetry, patches of n can be reflected (they need not have started in the same piece of the six-piece dissection). The n = 2 case, with all six pairs of pieces flipped is the second figure that Michael sent earlier; perhaps he had intended this more general situation. If this list is not complete, I'd be very interested in seeing more! On the other hand, if the list is complete, then it solves questions such as "can the center be in the interior of one of the pieces?" and "can the diameter of the pieces be less than 1 ?". Michael Reid
On Fri, May 18, 2012 at 9:16 PM, Michael Reid <reid@gauss.math.ucf.edu>wrote:
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?
Nah, ... the right question is to ask for a classification of congruent dissections.
Sure, of course this is the "real" question. All the "diameter 1" or "not all touching the center" modifiers are ways of asking whether there are more dissections than the "boring" ones. (In light of your examples, my next proposal for "boring" regarding sphere dissections is that every isometry of space that takes one piece of the dissection to another fixes the origin. This generalizes the notion of interesting from the 2D classification.) --Michael
I'll do this for the 2d case of a unit disc, but of course it only encompasses *known* congruent dissections. And, to be explicit, I'm only considering pieces that have connected interior.
First, there are the obvious, uninteresting dissections: use n equally spaced segments (possibly curved) from the center to the boundary. In this case, I think it's OK to describe the "uninteresting" cases as fundamental regions of a finite subgroup of SO(2) ; we don't even need all of O(2) .
Second, start by dissecting into six congruent pieces with arcs of unit circles. These pieces have mirror symmetry, so can be bisected with straight line segments. This is the figure in UPIG, and the first that Michael sent last month.
Third, start with the same six-piece dissection as before. Now further dissect each into n congruent pieces using arcs of unit circles from the center. Since n of these make a shape with reflective symmetry, patches of n can be reflected (they need not have started in the same piece of the six-piece dissection). The n = 2 case, with all six pairs of pieces flipped is the second figure that Michael sent earlier; perhaps he had intended this more general situation.
If this list is not complete, I'd be very interested in seeing more!
On the other hand, if the list is complete, then it solves questions such as "can the center be in the interior of one of the pieces?" and "can the diameter of the pieces be less than 1 ?".
Michael Reid
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Third, start with the same six-piece dissection as before. Now further dissect each into n congruent pieces using arcs of unit circles from the center. Since n of these make a shape with reflective symmetry, patches of n can be reflected (they need not have started in the same piece of the six-piece dissection). The n = 2 case, with all six pairs of pieces flipped is the second figure that Michael sent earlier; perhaps he had intended this more general situation.
Let's enumerate solutions of this form. Firstly, assume without loss of generality that the spirals are clockwise instead of anticlockwise. We can 'flip' between 1 and 6 patches of n pieces. For want of a better word, this will be the 'depth' of the solution. To enumerate them, it would be best to consider each depth separately: Depth = 1: solutions = 1. (Trivially one up to rotation.) Depth = 2: solutions = 2n + 1. (There can be between 0 and 2n pieces between the two flipped patches.) Depth = 6: solutions = 1. (Trivially one up to rotation.) For depth = 3, 4 and 5, it's probably best to apply Burnside's Lemma. In each case, we are effectively enumerating cyclic partitions of (6-d)n objects into d (possibly empty) sets. We have a preliminary value of (6-d)n+d-1 choose d-1, but we need to take into consideration that some of these are identical. Depth = 3: The partition is (a,b,c), and the cyclic group of order 3 has the following three elements: Identity -- fixes all (3n+2 choose 2) partitions; Rotation by 2pi/3 -- fixes only (n,n,n); Rotation by -2pi/3 -- fixes only (n,n,n); So, we have a total of [(3n+2 choose 2) + 2]/3 distinct solutions. Depth = 5: The cyclic group of order 5 has the following five elements: Identity -- fixes all (n+4 choose 4) partitions; 4 rotations -- fixes only (n/5,n/5,n/5,n/5,n/5), if it exists; So, we have a total of [(n+4 choose 4) + 4]/5 solutions if 5 divides n, and [n+4 choose 4]/5 otherwise. Depth = 4: This is the tricky one. We have four elements: Identity -- fixes all (2n+3 choose 3) partitions; Rotation by pi -- fixes n+1 partitions of the form (a,n-a,a,n-a); Rotation by pi/2 -- fixes only (n/2,n/2,n/2,n/2), if it exists; Rotation by pi/2 -- fixes only (n/2,n/2,n/2,n/2), if it exists; For this depth, there are [(2n+3 choose 3) + n + 3]/4 solutions if n is even, and [(2n+3 choose 3) + n + 1]/4 otherwise. To summarise, we have the following: Depth = 1, solutions = 1; Depth = 2, solutions = 2n+1; Depth = 3, solutions = [3n^2 + 3n + 2]/2; Depth = 4, solutions = [2n^3 + 6n^2 + 7n + 3]/6 (for n odd); or solutions = [2n^3 + 6n^2 + 7n + 6]/6 (for n even); Depth = 5, solutions = [n^4 + 10n^3 + 35n^2 + 50n + 24]/120 (for n not divisible by 5); or solutions = [n^4 + 10n^3 + 35n^2 + 50n + 120]/120 (for n divisible by 5). Of course, for the sequence of 'known partitions of the unit disk into p congruent pieces, at least one of which doesn't touch the centre', we have to add the curvilinear solution for p = 12. So, we get 0,0,0,0,0,0,0,0,0,0,0,30,0,0,0,0,0,56,0,0,0,0,0,98, ... The sequence for p = 6n is: 0, 30, 56, 98, 157, 236, 339, ... Mathematica code (for n >= 3): f[7] /. {f -> Function[{n}, 1 + 2 n + 1 + (3 n^2 + 3 n + 2)/2 + Floor[(2 n^3 + 6 n^2 + 7 n + 6)/6] + Floor[(n^4 + 10 n^3 + 35 n^2 + 50 n + 120)/120]]} New OEIS entry, perhaps? Sincerely, Adam P. Goucher
participants (3)
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Adam P. Goucher -
Michael Kleber -
Michael Reid