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Topic: Equidistant Spheres. (Read 3719 times) |
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rloginunix
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Equidistant Spheres.
« on: Dec 14th, 2013, 12:25pm » |
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Equidistant Spheres.
See. That's the problem with posting a solution before going to bed. Your brain keeps working quietly in the background while you're asleep and the next morning when you wake up - boom, another problem is born.
In the "Circles and straights" thread the answer to a more general problem for spheres turned out to be not interesting. My thought was how do I make it interesting? By adding the fifth point.
Problem:
how many spheres equidistant from five non-cospherical and non-coplanar points can be constructed?
Extra for experts: add planes to the question above - how many planes equidistant from five non-cospherical and non-coplanar points can be constructed?
I will post my line of reasoning later but if I've got my factorials right then the answer is 5 + 10 = 15.
If you think you are in trouble try these hints consecutively:
Hint 1: four in the bank
Hint 2: three on a line
Hint 3: two in a plane
If you think you are in real trouble work your way through the "Circles and straights" discussion first.
Good luck.
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rloginunix
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Re: Equidistant Spheres.
« Reply #1 on: Dec 24th, 2013, 8:53pm » |
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I hope that the moderators and the rest of the crowd aren't gonna kill me for posting lots of unhidden solution text and for solving my own problem. Let it be a fair warning for newcomers - stop reading now if you want to give it your own shot.
To sum up:
how many spheres and planes equidistant from five non-cospherical and non-coplanar points can be constructed?
This problem didn't just pop into my head all of a sudden one day. Well, it kinda did and that's the scary part but it happened gradually, in a series of steps.
I was working on the "Circles and straights" problem. I figure if you start digging - you start digging. You do it right or you don't do it at all. I first tried the 2-point 2D case. Then 3-point 2D case. While working on it I came to a certain conclusion. I then asked myself is that it? Is that the best you can do, mammal? I then realized that by switching to 3D I can come to a drastically different answer. From that point on 3D was on my mind.
I did the 4-point 2D case. Generalized it to 3D. The answer for spheres turned out to be not interesting. I thought how can I make it interesting? What will happen if I add the fifth point to the mix? I ran one case. It worked. Then the other. It also worked. At which point I decided that it can stand on its own.
So one way to unfold this solution is to mimic the process of its creation - gradually, over a series of steps which collectively I call Scope Reduction approach.
My solution path.
By working out the early cases I came to a conclusion that two patterns run through this problem. Combinatorial and geometric. Combinatorial pattern uncovers the why while geometric - the how. Both are need to obtain a thorough solution.
In the text that follows I refer to planes and spheres as just the Surface. Incidentally both surfaces have constant Gaussian curvature, zero for planes and one over radius squared for spheres.
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rloginunix
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Re: Equidistant Spheres.
« Reply #2 on: Dec 24th, 2013, 8:56pm » |
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Combinatorics.
If we keep all five points on one side of the Surface then to satisfy the requirement of equidistance they'll have to sit on a similar surface - plane for planes and sphere for spheres. This is a degenerate case excluded by the definition of the problem. It means that we have to group the points relative to the side of the Surface they are on in some other way. We only have two choices. We can group them as 1 + 4 by putting one point on one side of the Surface and the remaining four on the other or as 2 + 3 by putting two points on one side of the Surface and the remaining three on the other.
Whether we construct a line through A and B or B and A the line will remain the same. And if we pick three points A, B and C they will define a unique triangle whether we name/construct it as ABC or CBA, etc. It means that the order of items in the group doesn't matter. So from a purely combinatorial perspective we have our good old C(n, k) ways of picking them. It's C(5, 4) or C(5, 1) for the 1 + 4 case which is 5 and C(5, 3) or C(5, 2) for the 2 + 3 case which is 10. For a total of 15 objects.
So combinatorics and our gut are telling us the answer should be 15. We now have to back it up with a bit of geometry.
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rloginunix
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Re: Equidistant Spheres.
« Reply #3 on: Dec 24th, 2013, 9:00pm » |
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Geometry.
1 + 4.
Let's assume for convenience sake that E is on one side of the Surface and the remaining four points A, B, C and D are on its opposite side. It means that points A, B, C and D are neither collinear nor cocircular.
They are not collinear because otherwise we could construct a plane through the ABCD straight line and a point E not on it. That would mean that all five points are coplanar which is not allowed. Consequently for a unique plane-based solution to exist these points must be coplanar in such a way that they define a unique plane. A sphere-based solution then doesn't exist.
Points A, B, C and D are not cocircular because otherwise we could construct a sphere through all five points thus. Construct a line through the circle's center perpendicular to the circle's plane. Pick a point on the circle, say D. Construct a line segment DE. Bisect it with a plane. That plane and the perpendicular through the circle's center must intersect because if they don't they are parallel and if they are parallel all five points must be coplanar - not allowed. That intersection point is the center of the sphere that kisses all five points which is also not allowed. Consequently for a unique sphere-based solution to exist these four points must be non-coplanar. A plane-based solution then doesn't exist.
Spheres.
To construct a sphere equidistant from such 1 + 4 grouping we have to conceptually split the distance between two items in half. The first item is point E. The second item is an aggregate point ABCD. For that to work all four points A, B, C and D must sit on a sphere. Our task then is to locate it's center.
Let's say D is not in the ABC plane. Then three distinct points A, B and C define a unique plane. On it the same three points define a unique triangle. This unique triangle defines a unique circumcenter - a point equidistant from all three vertexes located at the intersection of three sides' bisectors (in practice just two is enough).
We can construct an infinite number of lines through that circumcenter. However we can construct only one line that passes through the circumcenter that is also perpendicular to ABC plane. Any point on that line is equidistant from the points in ABC plane. As such it can be used as a center of a sphere that kisses all three given points. So we can construct an infinite number of such spheres. See sketch.
[edit]
Moved the drawing file to this forum.
[/edit]
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| « Last Edit: Apr 5th, 2014, 8:12am by rloginunix » |
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rloginunix
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Re: Equidistant Spheres.
« Reply #4 on: Dec 24th, 2013, 9:01pm » |
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Next pick a point, say C. Construct a line segment CD. Bisect it with a plane. By definition any point on that plane is equidistant from C and D. That plane and the perpendicular through the triangle's circumcenter may or may not intersect. If they don't intersect they are parallel. And if they are parallel then it can only happen if D belongs to the ABC plane. It violates our original assumption that four points are non-coplanar so it can't be. They have to intersect. And their intersection point by definition is equidistant from all four points and we have our unique sphere centered at O. That was my first hint - four in the bank, meaning four non-coplanar points define a unique sphere.
On a separate note, if we treat these four non-coplanar points as vertexes of a tetrahedron then we just semi-proved that we can describe a sphere around any tetrahedron. Similar to its 2D counterpart - triangle. I'm risking deviating from the core problem too much but just for fun go ahead and figure out how to Inscribe a sphere into a tetrahedron.
Now that we have found the center O of a unique sphere all that remains to do is connect O and E, cut OE in half at F and construct a sphere centered at O with radius OF which is equidistant from all five points. Five spheres total.
Planes.
In this case all four points are coplanar. It means that these four points define a unique plane. If we drop a perpedicular from E on the ABCD plane and bisect it with a perpendicular plane then that plane will be equidistant from all five points. Five planes total.
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rloginunix
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Re: Equidistant Spheres.
« Reply #5 on: Dec 24th, 2013, 9:03pm » |
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2 + 3.
Let's assume for convenience sake that A and B are on one side of the Surface and C, D and E are on its opposite side. Through two points A and B we can construct only one unique line but an infinite number of circles, planes and spheres. Points C, D and E may either be collinear or not in which case they are cocircular. The inverse of the 1 + 4 case. If they are cocircular then they define a unique circumcenter and we can analyze this situation for spheres. If they are collinear we can analyze this situation for planes.
Spheres.
Points C, D and E are cocircular.
To construct a sphere equidistant from the 2 + 3 grouping we need to split the distance between two items in half. The first item is a 2-point aggregate and the second is a 3-point aggregate. Both must sit on concentric spheres and our task is to find that common center.
Borrow the 3-point sphere construction from our 1 + 4 deduction. Unique plane, triangle, circumcenter, perpendicular. That was my second hint for 3-point spheres - their centers sit on a special straight line which is a perpendicular through the triangle's circumcenter.
You can also borrow the 2-point sphere construction from our 1 + 4 deduction. Say A and B are two such points. Any point in the (perpendicular) plane bisecting the line segment AB is equidistant from A and B by definition. As such it can be used as a center of a sphere kissing both points. That was my final hint, two in a plane meaning for two points equidistant spheres are centered on the plane bisecting the segment connecting both points.
We already know that the 3-point perpendicular and the 2-point bisecting plane may or may not intersect. If they do then they locate a unique point in space O. Construct the 3-point CDE sphere and the 2-point AB sphere centered at O. Snap a line through O. That line will intersect both spheres locating two points, say F and G. Bisect FG to locate H. Construct a sphere equidistant from all five points centered at O with radius OH. Combinatorics tells us we can do 10 of those.
If the 3-point perpendicular and the 2-point bisecting plane do not intersect then they are parallel and we can't construct an equidistant sphere.
Planes.
Points C, D and E are collinear.
In that case we have points A and B in a different plane because otherwise all five points would've been coplanar.
These two sets of points (C, D, E) and (A, B) define two lines that cross over in space and by doing so define a unique pair of parallel planes constructed thus. Build a line through point C parallel to line AB. These two lines define a unique CDE plane. Build a line through point A parallel to line CDE. These two lines define a unique AB plane. This is similar to the second tetrahedron case in the "Circles and straights" discussion.
So now we have two distinct parallel planes. We construct a line perpendicular to both and its perpendicular bisecting plane is equidistant from all five points. 10 planes total.
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