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bonanova
Daedalian Member

 Posted: Thu Nov 04, 2010 5:25 pm    Post subject: 1 Is it true that the number ways a plane can slice a cube to get different regular polygons equals the number of ways a plane can slice a torus [doughnut] to get circles? Count only essentially different ways._________________ Vidi, vici, veni.
groza528
No Place Like Home

 Posted: Thu Nov 04, 2010 8:53 pm    Post subject: 2 I'm not certain I understand the question. When you refer to "essentially different" do you mean that a slice near the front face of the cube is identical to a slice near the top face? If I slice 1mm from the top face is that essentially different to slicing 2mm from the top face? Does this include a discussion of "how infinite" it is?
bonanova
Daedalian Member

 Posted: Thu Nov 04, 2010 9:40 pm    Post subject: 3 No, I was looking for answers that are much smaller than infinity. The first part asks about different regular polygons. A square is one regular polygon. Now consider other regular polygons. The second part asks about slicing a doughnut and getting circles. "Essentially different" was meant to say for example that slicing the doughnut with a plane containing the axis counts as [just] one way. You could rotate the slicing plane about the axis through an infinity of angles, but you'd get "essentially the same" result. If it helps in describing the results, consider the doughnut to be a perfect torus with inside diameter a and outside diameter 3a Sorry to make a deceptively easy puzzle sound unclear. _________________ Vidi, vici, veni.
Zag
Unintentionally offensive old coot

 Posted: Thu Nov 04, 2010 11:23 pm    Post subject: 4 I know you can slice a cube and make a regular hexagon. And, obviously, you can make a square. I don't know of any other regular polygons you can make. Slicing a torus to get circles, it seems that you could get two circles separate from each other, or two concentric circles, or a single circle. I think there are two different ways to get a single circle: one by the same way you get concentric circles, but just barely tangent to the top; the other as if you took a bite of the bagel. I don't see any other ways that arrive at circles, at least.
L'lanmal
Daedalian Member

Posted: Fri Nov 05, 2010 2:03 am    Post subject: 5

 Zag wrote: I know you can slice a cube and make a regular hexagon. And, obviously, you can make a square. I don't know of any other regular polygons you can make.

A triangle near the corner?
bonanova
Daedalian Member

 Posted: Fri Nov 05, 2010 2:08 am    Post subject: 6 I admit I hadn't thought of getting a single circle. Maybe it's a philosophical issue whether a vanishingly small slice ever becomes a line. For our purposes, I'll limit slices to operations that leave parts of the object being sliced on both sides of the slicing plane. That would eliminate the limiting case of concentric circles. So I guess with that definition, you'd always get two circles. Bites probably wouldn't count, either, unless ... no, I guess I can't visualize planar teeth. Maybe the guy in Dr. No? Nah. No biting._________________ Vidi, vici, veni.
Zag
Unintentionally offensive old coot

 Posted: Fri Nov 05, 2010 3:29 am    Post subject: 7 lol I didn't mean literally biting. I was just trying to describe the angle of the cut. If you draw a radius of the large circle, put the plane perpendicular to the radius, very near the edge of it. I'm guessing that there is a distance at which the section cut is a circle.
bonanova
Daedalian Member

 Posted: Fri Nov 05, 2010 3:48 am    Post subject: 8 Thanks for accepting my humor graciously. I think those cuts always make ovals. There is clearly a minor axis [parallel to the axis of the torus, the cross section can't be wider than a, the other axis has a limit of 3a altho it's never reached.] I don't think there is a way to get just one circle with a planar cut other than using a tangent plane, but I've defined that as not a cut. I think the original question hasn't been unanswered._________________ Vidi, vici, veni.
groza528
No Place Like Home

 Posted: Fri Nov 05, 2010 4:23 am    Post subject: 9 I thought it might be possible to slice the torus diagonally to create a circle, but the algebra isn't working out. I can't think of a third way to create circles with the torus, but as others have said, you can get a square, hexagon, and triangle out of the cube. My vote is for "no, it's not true."
mith
Pitbull of Truth

 Posted: Fri Nov 05, 2010 3:51 pm    Post subject: 10 I believe you can also cut a torus at an angle to get two intersecting circles.
bonanova
Daedalian Member

Posted: Fri Nov 05, 2010 5:47 pm    Post subject: 11

 mith wrote: I believe you can also cut a torus at an angle to get two intersecting circles.

The ones I knew about result from a slice through the center and tangent above and below on the sides. This gives the surprising result that every point on a torus is on four circles: the two that are obvious and two "Villarceau circles," of diameter 2a. Wikipedia shows a nice demo - and, for Groza, the somewhat involved mathematical proof.

So the puzzle answer is yes. But often these circles and the hexagon are missed, and yes is said for the wrong reason.

Along these lines, and this might belong in the chestnuts thread, is there a way to draw congruent closed curves on the surfaces of two potatoes? Conceptually at least? Well the question screams the answer, so the puzzle is, how?
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groza528
No Place Like Home

 Posted: Fri Nov 05, 2010 8:01 pm    Post subject: 12 Oh drat; that's exactly the cut I was picturing but I was imagining the wrong result. The reason my algebra wasn't working is because I was thinking it was an ellipse inside a circle rather than two intersecting circles.
groza528
No Place Like Home

 Posted: Fri Nov 05, 2010 8:07 pm    Post subject: 13 Ah no, that's not quite the cut I was picturing after all. I was going through the uppermost and lowermost points of the torus, rather than tangent to it. I was still probably visualizing it wrong though
bonanova
Daedalian Member

Posted: Tue Nov 09, 2010 7:12 am    Post subject: 14

 bonanova wrote: ... this might belong in the chestnuts thread, is there a way to draw congruent closed curves on the surfaces of two potatoes? Conceptually at least? Well the question screams the answer, so the puzzle is, how?

One on each that is. And ordinary potatoes. No tori, etc.
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Lepton*
Guest

 Posted: Tue Nov 09, 2010 1:29 pm    Post subject: 15 Choose two potatoes such that one is everywhere lumpier than the other, and both have the same volume. Draw an arbitrary closed curve on the less-lumpy potato. Next, completely cover the lumpier potato with your marker. By the intermediate value theorem, the curve on the less-lumpy potato corresponds to one of the many curves you've drawn on the lumpier potato. This proof is incorrect.
Jack_Ian
Big Endian

 Posted: Tue Nov 09, 2010 2:40 pm    Post subject: 16 Can lumpy mean spiky? If so then I'm dubious. Is a cube-shaped potato lumpier than a spherical one? Either potato, in that case, could be coloured in without using a curve that could be drawn on the other. This is one area of mathematics that I have absolutely no clue about. So pardon me if this is a ridiculous question.
Zahariel
Daedalian Member

 Posted: Tue Nov 09, 2010 5:11 pm    Post subject: 17 Jack_Ian, you could draw a circle entirely on one face of the cube, and a congruent circle somewhere on the sphere. The original theorem is easily enough proved: Identify the potatoes with their respective surfaces (any closed differentiable surface will do, although it's better if it can be embedded in 3 dimensions without self-intersection). Now move those surfaces so that they intersect nontrivially (this is always possible as a result of the intermediate value theorem). Their intersection (or more formally, each segment of their intersection) is a closed curve that can be drawn on both surfaces. QED. As to how you would do this construction with physical potatoes that cannot be moved to interpenetrate with each other, I have no idea.
bonanova
Daedalian Member

Posted: Tue Nov 09, 2010 5:51 pm    Post subject: 18

bonanova wrote:
 bonanova wrote: Conceptually at least?
And ordinary potatoes. No tori, etc.

Yeah, that's the idea. Merge a couple of ghost potatoes and note the intersection of their surfaces. You can even require one of them to be bisected. Sort of. If you can partition volume with a non-planar curve.
_________________
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Zag
Unintentionally offensive old coot

 Posted: Tue Nov 09, 2010 6:52 pm    Post subject: 19 Hi Zahariel. You haven't been around lately. Good to hear from you. Call your mother! Also, I bought myself the birthday present you "gave" me, so you owe me \$60 for it. I don't see how that demonstrates Lepton's situation. He stated that any closed curve on one potato can be reproduced on the other. You've shown that there are curves that are common between them, which I agree is clear, but not that any curve on one can be reproduced on the other.
ralphmerridew
Daedalian Member

 Posted: Tue Nov 09, 2010 8:05 pm    Post subject: 20 Lepton's can't always be done; have one be a perfect sphere. Choose a great circle for its curve. Have the second be a polyhedron with no face large enough to hold the circle.
Zahariel
Daedalian Member

 Posted: Tue Nov 09, 2010 8:54 pm    Post subject: 21 I'm sort of curious how well-behaved the potatoes have to be. Do they actually have to be differentiable? I'm imagining some kind of pathological continuous-but-nowhere-differentiable surface that refuses to form well-behaved intersections with other, similar, surfaces. Is this possible? I don't remember enough manifold calculus to be able to work it out in my head.
Lepton*
Guest

 Posted: Wed Nov 10, 2010 2:02 am    Post subject: 22 Just to be clear, for those that missed the empty space at the bottom of my "proof": it's not correct, and was intended as parody. Zahariel, it's (also) been a few years since my last differential geometry class, but I (also) remember derivatives as being the way to determine intersection on manifolds. Points where differentiability is lost would get "stuck", as it would be impossible to define a function that carries points from one manifold to the next. Maybe. Here's another not-really-a-solution, based on Zahariel's proof: slam the potatoes together at a velocity that is greater than the speed of sound in the potato cores (so that they crumple at the impact site, but stay rigid beyond that, as least momentarily). Then, after you separate the two potatoes, the impact will have left an impact area on each potato. The circumferences of these mush areas will be congruous.
Jack_Ian
Big Endian

Posted: Wed Nov 10, 2010 2:17 pm    Post subject: 23

 Zahariel wrote: Jack_Ian, you could draw a circle entirely on one face of the cube, and a congruent circle somewhere on the sphere. The original theorem is easily enough proved: Identify the potatoes with their respective surfaces (any closed differentiable surface will do, although it's better if it can be embedded in 3 dimensions without self-intersection). Now move those surfaces so that they intersect nontrivially (this is always possible as a result of the intermediate value theorem). Their intersection (or more formally, each segment of their intersection) is a closed curve that can be drawn on both surfaces. QED.
But wouldn't that only produce the subset of possible curves that are common to both?
Can you draw a triangle on a sphere?
Lepton*
Guest

 Posted: Wed Nov 10, 2010 2:19 pm    Post subject: 24 Define "triangle", first.
bonanova
Daedalian Member

Posted: Wed Nov 10, 2010 4:21 pm    Post subject: 25

 Lepton* wrote: Define "triangle", first.

Arcs of great circles joining three points. Which curves would not then lie on a plane. The answer has to be the equivalent of the intersection of the two surfaces.
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