Talk:Villarceau circles

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So let's try not copying mathworld.wolfram.com?

yeah come on guys geez


I wrote this page. I did not copy it from mathworld. In particular, note that mathworld does not say who Villarceau was. Michael Hardy 20:47, 11 November 2005 (UTC)
Correction: I did not write the original version. But any statement of the result must state the result. Mathworld's page is even stubbier than this page, and states the result. Michael Hardy 23:43, 11 November 2005 (UTC)
The article at MathWorld covers it, not in the Villarceau circles page, but in the torus page. ☢ Ҡi∊ff⌇↯ 18:33, 3 October 2006 (UTC)

Contents

[edit] Article expansion

We need some nice images and formulas... Anyone care to fill those in? I'll try to render something pretty in POV-Ray... ☢ Ҡi∊ff⌇↯ 11:56, 22 September 2006 (UTC)

And done! ☢ Ҡi∊ff⌇↯ 05:55, 1 October 2006 (UTC)

[edit] Why use an example of this kind?

The article now says:

Let the torus be given implicitly as the set of points on circles of radius three around points on a circle of radius five in the xy plane.
0 = (x^2+y^2+z^2 - 34)^2 - 100(9-z^2) \,\!

Why do that? No generality at all is lost by using a simpler and more symmetric example, so why complicate things with pointless trivia? If an exercise were posed for students, where the problem is to reduce such an equation to a simple and symmetrical form, that would make sense, but that seems off-topic for this article. Michael Hardy 17:34, 3 October 2006 (UTC)

... and I notice that someone has labeled this an "example" in a section heading. But it's really the most general case, and other such "examples" would differ only in different numbers, but everything that really matters to the present topic would be unchanged. If various examples differed in some way that is on-topic, that would make sense. But they don't. The so-called "example" is really the most general case. Michael Hardy 17:38, 3 October 2006 (UTC)
I reckon you're complaining to me, since I added the specific torus equations. This was a drive-by edit, and I did not put the article/talk on my watch list, so luckily the new animation brought me by for a second look and I saw your comment.
Yes, we could write an equation for a torus with major radius R and minor radius r. The natural form of such an equation is a quartic, not found in our Wikipedia article, nor among the links, but noted by Pat Hanrahan.
0 = (x^2+y^2+z^2 - (R^2+r^2))^2 - 4 R^2 (r^2-z^2) \,\!
And yes, we could write an equation for a specific cutting plane, tangent to the torus and passing through its center. In fact, such an equation is given at what is presently the sole external link.
Of course, that page uses a torus equation with a square root, and does not explain the essential criteria for the cutting plane. Still, it would seem to have some of the generality you seek for those who want it. Furthermore, given the demand for tangency, it should not be hard for a motivated reader to derive a plane equation for themselves. Working from information in the link, one suitable plane through the x-axis is
0 = x r - z \sqrt{R^2-r^2} . \,\!
It is perfectly reasonable to add these two equations to the article to make it more self-contained. However, this is hardly completely general, since the torus is in special position, as is the plane.
But I added the example because the article had no formulae, and I used carefully chosen R and r to make the equations work out simply (since R2−r2 is a perfect square). I felt it would help most readers to see an explicit example they could easily verify for themselves. This is, after all, a tradition in mathematics going back thousands of years. The ancients often never gave a general formula.
I would be delighted to see you add a new section with a proof that the Villarceau circles exist, and that the cutting plane must satisfy the conditions I have stated. I would object to removing or changing the chosen example, which is already the simplest possible (a fact that may not be obvious if you are not familiar with the relevant quartic and plane equations). --KSmrqT 08:07, 5 October 2006 (UTC)

[edit] Proof from projective geometry

There is a nice four-line proof of this result at [1], in french, and with a one-page explanation of this proof. It may be original research, though. --Bernard 04:31, 4 October 2006 (UTC)

If it was original research when it was put on the external web site, then it's no longer original research when it's explained here, since the external website can be cited. Michael Hardy 18:47, 4 October 2006 (UTC)
Nonsense. First, we should surely be competent to judge a four-line proof of a well-known fact, so I think worrying about "original research" in this instance is a peculiarly Wikipedian paranoia. Second, by Wikipedia WP:NOR standards the proof would have to appear in a peer-reviewed publication, not simply somebody's personal website, to be considered authoritative. But then, I happen to think Wikipedia's attempts at quality control are seriously broken, trying unsuccessfully to avoid the need for qualified reviewers and writers with good judgment.
In any event, we could cite this paper and this one for web-accessible proofs that are peer-reviewed. --KSmrqT 08:50, 5 October 2006 (UTC)
I am not sure I understand you. To me, the proof in question is clearly correct. I would like to know what you think of the proof (if you can read french...), if it should be included in the article and how? The papers you cite are somewhat beyond the scope of the article, in my opinion and after a quick reading. --Bernard 14:16, 5 October 2006 (UTC)
The gist of the proof in French, which I do not read, seems to be essentially the same as the proof in English by Hirsch. Essentially, we discover that the quartic intersection curve has four double points, and we conclude that it factors into a symmetric pair of conics. Babel Fish (with a little manual help) gives the following translation:
Théorème (Villarceau, 1838). L'intersection d'un tore avec un de ses plans bitangents est la réunion de deux cercles.
Démonstration : L'intersection du tore avec le plan à l'infini est l'ombilicale (d'équation X2+Y2+Z2 = 0), comptée deux fois. L'intersection avec le plan bitangent est donc une quartique possédant quatre points doubles, ce qui montre qu'elle est dégénérée en deux coniques, se recoupant aux points cycliques, et qui sont donc deux cercles.
Theorem (Villarceau, 1838). The intersection of a torus with one of its bitangential planes is the meeting of two circles.
Proof : The intersection of the torus with the plane at infinity is the umbilical point (with equation X2+Y2+Z2 = 0), counted twice. The intersection with the bitangential plane is thus a quartic having four double points, which shows that it degenerates into two conics, intersecting at the cyclic points, and which are thus two circles.
I have extended the article with further discussion, citing both Hirsch and Coxeter, and incorporating some addition enrichment involving a famous fiber bundle. --KSmrqT 03:18, 7 October 2006 (UTC)

[edit] Who's Bancroft?

The article now says:

The torus plays a central role in the Hopf fibration of the 3-sphere, S3, over the ordinary sphere, S2, which has circles, S1, as fibers. When the 3-sphere is visualized in Euclidean 3-space by stereographic projection, the inverse image of a great circle on S2 under the fiber map is a torus, and the fibers themselves are Villarceau circles. Bancroft has explored such a torus with computer graphics imagery.

But no one named Bancroft is listed in the References section, nor otherwise identified in the article. Who's Bancroft? Michael Hardy 20:17, 9 October 2006 (UTC)

I have no idea; ask my fingers! The name is Banchoff, and the relevant link is to the Flat Torus page at the Geometry Center. I'll fix it. (And thanks for catching this.) --KSmrqT 21:07, 9 October 2006 (UTC)