Talk:Ptolemy's theorem

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[edit] formatting

Am I the only one who finds the formatting on this page screwy in Firefox?

Probably got fixed some time in '05.

[edit] barycenter?

"Any square can be inscribed in a circle whose center is the barycenter of the square."

Why not simply the center of the square? If barycenter is indeed necessary, then it should be made a link to the barycenter article - the term is not common enough to be generally understood in my opinion.

[edit] moved

I've moved this from Ptolemaios's theorem to its much better known name, Ptolemy's theorem. Although I'm Greek myself, there's no real reason to use obscure Greek versions of names when there is a much more widely used form that's standard in English. --Delirium 08:43, 10 November 2005 (UTC)

[edit] ABCD?

Why not introduce the theorem with the much simpler ABCD notation used in the proof and its figure, instead of the complicated-looking point–subscript notation? OK if I change it?

Current way:

If the quadrilateral is given by its four vertices P1, P2, P3 and P4, then the relation is
\overline{P_1P_3}\cdot \overline{P_2P_4}=\overline{P_1P_2}\cdot \overline{P_3P_4}+\overline{P_1P_4}\cdot \overline{P_2P_3}
Here \overline{P_iP_{i+1}}=\overline{P_{i+1}P_{i}} for i=1,\ldots,4 denote the lengths of the four sides of the quadrilateral (with indices taken modulo 4), the two diagonals are then \overline{P_1P_3} and \overline{P_2P_4}.

Proposed new way:

If the quadrilateral is given by its four vertices A, B, C, and D in order, then the theorem states that:
\overline{AC}\cdot \overline{BD}=\overline{AB}\cdot \overline{CD}+\overline{BC}\cdot \overline{DA}
where the overbar denotes the lengths of the line segments between the named vertices.

And of course it could use an introductory illustration. Dicklyon 18:59, 28 August 2006 (UTC)

Hearing no objection, I'll go ahead. And I added a figure for the golden ratio application, too. Dicklyon 02:15, 29 August 2006 (UTC)

And I mangled the illustration to make one for the lead. These are the first two SVGs I've ever made, so I'm still running a bit clueless. Please feel free to improve, replace, or delete. Dicklyon 02:48, 29 August 2006 (UTC)

[edit] Merge?

When someone suggests a merge, it would be proper to open the merge discussion with the reasoning. But, since Michael Kinyon didn't, I will. I was not previously aware of Ptolemy's inequality, but since it has a tiny article and explains that equality is achieved only in the case the corresponds to Ptolemy's theorem, it seems like a no-brainer to merge it into here. Perhaps it's all part of the same theorem originally? Or posed as a corrolary? We should find out, to determine HOW to merge it exactly. Dicklyon 03:04, 28 September 2006 (UTC)

Sorry I didn't say something; after posting the merge tags, I got a bit busy in real life, and then it slipped my mind. Thanks for the Talk page reminder. In any case, I too consider this a no-brainer, and in fact, it is normally the sort of bold merge an editor does without bothering to post tags. However, a couple of folks other than myself, namely Bh3u4m and Charles Matthews, did some recent editing on Ptolemy's inequality, so I thought it better to tag the articles. The other difficulty is that Ptolemy's theorem is quite dense already and has multiple proofs in it, so I am not sure exactly how to work the inequality case into it. I was hoping that those of you who watch this page more closely than I do would have some ideas. I am not sure about the history of the two results. Michael Kinyon 12:55, 28 September 2006 (UTC)

PlanetMath calls Ptolemy's theorem the equality [1], and has and article about Ptolemy's inequality proof [2]. I would therefore conclude that the theorem is a special case of the inequality (when equality occurs), thus I would leave two distinct articles, specifying this feature. Bh3u4m 13:12, 28 September 2006 (UTC)

Most sources treat them together. Here's a book that treats the inequality as a strengthening of the converse of the theorem: [3] Very few books name the "Ptolemy's inequality;" most just include the inequality as an extension or corrolary of the theorem, but leave it unnamed, it appears. Dicklyon 13:58, 28 September 2006 (UTC)

That's a nice source. I think we could work the inequality into the Preview and into the Geometric Proof pretty easily. I am not enamored of the other proofs, but I suspect those who are could work out how to rewrite them appropriately. Michael Kinyon 15:32, 28 September 2006 (UTC)

If I'm not wrong, PlanetMath is published under the GFDL, so it would be possible to simply copy their proof (this has to be verified) Bh3u4m 16:16, 28 September 2006 (UTC)

Nice work with the merge, Dicklyon. Michael Kinyon 23:45, 7 October 2006 (UTC)

And thanks for fixing my typos, Michael Kinyon. Dicklyon 00:01, 8 October 2006 (UTC)
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