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I do happen to have access to Correspondance sur l'Ecole polytechnique. You needn't sell your soul when I've already done so. ;-) The Poncelet solution begins, "Mener un cercle tangent à trois cercles donnés?" But you could probably guess it was something like that. :-) I'll look it over (though I need to do real work for a while longer).
I don't know anything about the circle method myself, but I have access to a large library. If you have any ideas on where I should look, please say. Otherwise I can look randomly. Ozob ( talk) 20:17, 30 May 2008 (UTC)
I've asked the library to take it (as well as the first Correspondance volume) out of storage. It should show up sometime this week; you'll know when I add the titles.
Poncelet's solution looks like it might be different than the others, but I haven't understood it well enough to try to explain it yet. Again, sometime this week, I hope. Ozob ( talk) 20:45, 1 June 2008 (UTC)
OK, done with the references. I have some comments:
I still haven't looked at Poncelet's solution hard enough to see whether it's the same as Gergonne's or not. I don't think it's like any of the others. Ozob ( talk) 23:52, 4 June 2008 (UTC)
Honestly, I'm not really sure what it should look like, because I'm entirely ignorant of graphic design. What you've produced is a lot nicer than the raw scan, but you and your graphically literate (grapherate?) peers will have to decide what's appropriate. And yes, you're right, it ought to go on commons eventually. Ozob ( talk) 16:03, 5 June 2008 (UTC)
I do have Poncelet's article. That part came out better. The plate is actually on a fold-out sheet of paper (I guess you could call it a centerfold :-) which is why it didn't turn out so well: It wasn't covered by the other book pages, only by the librarian's hand as she made the photocopy for me. Using the copier cover would have crushed the binding and made the book fall apart. (And I'm not exaggerating, this volume was in terrible shape. The cover and first few pages had fallen entirely off; the whole thing came to me wrapped in a string.) I didn't get scans of all the other articles and solutions, but I could, and I bet they'll turn out okay; the first Correspondances volume was in reasonable shape, and the Nouvelles Annales volume, being 50 years younger, seemed quite healthy. I haven't read the Altshiller-Court article, so I don't know what's known and what's not; tell me where our information is lacking and I'll try to fill in as best as my library can.
As far as our article goes, we have a slight problem: Poncelet quotes prior work on the subject. As I said, I could get the first Correspondances volume back out of storage and see what it says, but I'm worried about how we could make it all fit in the article. For the moment, let me give you my best guess as to what he says. It doesn't make sense, reflecting my own lack of French fluency, but it gives an idea as to what's going on. Important: Any line or circle which is not named in the text is also not drawn in the picture! There are quite a few of these!
First, Poncelet resizes the smallest circle so that it is a point, thereby reducing to the CCP case. Call the point A and the two circles X and Y, respectively. There are two lines which are tangent to both X and Y, and Poncelet calls the intersection of these lines O. Next, Poncelet draws a secant line to X and Y through O. The secant line intersects X and Y in two points each, and Poncelet lets T and T' be the interior intersections of X and Y, respectively, with the secant line. The three points T, T' , and A determine a circle. The line from A to O meets this circle in two points: One of the points is A, and the other point Poncelet calls B. The secant-secant power theorem states that OA × OB = OT × OT' . Next, draw another secant line to X and Y having intersections t and t' with X and Y, respectively. By prior work on this problem ("voyez la page 20 du 1er vol. de la Correspondance"), we see that OT × OT' = Ot × Ot' , hence OA × OB = Ot × Ot' , which proves that A, B, t, and t' lie on a circle.
Poncelet now quotes the following fact from the same article: If a circle is tangent to X and Y at two points collinear with O, then either the circle contains both X and Y or it contains neither. So he will show that the circle tangent to X and Y and passing through A also passes through B. This reduces him to showing that through two points A and B, there is a circle which touches ("touche") X and Y. (???) Since there are two solutions, he says, he chooses the one which touches X and Y externally.
Next, let p and p' be the two other points of intersection of Ot with X and Y, respectively. It suffices to show that the circle through A, p, and p' passes through B, because a circle passing through A and touching X and Y exteriorly evidently passes through A and B. One verifies using the common tangents to X and Y that OT × OT' = Op × Op' , hence, by the equation above, Op × Op' = OA × OB. This proves that A, B, p, and p' lie on a circle.
If one lets O' be the intersection of the common interior tangents to X and Y, then one may repeat the process to get two more solutions. This lets him count the solutions, sort of. Let me give it to you in the original French: On peut voir facilement, en examinant les différentes circonstances du contact, que ce dernier probleme est susceptible de quatre solutions, et que par conséquent il se trouve entièrement résolu par ce que j'ai dit. ("One sees easily, by examining the different contact possibilities [external vs. internal], that this problem is susceptible of four solutions, and consequently it is entirely solved by what I have said.") Yes, that's correct for the CCP case; but he seems to have forgotten about his initial reduction. Oops! Ozob ( talk) 18:22, 5 June 2008 (UTC)
(Unindent.) I've thought about this more, and I think I understand better. I've transcribed a large part of the article so that you can see what I'm reading. (I didn't intend to do so much, but it really wouldn't have been worth it otherwise.) Thank you, Willow, for your comments; I think you've figured out, or at least pointed the way on, some of the tough parts. Poncelet begins with:
Mener par un point un cercle tangent à deux cercles donnés, en diminuant ou augmentant le rayon du cercle cherché du rayon du plus petit des trois cercles, suicant qu'il doit toucher ce dernier cercle extérieurement ou intérieuerment, ce qui revient à augmenter ou diminuer également les rayons des deux autres cercles d'après la nature de leur point de contact. I take this to mean that he's reducing to the CCP case.
Je vais d'abord démontrer la proposition suivante sur laquelle se fonde la solution du problême dont il est question : Si par le point O, fig. 4, pl. 4, où se coupent les tangentes extérieures communes aux cercles X et Y, et par le point A où doit passer le cercle tangent à ces deux cercles, on mène une droite AO, que l'on fasse passer ensuite par le point O une sécante quelconque OT, qui vient couper les cercles X et Y intérieurement en T et T' ; qu'enfin par ces deux points T et T' et par le point A on fasse passer un cercle, cette circonférence de cercle coupera A O en un point B qui sera le même, quelle que soit la sécante OT. Poncelet collapses several steps into this paragraph. First he looks at the common tangents to the two circles. He specifies that he wants the two tangents to meet externally in the point O. When Poncelet talks about internal versus external, he seems to be speaking loosely; I'm not sure that it's possible to make his notion precise. But his point is that you need to pick one pair of intersecting common tangents; later, to construct O' and another solution he uses a different pair. (How many of these points are there? I'm not convinced there are only two.) Then he takes any secant OT; that is, he draws any line through O that meets X and Y transversally, and he calls the points of intersection T and T' . Again, he has to make a choice; this time he chooses the "internal" intersections. I suspect that if you make the wrong choices, then his construction fails to give any solutions; this prevents him from getting too many solutions. Finally, using A, T, and T' , Poncelet constructs a circle. AO meets the circle twice, and the other point of intersection is B. If for some reason AO is tangent to the circle, then A = B; this must be some degenerate case.
En effet, OB et OT étant les sécantes d'un même cercle ABT, on a :
Mais si l'on mène une nouvelle sécante Ot, on a aussi ( voyez la page 20 du 1er vol. de la Correspondance ),
Il est évident, d'après cette dernière équation (1), que les quatre points T, T' , A et B sont placés sur une même circonférence de cercle. First we apply the secant-secant power theorem, as discussed above. Then Willow suggests that we invoke homology. I don't know anything about this kind of homology (I only know about the algebraic kind), so I'll defer to others.
Il est démontré aussi dans l'article cité, que tout cercle tangent aux cercles X et Y, a ses deux points de contact placés sur une droite qui passe par le point O, dans les deux cas où il laisse entièrement hors de sa circonférence, ou qu'il renferme à-la-fois les deux cercles X et Y. Here is the other fact that Poncelet is quoting: If a circle is tangent to X and Y and its points of contact with X and Y lie on a line passing through O, then there are two cases: The circle contains both X and Y, or it contains neither. Assuming this, he goes on to say, Il suit de là et de ce que j'ai démontré plus haut, que le cercle tangent aux cercles X et Y, et qui passe par le point A, passe aussi par le point B. Ainsi le problême dont il s'agit se trouve ramené à celui-ci: Par deux points A et B, mener un cercle qui touche le cercle X ou Y. This is what provoked the ??? from me when I read it initially. He says that it suffices to show that a circle tangent to X and Y and passing through A also passes through B. This might be true for the reason Willow gives, namely that the radical axis of two solutions passes through B. Then he says this will reduce him to the following problem: Through two points A and B, there is a circle which touches X or Y. I initially misread "et" for "ou", hence my confusion about the circle touching X and Y rather than X or Y. "Touche" here seems to mean "is tangent to". I don't see how this helps or where he's going, but I think he things this will solve the problem:
Comme ce dernier problême est susceptible de deux solutions, il est bon de faire voir que celle qui correspond au cas où le cercle est touché extérieurement, appartient aussi au cercle qui, passant par le point A, toucheroit extérieurement les cercles X et Y. Is he saying that we have a solution?
Pour le démontrer, il suffit de faire voir que toit cercle passant par le point A et par deux points p et p' , où une sécante quelconque Ot vient couper extérieurement les cercles X et Y, passera aussi par le même point B; car alors le cercle qui passe par le point A, et qui touche extérieurement les cercles X et Y, ayant ses points de contact dans la direction du point O, passera évidemment par les points A et B. Or, on voit sans peine (*) que OT × OT' = Op × Op' ; donc, d'après l'équation (1),
Cette équation prouve que les points A, B, p, p' , sont placés sur la même circonférence de cercle. As Willow says above, the important step (the "it suffices" part) seems to be something to do with homology. There is also a footnote at (*) where he comments that to get the equality, you use the common external tangents as I noted in my previous exposition way above. He finishes the article as I described above; I don't think we gain anything if I transcribe it. There's also a paragraph at the end where he talks about a generalization to cones circumscribing spheres; I haven't read it closely, but it seems to be a completely different kind of generalization from the ones we already have in the article.
A last comment. I should really put these scans somewhere where the rest of you can read them. Does anyone know if this is an appropriate thing for Wikisource? Ozob ( talk) 00:33, 6 June 2008 (UTC)
There's a lot of interesting material here, quite clearly described — well done! In view of any intended feature article nomination, I think it's worth taking a look first at the overall organization, to see whether it is appropriate or could be made better, rather than nitpicking the wording of individual sentences; that can come later.
First, I think it would be possible to use the algebraic solution to unify a lot of the seeming haphazardness of the numbers of solutions of the many different special cases shown. For instance, the case of three mutually tangent circles has either two solutions (the inner and outer Soddy circles) or five (those two plus the three original circles), not eight: why? Because the three original circles invert to themselves while the Soddy circles invert to each other. Thus, counting solutions according to their multiplicity, each of the Soddy circles counts once while the three original circles count two times each, for a total of eight. This even works when one circle separates the other two and there are no (real) solutions — in that case the eight solutions are all imaginary. Taking this point of view pervasively would, I think, make sense of what currently looks like a lot of ad hoc case analysis.
The other issue I have with the current organization is that it seems to be primarily organized around the mathematics of the problem, and not around its history. There's no obvious way to extract a clean timeline, partition the work on the problem into phases according to what types of mathematical knowledge were applied to it when, etc. I wouldn't want to get rid of the current mathematical content but maybe the history could be better integrated as well. — David Eppstein ( talk) 19:02, 1 June 2008 (UTC)
Hi all,
I have a little breather now, so I thought I'd write up my understanding of Poncelet's solution, and see whether you all agree? He seems to begin by reducing CCC to CCP by scaling; thanks, Ozob, for seeing that! :) The remaining two circles X and Y have two homothetic centers (centers of similitude), one external and one internal, which Poncelet denotes by O and O' , respectively. Any two rays emanating from such a center that intersect both circles define two pairs of antihomologous points, e.g., (T, T') and (t, t'); a simple proof involving similar triangles (which I'll add to homothetic center) shows that these four points always lie on a circle. By the secant-power theorem, the pairs of points (A, B) and (t, t') always lie on a common circle, where (t, t') is any pair of antihomologous points.
Here's the missing step, I think. Consider a solution circle that is tangent to X and Y at two points, call them U and U' . The two lines that are tangent to the solution circle at U and U' intersect the line through U and U' symmetrically at the same angle; you'll see a mirror image about the perpendicular bisector of the line segment U U' . Therefore, if a solution exists, its tangent points to the circles X and Y are antihomologous. Therefore (U, U') and (A, B) lie on a common circle, namely the solution circle. Thus, Poncelet reduces the problem to CPP, to find a solution circle that passes through A, B and is tangent to either X or (ou) Y. Poncelet chooses X and solves it as above, finding P and drawing the tangents to X to find the tangent points m and m' . Then , with three points, (A, B, m) and (A, B, m'), one can construct the two solutions for CPP. Using the interior homothetic center gives a different pair of solutions. Willow ( talk) 13:32, 7 June 2008 (UTC)
User:Nbarth added a comment [1] to the effect that the number of solutions to the Problem of Apollonius is the "oldest significant result" in enumerative geometry. This would be very interesting if we had a source for it, but I doubt that it's true. I'd say that "two points determine a line" is significant, so it would be better to say "oldest nontrivial result". Even then I'm not convinced: As Willow mentioned above, a pair of circles has exactly two homothetic centers. Does anyone have a reference for this fact? Ozob ( talk) 16:21, 10 June 2008 (UTC)
Hi Nbarth, You make a good point that Apollonius almost certainly knew that there were eight solutions in the general case. And counting the number of solutions in the general case is certainly a problem of enumerative geometry. But what I'm worried about is that we're hoping to eventually make this a Wikipedia:Featured article, so just about any claim we make will need an inline reference. That will include any claim about it being one of the oldest solved problems in enumerative geometry. I think it would be safe to say that it's old, but then the remark is pale and lifeless, don't you think? I don't think it's obvious that there are eight solutions for three general circles. If we started with three given conics in P2k, k algebraically closed, then I agree that it's obvious by Bezout's theorem. But instead we're working in R2. (By the way, here's an example of a nontrivial enumerative geometry problem solved earlier: How many regular polyhedra can be inscribed in a sphere? These are the Platonic solids and their classification (hence number) was known to Euclid.) Ozob ( talk) 17:09, 12 June 2008 (UTC)
To reply to above points:
Nbarth ( email) ( talk) 11:40, 22 June 2008 (UTC)
Those of you with this article on your watchlists have noticed this diff. It links to my transcription of Poncelet's article, now on the French wikisource. I'd like to do this for a number of the other sources in this article; as we noticed when I got some of the missing bibliographic information, it seems like the authors of the available historical surveys didn't look at the primary sources as hard as they should have. So if we're going to write a complete and accurate article, we'll all need to be able to study the primary sources. It turns out that there are a lot of those, so I'm proposing the following mini-project: I'll scan and OCR what I have access to, and whoever would like to can help proofread it and enter it on Wikisource. (This is not entirely trivial, because the OCR software I have access to is Adobe Acrobat, which gets confused sometimes. But it is much, much faster than just typing in everything by hand.) To transmit the images I can email them to you. If we do this then the online record will be complete, and we'll have a rock-solid foundation for our article.
Here are the articles which are public domain, not already available online, and which I think I have access to. Some of the bibliographic entries are not very specific, and hopefully we'd be able to fix those, too; but as a side effect of that inspecificity I might have made some errors when I looked at the library catalog: (Struck-through articles have been transcribed or scanned. Ozob ( talk) 23:22, 25 July 2008 (UTC))
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help)Some of these (the works by Gauss, Newton, Descartes, and Fermat) may already be online as part of project to scan a person's complete works; if that's the case, there should be a link to the complete works edition. At present the only person who has such links is Euler (which is why he's not listed above). Also, some of these are excerpts from books, and while I'm happy to scan a few pages, I'm not going to scan whole books. I don't know if such intentionally incomplete transcriptions can go on Wikisource or not. Lastly, all the above citations are copied straight from the article, and as you can see, some of them need to be converted to citation templates.
There are also the following works which are public domain and not already online but which I don't have access to:
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help) (The French National Library has the successor journal, Journal de mathématiques pures et appliquées, online, but not this one.)Anyone who can get at these is welcome to join in scanning.
So, is anyone up for helping me? Ozob ( talk) 01:34, 27 June 2008 (UTC)
Revisiting the page, I have just one idea concerning the images: if the colors of the three given circles would be equal, this would reflect the fact that they are interchangeable, and would ease the visual digestion of the images. Jakob.scholbach ( talk) 11:14, 27 July 2008 (UTC)
Let me be the first to welcome this one to GAN, Willow. It looks fabulous. - Dan Dank55 ( talk)( mistakes) 02:14, 19 August 2008 (UTC)
A few questions:
As a general note, I am not a specialist. I consider myself sharp enough to pick this sort of thing up after having it beaten over my head repeatedly, but this is not my forte nor my vocation. I won't attempt to check statements for validity or factual accuracy (except statements made outside the purview of geometry).
Overall this is an impressive article. I understood a good portion of it (more than I thought I would) and I felt that care was taken in writing it. I'll be happy to pass this article as soon as the quibbles above are dealt with. On hold. Protonk ( talk) 04:51, 19 August 2008 (UTC)
Hi Willow!
Seems like a very good article. I don't know anything about it, but here are a few comments:
Hope that is a bit helpful. I will continue reading later. Randomblue ( talk) 13:37, 22 September 2008 (UTC)
To everyone who has contributed, commented on or reviewed this article, congratulations and thank you for all your tremendous work.
Despite all our efforts, however, I think we all know that this article is an FA thanks almost entirely to the efforts of one editor: Willow. I therefore propose we celebrate on her talk page. She has not edited since 1 October, so she may not be able to join us in in actual edits, but her spirit infuses this article and the many editors she has touched, so she won't be far away or far from our thoughts! Geometry guy 19:58, 13 October 2008 (UTC)
Problem of Apollonius#Algebraic solutions says:
On my Windows XP Professional Version 2002 Service Pack 3 system, purchased from Hewlett-Packard June 2006, the second equation above appears as untranslated computerese instead of an equation. I can "fix" it by changing any of the 2's to another number (which would make the math wrong), but seven 2's in one expression is too many. One would think such a mistake wouldn't be in a featured article, so does the equation look OK on your systems? That is, is it the same as the other 2 equations, with a 2 instead of a 1 or a 3? One way to avoid seven 2's and thus fix it, is to separate the expression into 2 separate expressions, which looks better but still imperfect, like this:
Art LaPella ( talk) 02:21, 21 November 2008 (UTC)
If you get to this article via the link on the Main Page it redirects you to this page. The link on the Main Page should be changed. Tis the season to be jolly ( talk) 00:23, 22 November 2008 (UTC)
As soon as I saw this article on the front page, I added in the accents on the Greek word -- Ancient Greek, if possible, should never be written without accents unless it is in all-caps. So I changed Επαφαι to Ἐπαφαί. Problem is, I'm not an admin, so I can't fix the Today's featured article version, and so far no one has answered my request there. I figure more people read this page, so can someone with the authority please make that change for me? Thanks, Iustinus ( talk) 15:47, 22 November 2008 (UTC)
What if you have a small circle between two big ones? Then you can't have a solution that encloses all three while still touching the little one. —Preceding unsigned comment added by 134.153.216.129 ( talk) 16:37, 22 November 2008 (UTC)
This is the best treatment of the Problem of Apollonius that I have ever read. "Well Done" to all who've contributed to making such a fine article. htom ( talk) 17:04, 22 November 2008 (UTC)
Reference 48 goes to a broken link.-- Ssola ( talk) 12:18, 6 March 2010 (UTC)
I'm reading through es:Problema de Apolonio to check if it's a good article. The article came from here (through an intermediate version in Catalan), and the general quality is excellent. Kudos, by the way. The trouble is, although I am a scientist, I'm not a mathematician; moreover, our Catalan-Spanish translator is a (young!) teenager, and the user that did the English-Catalan translation says he has an en-2 level, plus he is not a mathematician either, more like an "older" teenager. I'm afraid that some content errors might have slipped in, and that should not happen in a good article. If someone that really understands all the math in this article (and reads some Spanish) can give our version a quick look to check for gross factual errors e.g. a wrong step in the solutions of the problem, it would be much appreciated. -- 4lex ( talk) 16:52, 6 August 2010 (UTC)
The article failed eswiki FA status because it was found that references aren't quite good, some differ from text statements, some are false/spurious, some are just copy from book indexes page numbers without really paying attention to where it actually states what's referenced, etc. Magister Mathematicae ( talk) 21:50, 13 November 2010 (UTC)
In this case I assume there is no Apollonius circle, so that the red and green circle are inside and the black circle is outside of that circle. Am I right? If yes: in which cases of 3 non-overlapping circles there are less than the promised 8 solutions? -- RokerHRO ( talk) 15:27, 16 February 2012 (UTC)
I find the article a bit confusing, is it about the unique Appolonius problem (the CCC problem or on all Appolonius problems? (the general versions are introduced much to late )
Also could there be sections that give compass-and-straightedge constructions to the problems solvable that way. (are they not all solvable this way?) WillemienH ( talk) — Preceding undated comment added 08:05, 8 June 2015 (UTC)
The following solution cannot be included in the article because the source does not meet Wikipedia standards. However, I wondered if it might encourage editors to investigate whether the Problem of Apollonius has ever been solved via vector algebra: The Problem of Apollonius as an Opportunity to Use Reflections and Rotations to Solve Geometry Problems via Geometric (Clifford) Algebra Xxyyzzyyxx ( talk) 20:09, 6 June 2016 (UTC)
Problem of Apollonius is a featured article; it (or a previous version of it) has been identified as one of the best articles produced by the Wikipedia community. Even so, if you can update or improve it, please do so. | ||||||||||||||||
This article appeared on Wikipedia's Main Page as Today's featured article on November 22, 2008. | ||||||||||||||||
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I do happen to have access to Correspondance sur l'Ecole polytechnique. You needn't sell your soul when I've already done so. ;-) The Poncelet solution begins, "Mener un cercle tangent à trois cercles donnés?" But you could probably guess it was something like that. :-) I'll look it over (though I need to do real work for a while longer).
I don't know anything about the circle method myself, but I have access to a large library. If you have any ideas on where I should look, please say. Otherwise I can look randomly. Ozob ( talk) 20:17, 30 May 2008 (UTC)
I've asked the library to take it (as well as the first Correspondance volume) out of storage. It should show up sometime this week; you'll know when I add the titles.
Poncelet's solution looks like it might be different than the others, but I haven't understood it well enough to try to explain it yet. Again, sometime this week, I hope. Ozob ( talk) 20:45, 1 June 2008 (UTC)
OK, done with the references. I have some comments:
I still haven't looked at Poncelet's solution hard enough to see whether it's the same as Gergonne's or not. I don't think it's like any of the others. Ozob ( talk) 23:52, 4 June 2008 (UTC)
Honestly, I'm not really sure what it should look like, because I'm entirely ignorant of graphic design. What you've produced is a lot nicer than the raw scan, but you and your graphically literate (grapherate?) peers will have to decide what's appropriate. And yes, you're right, it ought to go on commons eventually. Ozob ( talk) 16:03, 5 June 2008 (UTC)
I do have Poncelet's article. That part came out better. The plate is actually on a fold-out sheet of paper (I guess you could call it a centerfold :-) which is why it didn't turn out so well: It wasn't covered by the other book pages, only by the librarian's hand as she made the photocopy for me. Using the copier cover would have crushed the binding and made the book fall apart. (And I'm not exaggerating, this volume was in terrible shape. The cover and first few pages had fallen entirely off; the whole thing came to me wrapped in a string.) I didn't get scans of all the other articles and solutions, but I could, and I bet they'll turn out okay; the first Correspondances volume was in reasonable shape, and the Nouvelles Annales volume, being 50 years younger, seemed quite healthy. I haven't read the Altshiller-Court article, so I don't know what's known and what's not; tell me where our information is lacking and I'll try to fill in as best as my library can.
As far as our article goes, we have a slight problem: Poncelet quotes prior work on the subject. As I said, I could get the first Correspondances volume back out of storage and see what it says, but I'm worried about how we could make it all fit in the article. For the moment, let me give you my best guess as to what he says. It doesn't make sense, reflecting my own lack of French fluency, but it gives an idea as to what's going on. Important: Any line or circle which is not named in the text is also not drawn in the picture! There are quite a few of these!
First, Poncelet resizes the smallest circle so that it is a point, thereby reducing to the CCP case. Call the point A and the two circles X and Y, respectively. There are two lines which are tangent to both X and Y, and Poncelet calls the intersection of these lines O. Next, Poncelet draws a secant line to X and Y through O. The secant line intersects X and Y in two points each, and Poncelet lets T and T' be the interior intersections of X and Y, respectively, with the secant line. The three points T, T' , and A determine a circle. The line from A to O meets this circle in two points: One of the points is A, and the other point Poncelet calls B. The secant-secant power theorem states that OA × OB = OT × OT' . Next, draw another secant line to X and Y having intersections t and t' with X and Y, respectively. By prior work on this problem ("voyez la page 20 du 1er vol. de la Correspondance"), we see that OT × OT' = Ot × Ot' , hence OA × OB = Ot × Ot' , which proves that A, B, t, and t' lie on a circle.
Poncelet now quotes the following fact from the same article: If a circle is tangent to X and Y at two points collinear with O, then either the circle contains both X and Y or it contains neither. So he will show that the circle tangent to X and Y and passing through A also passes through B. This reduces him to showing that through two points A and B, there is a circle which touches ("touche") X and Y. (???) Since there are two solutions, he says, he chooses the one which touches X and Y externally.
Next, let p and p' be the two other points of intersection of Ot with X and Y, respectively. It suffices to show that the circle through A, p, and p' passes through B, because a circle passing through A and touching X and Y exteriorly evidently passes through A and B. One verifies using the common tangents to X and Y that OT × OT' = Op × Op' , hence, by the equation above, Op × Op' = OA × OB. This proves that A, B, p, and p' lie on a circle.
If one lets O' be the intersection of the common interior tangents to X and Y, then one may repeat the process to get two more solutions. This lets him count the solutions, sort of. Let me give it to you in the original French: On peut voir facilement, en examinant les différentes circonstances du contact, que ce dernier probleme est susceptible de quatre solutions, et que par conséquent il se trouve entièrement résolu par ce que j'ai dit. ("One sees easily, by examining the different contact possibilities [external vs. internal], that this problem is susceptible of four solutions, and consequently it is entirely solved by what I have said.") Yes, that's correct for the CCP case; but he seems to have forgotten about his initial reduction. Oops! Ozob ( talk) 18:22, 5 June 2008 (UTC)
(Unindent.) I've thought about this more, and I think I understand better. I've transcribed a large part of the article so that you can see what I'm reading. (I didn't intend to do so much, but it really wouldn't have been worth it otherwise.) Thank you, Willow, for your comments; I think you've figured out, or at least pointed the way on, some of the tough parts. Poncelet begins with:
Mener par un point un cercle tangent à deux cercles donnés, en diminuant ou augmentant le rayon du cercle cherché du rayon du plus petit des trois cercles, suicant qu'il doit toucher ce dernier cercle extérieurement ou intérieuerment, ce qui revient à augmenter ou diminuer également les rayons des deux autres cercles d'après la nature de leur point de contact. I take this to mean that he's reducing to the CCP case.
Je vais d'abord démontrer la proposition suivante sur laquelle se fonde la solution du problême dont il est question : Si par le point O, fig. 4, pl. 4, où se coupent les tangentes extérieures communes aux cercles X et Y, et par le point A où doit passer le cercle tangent à ces deux cercles, on mène une droite AO, que l'on fasse passer ensuite par le point O une sécante quelconque OT, qui vient couper les cercles X et Y intérieurement en T et T' ; qu'enfin par ces deux points T et T' et par le point A on fasse passer un cercle, cette circonférence de cercle coupera A O en un point B qui sera le même, quelle que soit la sécante OT. Poncelet collapses several steps into this paragraph. First he looks at the common tangents to the two circles. He specifies that he wants the two tangents to meet externally in the point O. When Poncelet talks about internal versus external, he seems to be speaking loosely; I'm not sure that it's possible to make his notion precise. But his point is that you need to pick one pair of intersecting common tangents; later, to construct O' and another solution he uses a different pair. (How many of these points are there? I'm not convinced there are only two.) Then he takes any secant OT; that is, he draws any line through O that meets X and Y transversally, and he calls the points of intersection T and T' . Again, he has to make a choice; this time he chooses the "internal" intersections. I suspect that if you make the wrong choices, then his construction fails to give any solutions; this prevents him from getting too many solutions. Finally, using A, T, and T' , Poncelet constructs a circle. AO meets the circle twice, and the other point of intersection is B. If for some reason AO is tangent to the circle, then A = B; this must be some degenerate case.
En effet, OB et OT étant les sécantes d'un même cercle ABT, on a :
Mais si l'on mène une nouvelle sécante Ot, on a aussi ( voyez la page 20 du 1er vol. de la Correspondance ),
Il est évident, d'après cette dernière équation (1), que les quatre points T, T' , A et B sont placés sur une même circonférence de cercle. First we apply the secant-secant power theorem, as discussed above. Then Willow suggests that we invoke homology. I don't know anything about this kind of homology (I only know about the algebraic kind), so I'll defer to others.
Il est démontré aussi dans l'article cité, que tout cercle tangent aux cercles X et Y, a ses deux points de contact placés sur une droite qui passe par le point O, dans les deux cas où il laisse entièrement hors de sa circonférence, ou qu'il renferme à-la-fois les deux cercles X et Y. Here is the other fact that Poncelet is quoting: If a circle is tangent to X and Y and its points of contact with X and Y lie on a line passing through O, then there are two cases: The circle contains both X and Y, or it contains neither. Assuming this, he goes on to say, Il suit de là et de ce que j'ai démontré plus haut, que le cercle tangent aux cercles X et Y, et qui passe par le point A, passe aussi par le point B. Ainsi le problême dont il s'agit se trouve ramené à celui-ci: Par deux points A et B, mener un cercle qui touche le cercle X ou Y. This is what provoked the ??? from me when I read it initially. He says that it suffices to show that a circle tangent to X and Y and passing through A also passes through B. This might be true for the reason Willow gives, namely that the radical axis of two solutions passes through B. Then he says this will reduce him to the following problem: Through two points A and B, there is a circle which touches X or Y. I initially misread "et" for "ou", hence my confusion about the circle touching X and Y rather than X or Y. "Touche" here seems to mean "is tangent to". I don't see how this helps or where he's going, but I think he things this will solve the problem:
Comme ce dernier problême est susceptible de deux solutions, il est bon de faire voir que celle qui correspond au cas où le cercle est touché extérieurement, appartient aussi au cercle qui, passant par le point A, toucheroit extérieurement les cercles X et Y. Is he saying that we have a solution?
Pour le démontrer, il suffit de faire voir que toit cercle passant par le point A et par deux points p et p' , où une sécante quelconque Ot vient couper extérieurement les cercles X et Y, passera aussi par le même point B; car alors le cercle qui passe par le point A, et qui touche extérieurement les cercles X et Y, ayant ses points de contact dans la direction du point O, passera évidemment par les points A et B. Or, on voit sans peine (*) que OT × OT' = Op × Op' ; donc, d'après l'équation (1),
Cette équation prouve que les points A, B, p, p' , sont placés sur la même circonférence de cercle. As Willow says above, the important step (the "it suffices" part) seems to be something to do with homology. There is also a footnote at (*) where he comments that to get the equality, you use the common external tangents as I noted in my previous exposition way above. He finishes the article as I described above; I don't think we gain anything if I transcribe it. There's also a paragraph at the end where he talks about a generalization to cones circumscribing spheres; I haven't read it closely, but it seems to be a completely different kind of generalization from the ones we already have in the article.
A last comment. I should really put these scans somewhere where the rest of you can read them. Does anyone know if this is an appropriate thing for Wikisource? Ozob ( talk) 00:33, 6 June 2008 (UTC)
There's a lot of interesting material here, quite clearly described — well done! In view of any intended feature article nomination, I think it's worth taking a look first at the overall organization, to see whether it is appropriate or could be made better, rather than nitpicking the wording of individual sentences; that can come later.
First, I think it would be possible to use the algebraic solution to unify a lot of the seeming haphazardness of the numbers of solutions of the many different special cases shown. For instance, the case of three mutually tangent circles has either two solutions (the inner and outer Soddy circles) or five (those two plus the three original circles), not eight: why? Because the three original circles invert to themselves while the Soddy circles invert to each other. Thus, counting solutions according to their multiplicity, each of the Soddy circles counts once while the three original circles count two times each, for a total of eight. This even works when one circle separates the other two and there are no (real) solutions — in that case the eight solutions are all imaginary. Taking this point of view pervasively would, I think, make sense of what currently looks like a lot of ad hoc case analysis.
The other issue I have with the current organization is that it seems to be primarily organized around the mathematics of the problem, and not around its history. There's no obvious way to extract a clean timeline, partition the work on the problem into phases according to what types of mathematical knowledge were applied to it when, etc. I wouldn't want to get rid of the current mathematical content but maybe the history could be better integrated as well. — David Eppstein ( talk) 19:02, 1 June 2008 (UTC)
Hi all,
I have a little breather now, so I thought I'd write up my understanding of Poncelet's solution, and see whether you all agree? He seems to begin by reducing CCC to CCP by scaling; thanks, Ozob, for seeing that! :) The remaining two circles X and Y have two homothetic centers (centers of similitude), one external and one internal, which Poncelet denotes by O and O' , respectively. Any two rays emanating from such a center that intersect both circles define two pairs of antihomologous points, e.g., (T, T') and (t, t'); a simple proof involving similar triangles (which I'll add to homothetic center) shows that these four points always lie on a circle. By the secant-power theorem, the pairs of points (A, B) and (t, t') always lie on a common circle, where (t, t') is any pair of antihomologous points.
Here's the missing step, I think. Consider a solution circle that is tangent to X and Y at two points, call them U and U' . The two lines that are tangent to the solution circle at U and U' intersect the line through U and U' symmetrically at the same angle; you'll see a mirror image about the perpendicular bisector of the line segment U U' . Therefore, if a solution exists, its tangent points to the circles X and Y are antihomologous. Therefore (U, U') and (A, B) lie on a common circle, namely the solution circle. Thus, Poncelet reduces the problem to CPP, to find a solution circle that passes through A, B and is tangent to either X or (ou) Y. Poncelet chooses X and solves it as above, finding P and drawing the tangents to X to find the tangent points m and m' . Then , with three points, (A, B, m) and (A, B, m'), one can construct the two solutions for CPP. Using the interior homothetic center gives a different pair of solutions. Willow ( talk) 13:32, 7 June 2008 (UTC)
User:Nbarth added a comment [1] to the effect that the number of solutions to the Problem of Apollonius is the "oldest significant result" in enumerative geometry. This would be very interesting if we had a source for it, but I doubt that it's true. I'd say that "two points determine a line" is significant, so it would be better to say "oldest nontrivial result". Even then I'm not convinced: As Willow mentioned above, a pair of circles has exactly two homothetic centers. Does anyone have a reference for this fact? Ozob ( talk) 16:21, 10 June 2008 (UTC)
Hi Nbarth, You make a good point that Apollonius almost certainly knew that there were eight solutions in the general case. And counting the number of solutions in the general case is certainly a problem of enumerative geometry. But what I'm worried about is that we're hoping to eventually make this a Wikipedia:Featured article, so just about any claim we make will need an inline reference. That will include any claim about it being one of the oldest solved problems in enumerative geometry. I think it would be safe to say that it's old, but then the remark is pale and lifeless, don't you think? I don't think it's obvious that there are eight solutions for three general circles. If we started with three given conics in P2k, k algebraically closed, then I agree that it's obvious by Bezout's theorem. But instead we're working in R2. (By the way, here's an example of a nontrivial enumerative geometry problem solved earlier: How many regular polyhedra can be inscribed in a sphere? These are the Platonic solids and their classification (hence number) was known to Euclid.) Ozob ( talk) 17:09, 12 June 2008 (UTC)
To reply to above points:
Nbarth ( email) ( talk) 11:40, 22 June 2008 (UTC)
Those of you with this article on your watchlists have noticed this diff. It links to my transcription of Poncelet's article, now on the French wikisource. I'd like to do this for a number of the other sources in this article; as we noticed when I got some of the missing bibliographic information, it seems like the authors of the available historical surveys didn't look at the primary sources as hard as they should have. So if we're going to write a complete and accurate article, we'll all need to be able to study the primary sources. It turns out that there are a lot of those, so I'm proposing the following mini-project: I'll scan and OCR what I have access to, and whoever would like to can help proofread it and enter it on Wikisource. (This is not entirely trivial, because the OCR software I have access to is Adobe Acrobat, which gets confused sometimes. But it is much, much faster than just typing in everything by hand.) To transmit the images I can email them to you. If we do this then the online record will be complete, and we'll have a rock-solid foundation for our article.
Here are the articles which are public domain, not already available online, and which I think I have access to. Some of the bibliographic entries are not very specific, and hopefully we'd be able to fix those, too; but as a side effect of that inspecificity I might have made some errors when I looked at the library catalog: (Struck-through articles have been transcribed or scanned. Ozob ( talk) 23:22, 25 July 2008 (UTC))
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help)Some of these (the works by Gauss, Newton, Descartes, and Fermat) may already be online as part of project to scan a person's complete works; if that's the case, there should be a link to the complete works edition. At present the only person who has such links is Euler (which is why he's not listed above). Also, some of these are excerpts from books, and while I'm happy to scan a few pages, I'm not going to scan whole books. I don't know if such intentionally incomplete transcriptions can go on Wikisource or not. Lastly, all the above citations are copied straight from the article, and as you can see, some of them need to be converted to citation templates.
There are also the following works which are public domain and not already online but which I don't have access to:
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help) (The French National Library has the successor journal, Journal de mathématiques pures et appliquées, online, but not this one.)Anyone who can get at these is welcome to join in scanning.
So, is anyone up for helping me? Ozob ( talk) 01:34, 27 June 2008 (UTC)
Revisiting the page, I have just one idea concerning the images: if the colors of the three given circles would be equal, this would reflect the fact that they are interchangeable, and would ease the visual digestion of the images. Jakob.scholbach ( talk) 11:14, 27 July 2008 (UTC)
Let me be the first to welcome this one to GAN, Willow. It looks fabulous. - Dan Dank55 ( talk)( mistakes) 02:14, 19 August 2008 (UTC)
A few questions:
As a general note, I am not a specialist. I consider myself sharp enough to pick this sort of thing up after having it beaten over my head repeatedly, but this is not my forte nor my vocation. I won't attempt to check statements for validity or factual accuracy (except statements made outside the purview of geometry).
Overall this is an impressive article. I understood a good portion of it (more than I thought I would) and I felt that care was taken in writing it. I'll be happy to pass this article as soon as the quibbles above are dealt with. On hold. Protonk ( talk) 04:51, 19 August 2008 (UTC)
Hi Willow!
Seems like a very good article. I don't know anything about it, but here are a few comments:
Hope that is a bit helpful. I will continue reading later. Randomblue ( talk) 13:37, 22 September 2008 (UTC)
To everyone who has contributed, commented on or reviewed this article, congratulations and thank you for all your tremendous work.
Despite all our efforts, however, I think we all know that this article is an FA thanks almost entirely to the efforts of one editor: Willow. I therefore propose we celebrate on her talk page. She has not edited since 1 October, so she may not be able to join us in in actual edits, but her spirit infuses this article and the many editors she has touched, so she won't be far away or far from our thoughts! Geometry guy 19:58, 13 October 2008 (UTC)
Problem of Apollonius#Algebraic solutions says:
On my Windows XP Professional Version 2002 Service Pack 3 system, purchased from Hewlett-Packard June 2006, the second equation above appears as untranslated computerese instead of an equation. I can "fix" it by changing any of the 2's to another number (which would make the math wrong), but seven 2's in one expression is too many. One would think such a mistake wouldn't be in a featured article, so does the equation look OK on your systems? That is, is it the same as the other 2 equations, with a 2 instead of a 1 or a 3? One way to avoid seven 2's and thus fix it, is to separate the expression into 2 separate expressions, which looks better but still imperfect, like this:
Art LaPella ( talk) 02:21, 21 November 2008 (UTC)
If you get to this article via the link on the Main Page it redirects you to this page. The link on the Main Page should be changed. Tis the season to be jolly ( talk) 00:23, 22 November 2008 (UTC)
As soon as I saw this article on the front page, I added in the accents on the Greek word -- Ancient Greek, if possible, should never be written without accents unless it is in all-caps. So I changed Επαφαι to Ἐπαφαί. Problem is, I'm not an admin, so I can't fix the Today's featured article version, and so far no one has answered my request there. I figure more people read this page, so can someone with the authority please make that change for me? Thanks, Iustinus ( talk) 15:47, 22 November 2008 (UTC)
What if you have a small circle between two big ones? Then you can't have a solution that encloses all three while still touching the little one. —Preceding unsigned comment added by 134.153.216.129 ( talk) 16:37, 22 November 2008 (UTC)
This is the best treatment of the Problem of Apollonius that I have ever read. "Well Done" to all who've contributed to making such a fine article. htom ( talk) 17:04, 22 November 2008 (UTC)
Reference 48 goes to a broken link.-- Ssola ( talk) 12:18, 6 March 2010 (UTC)
I'm reading through es:Problema de Apolonio to check if it's a good article. The article came from here (through an intermediate version in Catalan), and the general quality is excellent. Kudos, by the way. The trouble is, although I am a scientist, I'm not a mathematician; moreover, our Catalan-Spanish translator is a (young!) teenager, and the user that did the English-Catalan translation says he has an en-2 level, plus he is not a mathematician either, more like an "older" teenager. I'm afraid that some content errors might have slipped in, and that should not happen in a good article. If someone that really understands all the math in this article (and reads some Spanish) can give our version a quick look to check for gross factual errors e.g. a wrong step in the solutions of the problem, it would be much appreciated. -- 4lex ( talk) 16:52, 6 August 2010 (UTC)
The article failed eswiki FA status because it was found that references aren't quite good, some differ from text statements, some are false/spurious, some are just copy from book indexes page numbers without really paying attention to where it actually states what's referenced, etc. Magister Mathematicae ( talk) 21:50, 13 November 2010 (UTC)
In this case I assume there is no Apollonius circle, so that the red and green circle are inside and the black circle is outside of that circle. Am I right? If yes: in which cases of 3 non-overlapping circles there are less than the promised 8 solutions? -- RokerHRO ( talk) 15:27, 16 February 2012 (UTC)
I find the article a bit confusing, is it about the unique Appolonius problem (the CCC problem or on all Appolonius problems? (the general versions are introduced much to late )
Also could there be sections that give compass-and-straightedge constructions to the problems solvable that way. (are they not all solvable this way?) WillemienH ( talk) — Preceding undated comment added 08:05, 8 June 2015 (UTC)
The following solution cannot be included in the article because the source does not meet Wikipedia standards. However, I wondered if it might encourage editors to investigate whether the Problem of Apollonius has ever been solved via vector algebra: The Problem of Apollonius as an Opportunity to Use Reflections and Rotations to Solve Geometry Problems via Geometric (Clifford) Algebra Xxyyzzyyxx ( talk) 20:09, 6 June 2016 (UTC)