Talk:Problem of Apollonius

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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)[reply]

You've no idea how happy that makes me; thanks for adding those titles and, more fervently, for trying to understand the Poncelet solution! I totally empathize; it took me weeks to understand the Gergonne solution. :P Oh, and could you also find the title for the Hachette reference? I left a note for

circle method application; I trust that he's right, though. :) Willow (talk) 12:28, 1 June 2008 (UTC)[reply

]

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)[reply]

OK, done with the references. I have some comments:

• The volumes crumble when you touch them. By the time I was done there was a scattering of aged brown bits of paper over the desk I was working at. I've never seen a better argument for "ROOM USE ONLY".
• The first volume is Correspondance sur l'Ecole polytechnique whereas the second is Correspondance sur l'Ecole Impériale polytechnique.
• Some of the dates, volume numbers, and page numbers we had were wrong. I have corrected them.
• I have a scan of a figure that Poncelet uses in his solution. This might, might be appropriate for the article. I don't feel like the scan came out too well—it's entirely legible, but I wouldn't say that it's pretty. Someone with Photoshop expertise would have to touch it up before we could use it. The actual file is a 1.32MB png, but here's a thumbnail:

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)[reply]

Here's a quick and dirty cleanup of the image -- just a high-radius unsharp mask to enhance the contrast, radius-1 median filter to blur the background a little, and threshold to whiten the background. There's still a bit of noise left, making it obvious it's a scan. Is that what you had in mind or did you want a more careful cleanup that takes the background down to pure white and leaves a clean line drawing? And shouldn't this end up on commons rather than en? —David Eppstein (talk) 00:39, 5 June 2008 (UTC)[reply]

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)[reply]

A cleaner version. I think I like it better than the previous sloppier one. I'll let you take care of uploading it to commons with appropriate sourcing and integrating it into the article. —David Eppstein (talk) 17:22, 5 June 2008 (UTC)[reply]

You're the bee's knees, Ozob, for going to all that trouble; from millions of amateur geometers, thank you! Thank you, David, for making it look nice, too. :)

It's a beautiful image, and also excellent for referencing our discussion of Poncelet's solution. I can't imagine that anyone would {{

fact

}}-tag our discussion if we could present them with the original online. :) I suggest, though, that if we want the image for explaining Poncelet's solution (rather than for giving historical flavour), we should remake the figure as an SVG. I would be happy to do that, although if someone else wanted to do it, that'd be great, too. :)

I think I can interpret some of the image. The point O at the far right is the external center of similitude (homothetic center); the pairs of points (p, t) and (p^prime, t^prime) are (anti)homologous, and lie on the two middle circles centered on X and Y, respectively. The large encompassing circle above, and the complete circle below are a pair of solutions; they intersect at points A and B. The line passing through A and B is the radical axis of the two solutions, which also passes through the external center of similitude. Meanwhile, the two points of tangency on the given circle X to these two solutions are labeled by m and m^prime; the tangent lines at those points intersect on the radical axis of the solution pair (Figure 11 here). (Maybe m and m^prime represent the tangent lines themselves?) Thus far, it's similar to Gergonne's solution, but then it seems to diverge.

This is rather speculative, but I suspect that Poncelet's goal is to find the radical axis of the solution circles. Since the point O is known, he is seeking another point on the radical axis, possibly P. The points T and T^prime on the two given circles X and Y are (anti)homologous, since a line through them also passes through O. Somehow Poncelet knows that the four points (T, T^prime, A, and B) all lie on a circle (not a solution circle), and he also knows that the radical axis of the given circle X and that common circle also passes through P. (Some theorems about homothetic centers presumably help him to conclude these things.) Then perhaps by choosing two different pairs (T, T^prime) and, say, (U, U^prime) he can locate P by intersecting lines, and thence the tangent points m and m^prime, and thence the solutions.

I don't want the poor books to get damaged, but could you photocopy/photograph the whole article? And — I'm getting greedy— maybe those of the other authors, such as Cauchy? The problem is that there's only one historical article about Apollonius' problem, that of Altshiller-Court, and as you point out, he seems not to have looked at the original texts. So none of us has much of a clue about the solutions from that Monge era, except as to whether they were algebraic or geometric. I'll completely understand if you can't, but an online copy of those solutions would be a wonderful boon to the article and to other interested readers. Whatever you decide is fine! :)

Thanks a million an Apollonian gasket's worth of circles, Willow (talk) 17:44, 5 June 2008 (UTC)[reply]

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

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.

That theorem is great, isn't it? The OT × OT' = Ot × Ot' thing follows from the theorem that two pairs of antihomologous points (4 points in all) lie on a circle; I remember seeing the proof in Johnson (1929). Willow (talk) 19:03, 5 June 2008 (UTC)[reply]

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.

I think the gist of the argument is like this, although perhaps Poncelet wouldn't have worded it thus. The two solutions must intersect in A, the given point. Their radical axis, which passes through their two intersection points A and (call it) Q, must also pass through O, a center of similarity. Next Poncelet shows that Q=B? Willow (talk) 19:03, 5 June 2008 (UTC)[reply]

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.

The key step, and one that I don't really understand, is that the two antihomologous intersection points of any ray from the center of similarity O for the two circles X and Y must lie on a circle that passes through A and B, and vice versa. Then if the tangent points m and (say) n of the given circles X and Y with the solution circle lie on such a ray, then the solution circle must itself pass through B, which was to be proven Q=B. Willow (talk) 19:39, 5 June 2008 (UTC)[reply]

Now we get a solution. Draw the circle through A, T, and T' (again). Draw a chord lT, and let its point of intersection with the line AO be denoted by P. Through P, one draws the tangents Pm and Pm' to the circle X. A degenerate case of the secant-secant power theorem says that Pm² = Pl × PT, and since Pl × PT = PA × PB (power theorem again), we get Pm² = PA × PB. Therefore the circle through A, B, and m touches the line Pm at m. Similarly, the circle through A, B, and m' touches Pm' at m' . The end?

This final part I understand, I think. The line mP is the radical axis for the given circle and the chosen solution. One point on such a radical axis can always be found by drawing a circle (ABTT') that intersects both the given circle and the solution circle. The lines connecting their respective intersection points (T and l for the given circle, A and B for the solution circle) intersect on a point P that lies on the radical axis. Since the solution is known to be tangent, it suffices to draw tangent lines to the given circle from P, thus identifying m and m' , from which the two solutions can be found, using A and B. Willow (talk) 19:23, 5 June 2008 (UTC)[reply]

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)[reply]

Maybe Poncelet says something at the beginning of the article about how to get the other solutions? Maybe like shrinking on of the other circles to a point? Willow (talk) 19:23, 5 June 2008 (UTC)[reply]

Wow,

the inversive solution to CCP is so much easier! I think I need to digest this, too, but let me first finish my thing with the roots on the article... Willow (talk) 18:36, 5 June 2008 (UTC)[reply

]

(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 :

AO × OB = OT × OT' .

Mais si l'on mène une nouvelle sécante Ot, on a aussi ( voyez la page 20 du 1er vol. de la Correspondance ),

OT × OT' = Ot × Ot' ;

donc AO × OB = Ot × Ot'  (1).

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

, 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),

Op × Op' = AO × OB.

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)[reply]

Yes, the French transcriptions and the English translations would be perfect for their respective Wikisources. The scans themselves might be better on the Commons; they're definitely in the public domain, and there's a category there for the Problem of Apollonius.

I think I understand the proof. It's nicely complementary to the Gergonne solution, and a good lead-in to that. Unfortunately, I got distracted today with Emmy Noether — sorry about that! Willow (talk) 21:34, 6 June 2008 (UTC)[reply]

PS. You might like the stuff I added to homothetic center, although I forgot to label the exterior homothetic center in the first image.  :( For two circles, there are only two centers of similitude, one interior and one exterior, where their common tangents intersect. The homology thing is discussed there, briefly, but I can add some more this weekend. Willow (talk) 21:41, 6 June 2008 (UTC)[reply]

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)[reply]

Thank you, David! :) I hadn't clued in about how the given circles are solutions when the roots are real and degenerate — cool! :) I do appreciate the idea of starting with the algebraic solution, which for a while was indeed the organization of the article. In modern times, people do seem to know more about solving for roots than they do about geometry, so it'd be more familiar. On the other hand, it'd be an anachronism, since the geometrical solutions came first (Apollonius, van Roomen, Vieta, Newton, etc.) and then the algebraic ones.

A historical organization might be good — I'm always fond of a good story told well ;)— but I think that there wasn't a clean separation by time; for instance, the algebraic and "new" geometrical solutions of Monge's school overlapped in time. I also wanted to give special attention to the van Roomen/Newton and Vieta solutions, the former as foreshadowing for the trilateration application and the latter for looking backwards to Apollonius. But I'm not attached to the present organization; how would you recommend that we change it? Willow (talk) 18:07, 5 June 2008 (UTC)[reply]

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)[reply]

I added that proof about the antihomologous points lying on a circle to the homothetic center article. What do people think about this interpretation of Poncelet's method — have I overlooked something? Does it follow? If you agree with it, I'll add it to the article, and then maybe solicit a peer review before FAC. Thanks! :) Willow (talk) 22:45, 10 June 2008 (UTC)[reply]

PS. The FAC won't happen for some time yet; I'm engrossed in Emmy Noether! Any help there would be very welcome as well. Willow (talk) 22:45, 10 June 2008 (UTC)[reply]

I ASSUME THAT EVERYONE AGREES WITH THIS READING OF PONCELET'S PROOF. If not, now would be a good time to pipe up, before I add it to the article. ;)

As an aside, I was the one who added the link to

Bezout's theorem

. It seemed appropriate, but I had no citation for that, so you did well to delete it. I can't pretend to follow your discussion below, but maybe your ignoring of the signs is causing you the difficulty of 16 solutions? I see all too plainly how much I have left to learn! :)

Since I haven't heard any objections, I'm going to assume that it's OK if I put this article up for peer review? Thanks for everyone's help! :) Willow (talk) 18:32, 24 June 2008 (UTC)[reply]

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)[reply]

Hi Ozob,

Agreed, the original wording was a bit strong. I had dismissed “two points determine a line” as an axiom, though of course as an observation it’s certainly an enumerative result. I think it’s largely a judgment call which early result is the earliest significant result. I’ve changed the wording so that it’s less judgmental – how does it look?

Nbarth (email) (talk) 22:21, 10 June 2008 (UTC)[reply]

Hi Nbarth, and thanks for your contributions to the article! :)

I'm inclined to agree with Ozob that it would be good to have a reference? I'm concerned because the number of solutions wasn't really discussed until Muirhead's work in 1896, and even his work wasn't fully correct; the correct enumeration was given only in 1983. Of course, the general number of solutions — eight for the general CCC problem — was known earlier, but exactly how much earlier, I'm not sure; I'll have to re-read some of the old proofs. Where did you read about Apollonius' problem and enumerative geometry? Willow (talk) 22:40, 10 June 2008 (UTC)[reply]

Hi Willow,

I came upon this page because I was familiar with the geometric figure and name (thanks to XScreenSaver); in the past, I studied algebraic geometry (including enumerative geometry), so when I saw the result (on 8 circles), I added a link to enumerative geometry – I know nothing about Apollonius’s problem beyond what’s in the article.

The article claims that Apollonius’s original solution was lost, but it’s reasonable to believe that he did solve it – and at the time people didn’t give abstract existence proofs, only constructions.

That there are 8 solutions for 3 general circles is obvious (internally/externally tangent gives you a binary choice, so 3 choices yields ${\displaystyle 2^{3}}$, and in general these can all be realized), so I don’t doubt that this case is ancient – the point about not correctly enumerating the various cases recalls how Euclid’s axioms omitted order axioms, which was fixed in Hilbert's axioms.

Nbarth (email) (talk) 22:58, 11 June 2008 (UTC)[reply]

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

, 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 P²_k, 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 ]

Hey all! :)

I agree that Apollonius probably knew that there were eight solutions in general; I give him credit for being a genius! :) Unfortunately, the only surviving report we have of his work (by Pappus) doesn't mention that result, so we can't know for sure. Also, just because there are eight possible tangency conditions doesn't mean that there must be eight solutions; at least, that isn't so clear to me. For example, the algebraic equations appear to give two solutions for every tangency condition; the symmetry between r and −r eliminates half the solutions, although maybe you could show that the c coefficient was always negative (implying that one root of the quadratic had to be negative).

I kind of agree with Ozob that we can't mention the priority of teh Apollonius enumeration unless we can find a book somewhere that supports the assertion. Surely there must be a book that discusses the history of enumerative geometry?

That's cool about the

GPS. :) Willow (talk) 23:08, 12 June 2008 (UTC)[reply

]

Hi everyone! (I like greeting people, let me do it again.) Hi! Hi!!!!! :-)

OK, now I'm not convinced that having eight solutions is obvious. Here's what I was thinking: If we work in P²_k, k algebraically closed, then we ought to be able to forget about the signs because any two conics are projectively equivalent, and the equations that x_s, y_s, and r_s need to satisfy are quadratic. Then you just count by Bezout's theorem: 2×2×2=8, like the article says. But I don't think that applies to the problem: Those three variables aren't the homogenous coordinates of a point in the projective plane, and while (if you homogenized) you could eliminate one of the signs (you can switch it using a projective transformation) you can't do that to all three simultaneously. So you still have to consider at least four sign combinations, and that gives too many solutions. Anyone know where the Bezout's theorem statement in the article comes from?

Here's another idea (which, as we will see, does not quite work). Parameterize all (projective) conics as ${\displaystyle ax^{2}+bxy+cxz+dy^{2}+eyz+fz^{2}}$ for some constants a through f. Modulo scaling, we have a five dimensional space P⁵ of conics. We'll construct an incidence correspondence. We want to look at incidences between conics (i.e., points in P⁵) and tangent vectors in P²; TP² is non-projective, so instead we'll use X, its projective completion. (I.e., add points at infinity; more formally, X = Proj Sym (Ω⊕O). We can't work with the projectivized tangent bundle because that throws out the zero tangent vector, the one we're interested in.) Note that X is bigraded, so P⁵×X is trigraded. Inside P⁵×X, we'll look at the locus Y consisting of all points C (for "conic") of P⁵ and all tangent vectors (possibly infinitely long) to points of P² such that the tangent vector lies over a point lying on the conic represented by C. In other words, Y is the projective completion of the relative tangent bundle of the universal conic over P⁵. It's seven dimensional. The universal conic is degree 1 in the coefficients (the P⁵ variables) and degree 2 in the P² variables, so Y is also. Relative to P⁵×X, (that is, in the tangency direction) Y has degree one because each fiber is a linear subspace. Y will be our incidence correspondence.

To ensure that we get tangency, we let Z be the preimage in P⁵×X of the zero section of X. The zero section of X has degree zero with respect to the P⁵ variables and the P² variables and degree 1 in the tangency direction. It's seven dimensional (since it's P⁵×P²). And finally we need three given conics. Each of these is a point in P⁵; take the preimages of these points in P⁵×X to get three four dimensional varieties C₁, C₂, C₃, each of degree 1 in P⁵ and degree zero in the other directions.

Intersect all of these. We have a seven dimensional variety, a seven dimensional variety, and three four dimensional varieties, all inside a nine dimensional space. If we wanted to use this technique to prove existence, we'd be stuck: These varieties are codimension 2, 2, 5, 5, and 5, so the expected codimension is 19, giving an expected dimension of -10. So the varieties are meeting specially, which I guess is not really a surprise. Assuming that solutions exist and that there are only finitely many, however, then we only need to look at the degree: The tridegrees of these varieties are (1, 2, 1), (0, 0, 1), (1, 0, 0), (1, 0, 0), and (1, 0, 0). Degrees add when you intersect, so the total degree is (4, 2, 2). So we should get 16 solutions? Hmm. I'm not seeing why the number is too big at the moment. I might have screwed up, but I don't see anything wrong at the moment. If this could be made to work I suppose that it would qualify as another solution to Apollonius' problem. (It'd be a very, very 20th century solution.) Ozob (talk) 20:02, 13 June 2008 (UTC)[reply]

To reply to above points:

oldest…enumerative geometry

I’m mostly concerned that we cite enumerative geometry, and early on, as that’s a very useful context for the result, and I don’t think saying “this is a result in enumerative geometry” requires any citation. I’m not familiar with the chronology, but agree that it would be valuable

“obvious”

By “obvious” I mean in the Ancient Egyptian/Babylonian sense: it’s demonstrably true in many cases that come to mind (just try drawing circles), and a heuristic for why (oh, it’s either internally or externally tangent, so ${\displaystyle 2^{3}=8}$) is readily apparent, but agreed:

• a rigorous proof (in the nice case) is non-trivial (the greater glory for Apollonius), and
• a careful enumeration of cases is quite subtle and took until the 20th century.

XScreenSaver

The hack is named apollonian, though it illustrates Descartes' theorem. It is currently the 6th screenshot at the gallery: XScreenSaver: Screenshots (Written by Allan R. Wilks and David Bagley; 2002.) I’ve noted it on that page at Descartes' theorem#External links.

Nbarth (email) (talk) 11:40, 22 June 2008 (UTC)[reply]

Thanks for the screensaver link, Nbarth! :) I think it's showing the Apollonian gasket as well as Descartes' theorem.

I agree that the counting of solutions falls within enumerative geometry, although it would be nice to have a citation for the role, if any, that this problem played in its history. Unfortunately, the solutions of this problem don't seem "obvious" to me; at least, I don't see any simple way of iterating to a solution, as one can in, say, bisecting or trisecting an angle. There seem to be too many degrees of freedom (placement of solution center, plus radius of solution circle) to do a simple search. If you don't mind, I think we have to downplay the link to enumerative geometry in the lead. Willow (talk) 18:39, 24 June 2008 (UTC)[reply]

Hi Willow,

Agreed – the classical problem is more about the construction of solutions (as is classic) than the enumeration, and I’ve no problem with enumerative geometry (a topic dear to my heart but of admittedly more specialized interest) being only mentioned in a more relevant section. BTW, thanks for the fixes to enumerative geometry!

Nbarth (email) (talk) 21:21, 24 June 2008 (UTC)[reply]

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))[reply]

• Muirhead RF (1896). "On the Number and nature of the Solutions of the Apollonian Contact Problem". Proceedings of the Edinburgh Mathematical Society. 14: 135–147, attached figures 44–114.
• Descartes R, Oeuvres de Descartes, Correspondance IV, (C. Adam and P. Tannery, Eds.), Paris: Leopold Cert 1901.
• Beecroft H (1842). "Properties of Circles in Mutual Contact". Lady’s and Gentleman’s Diary. 139: 91–96. (the library says it has ibid., 1846, p. 51, but it's not available, I'd guess because it's in preservation)
• Pappus (1876). F Hultsch (ed.). Pappi Alexandrini collectionis quae supersunt (3 volumes ed.).
• ISBN 0-521-08719-8. {{cite book}}: |pages= has extra text (help); Cite has empty unknown parameter: |1= (help
  )
• Casey J (1881). A Sequel to Euclid. pp. p. 122. {{cite book}}: |pages= has extra text (help)
• Viète F (1970). Apollonius Gallus. Seu, Exsuscitata Apolloni Pergæi Περι Επαφων Geometria ((Paris, 1600) photographic reproduction of the Opera Mathematica (1646), Schooten FA, editor ed.). New York: Georg Olms. pp. pp. 325–346. {{cite book}}: |pages= has extra text (help)
• Robert Simson (1734) Mathematical Collection, volume VII, p. 117.
• Zeuthen HG (1886). Die Lehre von den Kegelschnitten im Altertum. Copenhagen: Unknown. pp. pp. 381–383. {{cite book}}: |pages= has extra text (help)
• Coaklay GW (1860). "Analytical Solutions of the Ten Problems in the Tangencies of Circles; and also of the Fifteen Problems in the Tangencies of Spheres". The Mathematical Monthly. 2: 116–126.
• ISBN 3-487-04636-9
  .
• Carnot L (1801). De la corrélation dans les figures de géométrie. Paris: Unknown publisher. pp. No. 158–159. (it's marked, "Special collections, by appointment only", which sounds like they won't let me scan it; but there's also "Oeuvres mathématiques du citoyen Carnot" which isn't so restricted)
• Carnot L (1803). Géométrie de position. Paris: Unknown publisher. pp. p. 390, &sect, 334. {{cite book}}: |pages= has extra text (help) (Special collections again)
• Cauchy AL (July 1806). "Du cercle tangent à trois cercles donnés". Correspondance sur l'Ecole Polytechnique. 1 (6): pp. 193–195. {{cite journal}}: |pages= has extra text (help
  ) (Should have scanned this before, but fortunately this volume of the Correspondance is in decent shape)
• Petersen J (1879). Methods and Theories for the Solution of Problems of Geometrical Constructions, Applied to 410 Problems. London: Sampson Low, Marston, Searle & Rivington. pp. pp. 94–95 (Example 403). {{cite book}}: |pages= has extra text (help) (turns out this is a translation, though oddly the catalog doesn't give the original title or language)
• Salmon G (1879). A Treatise on Conic Sections, Containing an Account of Some of the Most Important Modern Algebraic and Geometric Methods. London: Longmans, Green and Co. pp. 110–115, 291–292.
• Vannson (1855). "Contact des cercles sur la sphère, par la geométrie". Nouvelles Annales de Mathématiques. XIV: 55–71. (again, I should have scanned this one last time)
• Lachlan R (1893). An elementary treatise on modern pure geometry. London: Macmillan. pp. §383–396, pp. 244-251.
  ASIN B0008CQ720
  .
• de Fermat P, Varia opera mathematica, p. 74, Tolos, 1679.
• Hachette JNP (September 1808). "Sur le contact des sphères; sur la sphère tangente à quatre sphères données; sur le cercle tangent à trois cercles donnés". Correspondance sur l'Ecole Polytechnique. 1 (2): pp. 27–28. {{cite journal}}: |pages= has extra text (help)
• Francais, Jacques (January 1810). "De la sphère tangente à quatre sphères données". Correspondance sur l'Ecole Impériale Polytechnique. 2 (2): pp. 63–66. {{cite journal}}: |pages= has extra text (help)
• Francais, Jacques (January 1813). "Solution analytique du problème de la sphère tangente à quatre sphères données". Correspondance sur l'Ecole Impériale Polytechnique. 2 (5): pp. 409–410. {{cite journal}}: |pages= has extra text (help)
• Dupin C (January 1813). "Mémoire sur les sphères". Correspondance sur l'Ecole Impériale Polytechnique. 2 (5): p. 423. {{cite journal}}: |pages= has extra text (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:

• Beecroft H (1846). "Unknown title". Lady’s and Gentleman’s Diary. 139: 51. (as mentioned above, the library has this but says it's not available)
• Camerer JG (1795). Apollonii de Tactionibus, quae supersunt, ac maxime lemmata Pappi, in hos libros Graece nunc primum edita, e codicibus manuscriptis, cum Vietae librorum Apollonii restitutione, adjectis observationibus, computationibus, ac problematis Apolloniani historia. Gothae: Ettinger.
• van Roomen A (1596). Problema Apolloniacum quo datis tribus circulis, quaeritur quartus eos contingens, antea a...Francisco Vieta...omnibus mathematicis...ad construendum propositum, jam vero per Belgam...constructum. Würzburg.
• Gergonne J (1813–1814). "Recherche du cercle qui en touche trois autres sur une sphère". Ann. math. pures appl. 4. {{cite journal}}: Check date values in: |date= (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)[reply]

Hi Ozob,

I'm back from my sister's wedding! I'll be happy to help you with all that, although I need to warn you that I have very limited access to library materials, and I've never scanned or OCR'd anything in my whole life. Where does one go to find a scanner? Maybe we should also take care of the concerns on the peer review as well? Forgive me, but I'll be on Wikipedia only spottily as I dive back into the real world; I was away for a long time! Willow (talk) 20:25, 25 July 2008 (UTC)[reply]

Hi Willow,

The library I use has a scanner right there, so for me it's no problem. You could try asking. Also, it turns out that this work is not particularly fun (certainly less fun than writing articles). I typed in the Poncelet and Beecroft articles (they're both on Wikisource), but for Muirhead's article I discovered that Wikimedia Commons is happy to post a scan: Image:On the Number and Nature of Solutions to the Apollonian Contact Problem.djvu. I can't think of a good way to add this reference to the article, so I haven't done it yet. And since sitting in front of a scanner is boring, I ended up doing other things.

The peer review concerns are probably higher priority, since after all most FAs don't feature a full complement of historical sources! There are of course the inevitable copeyedits, but I think the biggest concern is structural: Balance, organization, etc. As you can see on the peer review page, I think that separating the history and organizing the remainder of article by solution type is a good idea—but I could be convinced otherwise. As

regular! Everyone else doesn't know what they're missing! :-) Ozob (talk) 23:22, 25 July 2008 (UTC)[reply

]

Images

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)[reply]

That's a good idea, Jakob! I've been troubled that something was wrong with the images, but I've also been loath to change them, since they were so much work to make. :P I'll try to fix one image, and we'll see how we like that; then maybe we can extend that to the other images? Cleonis also pointed out to me that my images tend to be too big, too many bytes, so this is also a good opportunity to shrink them. :)

One concern I have is that, in a few places, we need to distinguish between the three given circles. For example, we try to explain the difference between external and internal tangencies to the reader, as in the animation showing the circles growing and shrinking. We need to point out that the internally tangent given circles grow and shrink with the solution circle, whereas the externally tangent circles do the opposite, shrinking as the solution grows and vice versa.

Let me work on that. I'm also conscious that I owe you a group of reviews — or maybe it was the reverse? ;) Willow (talk) 17:01, 29 July 2008 (UTC)[reply]

Update: I changed the given circles to black, as you suggested. I hope you like them! :) Willow (talk) 19:29, 18 August 2008 (UTC)[reply]

Let me be the first to welcome this one to GAN, Willow. It looks fabulous. - Dan

]

Thank you, Dan! This is my first GA nomination, although I worked like crazy to save Catullus 2 way back when; so I was a little nervous. It's wonderfully reassuring to be welcomed by you. :) Willow (talk) 10:22, 19 August 2008 (UTC)[reply]

A few questions:

• "point in a place": point in a plane?

Yes, my bad; I missed that. Fixed.

• "generalizations to even higher dimensions are also possible": I have a gut feeling that some FAC reviewer is going to say that "even" is redundant, just in case I don't catch this at FAC.

I liked "even", since we go from two dimensions to three, and then "even higher". Onwards and upwards ;) Willow (talk) 10:22, 19 August 2008 (UTC)[reply]

I like "even" too. - Dan

talk)(mistakes) 16:17, 19 August 2008 (UTC)[reply

]

• If anyone gives you trouble over "A rich repertoire...have..." (not "has"), let me know. I'm armed (with style guidelines) and ready.

"Hey, I'm a poet and didn't even know it." ;) That grammatical infelicity was an accident, but I kind of like it. If anyone disses your reasoning, I'll send an unkindness of ravens after them — or even a murder of crows. ;) Willow (talk) 10:22, 19 August 2008 (UTC)[reply]

• "published in 1596 in Adriaan van Roomen": in him?

Ummm — oops.

• "van Roomen's method has a flaw": keep "flaw" if you want to, but I prefer "drawback", at least at this point in the article, since it hasn't been stated at this point that compass-and-straightedge is a requirement. Likewise, I might delete the "remedied this flaw" in the next paragraph and just say that Viète "developed a method..." - Dan
  talk)(mistakes) 02:51, 19 August 2008 (UTC)[reply
  ]

"Drawback" is excellent! My thesaurus gave me a few others: "shortcoming", "blemish", "imperfection", etc. but I like the sound of yours best. I liked "flaw" for its monosyllaby, and as an oblique diamond metaphor, but this is better. Willow (talk) 10:22, 19 August 2008 (UTC)[reply]

• I would wikilink the first occurrence of "foci".

Okey-dokey, that's smart. :)

• "by Apollonius' definition of the circle" Should this be "the solution circle"?

This needed to be clarified, so I tried to and added a small image. I hope that's better? Willow (talk) 10:22, 19 August 2008 (UTC)[reply]

• When you first mention LPP, I don't think you've said that L is for "line", so maybe add "(a line and two points)".

Gave more of an explanation.

• Per WP:MOSNUM, the minus sign is unspaced when it means "negative" (search WP:MOSNUM for "unary"). Of course, not everyone is impressed with everything MOSNUM says. (You have − r and − s).

How do you know what's lurking in the darkest corners of my heart? How do you know? But I'm not a rebel by nature, and dread the FAC conflict, so I amended those. Thanks for catching them! :) Willow (talk) 10:22, 19 August 2008 (UTC)[reply]

• How do you feel about giving "Lie sphere" a respelling (pronounced "Lee") or an IPA pronunciation at the first occurrence?

Added pronunciation and Sophus himself to the "History" section.

• I would wikilink the first occurrence of "resp." to either Wikipedia or Wiktionary.

On further reflection, I wrote out both clauses. The parallel construction had the virtues of brevity and flow, but elegant rhetoric is misplaced if your readers don't get you. Willow (talk) 10:22, 19 August 2008 (UTC)[reply]

• Okay, quick copyedit is done, but as usual, I didn't do "image issues" (although the captions are good) or endsections. Best of luck! - Dan
  talk)(mistakes) 03:51, 19 August 2008 (UTC)[reply
  ]

Thank you so much, Dan; that was way more pleasant than I was fearing. :) Please let me know if I can return the favour someday, Willow (talk) 10:22, 19 August 2008 (UTC)[reply]

Thanks Willow, it's always a pleasure. Something went wrong with the fix to "Lie", now we've got this sentence: "Another approach uses Lie sphere geometry, Nicolas Fuss, Carl Friedrich Gauss, Lazare Carnot, and Augustin Louis Cauchy." - Dan

talk)(mistakes) 14:40, 19 August 2008 (UTC)[reply

]

GA Review

This review is

transcluded from Talk:Problem of Apollonius/GA1

. The edit link for this section can be used to add comments to the review.

Protonk comments

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).

Thank you so much, Proton! Your review is wonderful and nicely complementary to Dan's just above it. You put your finger on some key things that needed improving. I'll try to fix these up today.  :) Willow (talk) 11:53, 19 August 2008 (UTC)[reply]

• Images Image tags all check out. Each image seems to provide some good illustration of this obviously geometric problem.
• Style/MOS

• Lead seems good as far as length and breadth are concerned
• "...are sometimes called Apollonius circles, although the term is ambiguous." Ambiguous links directly to Circles of Apollonius. I gather that I am to understand this to mean the sentence "These "circles of Apollonius" should not be confused with the Apollonian circles, which have a different technical definition." highlights the ambiguity. Is there a better way to make this notion clear to the reader?

I expanded this discussion and also improved the disambiguation page, Circles of Apollonius. Willow (talk) 16:25, 19 August 2008 (UTC)[reply]

• "These objects may be arranged in any way and may cross one another; however, they are usually taken to be distinct." Would this be improved if a link to Distinct were provided? That article doesn't really help the reader too much and it might be considered a dicdef, so YMMV.

Your suggestion of the link to distinct was great, so I added it and also fleshed this idea out more. Willow (talk) 16:25, 19 August 2008 (UTC)[reply]

• "In practice, two circles are tangent if they coincide or intersect at only one point; this is also true for a line and a circle" Couldn't this be better stated by showing the two plane counterexamples for circles?

Again, excellent suggest that I took over. I also expanded the discussion here as well. Willow (talk) 16:25, 19 August 2008 (UTC)[reply]

• "Apollonius' problem can also be formulated as the problem of locating one or more points such that the differences of its distances to three given points equal three known values" This should be explained as carefully as the geometric formulation of the problem. Links to the trilateration page don't seem immediately helpful.

Yes, it does seem as though

GPS. I also moved up the definitions of internal and external tangencies, since they seemed to flow nicely after the definition of tangency itself. Willow (talk) 22:53, 19 August 2008 (UTC)[reply

]

• Blue links. Several persons and concepts are linked multiple times throughout the piece. Unless you intend to link them multiple times (e.g. linking to Newton in the lead and again in the applications), following links should be removed.

Since people sometimes read articles out of order, or skip around to the sections they care about, I generally give a link every section or every few sections. I'll go through the article, though, and see if I went overboard somewhere, as I'm wont to do. Thank you! :) Willow (talk) 12:04, 19 August 2008 (UTC)[reply]

As a note, I really like how this article uses links for internal navigation. Don't touch those. Protonk (talk) 14:15, 19 August 2008 (UTC)[reply]

• Do the sources describe Van Roomen's method as flawed because of the inability to construct it with a compass and straightedge? I can certainly see that it could be judged as not a 'real' solution to the problem, but flawed brings to mind underlying error.

You and Dan both had the same association, so I changed "flaw" to "drawback". I chose "flaw" originally mainly because I liked its sound, a punchy monosyllable that wakes the reader up. I was also thinking of Van Roomen's proof as a diamond with a single imperfection. The idea the sources convey is just that his proof left something to be desired, since it wasn't yet known whether Apollonius' problem could be solved by ruler and straightedge, until Viète's proof. Willow (talk) 12:04, 19 August 2008 (UTC)[reply]

• Newton's and Van Roomen's solutions are written in past tense, but Viète's is written in present tense.

Again — oops.

OK, I think this has been fixed? Willow (talk) 22:53, 19 August 2008 (UTC)[reply]

• Is the bolded "point, point, point" notation (PPP) common to a student of this discipline? Should be quickly explained or (more drastically) dispensed with?

You make a very good point! I've tried now to explain it quickly at the beginning of that section. The notation does seem to be generally used in sources written after 1950, and maybe earlier; I need to check that. Originally, the "Special cases" section was above the History section, so that I didn't need to explain the notation. Willow (talk) 12:04, 19 August 2008 (UTC)[reply]

• Solution by inversion section: "...orange circle CG in Figure 6 crosses the red, blue and green given circles at right angles" I don't see red or green circles. Only black, pink and orange (in figure 6).

Thank you for catching that! At Jakob's suggestion, I'd changed the colors of several Figures, to help people understand them, but I missed a few color-specific references in the text. :( I think that I've caught them all now, though. :) Willow (talk) 16:35, 19 August 2008 (UTC)[reply]

• "If two of the three given circles are disjoint" Disjoint might benefit from a link.

I changed it to "If two of the three given circles do not intersect", which is consistent with the rest of the article. No need to introduce new words like "disjoint" if we don't have to. :) Willow (talk) 16:35, 19 August 2008 (UTC)[reply]

• "...Apollonius' problem degenerates to the CCP limiting cases," Perhaps a more direct link?
  limiting cases for CCP

Thank you, that was YAES (yet another excellent suggestion). ;) Willow (talk) 16:35, 19 August 2008 (UTC)[reply]

• The rest looks ok.

• Sourcing and claims

• Is there a source independent from John Casey (mathematician) that claims he refined Van Roomen's method independent from Newton almost 200 years after the publication of the Principia? That doesn't seem unbelievable, just unlikely.

I feel a little silly for not realizing that; I'm a little too trusting sometimes. :P Let me check my sources more carefully and get back to you with a more precise and verifiable statement. Willow (talk) 11:53, 19 August 2008 (UTC)[reply]

• Format

• I like the format and layout.
• POV No problems.

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)[reply]

• Updates: Once the Viète solution tense and differences of distances problem statement issues are fixed, I'll be happy to pass this article. As an aside, it is interesting to see how circle packing theorems intersect with Mathematics of paper folding, see this TED talk. Protonk (talk) 16:44, 19 August 2008 (UTC)[reply]
  • PS Another link related to your aside. This seems off-topic from the present article, though. —David Eppstein (talk) 17:28, 19 August 2008 (UTC)[reply]

Cool! :) I'd never heard of TED conferences or Robert Lang before, but I was fascinated; thank you, dead people! :) I've never learned origami, although I do like to make talking puppets out of paper for my nieces and god-daughter. :) You're motivating me to watch all those origami videos on YouTube, like this one. :) Willow (talk) 22:53, 19 August 2008 (UTC)[reply]

• Excellent job, Protonk. Even if you did point out all the things I missed. No, I don't mind! - Dan
  talk)(mistakes) 18:38, 19 August 2008 (UTC)[reply
  ]

Comments from Randomblue

Hi Willow!

Seems like a very good article. I don't know anything about it, but here are a few comments:

Hi Random! :) It's nice to be working together again. I'll try to address your concerns/questions/suggestions below. Willow (talk) 15:57, 23 September 2008 (UTC)[reply]

• 1) "lines (circles of infinite radius)" -> Why are lines circles of infinite radius?

To me, this seems hard to explain unless you see it intuitively? Still, I'll do my best to justify it. A line is a curve with zero curvature everywhere; a circle is a curve with a constant curvature everywhere that equals one over its radius; ergo, taking the limit of the circle's radius going to infinity produces a curve with zero curvature everywhere=a line. There's a sly little point missing here — namely the point at infinity ;) — but in my opinion, that proviso doesn't need to be mentioned in an article written at this level. Willow (talk) 15:57, 23 September 2008 (UTC)[reply]

Ok, I see. You take the curvature 'viewpoint', but I was thinking of the area 'viewpoint'. But I guess all is well when we consider the point at infinity. Randomblue (talk) 20:10, 23 September 2008 (UTC)[reply]

• 2) "fractals—curves of fractional dimension—" -> Is that a definition?
  Gosper islands
  have dimension log(8)/log(7).

I'm not sure if I get your point? I think you're saying that "fractional dimension" implies that "dimension=rational number", which clearly need not be the case. I was trying to say only that "dimension=non-integer" and take advantage of the acoustic/etymological fact that "fractal" and "fractional" sound alike and might make it easier for non-mathematicians to understand that more intuitively. Willow (talk) 15:57, 23 September 2008 (UTC)[reply]

Fixed up in the lead? Willow (talk) 16:10, 23 September 2008 (UTC)[reply]

Yes, that's it. It is true that they both sound alike :). Randomblue (talk) 20:10, 23 September 2008 (UTC)[reply]

• 3) 1/3 of the lead describes possible generalizations of the problem, however the actual "Generalizations" paragraph is about 1/15 of the whole text. This seems disproportionate.

I sympathize with that, but as I wrote at the FAC, I consider the lead to be a standalone entity with a possibly different audience than the main body. Therefore, it needs to represent the main body, but not necessarily proportionately; in particular, the lead needs to be accessible to a broader audience. Therefore I compressed the sundry technical methods for solving Apollonius' problem (and their history of development) into a single sentence. I dwelt on the applications and generalizations because they could be understood without much effort or prior training. Willow (talk) 15:57, 23 September 2008 (UTC)[reply]

That's fine with me. Randomblue (talk) 20:10, 23 September 2008 (UTC)[reply]

• 4) "If the angle between the objects at an intersection point is zero" -> How are angles defined in general for "objects"? What's the angle between a circle and a point lying on it?

I re-ordered the setences to take care of points first, and then lines and circles. Is it OK now? Willow (talk) 16:16, 23 September 2008 (UTC)[reply]

Theoretically that sounds great, I check it later. Randomblue (talk) 20:10, 23 September 2008 (UTC)[reply]

• 5) "the given circles equal d1=r1+rs" but "solution circle and C1 is either rs + r1 or rs − r1" -> Inconsistent spacing between +'s and -'s. I prefer spaces in between.

Me, too. I think I've fixed that now? Thanks for catching that! :) Willow (talk) 15:57, 23 September 2008 (UTC)[reply]

Great, it changed a few spacings with = signs. Maybe other spacing issues are lurking around. Randomblue (talk) 20:10, 23 September 2008 (UTC)[reply]

• 6) "Célèbres Problêmes mathématiques." The current spelling for 'problem' in French is 'problème', not 'problême'. However, the old spelling might have been 'problême', although I doubt it. Also, French titles usually only have a capital on the first word.
• 7) "Pappus d'Alexandrie: La Collection Mathématique" check capitalization.
• 8) "Correspondance sur l'Ecole Polytechnique" probably add accent on Ecole.
• 9) "Traité de Géométrie" capitalization
• 10) "Oeuvres de Descartes" 'Oeuvres' should be 'Œuvres'
• 11) "((pronounced [liː], as "Lee")" there are two opening brackets. But anyway, why is this information pertinent?

I forget why I added that? Consider it disparu. ;) Willow (talk) 15:57, 23 September 2008 (UTC)[reply]

• 12) "Eppstein D (2001). "Tangent Spheres and Triangle Centers"" You've linked David, but is it the right one?

Yes, that was the article I intended, since ti deals with mutually tangent circles in the plane, and with some of the special cases of Apollonius' problem. Is there another one you thought I should add? Willow (talk) 16:23, 23 September 2008 (UTC)[reply]

Slight misunderstanding here. I'm just pointing out that there are two mathematicians who are called David Ep(p)stein, namely David Eppstein and David Epstein, the second of which deals more with geometry. Randomblue (talk) 20:10, 23 September 2008 (UTC)[reply]

I can confirm that the "Tangent Spheres and Triangle Centers" paper is by me and not the one-p David E. There is much geometry in both of our researches. —David Eppstein (talk) 15:15, 26 September 2008 (UTC)[reply]

• 13) "Apollonius' problem is a pure problem" is that a common phrasing? Maybe "Apollonius' problem is a problem in pure mathematics"
• 14) "Apollonius himself was famously indifferent to the practical applications of his work" is the citation coming just after a proof of that? Because Apollonius said the problem was worthy of study for its own sake doesn't mean he was indifferent to the practical applications of his work. The citation seems a bit disconnected.
• 15) "Apollonius' problem has other applications as well." redundant wording.

Re 13–15, I've removed that quote and re-written that section.

• 16) "Correspondance sur l'Ecole Impériale Polytechnique" accent on Ecole, two occurrences
• 17) "Francais, Jacques" I expect it is Français.
• 18) Why is David Eppstein the only one linked in "Further reading"?

I couldn't find Wikipedia articles about the others, but I'll look again. Is there someone for whom you know an article? Willow (talk) 16:23, 23 September 2008 (UTC)[reply]

• 19) Inconsistent plurals. "hyperbolas" and "hyperbolae"

Sigh, yes. Being a Latin fan, I prefer hyperbolae, but people will have it otherwise, so I'm bowing to the modern era. Willow (talk) 15:57, 23 September 2008 (UTC)[reply]

• 20) "The problem of counting the number of feasible solutions" what's the difference between a 'feasible solution' and a 'solution'?

Hmmm, another example of overly salicine enthusiasm. ;) Willow (talk) 15:57, 23 September 2008 (UTC)[reply]

Hope that is a bit helpful. I will continue reading later. Randomblue (talk) 13:37, 22 September 2008 (UTC)[reply]

Thanks again! I always appreciate your keen eyes. :) Willow (talk) 15:57, 23 September 2008 (UTC)[reply]

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)[reply]

Problem of Apollonius#Algebraic solutions says:

${\displaystyle \left(x_{s}-x_{1}\right)^{2}+\left(y_{s}-y_{1}\right)^{2}=\left(r_{s}-s_{1}r_{2}\right)^{2}}$

${\displaystyle \left(x_{s}-x_{2}\right)^{2}+\left(y_{s}-y_{2}\right)^{2}=\left(r_{s}-s_{2}r_{2}\right)^{2}}$

${\displaystyle \left(x_{s}-x_{3}\right)^{2}+\left(y_{s}-y_{3}\right)^{2}=\left(r_{s}-s_{3}r_{3}\right)^{2}.}$

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:

${\displaystyle \left(x_{s}-x_{1}\right)^{2}+\left(y_{s}-y_{1}\right)^{2}=\left(r_{s}-s_{1}r_{2}\right)^{2}}$

${\displaystyle \left(x_{s}-x_{2}\right)^{2}+\left(y_{s}-y_{2}\right)^{2}=}$

${\displaystyle \left(r_{s}-s_{2}r_{2}\right)^{2}}$

${\displaystyle \left(x_{s}-x_{3}\right)^{2}+\left(y_{s}-y_{3}\right)^{2}=\left(r_{s}-s_{3}r_{3}\right)^{2}.}$

Art LaPella (talk) 02:21, 21 November 2008 (UTC)[reply]

Fixed. This is a known bug, caused by the server for some reason creating a corrupted image for a specific TeX input. It can be fixed by someone with access to the servers deleting the corrupted image, or by making any change whatsoever to the code. I've made a change that doesn't affect rendering at all. Algebraist 14:41, 21 November 2008 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

That's mentioned in the section 'number of solutions' and pictured in figure 11. Algebraist 16:56, 22 November 2008 (UTC)[reply]

That's one of the special cases: http://en.wikipedia.org/wiki/Problem_of_Apollonius#Number_of_solutions htom (talk) 17:04, 22 November 2008 (UTC)[reply]

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)[reply]

Reference 48 goes to a broken link.--Ssola (talk) 12:18, 6 March 2010 (UTC)[reply]

Fixed.--Ssola (talk) 17:23, 14 March 2010 (UTC)[reply]

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)[reply]

Being teenager doesn't discredits my translation, although it might contain errors as the translation of anybody else. In fact, I was already concerned about the correctness of the maths of the article at this age (nowadays I'm studying mathematics at undergraduate level). Furthermore, I contacted with the Catalan Service of Terminology to obtain accurate translations of some terms. The translation of ca:Problema d'Apol·loni was reviewed by experimented users with a better english level than mine. That's why it achieved the "Featured" (not featrued) status. I'm interested in the gaps you found in the Spanish review, some of which may be easy to correct in the other versions. --Ssola (talk) 20:02, 2 March 2012 (UTC)[reply]

Note

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)[reply]

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)[reply]

This question is more than a year old, but I want to address it anyway. In the image, the large circle appears to be centered on (17.5, 0) and it has a radius of 16. The other two are unit circles centered on (0, 2) and (0, -2). Eight distinct solutions do exist. No, it is not possible to have the red and green circles on the interior and the black on the exterior, but it is best not to look at them that way. Think of the circle as you would a line. Rather than having an inside and an outside, it simply has two sides. In this case there are two solutions having the black circle on one side and the red and green on the other. These are the approximate center coordinates of those two solutions: (-0.58, 1.05), (-22.99, 13.31) --Geometricks (talk) 14:10, 16 March 2013 (UTC)[reply]

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

]

A solution via Geometric (Clifford) Algebra

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)[reply]