Talk:Descartes' theorem

message: I can't help wondering of Descartes' theorem should be a disambiguation page?  Thoughts? Michael Hardy 22:35, 23 May 2005 (UTC)
thread_title:  Disambiguation 

message: Is "curvature" the right term for ±1/r? I've seen explanations with curvature as always positive, 1/r, and "bend" used for the positive or negative curvature. Daggerbox (talk) 14:29, 13 February 2008 (UTC)
thread_title:  Curvature 

message: This article takes entirely too long before it states even the gist of Descartes' Theorem.  Ideally, at least the gist should appear in the lede, and the full statement should have been given by the next section (if not within the lede). A reader should not, for example, have first to read of the history of the theorem. —SlamDiego←T 03:26, 2 November 2008 (UTC)
thread_title:  Organization 

message: The article mentions how the theorem has been extended to spheres and even arbitrary dimension hyperspheres. Is it possible anyone can add the equations for either of those? Even just 3D spheres for now would be nice.  —Preceding unsigned comment added by Skytopia (talk • contribs) 06:03, 26 February 2009 (UTC)
thread_title:  Equations for spheres too?  Quaternions? 

message: Testing the combination k_i = 1 1 1 gives  3 - 2*Sqrt(3) for the inner circle, which is negative. Is there supposed to be an initial +-? 121.45.179.128 (talk) 08:23, 16 December 2009 (UTC)
replies: text: It is outer circe. Inner will be 3+2*Sqrt(3). You can also look at wikibooks article. --Adam majewski (talk) 18:04, 15 August 2010 (UTC)
thread_title:  Equation needs clarifying 

message: In par 3 under Special Cases:
"...as it is not possible for three lines and one circle to be mutually tangent." - my immediate reaction was 'incircle of a triangle?' Probably should be clarified that mutual tangency of three lines makes no sense. But then the previous case of two parallel lines being considered mutually tangent to a pair of circles also becomes questionable, as tangency implies having common points. Thoughts?  —Preceding unsigned comment added by SwiftSurge (talk • contribs) 16:02, 13 August 2010 (UTC)
replies: text: It's not really true that "mutual tangency of three lines makes no sense". In the geometry of circles and lines it makes sense to think of lines as a limiting case of circles with infinite radius, and to think of parallel lines as being a limiting case of tangent circles. For instance, a Möbius transformation can transform tangent circles into parallel lines or vice versa. So with these conventions it is indeed possible for three lines to be mutually tangent: just make them all parallel. However in this case it would only be possible to make the fourth thing that is tangent to all of them be another line, not a circle. —David Eppstein (talk) 00:58, 14 August 2010 (UTC)
thread_title:  Clarify in Special Cases? 

message: There seems to be an implicit assertion that given three mutually tangent circles, there always exists a fourth—but this isn't clearly spelled out.  Given a formula such as  k_4 = k_1 + k_2 + k_3 \pm2 \sqrt{k_1 k_2 + k_2 k_3 + k_3 k_1}. \,, it's reasonable to ask if the expression under the square root can be negative.  Certainly if k_3 is zero, then the other circles must be external (you can't have a straight line fitting inside a circle), so k_1 and k_2 will be positive.  And if k_3 is strictly negative, the first two circles must be small enough to fit inside the third, so k_1 and k_2 are both positive and bigger in absolute value than k_3.  So I think Descartes' theorem does imply existence of a fourth circle.  But what I'm doing here is original research; is there a reference that clearly deals with the existence question? Jowa fan (talk) 04:19, 24 June 2011 (UTC)
replies: text: It's a lot easier than that if you ignore the formula and think about it more geometrically. Perform an inversion centered on one of the points of tangency; the resulting configuration will consist of two parallel lines with a circle sandwiched between them, at which point it should be obvious that there are two choices for the fourth circle. Actually, according to Problem of Apollonius#Mutually tangent given circles: Soddy's circles and Descartes' theorem there are three more solutions: each of the given three circles is itself mutually tangent to all three circles as well. So anyway, yes, existence is known. —David Eppstein (talk) 06:48, 24 June 2011 (UTC)
replies: text: That's all good info; this article should include enough of it to say that "such a circle must exist" and link where User:David Eppstein mentions.    yoyo (talk) 02:54, 25 January 2021 (UTC)
thread_title:  Existence? 

message: In its introduction the article refers to kissing circles.  Yet the standard definition of kissing, i.e. Osculating circle, implies 2nd order contact.  This is one order beyond tangency.  Two osculating circles would be the same circle, since their curvature would be the same. It seems kissing should be removed and only mutual tangency should be given as a condition.  — Preceding unsigned comment added by 66.188.89.180 (talk) 19:50, 20 March 2014 (UTC)
replies: text: They are certainly not osculating circles. But Soddy used the word "kissing" and for that reason I believe it should remain in the article. —David Eppstein (talk) 16:07, 21 March 2014 (UTC)
thread_title: Kissing?

message: I've been writing a program to generate Apollonian Gaskets, and I've come to a minor hiccup.  It seems the ± sign in Equation 2 does not correspond to the ± sign in the unnumbered equation immediately after Equation 4 (Complex Descartes' Theorem).  In some cases, using a + sign in both equations yields one solution while using a − sign in both equations yields the other.  In other cases, both of those are wrong; one solution is found using a + sign in one equation and a − sign in the other, while the other solution is found by swapping those signs.  I haven't yet determined what makes the difference.  Vid the Kid (t/c) Yeah, that guy. 21:03, 18 March 2015 (UTC)
thread_title:  Correspondence of ± signs 

message: An external resource bears the title "Descartes' circle theorem".  
It's one thing to follow historical norms (however horrible), it's another thing to have page titles that are immediately self-evident.  
In this matter, for myself. I'd choose the later.  Good grief, Descartes' theorem has it's own disambiguation page.  How sucky is that?  If he'd been any more productive, perhaps the thing to do would be to assign his theorems Bach-Werke-Verzeichnis numbers.  At least then we could tell them apart on surface inspection.  — MaxEnt 18:45, 15 January 2017 (UTC)
thread_title:  Possible title upgrade 

message: Is it error? 
May be that for every three kissing...? Jumpow (talk) 07:33, 19 February 2018 (UTC)
replies: text: No. The quadratic equation is the one labeled (1). It involves the radii of four circles, not three. —David Eppstein (talk) 07:51, 19 February 2018 (UTC)
replies: text: Yes, of course. It was request from one of russian users and I did not recognize answer immeiately... Jumpow (talk) 13:31, 5 March 2018 (UTC)
replies: text: It looks like an error: you start with three circles, solve the equation and reach a fourth. The way it's stated now reads like "you start with four circles, solve and equation, and reach a fourth circle", which doesn't make sense.Fernandohbc (talk) 02:41, 14 October 2018 (UTC)
replies: text: Answering myself: After re-reading, I realize that my paraphrasing is wrong too. My apologies. I think that the clarification needs to be made in the following sentence, though. Something along the lines of "By solving this equation, one can start with three given mutually tangent circles and construct a fourth circle tangent to all of them.", maybe?Fernandohbc (talk) 02:46, 14 October 2018 (UTC)
replies: text: If readers don't understand that an equation involving four things can be solved to find one of the things from the other three, then perhaps their level of mathematical sophistication does not reach that of the target audience for this article. —David Eppstein (talk) 03:26, 14 October 2018 (UTC)
replies: text: This is not what I claimed readers would misunderstand. My comment was regarding the phrasing in plain English, which is a bit confusing, not the mathematical sophistication of the theorem itself. Fernandohbc (talk) 04:01, 14 October 2018 (UTC)
thread_title:  that for every four kissing 

message: The article could usefully tell readers where they may find proofs of the given facts, solutions and equations.  Not implying that the article needs to contain proofs, since Wikipedia doesn't do that.  yoyo (talk) 02:58, 25 January 2021 (UTC)
replies: text: I agree with this. An article about a theorem should do a bit better job showing or at least explicitly linking to proofs. The history section mentions that "Multiple proofs of the theorem have been published" but does not directly cite any of those inline, and the few-words summaries are unhelpfully vague. The "Proof from Heron's formula" is algebraic and not especially satisfying. @David Eppstein have you done any survey of the published proofs of Descartes' theorem? Can we make a section including a couple of them or at least providing a sketch? I feel like a section about proofs of this theorem (and maybe some more images for the history section) is the only thing really missing to make this "good article" quality. –jacobolus (t) 18:15, 8 July 2023 (UTC)
replies: text: To be honest, the Heron's formula proof is the least algebraic and unsatisfying proof that I encountered. They all seem to have a similar flavor: take a relatively simple piece of geometry (Heron's formula, Pappus chains, etc) and then elaborate it into a page or two of algebra to get the formula. —David Eppstein (talk) 18:32, 8 July 2023 (UTC)
replies: text: That's a bummer. We should perhaps at least track down some citations to the original sources then, so readers can save some effort on their path to disappointment. :-) –jacobolus (t) 18:59, 8 July 2023 (UTC)
thread_title:  Help those who would prove it 

message: I'm not able to edit the article right now due to Wikipedia technical problems (keeps timing out), but I made an image showing the 4 radii in the Heron's formula proof. If I can't successfully add it to the article, feel free to. –jacobolus (t) 00:20, 21 July 2023 (UTC)
replies: text: I wonder if it would be better to show the whole sides of the triangles labeled r_1 + r_2 etc. instead of splitting each side up into two parts and labeling them separately. –jacobolus (t) 00:29, 21 July 2023 (UTC)
replies: text: Okay I made that change, which I think makes the picture a bit more legible. –jacobolus (t) 00:46, 21 July 2023 (UTC)
replies: text: New picture looks helpful to me. —David Eppstein (talk) 07:28, 21 July 2023 (UTC)

text: I think this article should be readable by people who don't care to understand signed areas. And they are completely unnecessary here. See WP:TECHNICAL (which by the way is one of the GA requirements): We should not make the article more technical than it needs to be. Signed areas are an unnecessary technicality. As for P_4 at infinity: let's not get into trying to set up equations of equal area of triangles that have a vertex at infinity. That way lies original research and dubious definitions and confusion. —David Eppstein (talk) 20:23, 21 July 2023 (UTC)
replies: text: To be clear: I am not suggesting we should try to get fancy about areas in the straight line case. It just falls between the other two (I guess I might also mention one additional case when the center is on the triangle and one of the triangles degenerates, with area 0). Maybe the clearest for this proof is to just enumerate the possibilities then. –jacobolus (t) 20:48, 21 July 2023 (UTC)

text: I guess your alternative is okay. Let me read more carefully. –jacobolus (t) 20:49, 21 July 2023 (UTC)
thread_title:  Heron's formula proof 

message: Is there a solution (straightedge and compass construction) using any right-angled triangle without first having to calculate the position of the Descartes circles? Petrus3743 (talk) 21:49, 13 December 2023 (UTC)
replies: text: Meaning, is there a straightedge and compass construction of the fourth circle tangent to three given circles? Yes, but this is a different topic (and a much older one) than Descartes' theorem. It is the Problem of Apollonius, which we describe early in the history of this section. It is the attempted reformulation of this problem using algebra rather than compass and straightedge that led Descartes to Descartes' theorem, as we also describe in the history section. —David Eppstein (talk) 21:59, 13 December 2023 (UTC)
replies: text: Thanks for your quick answer. The question was, is there a construction with any right-angled triangle without first calculating the position of the 4th circle?--Petrus3743 (talk) 22:21, 13 December 2023 (UTC)
replies: text: A construction of what, from what givens? If you are just going to make me guess what you mean by repeating the same question in the same words that you already asked, I am not going to be able to give you a better answer than the one I already gave. —David Eppstein (talk) 02:08, 14 December 2023 (UTC)
replies: text: Maybe the following picture is better than my bad English. Is there a reference for this: GeoGebra? — Preceding unsigned comment added by Petrus3743 (talk • contribs) 06:04, 14 December 2023 (UTC)
replies: text:  Yes, see Problem of Apollonius. This is a particular special case. The three circles at the vertices have diameters -a+b+c, a-b+c, and a+b-c, (for a triangle of sides a,b,c), so those are not hard to construct. –jacobolus (t) 06:36, 14 December 2023 (UTC)
replies: text: The special case in which the three initial circles are tangent, as in the GeoGebra link, is easier than the general problem. See my 1999 web page https://ics.uci.edu/~eppstein/junkyard/tangencies/apollonian.html and related paper "Tangent Spheres and Triangle Centers" (Amer. Math. Monthly 2001). The right angle helps a tiny bit more, as two of the first three lines of the construction are already sides of the right triangle. —David Eppstein (talk) 06:42, 14 December 2023 (UTC)
replies: text: Thanks for the good solution! Maybe I will also draw your construction and enter it into the article Apollonisches Problem with the two links as references.--Petrus3743 (talk) 15:19, 14 December 2023 (UTC)

text: Thank you again for your references! I was able to use it well in the article Satz von Descartes. With regards--Petrus3743 (talk) 11:46, 18 December 2023 (UTC)
thread_title:  Straightedge and compass construction