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Bui Quang Tuan
Member since Jun2307

Jul2310, 09:49 AM (EST) 

"Cyclic Touching Circles Around Triangle"

Dear All My Friends, In the triangle ABC we select one direction, say A to B to C to A to B... At vertex A: construct a circle (A1) touching 2 lines of vertex A At vertex B: construct a circle (B1) touching 2 lines of vertex B and sharing with (A1) one touching point. At vertex C: construct a circle (C1) touching 2 lines of vertex C and sharing with (B1) one touching point. Again at vertex A... Generally, at one vertex, we construct a circle touching 2 lines of the vertex and sharing with previous constructed circle one touching point. Please prove that the process is stop after construction of six circles and all touching point are concyclic! Best regards, Bui Quang Tuan 

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Bui Quang Tuan
Member since Jun2307

Jul2410, 11:24 AM (EST) 

5. "RE: Cyclic Touching Circles Around Triangle"
In response to message #4

Dear Alex, Thank you for putting my name in the theorem! This interesting fact based on very wellknown geometrical essence and I only have found another configuration from another view point for it. I do not sure that it is new one. If any one known another reference before, please give us so we can have full history! Dear Alex, please correct in your explaining text: "Next inscribe circle C1 into <A1A2A3 so that it is to A1A2 at T12" I think it must be: "Next inscribe circle C2 into <A1A2A3 so that it is tangent to A1A2 at T12" Please put also all names: A1, A2, A3, C1, C2, C3, C4, C5, C6, T12, T23,... in the applet so we can use in the future! Best regards, Bui Quang Tuan 

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alexb
Charter Member
2550 posts 
Jul2410, 11:54 AM (EST) 

6. "RE: Cyclic Touching Circles Around Triangle"
In response to message #5

>Dear Alex, > >Thank you for putting my name in the theorem! This >interesting fact based on very wellknown geometrical >essence and I only have found another configuration from >another view point for it. You certainly brought it up. I made a reference to a better known fact to which your statement is equivalent. But you observation sheds extra light on the configuration. >I do not sure that it is new one. >If any one known another reference before, please give us so >we can have full history! Should I run into anything related I shal upgrade the page. Certainly. > >Dear Alex, please correct in your explaining text: >"Next inscribe circle C1 into <A1A2A3 so that it is to A1A2 >at T12" >I think it must be: >"Next inscribe circle C2 into <A1A2A3 so that it is tangent >to A1A2 at T12" Yes, thank you. >Please put also all names: A1, A2, A3, C1, C2, C3, C4, C5, >C6, T12, T23,... in the applet so we can use in the future! > OK 

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Bui Quang Tuan
Member since Jun2307

Jul2410, 07:52 PM (EST) 

7. "RE: Cyclic Touching Circles Around Triangle"
In response to message #0

Dear Alex, We can construct similar closed chain of six concyclic points as following: A1A2A3 is a triangle and O is circumcenter. L1 = line OA1 L2 = line OA2 L3 = line OA3 We start with any point P1 on the line L1 Perpendicular from P1 to A1A2 cuts L2 at P2 Perpendicular from P2 to A2A3 cuts L3 at P3 Perpendicular from P3 to A3A1 cuts L1 at P4 ... At the end P7=P1 and six points P1, P2, P3, P4, P5, P6 are on one circle centered at O. If instead of circumcenter O we take orthocenter H then P7=P1 but six points are not concyclic. I think the fact P7=P1 is true if instead of O we take any point on Darboux cubic. This cubic contains: incenter, circumcenter, orthocenter... But this fact need a lot of calculation. Only in circumcenter case six points are concyclic. In the incenter case: P1, P2, P3, P4, P5, P6 are centers of touching circles. They are not concyclic but we can construct touching circles and touching points are concyclic. Best regards, Bui Quang Tuan 

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Bui Quang Tuan
Member since Jun2307

Jul2410, 10:12 PM (EST) 

9. "RE: Cyclic Touching Circles Around Triangle"
In response to message #7

Dear Alex, I have calculated and the locus of points such that we can make closed chain of six points is union of circumcircle and Darboux cubic which contains: incenter, circumcenter, orthocenter... Two cases circumcenter, orthocenter are very trivial so we have here only interesting case with incenter. Best regards, Bui Quang Tuan >Dear Alex, > >We can construct similar closed chain of six concyclic >points as following: > >A1A2A3 is a triangle and O is circumcenter. >L1 = line OA1 >L2 = line OA2 >L3 = line OA3 > >We start with any point P1 on the line L1 >Perpendicular from P1 to A1A2 cuts L2 at P2 >Perpendicular from P2 to A2A3 cuts L3 at P3 >Perpendicular from P3 to A3A1 cuts L1 at P4 >... >At the end P7=P1 and six points P1, P2, P3, P4, P5, P6 are >on one circle centered at O. > >If instead of circumcenter O we take orthocenter H then >P7=P1 but six points are not concyclic. > >I think the fact P7=P1 is true if instead of O we take any >point on Darboux cubic. This cubic contains: incenter, >circumcenter, orthocenter... But this fact need a lot of >calculation. > >Only in circumcenter case six points are concyclic. > >In the incenter case: P1, P2, P3, P4, P5, P6 are centers of >touching circles. They are not concyclic but we can >construct touching circles and touching points are >concyclic. > >Best regards, >Bui Quang Tuan


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alexb
Charter Member
2550 posts 
Jul2410, 10:15 PM (EST) 

10. "RE: Cyclic Touching Circles Around Triangle"
In response to message #9

>I have calculated and the locus of points such that we can >make closed chain of six points is union of circumcircle and >Darboux cubic which contains: incenter, circumcenter, >orthocenter... Have you tried drawing the lines which are not perpendicular but fall under the same inclination? >Two cases circumcenter, orthocenter are very trivial so we >have here only interesting case with incenter. Yes, I think I can handle the circumcenter and the orthocenter. These come naturally.


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Bui Quang Tuan
Member since Jun2307

Jul2410, 07:40 AM (EST) 

11. "RE: Cyclic Touching Circles Around Triangle"
In response to message #10

Dear Alex, I just finished calculation for general case, say "parallel with one cevians and cutting another cevians". Our perpendicular case is one when "parallel with altitudes" and cutting another cevians (of incenter, circumcenter, orthocenter...). The locus of all points such that we can make closed chain of six points is union of:  One circumconic  One circumcubic When "parallel with altitudes" the circumconic is circumcircle and the circumcubic is Darboux cubic. I will collect some special cases with simple points so may be we can find interesting elementar proof. Best regards, Bui Quang Tuan >Have you tried drawing the lines which are not perpendicular >but fall under the same inclination? 

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Bui Quang Tuan
Member since Jun2307

Jul2610, 08:36 AM (EST) 

12. "RE: Cyclic Touching Circles Around Triangle"
In response to message #11

Dear Alex, I generalize the fact as following:Given triangle ABC with fixed point (not infinite) U with barycentrics (u:v:w) and variable point X. Take any point A1 on Cevian line XA Parallel line from A1 with UC cuts XB at B1 Parallel line from B1 with UA cuts XC at C1 Parallel line from C1 with UB cuts XA at A2 Parallel line from A2 with UC cuts XB at B2 Parallel line from B2 with UA cuts XC at C2 Parallel line from C2 with UB cuts XA at A3 The locus of X such that A3=A1 with any A1 on XA is union of: 1. One circumconic: w*(u + v)*x*y + u*(v + w)*y*z + v*(w + u)*z*x = 0 2. One circumcubic: CyclicSum=0 Some special triangle centers on this locus: U = Incenter, X= (Incenter, Orthocenter, Nagel Point, Spieker Center... ) U = Centroid, X= (Centroid, it is special case: there is only Centroid is the triangle center on the locus) U = Circumcenter, X = (Circumcenter, Orthocenter, Nine Point Center... ) U = Orthocenter, X = (Orthocenter, Incenter, Circumcenter... ) U = Nine Point Center, X = (Nine Point Center, Circumcenter, Midpoint of Nine Point Center and Circumcenter... ) U = Symmedian Point, X = (Symmedian Point... ) U = Gergonne Point, X = (Gergonne Point... ) U = Nagel Point, X = (Nagel Point, Incenter,... ) I propose one problem when U = Incenter and X = Nagel Point: in this case, A1,B1, C1, A2, B2, C2 are not concyclic, but they bound another six concylic points. We construct six following intersections: A12 = A1B1, A2C1 B12 = B1C1, B2A2 C12 = C1A2, C2B2 A21 = A2B2, A1C2 B21 = B2C2, B1A1 C21 = C2A1, C1B1 Please prove that six points A12, B12, C12, A21, B21, C21 are concyclic. This problem seems very hard! Best regards, Bui Quang Tuan > >I will collect some special cases with simple points so may >be we can find interesting elementar proof. > >Best regards, >Bui Quang Tuan 

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alexb
Charter Member
2550 posts 
Jul2610, 08:40 AM (EST) 

13. "RE: Cyclic Touching Circles Around Triangle"
In response to message #12

Dear Bui Quang Tuan: I admire your perseverance. I am afraid, though, that the whole thing is beyond my abilities. This is to be regretted as what you say is very interesting. Alex 

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alexb
Charter Member
2550 posts 
Jul2610, 10:07 AM (EST) 

15. "RE: Cyclic Touching Circles Around Triangle"
In response to message #14

>I don't think so! I disagree with you :) >Let's us come back to Six Circles Theorem configuration. I >have some things interesting and will send you when I finish >all. I would appreciate that very much. 

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Bui Quang Tuan
Member since Jun2307

Jul2610, 03:07 PM (EST) 

16. "RE: Cyclic Touching Circles Around Triangle"
In response to message #14

>Let's us come back to Six Circles Theorem configuration. I >have some things interesting and will send you when I finish >all. >Best regards, >Bui Quang TuanDear Alex, When playing the applet of Six Circles Theorem configuration we can see one special point when three circles C1, C3, C5 or C2, C4, C6 are concurrent at it. Now we try to find what is this point. We denote C(I) as a concyclic circle of six touching points. Incenter I is center of C(I). Take a circle C1 with touching points T12 on A1A2 and T13 on A1A3. A circle C2 shares with C1 touching point T12. T23 is touching point of C2 with A2A3 A circle C6 shares with C1 touching point T13. T32 is touching point of C6 with A2A3 Radical center of three circles C1, C2, C6 is A1 since A1 is intersection of two tangent lines. Hence radical axis of two circles C2, C6 is a line connected A1 and midpoint M1 of T23T32 (radical axis of two circles passes midpoint of two touching points of common tangent line). In the circle C(I), T23T32 is one chord, hence IM1 is perpendicular with A2A3, so M1 is touching point of incircle with A2A3. It means radical axis of C2, C6 (line A1M1) passes through Gergonne point Ge of A1A2A3. Similarly at the end we can prove that radical center of C2, C4, C6 is Gergonne point, and radical center of C1, C3, C5 is also Gergonne point. When changing the configuration, the radical center of triple of circles is fixed, hence when they concurrent, the concurrent point is Gergonne point. There are two triples but only one concurrent point Ge. So our configuration can be used as one solution for following problem: Given a triangle, to construct three circles through a common point, each tangent to two sides of the triangle, such that the 6 points of contact are concyclic. This is problem of Thebault  Eves, AMM E457. I do not have original text of the problem and read it from Paul Yiu book (page 135): https://math.fau.edu/Yiu/EuclideanGeometryNotes.pdf We can see also one another six concyclic points when there is one triple of circles cutting side lines of A1A2A3. In this case, six cutting points are concyclic. When three circles intersect side lines of A1A2A3, they also intersect each other on three radical axis A1Ge, A2Ge, A3Ge. Six cutting points are concyclic by the theorem of Six Concyclic Points: https://www.cuttheknot.org/Curriculum/Geometry/SixConcyclicPoints.shtml Best regards, Bui Quang Tuan 

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mpdlc
guest

Jul2910, 05:24 PM (EST) 

19. "RE%3A Cyclic Touching Circles Around Triangle"
In response to message #0

I must congratulate you both Bui Quam and Alex for the rich and interesting research on the subject "RE: Cyclic Touching Circles Around Triangle" posed by our friend Bui I outline below an "engineering solution" to the original challenge , of course not rigorous as the one exhibited in CTK but I belive is valid so I kindly ask your expertise in checking if any flaws on it. Let us asume that we have twelve travellers as follow 4 located on vertex A we shall call them A1b , A2c, A2b and A3c 4 located on vertex B we shall call them B1c, B1c, B2a and B2c 4 located on vertex C we shall call them C1b, C1a, C2b and C2a The first subindex take into account the generation they belong and the second subindex for the destination vertex At the instant T=0 all the travellers start to depart to the each destination vertex, The speed of each traveller will be such that, after elapsed a given unit time U, all Travellers simultaneously will meet by pairs as follows
A1b meet with B1a ; B1c meet with C1b ; C1a meet with A2c ; A2b meet with B2a ; B2c meet with C2b ; C2a meet with A3c We name the speed of the travellers as follows For A1b equals Va1 For B1a and B1c equal Vb1 For C1b and C1a equal Vc1 For A2c and A2b equal Va2 For B2a and B2c equal Vb2 For C2b and C2a equal Vc2 After a time U he distance walked by each of above the pairs of travellers will amount respectively one side of the triangle ( La, Lb, Lc) So we can write the following equations (Va1+Vb1) U = Lc ; (1) (Vb1+Vc1) U = La ; (2) (Vc1+Va2) U = Lb ; (3) (Va2+Vb2) U = Lc ; (4) (Va2+Vc2) U = La ; (5) Substracting Eq (4) from (1) and Eq (5) from (2) We get U( Va1Va2) = U( Vb2 Vb1) = U( Vc1Vc2) what means the difference between the radius of circunferences drawn from vertices A, B and C are equals and will keep equals forever. Now to the second part, once is proved that distance of the two radius drawn from the same vertex is constant, is obvious that midpoint of this two radius is the tangency point of the incircle since two consecutive midpoints are equidistant from the common vertex of its side. Being the tangency point the mentioned midpoint if from the incenter we draw a circunference passing intersection of side Lc with Ra1 (UVa1) it will pass also over the other intersection with Ra2 (UVa2) 

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Bui Quang Tuan
Member since Jun2307

Jul3010, 06:54 AM (EST) 

20. "RE: RE%3A Cyclic Touching Circles Around Triangle"
In response to message #19

Dear Mariano Perez de la Cruz, Thank you for your kind words and interesting with our problem! It is very interesting if we can create some more mechanisms from this geometric essence. After careful studying your idea, I am not sure that I understand you well. We are talking about two configurations: Eves' configuration: https://www.cuttheknot.org/Curriculum/Geometry/ExpandedIncircle.shtml and my configuration: https://www.cuttheknot.org/Curriculum/Geometry/BuiQuangTuan.shtml Both configurations generate six concyclic points on sidelines of a triangle. Eves created these six points as common intersection points of two circles centered at two vertices of the triangle. I created these six points as common tangent points of two circles each touching two sidelines of the triangle. If you use symbol names (points, lines, circle...) in one of two above configurations then it is better to understand. May be you can also create your image and send us please! In any case, I am worry about two matters: 1. Twelve travellers (four at each vertex) are may be too much? 2. Where their derpart positions? As you can play in two configurations (of Eves and of mine), derpart positions are at three touching points of incircle, so not at the vertices as you design. I am very sorry if I understand you not very well! Best regards, Bui Quang Tuan > >Let us asume that we have twelve travellers as follow >4 located on vertex A we shall call them A1b , A2c, A2b and >A3c >4 located on vertex B we shall call them B1c, B1c, B2a and >B2c >4 located on vertex C we shall call them C1b, C1a, C2b and >C2a > >The first subindex take into account the generation they >belong and the second subindex for the destination vertex > > >At the instant T=0 all the travellers start to depart to the >each destination vertex,


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mpdlc
Member since Mar1207

Jul3010, 05:15 PM (EST) 

21. "RE: RE%3A Cyclic Touching Circles Around Triangle"
In response to message #20

Not having paper nor pencil at hand, the unconventional approach of using travelers came to me meanwhile seating at dentist office trying to entertain my mind in some absorbing subject and forget the upcoming procedures. Now I realize my solution It basically is coincident with H. Eve , I enclose an sketch of it. Certainly I need eleven travelers four for each vertex except for vertex A since I only have used five equations.In my opinion for the original problem , there are two interesting facts as you point out no matter what size the first radius is, the midpoint of the two consecutive radii is always the tangency point of the incircle to the given triangle ABC and of course the length between the consecutive radii is invariant The second is using the same sequence drawing circles in any polygon after the second round the value of the radii keep repeating and also the length between the consecutive radii is invariant. Of course there are more interesting facts from the deep research you made like Gergonne Adams Line and obviously all your second findings is a whole new prospective mpdlc 
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