>Bui Quang Tuan,
>
>Is this what you have in mind
>
>https://www.cut-the-knot.org/Curriculum/Geometry/Evelyn.shtml
>
>?

https://www.cut-the-knot.org/Curriculum/Geometry/ExpandedIncircle.shtml

Thank you.

Thank you for putting my name in the theorem! This interesting fact based on very well-known 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

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

I am now preoccupied with trying to answer the Why? question at the end of

https://www.cut-the-knot.org/Curriculum/Arithmetic/FlipThem.shtml

The case of odd numbers of triangles and the ones flipped on a move is simple. But when the total is even I am stuck with particular cases.

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

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.

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?

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

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

Dear Alex,
I don't think so! But I am agreed with you that your site is not necessary to include completed drawings with tired calculations.
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 Tuan

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.

Dear 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.cut-the-knot.org/Curriculum/Geometry/SixConcyclicPoints.shtml

Best regards,
Bui Quang Tuan

All the best,
Alex

https://www.cut-the-knot.org/Curriculum/Geometry/BuiQuangTuan2.shtml

Thank you

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.cut-the-knot.org/Curriculum/Geometry/ExpandedIncircle.shtml

and my configuration:
https://www.cut-the-knot.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,

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