## New circle related to a triangle

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• Dear friends! Last summer I ve found one new (as I hope) circle related to a triangle. The construction is as following: Let ABC be an arbitrary triangle and
Message 1 of 14 , Nov 9 12:01 AM
Dear friends!
Last summer I've found one new (as I hope) circle related to a triangle.
The construction is as following:
Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
proof.
May be somebody be more successful in search of such proof?
Best regards,
Alexei Myakishev
• ... Dear Alexei Is the center of the circle a known point? (ie already listed in ETC) Coordinates? APH [Non-text portions of this message have been removed]
Message 2 of 14 , Nov 9 12:24 AM
On Wed, Nov 9, 2011 at 10:01 AM, amyakishev <amyakishev@...> wrote:

> **
>
>
> Dear friends!
> Last summer I've found one new (as I hope) circle related to a triangle.
> The construction is as following:
> Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle
> of ABC.
> Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine
> point K' of A'B'C'.
> By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots
> are with respect to ABC).
> Let Ab is the point of intersection of the parallel line to AB (trough A1)
> with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB -
> sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
> So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
> I can proof that fact by rather easy barycentric calculations but have
> failed to get synthetical (pure geometrical)
> proof.
> May be somebody be more successful in search of such proof?
> Best regards,
> Alexei Myakishev
>
>
>

Dear Alexei

Is the center of the circle a known point? (ie already listed in ETC)
Coordinates?

APH

[Non-text portions of this message have been removed]
• Dear Antreas! The coordinates of the center are rather ugly, smth. like that (if Mathematica 5.1 tells the truth^)):
Message 3 of 14 , Nov 9 1:06 AM
Dear Antreas!
The coordinates of the center are rather ugly, smth. like that (if Mathematica 5.1 tells the truth^)):
8a^4b^2c^2(-a^2(b^2+c^2)(b^4-5b^2c^2+c^4)+a^4(b^4-4b^2c^2+c^4)+b^2c^2(b^4-4b^2c^2+c^4)):...:...
This point is absent in ETC (was checked by magic triangle 6,9,13:))
I've posted about it to Kimberling in ceptember but still have no reply:)
Best regards,
Alex

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> On Wed, Nov 9, 2011 at 10:01 AM, amyakishev <amyakishev@...> wrote:
>
> > **
> >
> >
> > Dear friends!
> > Last summer I've found one new (as I hope) circle related to a triangle.
> > The construction is as following:
> > Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle
> > of ABC.
> > Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine
> > point K' of A'B'C'.
> > By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots
> > are with respect to ABC).
> > Let Ab is the point of intersection of the parallel line to AB (trough A1)
> > with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB -
> > sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
> > So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
> > I can proof that fact by rather easy barycentric calculations but have
> > failed to get synthetical (pure geometrical)
> > proof.
> > May be somebody be more successful in search of such proof?
> > Best regards,
> > Alexei Myakishev
> >
> >
> >
>
> Dear Alexei
>
> Is the center of the circle a known point? (ie already listed in ETC)
> Coordinates?
>
> APH
>
>
> [Non-text portions of this message have been removed]
>
• Dear Alexei ... I don t know if the following can help you but If M,N are the isotomic conjugates (wrt ABC) of the Brocard points of A B C , it seems that your
Message 4 of 14 , Nov 9 2:08 AM
Dear Alexei

> Last summer I've found one new (as I hope) circle related to a triangle.
> The construction is as following:
> Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
> Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
> By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
> Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
> So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
> I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
> proof.
> May be somebody be more successful in search of such proof?

If M,N are the isotomic conjugates (wrt ABC) of the Brocard points of A'B'C', it seems that your circle is the common cevian circle (wrt ABC) of M and N.
Kind regards. Jean-Pierre
• Dear Jean-Pierre! Thank you for that wonderful observation. The next question is: Is this fact about isotomic Brocard s points and the corresponding circle
Message 5 of 14 , Nov 9 3:00 AM
Dear Jean-Pierre!
Thank you for that wonderful observation.
The next question is:
Is this fact about isotomic Brocard's points and the corresponding circle well known?
Or you have established it righn now?:)
Best regards,
Alex

--- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...> wrote:
>
> Dear Alexei
>
> > Last summer I've found one new (as I hope) circle related to a triangle.
> > The construction is as following:
> > Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
> > Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
> > By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
> > Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
> > So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
> > I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
> > proof.
> > May be somebody be more successful in search of such proof?
>
> If M,N are the isotomic conjugates (wrt ABC) of the Brocard points of A'B'C', it seems that your circle is the common cevian circle (wrt ABC) of M and N.
> Kind regards. Jean-Pierre
>
• Dear Alexei ... In my case, I didn t know this result before your message. In fact, if you put u=1/bb+1/cc-1/aa, v,w cyclically, we have
Message 6 of 14 , Nov 9 6:21 AM
Dear Alexei

> Thank you for that wonderful observation.
> The next question is:
> Is this fact about isotomic Brocard's points and the corresponding circle well known?
> Or you have established it righn now?:)

In my case, I didn't know this result before your message.
In fact, if you put u=1/bb+1/cc-1/aa, v,w cyclically, we have
Ba=(0,u,w);Cb=(u,0,v);Ac=(w,v,0)
Ca=(0,v,u);Ab=(v,0,w);Bc=(u,w,0)
Thus BaCbAc is the cevian triangle of (1/v,1/w,1/u) and CaAbBc is the cevian triangle of (1/w,1/u,1/v)
As (v,w,u) and (w,u,v) are the anticomplements of the Brocard points of ABC, my assertion follows.
Now AAb.ACb = AAc.ABc = aabbccvw/4 and two similar identities shows that the six points are concyclic
Kind regards. Jean-Pierre
• Dear Alexei, Bernard has found this circle before in Hyacinthos 5710: [BG, 5710]: a new (?) circle Dear friends, P a point, PaPbPc its cevian triangle. the
Message 7 of 14 , Nov 9 6:36 AM
Dear Alexei,

Bernard has found this circle before in Hyacinthos 5710:

[BG, 5710]: a new (?) circle

Dear friends,

P a point, PaPbPc its cevian triangle.
the parallel at Pa to AC meets AB at Pac.
the parallel at Pa to AB meets AC at Pab.
Similarly define Pba, Pbc, Pcb, Pca.

Those six points always lie on the same conic.

This conic is a circle iff P=X194 ie the perspector of the cevian triangle
of K=X6 and the antimedial triangle.

Its center is unknown.

Maybe there are some other interesting properties.

=============
Best regards
Sincerely
Paul

--- In Hyacinthos@yahoogroups.com, "amyakishev" <amyakishev@...> wrote:
>
> Dear friends!
> Last summer I've found one new (as I hope) circle related to a triangle.
> The construction is as following:
> Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
> Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
> By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
> Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
> So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
> I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
> proof.
> May be somebody be more successful in search of such proof?
> Best regards,
> Alexei Myakishev
>
• Dear Paul and friends, I had forgotten all this which also appears in §2.3 of http://bernard.gibert.pagesperso-orange.fr/files/tuckercubics.html Thank you and
Message 8 of 14 , Nov 9 6:45 AM
Dear Paul and friends,

I had forgotten all this which also appears in §2.3 of

http://bernard.gibert.pagesperso-orange.fr/files/tuckercubics.html

Thank you and best regards

Bernard

> [PY] Bernard has found this circle before in Hyacinthos 5710:

[Non-text portions of this message have been removed]
• Dear Paul, Bernard, Alexei and other Hyacinthists a good advice : before posting a message, we must read carefully the complete works of Bernard. Friendly.
Message 9 of 14 , Nov 9 9:14 AM
Dear Paul, Bernard, Alexei and other Hyacinthists
a good advice : before posting a message, we must read carefully the complete works of Bernard.
Friendly. Jean-Pierre

> Bernard has found this circle before in Hyacinthos 5710:
>
> [BG, 5710]: a new (?) circle
>
> Dear friends,
>
> P a point, PaPbPc its cevian triangle.
> the parallel at Pa to AC meets AB at Pac.
> the parallel at Pa to AB meets AC at Pab.
> Similarly define Pba, Pbc, Pcb, Pca.
>
> Those six points always lie on the same conic.
>
> This conic is a circle iff P=X194 ie the perspector of the cevian triangle
> of K=X6 and the antimedial triangle.
>
> Its center is unknown.
>
> Maybe there are some other interesting properties.
• Sound advice from Pope J.-P. Best regards Sincerely Paul ... From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On Behalf Of jpehrmfr Sent:
Message 10 of 14 , Nov 9 9:53 AM

Best regards
Sincerely
Paul

-----Original Message-----
From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On Behalf Of jpehrmfr
Sent: Wednesday, November 09, 2011 12:15 PM
To: Hyacinthos@yahoogroups.com
Subject: [EMHL] Re: New circle related to a triangle

Dear Paul, Bernard, Alexei and other Hyacinthists a good advice : before posting a message, we must read carefully the complete works of Bernard.
Friendly. Jean-Pierre

> Bernard has found this circle before in Hyacinthos 5710:
>
> [BG, 5710]: a new (?) circle
>
> Dear friends,
>
> P a point, PaPbPc its cevian triangle.
> the parallel at Pa to AC meets AB at Pac.
> the parallel at Pa to AB meets AC at Pab.
> Similarly define Pba, Pbc, Pcb, Pca.
>
> Those six points always lie on the same conic.
>
> This conic is a circle iff P=X194 ie the perspector of the cevian
> triangle of K=X6 and the antimedial triangle.
>
> Its center is unknown.
>
> Maybe there are some other interesting properties.

------------------------------------

• Dear Paul, thank you. It s really so hard tj invent something new. But the question about geometrical proof is still remain? Best regards, Sincerely yours,
Message 11 of 14 , Nov 9 12:28 PM
Dear Paul,
thank you.
It's really so hard tj invent something new.
But the question about geometrical proof is still remain?
Best regards,
Sincerely yours,
Alex

--- In Hyacinthos@yahoogroups.com, "yiuatfauedu" <yiu@...> wrote:
>
> Dear Alexei,
>
> Bernard has found this circle before in Hyacinthos 5710:
>
> [BG, 5710]: a new (?) circle
>
> Dear friends,
>
> P a point, PaPbPc its cevian triangle.
> the parallel at Pa to AC meets AB at Pac.
> the parallel at Pa to AB meets AC at Pab.
> Similarly define Pba, Pbc, Pcb, Pca.
>
> Those six points always lie on the same conic.
>
> This conic is a circle iff P=X194 ie the perspector of the cevian triangle
> of K=X6 and the antimedial triangle.
>
> Its center is unknown.
>
> Maybe there are some other interesting properties.
>
> =============
> Best regards
> Sincerely
> Paul
>
> --- In Hyacinthos@yahoogroups.com, "amyakishev" <amyakishev@> wrote:
> >
> > Dear friends!
> > Last summer I've found one new (as I hope) circle related to a triangle.
> > The construction is as following:
> > Let ABC be an arbitrary triangle and A'B'C' - anticomplementary triangle of ABC.
> > Let Km' be the isotomic conjugate (with respect to A'B'C') of Lemoine point K' of A'B'C'.
> > By A1,B1,C1 let us mark the foots of the cevians trough Km' (now the foots are with respect to ABC).
> > Let Ab is the point of intersection of the parallel line to AB (trough A1) with AC-sideline and Ac - of the parallel line to AC (trough A1) with AB - sideline. The points Bc,Ba, Cb,Ca are defined cyclically in the same way.
> > So all six points Ab,Ac,Ba,Bc,Ca,Cb lie on a circle.
> > I can proof that fact by rather easy barycentric calculations but have failed to get synthetical (pure geometrical)
> > proof.
> > May be somebody be more successful in search of such proof?
> > Best regards,
> > Alexei Myakishev
> >
>
• Dear Bernard, sorry for the new circle. Newertheless, it would be interesting to know (for me) how you have proved it. Best regards, Alex
Message 12 of 14 , Nov 9 12:58 PM
Dear Bernard, sorry for the "new" circle.
Newertheless, it would be interesting to know (for me) how you have proved it.
Best regards,
Alex

--- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@...> wrote:
>
> Dear Paul and friends,
>
> I had forgotten all this which also appears in §2.3 of
>
> http://bernard.gibert.pagesperso-orange.fr/files/tuckercubics.html
>
> Thank you and best regards
>
> Bernard
>
>
>
> > [PY] Bernard has found this circle before in Hyacinthos 5710:
>
>
>
> [Non-text portions of this message have been removed]
>
• Dear Alex, ... That s OK. I m pretty sure one can find it in older literature ! ... through barycentric calculations. Best regards Bernard [Non-text portions
Message 13 of 14 , Nov 9 10:37 PM
Dear Alex,

> Dear Bernard, sorry for the "new" circle.

That's OK. I'm pretty sure one can find it in older literature !

> Newertheless, it would be interesting to know (for me) how you have proved it.

through barycentric calculations.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard, in fact, nominal by barycentric, I ve get it:) Now I m prepareing the article about this circle. Of course, you would be menthion in it.:) But
Message 14 of 14 , Nov 10 12:41 AM
Dear Bernard,
in fact, nominal by barycentric, I've get it:)
Of course, you would be menthion in it.:)
But can you help me with "older literature", i.e. some sourses?:)
Best regards,
Alex

--- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@...> wrote:
>
> Dear Alex,
>
> > Dear Bernard, sorry for the "new" circle.
>
> That's OK. I'm pretty sure one can find it in older literature !
>
> > Newertheless, it would be interesting to know (for me) how you have proved it.
>
> through barycentric calculations.
>
> Best regards
>
> Bernard
>
>
>
> [Non-text portions of this message have been removed]
>
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