## loci related to Taylor circle

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• Dear friends, Let ABC be a triangle, and P a point. Let A B C be the pedal triangle of P. Let Ba, Ca be the orthogonal projections of A onto lines CA, AB,
Message 1 of 5 , Apr 16, 2013
Dear friends,

Let ABC be a triangle, and P a point.
Let A'B'C' be the pedal triangle of P.
Let Ba, Ca be the orthogonal projections of A' onto lines CA, AB, resp.
Define Cb, Ab, Ac, Bc cyclically.

What is the locus of P such that Ba, Ca, Cb, Ab, Ac, Bc lie on a common conic? The locus would include H, for which the conic is the Taylor circle.

What is the locus of the centers of the conics for P in the above locus? It would include X(394), the center of the Taylor circle.

Best regards,
Randy
• Dear Randy, ... a quintic with many simple points but only two (I think) ETC centers : X4, X1498. ... seems very difficult... Best regards Bernard [Non-text
Message 2 of 5 , Apr 17, 2013
Dear Randy,

> [RH] Let ABC be a triangle, and P a point.
> Let A'B'C' be the pedal triangle of P.
> Let Ba, Ca be the orthogonal projections of A' onto lines CA, AB, resp.
> Define Cb, Ab, Ac, Bc cyclically.
>
> What is the locus of P such that Ba, Ca, Cb, Ab, Ac, Bc lie on a common conic? The locus would include H, for which the conic is the Taylor circle.

a quintic with many simple points but only two (I think) ETC centers : X4, X1498.

> What is the locus of the centers of the conics for P in the above locus? It would include X(394), the center of the Taylor circle.

seems very difficult...

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Randy and Bernard, [RH] Let ABC be a triangle, and P a point. Let A B C be the pedal triangle of P. Let Ba, Ca be the orthogonal projections of A onto
Message 3 of 5 , Apr 17, 2013
Dear Randy and Bernard,

[RH] Let ABC be a triangle, and P a point.
Let A'B'C' be the pedal triangle of P.
Let Ba, Ca be the orthogonal projections of A' onto lines CA, AB, resp.
Define Cb, Ab, Ac, Bc cyclically.
What is the locus of P such that Ba, Ca, Cb, Ab, Ac, Bc lie on a common conic? The locus would include H, for which the conic is the Taylor circle.

[BG]: a quintic with many simple points but only two (I think) ETC centers : X4, X1498.

*** X(1498) is the Nagel point of the tangential triangle.
Ab = (0 : S_{AB}+S_{AC}-S_{BC} : S_{AB}+S_{AC}+S_{BC}) and
Ac = (0 : S_{AB}+S_{AC}+S_{BC} : S_{AB}+S_{AC}+S_{BC})
are isotomic points on BC, so are Bc, Ba, and Ca, Cb.
X(1498) is the unique point with this property. The conic is

(4/S^2)cyclic sum ((a^4S_{AA})/(S_{AB}+S_AC}-S_{BC}))yz - (x+y+z)^2 = 0,
concentric (and homothetic) with the circumconic with perspector
((a^4S_{AA}/(S_{AB}+S_{AC}-S_{BC}):...:...)
[with (6-9-13)-search number 5.10435062529...]
and has center
(a^4(S_{AAAB}+S_{AAAC}+S_{AABB}-S_{AABC}+S_{AACC}-S_{BBCC}/
(S_{AB}+S_{AC}-S_{BC}) :...:...)
with (6-9-13) search number 1.09478783248....

Best regards
Sincerely
Paul
• Paul, This is interesting: the perspector you mention with ETC search value 5.10435062529 matches the isogonal conjugate of the polar conjugate of X(1073), and
Message 4 of 5 , Apr 17, 2013
Paul,

This is interesting: the perspector you mention with ETC search value 5.10435062529 matches the isogonal conjugate of the polar conjugate of X(1073), and as such would have trilinears (cos^2 A)/(cos A - cos B cos C) : :.

What do you mean by the notations S_{AB}, etc.? I assume it has to do with Conway notation. Also, are your coordinates trilinears or barycentrics?

Randy

--- In Hyacinthos@yahoogroups.com, "yiuatfauedu" <yiu@...> wrote:
>
> Dear Randy and Bernard,
>
> [RH] Let ABC be a triangle, and P a point.
> Let A'B'C' be the pedal triangle of P.
> Let Ba, Ca be the orthogonal projections of A' onto lines CA, AB, resp.
> Define Cb, Ab, Ac, Bc cyclically.
> What is the locus of P such that Ba, Ca, Cb, Ab, Ac, Bc lie on a common conic? The locus would include H, for which the conic is the Taylor circle.
>
> [BG]: a quintic with many simple points but only two (I think) ETC centers : X4, X1498.
>
> *** X(1498) is the Nagel point of the tangential triangle.
> Ab = (0 : S_{AB}+S_{AC}-S_{BC} : S_{AB}+S_{AC}+S_{BC}) and
> Ac = (0 : S_{AB}+S_{AC}+S_{BC} : S_{AB}+S_{AC}+S_{BC})
> are isotomic points on BC, so are Bc, Ba, and Ca, Cb.
> X(1498) is the unique point with this property. The conic is
>
> (4/S^2)cyclic sum ((a^4S_{AA})/(S_{AB}+S_AC}-S_{BC}))yz - (x+y+z)^2 = 0,
> concentric (and homothetic) with the circumconic with perspector
> ((a^4S_{AA}/(S_{AB}+S_{AC}-S_{BC}):...:...)
> [with (6-9-13)-search number 5.10435062529...]
> and has center
> (a^4(S_{AAAB}+S_{AAAC}+S_{AABB}-S_{AABC}+S_{AACC}-S_{BBCC}/
> (S_{AB}+S_{AC}-S_{BC}) :...:...)
> with (6-9-13) search number 1.09478783248....
>
> Best regards
> Sincerely
> Paul
>
• Dear Randy, Yes, S_{AB} is shorthand for (S_A)(S_B) etc. I always use barycentric coordinates unless the context clearly favors trilinear coordinates. Let s
Message 5 of 5 , Apr 17, 2013
Dear Randy,

Yes, S_{AB} is shorthand for (S_A)(S_B) etc. I always use barycentric coordinates unless the context clearly favors trilinear coordinates.

Let's check with the trilinear coordinates you gave. With an extra factor a (in the first component), the barycentric coordinates are

a cos^2 A/(cos A - cos B cos C) : ... : ...
= a(S_{AA}/(bc)^2) /(S_A/(bc) - S_{BC}/(a^2bc)) : ... : ...
= a^3S_{AA}/(a^2bc S_A - bcS_{BC}) : ... : ...
= a^4S_{AA}/(a^2S_A - S_{BC}) : ... : ...

Yes, it is the same as the one I gave!

Best regards
Sincerely
Paul
________________________________________
From: Hyacinthos@yahoogroups.com [Hyacinthos@yahoogroups.com] on behalf of rhutson2 [rhutson2@...]
Sent: Wednesday, April 17, 2013 3:37 PM
To: Hyacinthos@yahoogroups.com
Subject: [EMHL] Re: loci related to Taylor circle

Paul,

This is interesting: the perspector you mention with ETC search value 5.10435062529 matches the isogonal conjugate of the polar conjugate of X(1073), and as such would have trilinears (cos^2 A)/(cos A - cos B cos C) : :.

What do you mean by the notations S_{AB}, etc.? I assume it has to do with Conway notation. Also, are your coordinates trilinears or barycentrics?

Randy

--- In Hyacinthos@yahoogroups.com, "yiuatfauedu" <yiu@...> wrote:
>
> Dear Randy and Bernard,
>
> [RH] Let ABC be a triangle, and P a point.
> Let A'B'C' be the pedal triangle of P.
> Let Ba, Ca be the orthogonal projections of A' onto lines CA, AB, resp.
> Define Cb, Ab, Ac, Bc cyclically.
> What is the locus of P such that Ba, Ca, Cb, Ab, Ac, Bc lie on a common conic? The locus would include H, for which the conic is the Taylor circle.
>
> [BG]: a quintic with many simple points but only two (I think) ETC centers : X4, X1498.
>
> *** X(1498) is the Nagel point of the tangential triangle.
> Ab = (0 : S_{AB}+S_{AC}-S_{BC} : S_{AB}+S_{AC}+S_{BC}) and
> Ac = (0 : S_{AB}+S_{AC}+S_{BC} : S_{AB}+S_{AC}+S_{BC})
> are isotomic points on BC, so are Bc, Ba, and Ca, Cb.
> X(1498) is the unique point with this property. The conic is
>
> (4/S^2)cyclic sum ((a^4S_{AA})/(S_{AB}+S_AC}-S_{BC}))yz - (x+y+z)^2 = 0,
> concentric (and homothetic) with the circumconic with perspector
> ((a^4S_{AA}/(S_{AB}+S_{AC}-S_{BC}):...:...)
> [with (6-9-13)-search number 5.10435062529...]
> and has center
> (a^4(S_{AAAB}+S_{AAAC}+S_{AABB}-S_{AABC}+S_{AACC}-S_{BBCC}/
> (S_{AB}+S_{AC}-S_{BC}) :...:...)
> with (6-9-13) search number 1.09478783248....
>
> Best regards
> Sincerely
> Paul
>

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