## Re: NPC-related conic

Expand Messages
• Dear Randy, you can take A B C the medial triangle and A B C the cevian triangle of any point (p:q:r) and this gives a conic with center {(p + q) (p + r) (10
Message 1 of 3 , Feb 24, 2013
Dear Randy, you can take A'B'C' the medial triangle and A"B"C" the cevian triangle of any point (p:q:r) and this gives a conic with center

{(p + q) (p + r) (10 p^3 q^2 + 25 p^2 q^3 + 5 p q^4 + 12 p^3 q r +
87 p^2 q^2 r + 58 p q^3 r + 3 q^4 r + 10 p^3 r^2 + 87 p^2 q r^2 +
98 p q^2 r^2 + 13 q^3 r^2 + 25 p^2 r^3 + 58 p q r^3 +
13 q^2 r^3 + 5 p r^4 + 3 q r^4), (p + q) (q + r) (5 p^4 q +
25 p^3 q^2 + 10 p^2 q^3 + 3 p^4 r + 58 p^3 q r + 87 p^2 q^2 r +
12 p q^3 r + 13 p^3 r^2 + 98 p^2 q r^2 + 87 p q^2 r^2 +
10 q^3 r^2 + 13 p^2 r^3 + 58 p q r^3 + 25 q^2 r^3 + 3 p r^4 +
5 q r^4), (p + r) (q + r) (3 p^4 q + 13 p^3 q^2 + 13 p^2 q^3 +
3 p q^4 + 5 p^4 r + 58 p^3 q r + 98 p^2 q^2 r + 58 p q^3 r +
5 q^4 r + 25 p^3 r^2 + 87 p^2 q r^2 + 87 p q^2 r^2 + 25 q^3 r^2 +
10 p^2 r^3 + 12 p q r^3 + 10 q^2 r^3)}

that is homothetic to the circumconic

(3 p + q) (p + 3 q) r (p + r) (q + r) x y +
q (p + q) (3 p + r) (q + r) (p + 3 r) x z +
p (p + q) (p + r) (3 q + r) (q + 3 r) y z = 0.

Best regards,

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
>
> Let A'B'C' be the medial triangle and A"B"C" be the orthic triangle.
> (So A', A" are the intersections of the NPC with BC, etc.)
> Let Ab be the {A',A"}-Harmonic conjugate of B, and let Ac be the {A',A"}-Harmonic conjugate of C. Define Bc, Ba, Ca, Cb cyclically.
>
> Ab, Ac, Bc, Ba, Ca, Cb lie on a common conic.
>
> What are the coordinates of the conic center?
>
> Best regards,
> Randy
>
• Thank you, Francisco!  This could be further generalized by letting A B C and A B C be any 2 cevian triangles. Rsndy ... [Non-text portions of this message
Message 2 of 3 , Feb 25, 2013
Thank you, Francisco!  This could be further generalized by letting A'B'C' and A"B"C" be any 2 cevian triangles.

Rsndy

>________________________________
> From: Francisco Javier <garciacapitan@...>
>To: Hyacinthos@yahoogroups.com
>Sent: Sunday, February 24, 2013 2:14 AM
>Subject: [EMHL] Re: NPC-related conic
>
>

>Dear Randy, you can take A'B'C' the medial triangle and A"B"C" the cevian triangle of any point (p:q:r) and this gives a conic with center
>
>{(p + q) (p + r) (10 p^3 q^2 + 25 p^2 q^3 + 5 p q^4 + 12 p^3 q r +
>87 p^2 q^2 r + 58 p q^3 r + 3 q^4 r + 10 p^3 r^2 + 87 p^2 q r^2 +
>98 p q^2 r^2 + 13 q^3 r^2 + 25 p^2 r^3 + 58 p q r^3 +
>13 q^2 r^3 + 5 p r^4 + 3 q r^4), (p + q) (q + r) (5 p^4 q +
>25 p^3 q^2 + 10 p^2 q^3 + 3 p^4 r + 58 p^3 q r + 87 p^2 q^2 r +
>12 p q^3 r + 13 p^3 r^2 + 98 p^2 q r^2 + 87 p q^2 r^2 +
>10 q^3 r^2 + 13 p^2 r^3 + 58 p q r^3 + 25 q^2 r^3 + 3 p r^4 +
>5 q r^4), (p + r) (q + r) (3 p^4 q + 13 p^3 q^2 + 13 p^2 q^3 +
>3 p q^4 + 5 p^4 r + 58 p^3 q r + 98 p^2 q^2 r + 58 p q^3 r +
>5 q^4 r + 25 p^3 r^2 + 87 p^2 q r^2 + 87 p q^2 r^2 + 25 q^3 r^2 +
>10 p^2 r^3 + 12 p q r^3 + 10 q^2 r^3)}
>
>that is homothetic to the circumconic
>
>(3 p + q) (p + 3 q) r (p + r) (q + r) x y +
>q (p + q) (3 p + r) (q + r) (p + 3 r) x z +
>p (p + q) (p + r) (3 q + r) (q + 3 r) y z = 0.
>
>Best regards,
>
>Francisco Javier.
>
>
>--- In Hyacinthos@yahoogroups.com, "rhutson2" wrote:
>>
>> Let A'B'C' be the medial triangle and A"B"C" be the orthic triangle.
>> (So A', A" are the intersections of the NPC with BC, etc.)
>> Let Ab be the {A',A"}-Harmonic conjugate of B, and let Ac be the {A',A"}-Harmonic conjugate of C. Define Bc, Ba, Ca, Cb cyclically.
>>
>> Ab, Ac, Bc, Ba, Ca, Cb lie on a common conic.
>>
>> What are the coordinates of the conic center?
>>
>> Best regards,
>> Randy
>>
>
>
>
>
>

[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.