## Another locus

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• let ABC be a triangle and A B C the pedal triangle of a point P. Let A be a point on PA (see figure) A / / B C / /
Message 1 of 6 , Mar 1, 2001
let ABC be a triangle and A'B'C' the pedal triangle of a point P.

Let A" be a point on PA' (see figure)

A
/\
/ \ B"
C" / \
/ \
/ \
/ \
/ \
/ \
/ P \
/ | \
/ | \
B--------A'------------C
|
|
|
|
|
|
A"

such that PA" = a.

Similarly we define the points B", C" on PB', PC', respectively
(PB" = b, PC" = c)

Which is the locus of P such that AA", BB", CC" are concurrent?

APH
• Let ABC be a triangle, P a point and A B C the pedal triangle of P. The parallels through A,B,C to B C , C A , A B , resp. bound a triangle A B C . Which is
Message 2 of 6 , Aug 16, 2009
Let ABC be a triangle, P a point and A'B'C' the
pedal triangle of P.

The parallels through A,B,C to B'C', C'A', A'B', resp.
bound a triangle A"B"C".

Which is the locus of P such that the Euler Lines of
A"BC, B"CA, C"AB are concurrent?

Variation: A'B'C' : cevian triangle of P.

APH
• Another unanswered post. In this case, the locus is a cubic with long equation (shown below) through X3 and X4. Best regards, Francisco Javier. ... S stands
Message 3 of 6 , Oct 23, 2011

In this case, the locus is a cubic with long equation (shown below) through X3 and X4.

Best regards,

Francisco Javier.

----

S stands for twice the area of ABC.

-8 a^4 b^4 c^2 x^3 + 8 a^2 b^6 c^2 x^3 + 8 a^4 b^2 c^4 x^3 -
8 a^2 b^2 c^6 x^3 - 8 a^2 b^4 c^2 S x^3 + 8 b^6 c^2 S x^3 +
8 a^2 b^2 c^4 S x^3 - 8 b^2 c^6 S x^3 + a^8 c^2 x^2 y -
8 a^6 b^2 c^2 x^2 y - 14 a^4 b^4 c^2 x^2 y + 24 a^2 b^6 c^2 x^2 y -
3 b^8 c^2 x^2 y - 4 a^6 c^4 x^2 y + 16 a^4 b^2 c^4 x^2 y -
12 a^2 b^4 c^4 x^2 y + 8 b^6 c^4 x^2 y + 6 a^4 c^6 x^2 y -
8 a^2 b^2 c^6 x^2 y - 6 b^4 c^6 x^2 y - 4 a^2 c^8 x^2 y +
c^10 x^2 y - 4 a^6 c^2 S x^2 y - 20 a^4 b^2 c^2 S x^2 y +
12 a^2 b^4 c^2 S x^2 y + 12 b^6 c^2 S x^2 y + 8 a^4 c^4 S x^2 y +
8 a^2 b^2 c^4 S x^2 y - 8 b^4 c^4 S x^2 y - 4 a^2 c^6 S x^2 y -
4 b^2 c^6 S x^2 y + 3 a^8 c^2 x y^2 - 24 a^6 b^2 c^2 x y^2 +
14 a^4 b^4 c^2 x y^2 + 8 a^2 b^6 c^2 x y^2 - b^8 c^2 x y^2 -
8 a^6 c^4 x y^2 + 12 a^4 b^2 c^4 x y^2 - 16 a^2 b^4 c^4 x y^2 +
4 b^6 c^4 x y^2 + 6 a^4 c^6 x y^2 + 8 a^2 b^2 c^6 x y^2 -
6 b^4 c^6 x y^2 + 4 b^2 c^8 x y^2 - c^10 x y^2 -
12 a^6 c^2 S x y^2 - 12 a^4 b^2 c^2 S x y^2 +
20 a^2 b^4 c^2 S x y^2 + 4 b^6 c^2 S x y^2 + 8 a^4 c^4 S x y^2 -
8 a^2 b^2 c^4 S x y^2 - 8 b^4 c^4 S x y^2 + 4 a^2 c^6 S x y^2 +
4 b^2 c^6 S x y^2 - 8 a^6 b^2 c^2 y^3 + 8 a^4 b^4 c^2 y^3 -
8 a^2 b^4 c^4 y^3 + 8 a^2 b^2 c^6 y^3 - 8 a^6 c^2 S y^3 +
8 a^4 b^2 c^2 S y^3 - 8 a^2 b^2 c^4 S y^3 + 8 a^2 c^6 S y^3 -
a^8 b^2 x^2 z + 4 a^6 b^4 x^2 z - 6 a^4 b^6 x^2 z + 4 a^2 b^8 x^2 z -
b^10 x^2 z + 8 a^6 b^2 c^2 x^2 z - 16 a^4 b^4 c^2 x^2 z +
8 a^2 b^6 c^2 x^2 z + 14 a^4 b^2 c^4 x^2 z + 12 a^2 b^4 c^4 x^2 z +
6 b^6 c^4 x^2 z - 24 a^2 b^2 c^6 x^2 z - 8 b^4 c^6 x^2 z +
3 b^2 c^8 x^2 z + 4 a^6 b^2 S x^2 z - 8 a^4 b^4 S x^2 z +
4 a^2 b^6 S x^2 z + 20 a^4 b^2 c^2 S x^2 z - 8 a^2 b^4 c^2 S x^2 z +
4 b^6 c^2 S x^2 z - 12 a^2 b^2 c^4 S x^2 z + 8 b^4 c^4 S x^2 z -
12 b^2 c^6 S x^2 z + a^10 y^2 z - 4 a^8 b^2 y^2 z +
6 a^6 b^4 y^2 z - 4 a^4 b^6 y^2 z + a^2 b^8 y^2 z -
8 a^6 b^2 c^2 y^2 z + 16 a^4 b^4 c^2 y^2 z - 8 a^2 b^6 c^2 y^2 z -
6 a^6 c^4 y^2 z - 12 a^4 b^2 c^4 y^2 z - 14 a^2 b^4 c^4 y^2 z +
8 a^4 c^6 y^2 z + 24 a^2 b^2 c^6 y^2 z - 3 a^2 c^8 y^2 z -
4 a^6 b^2 S y^2 z + 8 a^4 b^4 S y^2 z - 4 a^2 b^6 S y^2 z -
4 a^6 c^2 S y^2 z + 8 a^4 b^2 c^2 S y^2 z - 20 a^2 b^4 c^2 S y^2 z -
8 a^4 c^4 S y^2 z + 12 a^2 b^2 c^4 S y^2 z + 12 a^2 c^6 S y^2 z -
3 a^8 b^2 x z^2 + 8 a^6 b^4 x z^2 - 6 a^4 b^6 x z^2 + b^10 x z^2 +
24 a^6 b^2 c^2 x z^2 - 12 a^4 b^4 c^2 x z^2 - 8 a^2 b^6 c^2 x z^2 -
4 b^8 c^2 x z^2 - 14 a^4 b^2 c^4 x z^2 + 16 a^2 b^4 c^4 x z^2 +
6 b^6 c^4 x z^2 - 8 a^2 b^2 c^6 x z^2 - 4 b^4 c^6 x z^2 +
b^2 c^8 x z^2 + 12 a^6 b^2 S x z^2 - 8 a^4 b^4 S x z^2 -
4 a^2 b^6 S x z^2 + 12 a^4 b^2 c^2 S x z^2 + 8 a^2 b^4 c^2 S x z^2 -
4 b^6 c^2 S x z^2 - 20 a^2 b^2 c^4 S x z^2 + 8 b^4 c^4 S x z^2 -
4 b^2 c^6 S x z^2 - a^10 y z^2 + 6 a^6 b^4 y z^2 - 8 a^4 b^6 y z^2 +
3 a^2 b^8 y z^2 + 4 a^8 c^2 y z^2 + 8 a^6 b^2 c^2 y z^2 +
12 a^4 b^4 c^2 y z^2 - 24 a^2 b^6 c^2 y z^2 - 6 a^6 c^4 y z^2 -
16 a^4 b^2 c^4 y z^2 + 14 a^2 b^4 c^4 y z^2 + 4 a^4 c^6 y z^2 +
8 a^2 b^2 c^6 y z^2 - a^2 c^8 y z^2 + 4 a^6 b^2 S y z^2 +
8 a^4 b^4 S y z^2 - 12 a^2 b^6 S y z^2 + 4 a^6 c^2 S y z^2 -
8 a^4 b^2 c^2 S y z^2 - 12 a^2 b^4 c^2 S y z^2 - 8 a^4 c^4 S y z^2 +
20 a^2 b^2 c^4 S y z^2 + 4 a^2 c^6 S y z^2 + 8 a^6 b^2 c^2 z^3 -
8 a^2 b^6 c^2 z^3 - 8 a^4 b^2 c^4 z^3 + 8 a^2 b^4 c^4 z^3 +
8 a^6 b^2 S z^3 - 8 a^2 b^6 S z^3 - 8 a^4 b^2 c^2 S z^3 +
8 a^2 b^4 c^2 S z^3 ==0.

--- In Hyacinthos@yahoogroups.com, xpolakis@... wrote:
>
> let ABC be a triangle and A'B'C' the pedal triangle of a point P.
>
> Let A" be a point on PA' (see figure)
>
> A
> /\
> / \ B"
> C" / \
> / \
> / \
> / \
> / \
> / \
> / P \
> / | \
> / | \
> B--------A'------------C
> |
> |
> |
> |
> |
> |
> A"
>
> such that PA" = a.
>
> Similarly we define the points B", C" on PB', PC', respectively
> (PB" = b, PC" = c)
>
> Which is the locus of P such that AA", BB", CC" are concurrent?
>
>
> APH
>
• In the pedal case, the locus is (line at infinity) + (circumcircle) + (napoleon cubic). In the cevian case, the locus is a 15th degree curve through X2, X4 and
Message 4 of 6 , Oct 23, 2011
In the pedal case, the locus is (line at infinity) + (circumcircle) + (napoleon cubic).

In the cevian case, the locus is a 15th degree curve through X2, X4 and X7.

Best regards,

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, "xpolakis" <anopolis72@...> wrote:
>
> Let ABC be a triangle, P a point and A'B'C' the
> pedal triangle of P.
>
> The parallels through A,B,C to B'C', C'A', A'B', resp.
> bound a triangle A"B"C".
>
> Which is the locus of P such that the Euler Lines of
> A"BC, B"CA, C"AB are concurrent?
>
> Variation: A'B'C' : cevian triangle of P.
>
>
> APH
>
• ... [long equation snipped -- BW] ... Since points at infinity are extraneous roots, this becomes a conic, CyclicSum( (SB-SC) (SA x (x+y+z) + S y z) = 0. --
Message 5 of 6 , Oct 27, 2011
>
> In this case, the locus is a cubic with long equation (shown below) through X3
> and X4.
>
> Best regards,
>
> Francisco Javier.
>
> ----
>
> S stands for twice the area of ABC.
>

[long equation snipped -- BW]

> --- In Hyacinthos@yahoogroups.com, xpolakis@... wrote:
>>
>> let ABC be a triangle and A'B'C' the pedal triangle of a point
> P.
>>
>> Let A" be a point on PA' (see figure)
>>
>>                               A
>>                               /\
>>                               /  \            B"
>>               C"            /    \
>>                             /      \
>>                           /        \
>>                           /          \
>>                         /            \
>>                         /              \
>>                       /    P          \
>>                       /      |          \
>>                     /      |            \
>>                     B--------A'------------C
>>                             |
>>                             |
>>                             |
>>                             |
>>                             |
>>                             |
>>                             A"
>>
>> such that PA" = a.
>>
>> Similarly we define the points B", C" on PB', PC',
> respectively
>> (PB" = b, PC" = c)
>>
>> Which is the locus of P such that AA", BB", CC" are
> concurrent?
>>
>>
>> APH

Since points at infinity are extraneous roots, this becomes a conic,
CyclicSum( (SB-SC) (SA x (x+y+z) + S y z) = 0.
--
Barry Wolk
• Dear Barry Wolk: Yes, you are right. I misunderstood the statement, since I toke A satisfying A A = a, etc,instead of PA =a, etc, then A lying on the
Message 6 of 6 , Oct 27, 2011
Dear Barry Wolk:

Yes, you are right. I misunderstood the statement, since I toke A'' satisfying A'A'' = a, etc,instead of PA''=a, etc, then A'' lying on the square constructed externally on BC.

Best regards,

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, Barry Wolk <wolkbarry@...> wrote:
>
> >
> > In this case, the locus is a cubic with long equation (shown below) through X3
> > and X4.
> >
> > Best regards,
> >
> > Francisco Javier.
> >
> > ----
> >
> > S stands for twice the area of ABC.
> >
>
> [long equation snipped -- BW]
>
>
> >ï¿½
> > --- In Hyacinthos@yahoogroups.com, xpolakis@ wrote:
> >>
> >> let ABC be a triangle and A'B'C' the pedal triangle of a point
> > P.
> >>
> >> Let A" be a point on PA' (see figure)
> >>
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ A
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /\
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ \ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ B"
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ C"ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ ï¿½ \
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ ï¿½ ï¿½ \
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ ï¿½ ï¿½ ï¿½ \
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ \
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ \
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ \
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ ï¿½ Pï¿½ ï¿½ ï¿½ ï¿½ ï¿½ \
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ ï¿½ ï¿½ |ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ \
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ /ï¿½ ï¿½ ï¿½ |ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ \
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ B--------A'------------C
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ |
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ |
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ |
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ |
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ |
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ |
> >> ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ ï¿½ A"
> >>
> >> such that PA" = a.
> >>
> >> Similarly we define the points B", C" on PB', PC',
> > respectively
> >> (PB" = b, PC" = c)
> >>
> >> Which is the locus of P such that AA", BB", CC" are
> > concurrent?
> >>
> >>
> >> APH
>
> Since points at infinity are extraneous roots, this becomes a conic,
> CyclicSum( (SB-SC) (SA x (x+y+z) + S y z) = 0.
> --
> Barry Wolk
>
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