## Re: [EMHL] LOCUS

Expand Messages
• Dear Antreas What do you mean by circle (Pb, PbP)? Friendly Francois ... [Non-text portions of this message have been removed]
Message 1 of 26 , Apr 13, 2010
Dear Antreas
What do you mean by circle (Pb, PbP)?
Friendly
Francois

On Wed, Apr 7, 2010 at 1:11 PM, Antreas <anopolis72@...> wrote:

>
>
> Let ABC be a triangle, P a point and
> PaPbPc the pedal triangle of P.
>
> Let A' be the other than P intersection of
> the circles (Pb, PbP) and (Pc, PcP)
> and similarly B', C'.
>
> Which is the locus of P such that:
>
> 1. ABC, A'B'C' are perspective
>
> 2. PaPbPc, A'B'C' are perspective
>
> 3. ABC, A'B'C' are orthologic.
>
> The triangles PaPbPc, A'B'C' are orthologic.
> One orthologic center is P. The other one?
>
> Antreas
>
>
>

[Non-text portions of this message have been removed]
• Dear Francois Circle with center Pb and radius PbP APH On Tue, Apr 13, 2010 at 6:17 PM, Francois Rideau ... -- http://anopolis72000.blogspot.com/
Message 2 of 26 , Apr 13, 2010
Dear Francois

Circle with center Pb and radius PbP

APH

On Tue, Apr 13, 2010 at 6:17 PM, Francois Rideau
<francois.rideau@...> wrote:
> Dear Antreas
> What do you mean by circle (Pb, PbP)?
> Friendly
> Francois
>
> On Wed, Apr 7, 2010 at 1:11 PM, Antreas <anopolis72@...> wrote:
>
>>
>>
>> Let ABC be a triangle, P a point and
>> PaPbPc the pedal triangle of P.
>>
>> Let A' be the other than P intersection of
>> the circles (Pb, PbP) and (Pc, PcP)
>> and similarly B', C'.
>>
>> Which is the locus of P such that:
>>
>> 1. ABC, A'B'C' are perspective
>>
>> 2. PaPbPc, A'B'C' are perspective
>>
>> 3. ABC, A'B'C' are orthologic.
>>
>> The triangles PaPbPc, A'B'C' are orthologic.
>> One orthologic center is P. The other one?
>>
>> Antreas
>>
>>
>>
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
>
>

--
http://anopolis72000.blogspot.com/
• Let ABC be a triangle, P a point, A B C the pedal triangle of P and L the Euler line of ABC. Let La,Lb,Lc be the reflections of L in the PA ,PB ,PC , resp.
Message 3 of 26 , Aug 3, 2010
Let ABC be a triangle, P a point, A'B'C'
the pedal triangle of P and L the Euler
line of ABC.

Let La,Lb,Lc be the reflections of L in the
PA',PB',PC', resp.

Which is the locus of P such that
ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

Antiedal triangle version:

Let ABC be a triangle, P a point, A'B'C'
the antipedal triangle of P and L the Euler
line of A'B'C'. Let La,Lb,Lc be the reflections
of L in the PA,PB,PC, resp.

Which is the locus of P such that
A'B'C', Triangle bounded by (La,Lb,Lc) are parallelogic?

Are the simple points O,H,I lying on the locus?

APH
• ... ABC and Triangle bounded by (La,Lb,Lc) are parallelogic = the parallels through A,B,C to La,Lb,Lc, resp. concur on a point U U=X(265) for every point P of
Message 4 of 26 , Aug 4, 2010
--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle, P a point, A'B'C'
> the pedal triangle of P and L the Euler
> line of ABC.
>
> Let La,Lb,Lc be the reflections of L in the
> PA',PB',PC', resp.
>
> Which is the locus of P such that
> ABC, Triangle bounded by (La,Lb,Lc) are parallelogic?

> APH
>

ABC and Triangle bounded by (La,Lb,Lc) are parallelogic =
the parallels through A,B,C to La,Lb,Lc, resp. concur on a point U

U=X(265) for every point P of the plane

Angel
• Let ABC be a triangle, P a point, PaPbPc the pedal (or cevian) triangle of P, and (Ia),(Ib),(Ic) the excircles. Let (Oa) be the circle passing through Pb, Pc
Message 5 of 26 , Sep 7, 2010
Let ABC be a triangle, P a point, PaPbPc the pedal (or cevian)
triangle of P, and (Ia),(Ib),(Ic) the excircles.

Let (Oa) be the circle passing through Pb, Pc and touching
externally (or internally) the circle (Ia) at Qa.

Similarly (Ob),(Oc) and Qb,Qc.

Which is the locus of P such that

1. ABC, OaObOc

2. ABC, QaQbQc

are perspective?

(There are four variations:
pedal/cevian - externally/internally)

APH
• Dear Antreas TWO PARTICULAR CASES ... - PaPbPc the cevian triangle of G (PaPbPc the medial triangle). Let (Oa) be the circle passing through Pb, Pc and
Message 6 of 26 , Sep 9, 2010
Dear Antreas

TWO PARTICULAR CASES
----------------------------------------

- PaPbPc the cevian triangle of G (PaPbPc the medial triangle).

Let (Oa) be the circle passing through Pb, Pc and touching
internally the circle (Ia) at Qa. Similarly (Ob),(Oc) and Qb,Qc.

Perspector of ABC and QaQbQc:

( (b+c-a)(b+c)^2 : (c+a-b)(c+a)^2 : (a+b-c)(a+b)^2 )

Radical center of (Oa), (Ob), (Oc):

(a (a^2(b + c) + 2 a (b^2 + 3 b c + c^2) + (b + c)^3): ... : ...)

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

- PaPbPc the cevian triangle of H (PaPbPc the orthic triangle)
Let (Oa) be the circle passing through Pb, Pc and touching
internally the circle (Ia) at Qa. Similarly (Ob),(Oc) and Qb,Qc.

Perspector of ABC and QaQbQc:

( (b+c)^2/(b+c-a)^3 : (c+a)^2/(c+a-b)^3 : (a+b)^2/(a+b-c)^3 )

Radical center of (Oa), (Ob), (Oc): X(1829) = ZOSMA TRANSFORM OF X(10)

(a (a^5(b + c) + a^4(b^2 + c^2) - a(b - c)^2 (b + c)^3 - (b^2-c^2)^2 (b^2 + c^2)): ... : ...)

Best regards,

Angel

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle, P a point, PaPbPc the pedal (or cevian)
> triangle of P, and (Ia),(Ib),(Ic) the excircles.
>
> Let (Oa) be the circle passing through Pb, Pc and touching
> externally (or internally) the circle (Ia) at Qa.
>
> Similarly (Ob),(Oc) and Qb,Qc.
>
> Which is the locus of P such that
>
> 1. ABC, OaObOc
>
> 2. ABC, QaQbQc
>
> are perspective?
>
> (There are four variations:
> pedal/cevian - externally/internally)
>
> APH
>
• Let ABC be a triangle, P a point and Ab,Ac points on AB,AC, resp. (on the same size of BC) such that: area(PBC) = area(PBAb) = area(PCAc). Similarly the points
Message 7 of 26 , May 5, 2011
Let ABC be a triangle, P a point and Ab,Ac points on AB,AC, resp.
(on the same size of BC) such that:
area(PBC) = area(PBAb) = area(PCAc).

Similarly the points Bc,Ba and Ca,Cb.

Which is the locus of P such that the triangles: ABC, Triangle
bounded by AbAc,BcBa,CaCb are perspective?
(Special case: AbAc,BcBa,CaCb are concurrent)

APH
• Dear Antreas: The lines AbAc,BcBa,CaCb are always parallel to the trilinear polar of P. Best regards, Francisco Javier.
Message 8 of 26 , May 5, 2011
Dear Antreas:

The lines AbAc,BcBa,CaCb are always parallel to the trilinear polar of P.

Best regards,

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle, P a point and Ab,Ac points on AB,AC, resp.
> (on the same size of BC) such that:
> area(PBC) = area(PBAb) = area(PCAc).
>
> Similarly the points Bc,Ba and Ca,Cb.
>
> Which is the locus of P such that the triangles: ABC, Triangle
> bounded by AbAc,BcBa,CaCb are perspective?
> (Special case: AbAc,BcBa,CaCb are concurrent)
>
> APH
>
• And, if lines AbAc, BcBa, CaCb intersect the lines BC, CA, AB at A , B , C , then lines AA , BB , CC are parallel to the polar trilineal of isotomic conjugate
Message 9 of 26 , May 6, 2011
And, if lines AbAc, BcBa, CaCb intersect the lines BC, CA, AB at A', B', C', then lines AA', BB', CC' are parallel to the polar trilineal of isotomic conjugate of P.

--- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
>
> Dear Antreas:
>
> The lines AbAc,BcBa,CaCb are always parallel to the trilinear polar of P.
>
> Best regards,
>
> Francisco Javier.
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> >
> > Let ABC be a triangle, P a point and Ab,Ac points on AB,AC, resp.
> > (on the same size of BC) such that:
> > area(PBC) = area(PBAb) = area(PCAc).
> >
> > Similarly the points Bc,Ba and Ca,Cb.
> >
> > Which is the locus of P such that the triangles: ABC, Triangle
> > bounded by AbAc,BcBa,CaCb are perspective?
> > (Special case: AbAc,BcBa,CaCb are concurrent)
> >
> > APH
> >
>
• Let ABC be a triangle and A B C , A B C the cevian, pedal triangles of P, resp. Denote: Ab, Ac = the reflections of A in BB , CC Bc, Ba = the reflections of
Message 10 of 26 , Feb 25, 2013
Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
triangles of P, resp.

Denote:

Ab, Ac = the reflections of A' in BB', CC'

Bc, Ba = the reflections of B' in CC', AA'

Ca, Cb = the reflections of C' in AA', BB'

Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at P = common circumcenter of the triangles)

La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.

For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
(antipode of Feuerbach point)

For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
on the pedal circle of H (=NPC).

Which is the point of concurrence?

In general:
Which is the locus of P such that the lines La,Lb,Lc
are concurrent?

Antreas
• Dear Antreas, If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle. If P=H the lines La,Lb,Lc concur in X(1986)=
Message 11 of 26 , Feb 26, 2013
Dear Antreas,

If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle.

If P=H the lines La,Lb,Lc concur in X(1986)= HATZIPOLAKIS REFLECTION POINT (Antreas Hatzipolakis,Hyacinthos 7868,9/12/03;coordinates by Barry Wolk,Hyacinthos 7876,9/13/03)

X(1986)=(a^2(4SA^2-b^2c^2)(a^2(SA^2-SB SC)-SA(c^2-b^2)^2)/SA: ... : ...)

In general:
The lines La,Lb,Lc are concurrent if P is on the algebraic curve (Gamma) of degree nine (SA, SB, SC usual Conway notation):

(SA+SB)^3(SA+SC)(SA*SB-2SA*SC+SB*SC)x^6y^3-

(SA+SB)^3(SA+SC)(5SA*SB+5SA*SC-4SB*SC)x^5y^4+

(SA+SB)^3(SB+SC)(5SA*SB-4SA*SC+5SB*SC)x^4y^5-

(SA+SB)^3(SB+SC)(SA*SB+SA*SC-2SB*SC)x^3y^6+

(SA+SB)^2x^2y^2z((-SA-SC)(-4SA^2SB+SA(5SA-3SB)SC+(SA+SB)SC^2)x^4-
2(SA+SC)(SA(SB-SC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y-
4(SA-SB)SC(SB*SC+SA(SB+SC))x^2y^2+
2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^2-2SB*SC+SC^2))x*y^3+
(SB+SC)(-SA(4SB-SC)(SB+SC)+SB*SC(5SB+SC))y^4)+

(SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)-(4SA-SB)(SA+SB)SC)x^5-
(SA+SC)(SA^2(19SB-14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
SA*SB(7SB^2-8SB*SC-3SC^2))x^3y^2+(SB+SC)*
(SA^2SB(7SA+19SB)+SA(7SA^2-8SA*SB+5SB^2)SC+
(11SA-14SB)(SA+SB)SC^2)x^2y^3-(SB+SC)^2(SA*SB(SA+5SB)+
(SA-4SB)(SA+SB)SC)y^5)+

(SA+SB)z^3((-(SA+SC)^3)(-2SA*SB+
(SA+SB)SC)x^6+2(SA+SC)^2(SA(SB-SC)^2+5SA^2(SB+SC)+
SB*SC(SB+SC))x^5y-(SA+SC)(SA^2(14SB-19SC)*

(SB+SC)-SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC-7SC^2))*
x^4y^2+(SB+SC)(SA^2(14SB-11SC)(SB+SC)-
SB*SC^2(19SB+7SC)+SA*SC(-5SB^2+8SB*SC-7SC^2))*
x^2y^4-2(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
SA(5SB^2-2SB*SC+SC^2))x*y^5+(SB+SC)^3(-2SA*SB+(SA+SB)SC)*y^6)+

((SA+SB)(SA+SC)^3(-4SB*SC+5SA(SB+SC))x^5+

4SB(SA-SC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y-

(SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
SA(7SA^2-8SA*SB-3SB^2)SC+(19SA-14SB)(SA+SB)SC^2)x^3y^2+

(SA+SC)(SB+SC)(SA^2(11SB-14SC)(SB+SC)+
SB^2SC(7SB+19SC)+SA*SB(7SB^2-8SB*SC+5SC^2))x^2y^3-

4SA(SB-SC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4-

(SA+SB)(SB+SC)^3(5SA*SB-4SA*SC+5SB*SC)y^5)z^4-

(SA+SC)(SB+SC)((SA+SC)x^2-(SB+SC)y^2)((SA+SC)(-4SA*SB+5(SA+SB)SC)x^2+
2(SA*SB(SA+SB)+(SA-SB)^2SC+5(SA+SB)SC^2)x*y+
(SB+SC)(-4SA*SB+5(SA+SB)SC)y^2)z^5+

(SA+SC)(SB+SC)((SA+SC)(-2SB*SC+SA(SB+SC))x-(SB+SC)(SA(SB-2SC)+SB*SC)y)*(SA*x^2+SB*y^2+SC(x+y)^2)z^6 =0

(There must be a better simplification of this equation!)

This curve (Gamma) contains points isodynamic (X13, X14) but the corresponding triangles AAbAc, BBcBa and CCaCb are equilateral.

The triangle center X(74) -isogonal conjugate of Euler infinity point- is on the curve (Gamma) and the Euler Lines of A'AbAc, B'BcBa, C'CaCb
passing through A'', B'', C'' resp. (concurrent at X74 = common circumcenter of the triangles).

Best regards
Angel Montesdeoca

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
> triangles of P, resp.
>
> Denote:
>
> Ab, Ac = the reflections of A' in BB', CC'
>
> Bc, Ba = the reflections of B' in CC', AA'
>
> Ca, Cb = the reflections of C' in AA', BB'
>
> Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at P = common circumcenter of the triangles)
>
> La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
>
> For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
> (antipode of Feuerbach point)
>
> For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
> on the pedal circle of H (=NPC).
>
> Which is the point of concurrence?
>
> In general:
> Which is the locus of P such that the lines La,Lb,Lc
> are concurrent?
>
> Antreas
>
• More information on the algebraic curve of degree nine (Gamma): - Passes through the vertices of the triangle ABC. - The vertices are multiple points of order
Message 12 of 26 , Feb 26, 2013

- Passes through the vertices of the triangle ABC.

- The vertices are multiple points of order 3.

- The real tangents at the vertices of ABC intersect at the point X(74).

Angel M.

--- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
>
>
> Dear Antreas,
>
> If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of Feuerbach point on the incircle.
>
> If P=H the lines La,Lb,Lc concur in X(1986)= HATZIPOLAKIS REFLECTION POINT (Antreas Hatzipolakis,Hyacinthos 7868,9/12/03;coordinates by Barry Wolk,Hyacinthos 7876,9/13/03)
>
> X(1986)=(a^2(4SA^2-b^2c^2)(a^2(SA^2-SB SC)-SA(c^2-b^2)^2)/SA: ... : ...)
>
>
> In general:
> The lines La,Lb,Lc are concurrent if P is on the algebraic curve (Gamma) of degree nine (SA, SB, SC usual Conway notation):
>
>
> (SA+SB)^3(SA+SC)(SA*SB-2SA*SC+SB*SC)x^6y^3-
>
> (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC-4SB*SC)x^5y^4+
>
> (SA+SB)^3(SB+SC)(5SA*SB-4SA*SC+5SB*SC)x^4y^5-
>
> (SA+SB)^3(SB+SC)(SA*SB+SA*SC-2SB*SC)x^3y^6+
>
> (SA+SB)^2x^2y^2z((-SA-SC)(-4SA^2SB+SA(5SA-3SB)SC+(SA+SB)SC^2)x^4-
> 2(SA+SC)(SA(SB-SC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y-
> 4(SA-SB)SC(SB*SC+SA(SB+SC))x^2y^2+
> 2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^2-2SB*SC+SC^2))x*y^3+
> (SB+SC)(-SA(4SB-SC)(SB+SC)+SB*SC(5SB+SC))y^4)+
>
> (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)-(4SA-SB)(SA+SB)SC)x^5-
> (SA+SC)(SA^2(19SB-14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
> SA*SB(7SB^2-8SB*SC-3SC^2))x^3y^2+(SB+SC)*
> (SA^2SB(7SA+19SB)+SA(7SA^2-8SA*SB+5SB^2)SC+
> (11SA-14SB)(SA+SB)SC^2)x^2y^3-(SB+SC)^2(SA*SB(SA+5SB)+
> (SA-4SB)(SA+SB)SC)y^5)+
>
> (SA+SB)z^3((-(SA+SC)^3)(-2SA*SB+
> (SA+SB)SC)x^6+2(SA+SC)^2(SA(SB-SC)^2+5SA^2(SB+SC)+
> SB*SC(SB+SC))x^5y-(SA+SC)(SA^2(14SB-19SC)*
>
> (SB+SC)-SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC-7SC^2))*
> x^4y^2+(SB+SC)(SA^2(14SB-11SC)(SB+SC)-
> SB*SC^2(19SB+7SC)+SA*SC(-5SB^2+8SB*SC-7SC^2))*
> x^2y^4-2(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
> SA(5SB^2-2SB*SC+SC^2))x*y^5+(SB+SC)^3(-2SA*SB+(SA+SB)SC)*y^6)+
>
> ((SA+SB)(SA+SC)^3(-4SB*SC+5SA(SB+SC))x^5+
>
> 4SB(SA-SC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y-
>
> (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
> SA(7SA^2-8SA*SB-3SB^2)SC+(19SA-14SB)(SA+SB)SC^2)x^3y^2+
>
> (SA+SC)(SB+SC)(SA^2(11SB-14SC)(SB+SC)+
> SB^2SC(7SB+19SC)+SA*SB(7SB^2-8SB*SC+5SC^2))x^2y^3-
>
> 4SA(SB-SC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4-
>
> (SA+SB)(SB+SC)^3(5SA*SB-4SA*SC+5SB*SC)y^5)z^4-
>
> (SA+SC)(SB+SC)((SA+SC)x^2-(SB+SC)y^2)((SA+SC)(-4SA*SB+5(SA+SB)SC)x^2+
> 2(SA*SB(SA+SB)+(SA-SB)^2SC+5(SA+SB)SC^2)x*y+
> (SB+SC)(-4SA*SB+5(SA+SB)SC)y^2)z^5+
>
> (SA+SC)(SB+SC)((SA+SC)(-2SB*SC+SA(SB+SC))x-(SB+SC)(SA(SB-2SC)+SB*SC)y)*(SA*x^2+SB*y^2+SC(x+y)^2)z^6 =0
>
>
> (There must be a better simplification of this equation!)
>
>
> This curve (Gamma) contains points isodynamic (X13, X14) but the corresponding triangles AAbAc, BBcBa and CCaCb are equilateral.
>
> The triangle center X(74) -isogonal conjugate of Euler infinity point- is on the curve (Gamma) and the Euler Lines of A'AbAc, B'BcBa, C'CaCb
> passing through A'', B'', C'' resp. (concurrent at X74 = common circumcenter of the triangles).
>
>
> Best regards
> Angel Montesdeoca
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> >
> > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
> > triangles of P, resp.
> >
> > Denote:
> >
> > Ab, Ac = the reflections of A' in BB', CC'
> >
> > Bc, Ba = the reflections of B' in CC', AA'
> >
> > Ca, Cb = the reflections of C' in AA', BB'
> >
> > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent at P = common circumcenter of the triangles)
> >
> > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
> >
> > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
> > (antipode of Feuerbach point)
> >
> > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
> > on the pedal circle of H (=NPC).
> >
> > Which is the point of concurrence?
> >
> > In general:
> > Which is the locus of P such that the lines La,Lb,Lc
> > are concurrent?
> >
> > Antreas
> >
>
• Dear Angel Thank you. I came to this configuration trying to find three homocentric (concentric) circles but not by construction (ie not by taking a point as
Message 13 of 26 , Feb 26, 2013
Dear Angel

Thank you.

I came to this configuration trying to find three homocentric (concentric)
circles
but not by construction (ie not by taking a point as center).

Antreas

On Tue, Feb 26, 2013 at 2:57 PM, Angel <amontes1949@...> wrote:

> **
>
>
>
> Dear Antreas,
>
> If P=I the lines La,Lb,Lc intersect at X(1317) is the antipode of
> Feuerbach point on the incircle.
>
> If P=H the lines La,Lb,Lc concur in X(1986)= HATZIPOLAKIS REFLECTION POINT
> (Antreas Hatzipolakis,Hyacinthos 7868,9/12/03;coordinates by Barry
> Wolk,Hyacinthos 7876,9/13/03)
>
> X(1986)=(a^2(4SA^2-b^2c^2)(a^2(SA^2-SB SC)-SA(c^2-b^2)^2)/SA: ... : ...)
>
> In general:
> The lines La,Lb,Lc are concurrent if P is on the algebraic curve (Gamma)
> of degree nine (SA, SB, SC usual Conway notation):
>
> (SA+SB)^3(SA+SC)(SA*SB-2SA*SC+SB*SC)x^6y^3-
>
> (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC-4SB*SC)x^5y^4+
>
> (SA+SB)^3(SB+SC)(5SA*SB-4SA*SC+5SB*SC)x^4y^5-
>
> (SA+SB)^3(SB+SC)(SA*SB+SA*SC-2SB*SC)x^3y^6+
>
> (SA+SB)^2x^2y^2z((-SA-SC)(-4SA^2SB+SA(5SA-3SB)SC+(SA+SB)SC^2)x^4-
> 2(SA+SC)(SA(SB-SC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y-
> 4(SA-SB)SC(SB*SC+SA(SB+SC))x^2y^2+
> 2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^2-2SB*SC+SC^2))x*y^3+
> (SB+SC)(-SA(4SB-SC)(SB+SC)+SB*SC(5SB+SC))y^4)+
>
> (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)-(4SA-SB)(SA+SB)SC)x^5-
> (SA+SC)(SA^2(19SB-14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
> SA*SB(7SB^2-8SB*SC-3SC^2))x^3y^2+(SB+SC)*
> (SA^2SB(7SA+19SB)+SA(7SA^2-8SA*SB+5SB^2)SC+
> (11SA-14SB)(SA+SB)SC^2)x^2y^3-(SB+SC)^2(SA*SB(SA+5SB)+
> (SA-4SB)(SA+SB)SC)y^5)+
>
> (SA+SB)z^3((-(SA+SC)^3)(-2SA*SB+
> (SA+SB)SC)x^6+2(SA+SC)^2(SA(SB-SC)^2+5SA^2(SB+SC)+
> SB*SC(SB+SC))x^5y-(SA+SC)(SA^2(14SB-19SC)*
>
> (SB+SC)-SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC-7SC^2))*
> x^4y^2+(SB+SC)(SA^2(14SB-11SC)(SB+SC)-
> SB*SC^2(19SB+7SC)+SA*SC(-5SB^2+8SB*SC-7SC^2))*
> x^2y^4-2(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
> SA(5SB^2-2SB*SC+SC^2))x*y^5+(SB+SC)^3(-2SA*SB+(SA+SB)SC)*y^6)+
>
> ((SA+SB)(SA+SC)^3(-4SB*SC+5SA(SB+SC))x^5+
>
> 4SB(SA-SC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y-
>
> (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
> SA(7SA^2-8SA*SB-3SB^2)SC+(19SA-14SB)(SA+SB)SC^2)x^3y^2+
>
> (SA+SC)(SB+SC)(SA^2(11SB-14SC)(SB+SC)+
> SB^2SC(7SB+19SC)+SA*SB(7SB^2-8SB*SC+5SC^2))x^2y^3-
>
> 4SA(SB-SC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4-
>
> (SA+SB)(SB+SC)^3(5SA*SB-4SA*SC+5SB*SC)y^5)z^4-
>
> (SA+SC)(SB+SC)((SA+SC)x^2-(SB+SC)y^2)((SA+SC)(-4SA*SB+5(SA+SB)SC)x^2+
> 2(SA*SB(SA+SB)+(SA-SB)^2SC+5(SA+SB)SC^2)x*y+
> (SB+SC)(-4SA*SB+5(SA+SB)SC)y^2)z^5+
>
> (SA+SC)(SB+SC)((SA+SC)(-2SB*SC+SA(SB+SC))x-(SB+SC)(SA(SB-2SC)+SB*SC)y)*(SA*x^2+SB*y^2+SC(x+y)^2)z^6
> =0
>
> (There must be a better simplification of this equation!)
>
> This curve (Gamma) contains points isodynamic (X13, X14) but the
> corresponding triangles AAbAc, BBcBa and CCaCb are equilateral.
>
> The triangle center X(74) -isogonal conjugate of Euler infinity point- is
> on the curve (Gamma) and the Euler Lines of A'AbAc, B'BcBa, C'CaCb
> passing through A'', B'', C'' resp. (concurrent at X74 = common
> circumcenter of the triangles).
>
> Best regards
> Angel Montesdeoca
>
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
> >
> > Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
> > triangles of P, resp.
> >
> > Denote:
> >
> > Ab, Ac = the reflections of A' in BB', CC'
> >
> > Bc, Ba = the reflections of B' in CC', AA'
> >
> > Ca, Cb = the reflections of C' in AA', BB'
> >
> > Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp.
> (concurrent at P = common circumcenter of the triangles)
> >
> > La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
> >
> > For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
> > (antipode of Feuerbach point)
> >
> > For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
> > on the pedal circle of H (=NPC).
> >
> > Which is the point of concurrence?
> >
> > In general:
> > Which is the locus of P such that the lines La,Lb,Lc
> > are concurrent?
> >
> > Antreas
> >
>
>
>

--
http://anopolis72000.blogspot.com/

[Non-text portions of this message have been removed]
• [APH] ... More for P = H: The NPCs of A AbAc, B BcBa, C CaCb are concurrent on the NPC of A B C (On the Poncelet point of H with respect A B C ie the point
Message 14 of 26 , Feb 26, 2013
[APH]

>
> Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal
> triangles of P, resp.
>
> Denote:
>
> Ab, Ac = the reflections of A' in BB', CC'
>
> Bc, Ba = the reflections of B' in CC', AA'
>
> Ca, Cb = the reflections of C' in AA', BB'
>
> Ea, Eb, Ec = the Euler Lines of A'AbAc, B'BcBa, C'CaCb, resp. (concurrent
> at P = common circumcenter of the triangles)
>
> La, Lb, Lc = the parallels through A",B",C" to Ea,Eb,Ec, resp.
>
> For P = I, the lines La,Lb,Lc concur on the pedal circle of I.
> (antipode of Feuerbach point)
>
> For P = H (A'B'C' = A"B"C") the lines La,Lb,Lc concur
> on the pedal circle of H (=NPC).
>
>
>
More for P = H:

The NPCs of A'AbAc, B'BcBa, C'CaCb are concurrent on
the NPC of A'B'C' (On the Poncelet point of H with respect
A'B'C' ie the point of concurrence of the NPCs of
A'B'C', HB'C', HC'A', HA'B'. So we have seven concurrent NPCs)

The parallels through A,B,C to the (concurrent at H)
Euler lines Ea, Eb, Ec of A'AbAc, B'BcBa, C'CaCb, resp. concur on the
circumcircle of ABC on the antipode of the Euler line
reflection point. And the perpendiculars, on the Euler line reflection
point.

APH

[Non-text portions of this message have been removed]
• Let ABC be a triangle and P a point. Denote: P1,P2,P3 = the P-points of PBC,PCA,PAB resp. P12,P13 = the reflections of P1 in AC,AB, resp. P23,P21 = the
Message 15 of 26 , May 29, 2016
Let ABC be a triangle and P a point.

Denote:

P1,P2,P3 = the P-points of PBC,PCA,PAB resp.

P12,P13 = the reflections of P1 in AC,AB, resp.
P23,P21 = the reflections of P2 in BA,BC, resp.
P31,P32 = the reflections of P3 in CB,CA, resp.

Which is the locus of P such that the perpendicular bisectors of

P12P13, P23P21, P31P32 are concurrent ?

(equivalently: Let M1,M2,M3 be the midpoints of P12P13, P23P21, P31P32, resp.

Which is the locus of P such that ABC, M1M2M3 are perspective?)

1. P = O

O1,O2,O3 = the circumcenters of OBC,OCA,OAB resp.
O12,O13 = the reflections of O1 in AC,AB, resp.
O23,O21 = the reflections of O2 in BA,BC, resp.
O31,O32 = the reflections of O3 in CB,CA, resp.

The perpendicular bisectors of O12O13,O23O21,O31O32 are concurrent at N.

2. P = I

I1,I2,I3 = the Incenters of IBC,ICA,IAB resp.
I12,I13 = the reflections of I1 in AC,AB, resp.
I23,I21 = the reflections of I2 in BA,BC, resp.
I31,I32 = the reflections of I3 in CB,CA, resp.

The perpendicular bisectors of I12I13,I23I21,I31I32 are concurrent at ??? (on
the OI line)

3. P = N

N1, N2, N3 = the NPC centers of NBC,NCA,NAB, resp.

N12,N13 = the reflections of N1 in AC,AB, resp.
N23,N21 = the reflections of N2 in BA,BC, resp.
N31,N32 = the reflections of N3 in CB,CA, resp.

The perpendicular bisectors of N12N13, N23N21, N31N32 are concurrent at ??.

Antreas

• [APH]: 3. P = N N1, N2, N3 = the NPC centers of NBC,NCA,NAB, resp. N12,N13 = the reflections of N1 in AC,AB, resp. N23,N21 = the reflections of N2 in BA,BC,
Message 16 of 26 , May 29, 2016

[APH]:

3. P = N

N1, N2, N3 = the NPC centers of NBC,NCA,NAB, resp.

N12,N13 = the reflections of N1 in AC,AB, resp.
N23,N21 = the reflections of N2 in BA,BC, resp.
N31,N32 = the reflections of N3 in CB,CA, resp.

The perpendicular bisectors of N12N13, N23N21, N31N32 are concurrent at ??.

Antreas

[Angel Montesdeoca]:

The perpendicular bisectors of N12N13, N23N21, N31N32 are cevians of  X(1493) = NAPOLEON CROSSSUM

Best regards,
Angel M.

• [APH] Let ABC be a triangle and P a point. Denote: P1,P2,P3 = the P-points of PBC,PCA,PAB resp. P12,P13 = the reflections of P1 in AC,AB, resp. P23,P21 = the
Message 17 of 26 , May 30, 2016
[APH]

Let ABC be a triangle and P a point.

Denote:

P1,P2,P3 = the P-points of PBC,PCA,PAB resp.

P12,P13 = the reflections of P1 in AC,AB, resp.
P23,P21 = the reflections of P2 in BA,BC, resp.
P31,P32 = the reflections of P3 in CB,CA, resp.

Which is the locus of P such that the perpendicular bisectors of

P12P13, P23P21, P31P32 are concurrent ?

(equivalently: Let M1,M2,M3 be the midpoints of P12P13, P23P21, P31P32, resp.

Which is the locus of P such that ABC, M1M2M3 are perspective?)

1. P = O

O1,O2,O3 = the circumcenters of OBC,OCA,OAB resp.
O12,O13 = the reflections of O1 in AC,AB, resp.
O23,O21 = the reflections of O2 in BA,BC, resp.
O31,O32 = the reflections of O3 in CB,CA, resp.

The perpendicular bisectors of O12O13,O23O21,O31O32 are concurrent at N.

2. P = I

I1,I2,I3 = the Incenters of IBC,ICA,IAB resp.
I12,I13 = the reflections of I1 in AC,AB, resp.
I23,I21 = the reflections of I2 in BA,BC, resp.
I31,I32 = the reflections of I3 in CB,CA, resp.

The perpendicular bisectors of I12I13,I23I21,I31I32 are concurrent at ??? (on

the OI line)

3. P = N

N1, N2, N3 = the NPC centers of NBC,NCA,NAB, resp.

N12,N13 = the reflections of N1 in AC,AB, resp.
N23,N21 = the reflections of N2 in BA,BC, resp.
N31,N32 = the reflections of N3 in CB,CA, resp.

The perpendicular bisectors of N12N13, N23N21, N31N32 are concurrent at ??.

Antreas

Dear Antreas,

The list is longer.

The appearance of i(j) in the following list means that the perpendicular bisectors for P=X(i) concur at X(j). If no j is given it means that they concur at a non-ETC center:

1(1129), 3(5), 5(1493), 6(141), 13(15), 15(17), 17(61), 20(3), 31, 32, 75, 76, 140, 176, 365, 366, 376(4550), 381, 382, 399(5671), 485(371), 546, 547, 548, 549, 550, 560, 561, 631, 632, 1501, 1502, 1656, 1657, 1917, 1928, 2042, 2043, 2044, 2045, 2046, 2671, 2675, 2676

Note: Found with i<=3054.

Regards

César
• ... Dear Angel and Cesar, Thank you for your responses. Now, I think the same is true if we replace the Incenters of the component triangles with excenters.
Message 18 of 26 , May 30, 2016

[APH]:

2. P = I

I1,I2,I3 = the Incenters of IBC,ICA,IAB resp.
I12,I13 = the reflections of I1 in AC,AB, resp.
I23,I21 = the reflections of I2 in BA,BC, resp.
I31,I32 = the reflections of I3 in CB,CA, resp.

The perpendicular bisectors of I12I13,I23I21,I31I32 are concurrent at ??? (on
the OI line)

Dear Angel and Cesar,

Now, I think the same is true if we replace the Incenters of the component triangles
with excenters. That is:

Denote:

J1 = the BC-excenter of IBC
J2 = the CA-excenter of ICA
J3 = the AB-excenter of IAB

J12,J13 = the reflections of J1 in AC,AB, resp.
J23,J21 = the reflections of J2 in BA,BC, resp.
J31,J32 = the reflections of J3 in CB,CA, resp.

The perpendicular bisectors of J12J13,J23IJ21,J31J32 are concurrent.

Antreas

Your message has been successfully submitted and would be delivered to recipients shortly.