## Re: LOCUS

Expand Messages
• Dear Antreas, ... Here, examples are : any, except from circumcircle. 0. OaObOc, HaHbHc : circumcircle and K001, Neuberg cubic. Examples : 1, 3, 4, 13, 14, 15,
Message 1 of 22 , Jan 24, 2010
• 0 Attachment
Dear Antreas,

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
> Let ABC be a triangle, P a point, Oa,Ob,Oc and Ha,Hb,Hc the
> circumcenters, orthocenters of PBC,PCA,PAB, resp.
> Let A'B'C', A"B"C" be the cevian, pedal triangles of P, resp.
> Which is the locus of P such that:
> 0. OaObOc, HaHbHc ... WNR
> 1. cevian, OaObOc
> 2. cevian, HaHbHc
> 3. pedal , OaObOc
> 4. pedal , HaHbHc
> are perspective?

Here, examples are : any, except from circumcircle.

0. OaObOc, HaHbHc : circumcircle and K001, Neuberg cubic. Examples : 1, 3, 4, 13, 14, 15, 16, 30, 74, 370, 399, 484, 616, 617, 1138, 1157, 1263, 1276, 1277, 1337, 1338, 2132, 2133, 3065, 3440, 3441, 3464, 3465, 3466, 3479, 3480, 3481, 3482, 3483, 3484

1. cevian, OaObOc : circumcircle and 5th degree curve
(q^2*c^2-r^2*b^2)*(b^2+c^2-a^2)*p^3+(r^2*a^2-p^2*c^2)*(c^2+a^2-b^2)*q^3+(p^2*b^2-q^2*a^2)*(a^2+b^2-c^2)*r^3=0. Examples : 1, 2, 4, 13, 14, 357, 1113, 1114, 1134, 1136, 1156

2. cevian, HaHbHc : circumcircle and 7th degree curve. Examples : 4.

3. pedal , OaObOc : circumcircle and K006 ORTHOCUBIC, pK(X6, X4). Examples : 1, 3, 4, 46, 90, 155, 254, 371, 372, 485, 486, 487, 488.

4. pedal , HaHbHc : ever. Perspector : P.

5. anticev , OaObOc : circumcircle and 7th degree. Examples : 1, 6, 13, 14, 399 (1->1, 6->3)

6. anticev, HaHbHc : 8th degree. Examples : 1,4 (1->9, 4->4)

7. circumcev , OaObOc : ever. Examples : (P, persp) : [2, 1296], [3, 110], [4, 110], [5, 1291], [6, 1296], [13, 110], [14, 110], [15, 110], [16, 110], [20, 112]...
In fact, persp= X(110) when P belongs to K001

Formula : a^2/((q^2*a^2-p^2*b^2)*S[b]*r+(-r^2*a^2+p^2*c^2)*S[c]*q+(q^2*c^2-r^2*b^2)*a^2*p), etc with S[a]=Conway symbol.

8. circumcev, HaHbHc : circumcircle and 9th degree. Examples : 1.

Best regards,
Pierre.
• Dear Pierre Thanks! ... This quintic looks interesting. Is it already listed by Bernard? APH [Non-text portions of this message have been removed]
Message 2 of 22 , Jan 25, 2010
• 0 Attachment
Dear Pierre

Thanks!

> --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
> "Antreas" <anopolis72@...> wrote:
> > Let ABC be a triangle, P a point, Oa,Ob,Oc and Ha,Hb,Hc the
> > circumcenters, orthocenters of PBC,PCA,PAB, resp.
> > Let A'B'C', A"B"C" be the cevian, pedal triangles of P, resp.
> > Which is the locus of P such that:
> > 0. OaObOc, HaHbHc ... WNR
> > 1. cevian, OaObOc
> > are perspective?
>
> Here, examples are : any, except from circumcircle.
>
> 0. OaObOc, HaHbHc : circumcircle and K001, Neuberg cubic. Examples : 1, 3,
> 4, 13, 14, 15, 16, 30, 74, 370, 399, 484, 616, 617, 1138, 1157, 1263, 1276,
> 1277, 1337, 1338, 2132, 2133, 3065, 3440, 3441, 3464, 3465, 3466, 3479,
> 3480, 3481, 3482, 3483, 3484
>
> 1. cevian, OaObOc : circumcircle and 5th degree curve
> (q^2*c^2-r^2*b^2)*(b^2+c^2-a^2)*p^3+(r^2*a^2-p^2*c^2)*(c^2+a^2-b^2)*q^3+(p^2*b^2-q^2*a^2)*(a^2+b^2-c^2)*r^3=0.
> Examples : 1, 2, 4, 13, 14, 357, 1113, 1114, 1134, 1136, 1156
>
>
This quintic looks interesting. Is it already listed by Bernard?

APH

[Non-text portions of this message have been removed]
• Dear Antreas, ... Yes, you are right ! Q003, the Euler-Morley Quintic. Since there are 95 identified points on Q003, it remains only to obtain 95
Message 3 of 22 , Jan 25, 2010
• 0 Attachment
Dear Antreas,
--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:

> > 1. cevian, OaObOc : circumcircle and 5th degree curve
> > (q^2*c^2-r^2*b^2)*(b^2+c^2-a^2)*p^3+(r^2*a^2-p^2*c^2)*(c^2+a^2-b^2)*q^3+(p^2*b^2-q^2*a^2)*(a^2+b^2-c^2)*r^3=0.
> > Examples : 1, 2, 4, 13, 14, 357, 1113, 1114, 1134, 1136, 1156
> >
> This quintic looks interesting. Is it already listed by Bernard?

Yes, you are right ! Q003, the Euler-Morley Quintic.

Since there are 95 "identified" points on Q003, it remains "only" to obtain 95 interpretations of the property persp(cevian, OaObOc)...

Best regards.
• Dear Pierre ... Probably are interesting the variations with Orthologic instead of Perspective triangles. Most likely the loci are high degree curves, but it
Message 4 of 22 , Jan 25, 2010
• 0 Attachment
Dear Pierre

> --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
> "Antreas" <anopolis72@...> wrote:
> > Let ABC be a triangle, P a point, Oa,Ob,Oc and Ha,Hb,Hc the
> > circumcenters, orthocenters of PBC,PCA,PAB, resp.
> > Let A'B'C', A"B"C" be the cevian, pedal triangles of P, resp.
> > Which is the locus of P such that:
> > 0. OaObOc, HaHbHc ... WNR
> > 1. cevian, OaObOc
> > 2. cevian, HaHbHc
> > 3. pedal , OaObOc
> > 4. pedal , HaHbHc
> > are perspective?
>
> Here, examples are : any, except from circumcircle.
>
> 0. OaObOc, HaHbHc : circumcircle and K001, Neuberg cubic. Examples : 1, 3,
> 4, 13, 14, 15, 16, 30, 74, 370, 399, 484, 616, 617, 1138, 1157, 1263, 1276,
> 1277, 1337, 1338, 2132, 2133, 3065, 3440, 3441, 3464, 3465, 3466, 3479,
> 3480, 3481, 3482, 3483, 3484
>
> 1. cevian, OaObOc : circumcircle and 5th degree curve
> (q^2*c^2-r^2*b^2)*(b^2+c^2-a^2)*p^3+(r^2*a^2-p^2*c^2)*(c^2+a^2-b^2)*q^3+(p^2*b^2-q^2*a^2)*(a^2+b^2-c^2)*r^3=0.
> Examples : 1, 2, 4, 13, 14, 357, 1113, 1114, 1134, 1136, 1156
>
> 2. cevian, HaHbHc : circumcircle and 7th degree curve. Examples : 4.
>
> 3. pedal , OaObOc : circumcircle and K006 ORTHOCUBIC, pK(X6, X4). Examples
> : 1, 3, 4, 46, 90, 155, 254, 371, 372, 485, 486, 487, 488.
>
> 4. pedal , HaHbHc : ever. Perspector : P.
>
> 5. anticev , OaObOc : circumcircle and 7th degree. Examples : 1, 6, 13, 14,
> 399 (1->1, 6->3)
>
> 6. anticev, HaHbHc : 8th degree. Examples : 1,4 (1->9, 4->4)
>
> 7. circumcev , OaObOc : ever. Examples : (P, persp) : [2, 1296], [3, 110],
> [4, 110], [5, 1291], [6, 1296], [13, 110], [14, 110], [15, 110], [16, 110],
> [20, 112]...
> In fact, persp= X(110) when P belongs to K001
>
> Formula :
> a^2/((q^2*a^2-p^2*b^2)*S[b]*r+(-r^2*a^2+p^2*c^2)*S[c]*q+(q^2*c^2-r^2*b^2)*a^2*p),
> etc with S[a]=Conway symbol.
>
> 8. circumcev, HaHbHc : circumcircle and 9th degree. Examples : 1.
>
>

Probably are interesting the variations with Orthologic instead of
Perspective triangles.

Most likely the loci are high degree curves, but it will be interesting to
find simple points
lying on these loci.

For example:
I think that in the the above case 0 [ ie OaObOc, HaHbHc be orthologic] the
point I = Incenter
is on the locus. If so we have a line passing through the two orthologic
centers and the Schiffler point.

Greetings

Antreas

[Non-text portions of this message have been removed]
• 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
Message 5 of 22 , Apr 7, 2010
• 0 Attachment
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
• ... A B C is the reflection triangle of P(x:y:z) in its pedal triangle. 1. ABC, A B C are perspective if and only if P lies on the orthocubic cubic (Antreas
Message 6 of 22 , Apr 7, 2010
• 0 Attachment
--- In Hyacinthos@yahoogroups.com, "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
>

A'B'C' is the reflection triangle of P(x:y:z) in its pedal triangle.

1. ABC, A'B'C' are perspective if and only if P lies on the orthocubic cubic (Antreas P Hatzipolakis and Paul Yiu, Reflections in triangle geometry, Forum Geometricorum, 9 (2009) 301--348. Proposition 26)

3. ABC, A'B'C' are orthologic if and only if P lies (infinity lines, circumcircle, McCay Cubic).

The triangles PaPbPc, A'B'C' are orthologic.
One orthologic center is P(x:y:z). The other one:
a^2(2b^2c^2x^2 + c^2(a^2+b^2-c^2)x y + b^2(a^2-b^2+c^2)x z + (a^2b^2-b^4+a^2c^2+2b^2c^2-c^4)y z): ... : ....

Angel
• Dear Antreas What do you mean by circle (Pb, PbP)? Friendly Francois ... [Non-text portions of this message have been removed]
Message 7 of 22 , Apr 13, 2010
• 0 Attachment
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 8 of 22 , Apr 13, 2010
• 0 Attachment
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 9 of 22 , Aug 3, 2010
• 0 Attachment
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 10 of 22 , Aug 4, 2010
• 0 Attachment
--- 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 11 of 22 , Sep 7, 2010
• 0 Attachment
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 12 of 22 , Sep 9, 2010
• 0 Attachment
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 13 of 22 , May 5, 2011
• 0 Attachment
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 14 of 22 , May 5, 2011
• 0 Attachment
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 15 of 22 , May 6, 2011
• 0 Attachment
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 16 of 22 , Feb 25, 2013
• 0 Attachment
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 17 of 22 , Feb 26, 2013
• 0 Attachment
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 18 of 22 , Feb 26, 2013
• 0 Attachment

- 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 19 of 22 , Feb 26, 2013
• 0 Attachment
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 20 of 22 , Feb 26, 2013
• 0 Attachment
[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]
Your message has been successfully submitted and would be delivered to recipients shortly.