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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^2r^2*b^2)*(b^2+c^2a^2)*p^3+(r^2*a^2p^2*c^2)*(c^2+a^2b^2)*q^3+(p^2*b^2q^2*a^2)*(a^2+b^2c^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^2p^2*b^2)*S[b]*r+(r^2*a^2+p^2*c^2)*S[c]*q+(q^2*c^2r^2*b^2)*a^2*p), etc with S[a]=Conway symbol.
8. circumcev, HaHbHc : circumcircle and 9th degree. Examples : 1.
Best regards,
Pierre. 0 Attachment
Dear Pierre
Thanks!
>  In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
This quintic looks interesting. Is it already listed by Bernard?
> "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^2r^2*b^2)*(b^2+c^2a^2)*p^3+(r^2*a^2p^2*c^2)*(c^2+a^2b^2)*q^3+(p^2*b^2q^2*a^2)*(a^2+b^2c^2)*r^3=0.
> Examples : 1, 2, 4, 13, 14, 357, 1113, 1114, 1134, 1136, 1156
>
>
APH
[Nontext portions of this message have been removed] 0 Attachment
Dear Antreas, In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
> > 1. cevian, OaObOc : circumcircle and 5th degree curve
> > (q^2*c^2r^2*b^2)*(b^2+c^2a^2)*p^3+(r^2*a^2p^2*c^2)*(c^2+a^2b^2)*q^3+(p^2*b^2q^2*a^2)*(a^2+b^2c^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 EulerMorley Quintic.
Since there are 95 "identified" points on Q003, it remains "only" to obtain 95 interpretations of the property persp(cevian, OaObOc)...
Best regards. 0 Attachment
Dear Pierre
>  In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
Probably are interesting the variations with Orthologic instead of
> "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^2r^2*b^2)*(b^2+c^2a^2)*p^3+(r^2*a^2p^2*c^2)*(c^2+a^2b^2)*q^3+(p^2*b^2q^2*a^2)*(a^2+b^2c^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^2p^2*b^2)*S[b]*r+(r^2*a^2+p^2*c^2)*S[c]*q+(q^2*c^2r^2*b^2)*a^2*p),
> etc with S[a]=Conway symbol.
>
> 8. circumcev, HaHbHc : circumcircle and 9th degree. Examples : 1.
>
>
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
[Nontext portions of this message have been removed] 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 0 Attachment
 In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:>
A'B'C' is the reflection triangle of P(x:y:z) in its pedal triangle.
> 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
>
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) 301348. 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^2c^2)x y + b^2(a^2b^2+c^2)x z + (a^2b^2b^4+a^2c^2+2b^2c^2c^4)y z): ... : ....
Angel 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
>
>
>
[Nontext portions of this message have been removed] 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
>>
>>
>>
>
>
> [Nontext portions of this message have been removed]
>
>
>
> 
>
> Yahoo! Groups Links
>
>
>
>
http://anopolis72000.blogspot.com/ 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 0 Attachment
 In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:>
ABC and Triangle bounded by (La,Lb,Lc) are parallelogic =
> 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
>
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 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 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+ca)(b+c)^2 : (c+ab)(c+a)^2 : (a+bc)(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+ca)^3 : (c+a)^2/(c+ab)^3 : (a+b)^2/(a+bc)^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^2c^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
> 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 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
> 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
> >
> 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 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^2b^2c^2)(a^2(SA^2SB SC)SA(c^2b^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*SB2SA*SC+SB*SC)x^6y^3
(SA+SB)^3(SA+SC)(5SA*SB+5SA*SC4SB*SC)x^5y^4+
(SA+SB)^3(SB+SC)(5SA*SB4SA*SC+5SB*SC)x^4y^5
(SA+SB)^3(SB+SC)(SA*SB+SA*SC2SB*SC)x^3y^6+
(SA+SB)^2x^2y^2z((SASC)(4SA^2SB+SA(5SA3SB)SC+(SA+SB)SC^2)x^4
2(SA+SC)(SA(SBSC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y
4(SASB)SC(SB*SC+SA(SB+SC))x^2y^2+
2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^22SB*SC+SC^2))x*y^3+
(SB+SC)(SA(4SBSC)(SB+SC)+SB*SC(5SB+SC))y^4)+
(SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)(4SASB)(SA+SB)SC)x^5
(SA+SC)(SA^2(19SB14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
SA*SB(7SB^28SB*SC3SC^2))x^3y^2+(SB+SC)*
(SA^2SB(7SA+19SB)+SA(7SA^28SA*SB+5SB^2)SC+
(11SA14SB)(SA+SB)SC^2)x^2y^3(SB+SC)^2(SA*SB(SA+5SB)+
(SA4SB)(SA+SB)SC)y^5)+
(SA+SB)z^3(((SA+SC)^3)(2SA*SB+
(SA+SB)SC)x^6+2(SA+SC)^2(SA(SBSC)^2+5SA^2(SB+SC)+
SB*SC(SB+SC))x^5y(SA+SC)(SA^2(14SB19SC)*
(SB+SC)SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC7SC^2))*
x^4y^2+(SB+SC)(SA^2(14SB11SC)(SB+SC)
SB*SC^2(19SB+7SC)+SA*SC(5SB^2+8SB*SC7SC^2))*
x^2y^42(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
SA(5SB^22SB*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(SASC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y
(SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
SA(7SA^28SA*SB3SB^2)SC+(19SA14SB)(SA+SB)SC^2)x^3y^2+
(SA+SC)(SB+SC)(SA^2(11SB14SC)(SB+SC)+
SB^2SC(7SB+19SC)+SA*SB(7SB^28SB*SC+5SC^2))x^2y^3
4SA(SBSC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4
(SA+SB)(SB+SC)^3(5SA*SB4SA*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)+(SASB)^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(SB2SC)+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
> 0 Attachment
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 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^2b^2c^2)(a^2(SA^2SB SC)SA(c^2b^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*SB2SA*SC+SB*SC)x^6y^3
>
> (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC4SB*SC)x^5y^4+
>
> (SA+SB)^3(SB+SC)(5SA*SB4SA*SC+5SB*SC)x^4y^5
>
> (SA+SB)^3(SB+SC)(SA*SB+SA*SC2SB*SC)x^3y^6+
>
> (SA+SB)^2x^2y^2z((SASC)(4SA^2SB+SA(5SA3SB)SC+(SA+SB)SC^2)x^4
> 2(SA+SC)(SA(SBSC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y
> 4(SASB)SC(SB*SC+SA(SB+SC))x^2y^2+
> 2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^22SB*SC+SC^2))x*y^3+
> (SB+SC)(SA(4SBSC)(SB+SC)+SB*SC(5SB+SC))y^4)+
>
> (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)(4SASB)(SA+SB)SC)x^5
> (SA+SC)(SA^2(19SB14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
> SA*SB(7SB^28SB*SC3SC^2))x^3y^2+(SB+SC)*
> (SA^2SB(7SA+19SB)+SA(7SA^28SA*SB+5SB^2)SC+
> (11SA14SB)(SA+SB)SC^2)x^2y^3(SB+SC)^2(SA*SB(SA+5SB)+
> (SA4SB)(SA+SB)SC)y^5)+
>
> (SA+SB)z^3(((SA+SC)^3)(2SA*SB+
> (SA+SB)SC)x^6+2(SA+SC)^2(SA(SBSC)^2+5SA^2(SB+SC)+
> SB*SC(SB+SC))x^5y(SA+SC)(SA^2(14SB19SC)*
>
> (SB+SC)SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC7SC^2))*
> x^4y^2+(SB+SC)(SA^2(14SB11SC)(SB+SC)
> SB*SC^2(19SB+7SC)+SA*SC(5SB^2+8SB*SC7SC^2))*
> x^2y^42(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
> SA(5SB^22SB*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(SASC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y
>
> (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
> SA(7SA^28SA*SB3SB^2)SC+(19SA14SB)(SA+SB)SC^2)x^3y^2+
>
> (SA+SC)(SB+SC)(SA^2(11SB14SC)(SB+SC)+
> SB^2SC(7SB+19SC)+SA*SB(7SB^28SB*SC+5SC^2))x^2y^3
>
> 4SA(SBSC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4
>
> (SA+SB)(SB+SC)^3(5SA*SB4SA*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)+(SASB)^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(SB2SC)+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
> >
> 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^2b^2c^2)(a^2(SA^2SB SC)SA(c^2b^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*SB2SA*SC+SB*SC)x^6y^3
>
> (SA+SB)^3(SA+SC)(5SA*SB+5SA*SC4SB*SC)x^5y^4+
>
> (SA+SB)^3(SB+SC)(5SA*SB4SA*SC+5SB*SC)x^4y^5
>
> (SA+SB)^3(SB+SC)(SA*SB+SA*SC2SB*SC)x^3y^6+
>
> (SA+SB)^2x^2y^2z((SASC)(4SA^2SB+SA(5SA3SB)SC+(SA+SB)SC^2)x^4
> 2(SA+SC)(SA(SBSC)^2+5SA^2(SB+SC)+SB*SC(SB+SC))x^3y
> 4(SASB)SC(SB*SC+SA(SB+SC))x^2y^2+
> 2(SB+SC)(SA^2(SB+SC)+SB*SC(5SB+SC)+SA(5SB^22SB*SC+SC^2))x*y^3+
> (SB+SC)(SA(4SBSC)(SB+SC)+SB*SC(5SB+SC))y^4)+
>
> (SA+SB)x*y*z^2((SA+SC)^2(SA*SB(5SA+SB)(4SASB)(SA+SB)SC)x^5
> (SA+SC)(SA^2(19SB14SC)(SB+SC)+SB^2SC(7SB+11*SC)+
> SA*SB(7SB^28SB*SC3SC^2))x^3y^2+(SB+SC)*
> (SA^2SB(7SA+19SB)+SA(7SA^28SA*SB+5SB^2)SC+
> (11SA14SB)(SA+SB)SC^2)x^2y^3(SB+SC)^2(SA*SB(SA+5SB)+
> (SA4SB)(SA+SB)SC)y^5)+
>
> (SA+SB)z^3(((SA+SC)^3)(2SA*SB+
> (SA+SB)SC)x^6+2(SA+SC)^2(SA(SBSC)^2+5SA^2(SB+SC)+
> SB*SC(SB+SC))x^5y(SA+SC)(SA^2(14SB19SC)*
>
> (SB+SC)SB*SC^2(11SB+7SC)+SA*SC(3SB^2+8SB*SC7SC^2))*
> x^4y^2+(SB+SC)(SA^2(14SB11SC)(SB+SC)
> SB*SC^2(19SB+7SC)+SA*SC(5SB^2+8SB*SC7SC^2))*
> x^2y^42(SB+SC)^2(SA^2(SB+SC)+SB*SC(5SB+SC)+
> SA(5SB^22SB*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(SASC)(SA+SC)^2(SB*SC+SA(SB+SC))x^4y
>
> (SA+SC)(SB+SC)(SA^2SB(7SA+11SB)+
> SA(7SA^28SA*SB3SB^2)SC+(19SA14SB)(SA+SB)SC^2)x^3y^2+
>
> (SA+SC)(SB+SC)(SA^2(11SB14SC)(SB+SC)+
> SB^2SC(7SB+19SC)+SA*SB(7SB^28SB*SC+5SC^2))x^2y^3
>
> 4SA(SBSC)(SB+SC)^2(SB*SC+SA(SB+SC))x*y^4
>
> (SA+SB)(SB+SC)^3(5SA*SB4SA*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)+(SASB)^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(SB2SC)+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/
[Nontext portions of this message have been removed] 0 Attachment
[APH]
>
More for P = H:
> 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).
>
>
>
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
[Nontext portions of this message have been removed]
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