- Well... let's take three other points on the circumcircle....

The perpendicular bisectors of the sides of ABC intersect

the circumcircle at A',B',C' on the positive sides of BC,CA,AB, resp.

(ie both A,A' are on the same side of BC etc)

The Simson lines of A',B',C' are concurrent at a point P.

Note: If A",B",C" are the other intersections of the perp. bisectors

withthe circumcircle, then the Simson Lines of A",B",C" bound a

triangle which is perspective with the pedal triangle of O (medial

triangle) at P.

Generalization (Locus):

Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P.

The line PPa intersects the circumcircle at A' on the positive side

of BC (ie both A,A' are on the same side of BC). Similarly B',C'.

Which is the locus of P such that the Simson lines of A',B',C' are

concurrent?

APH

On Thu, Mar 7, 2013 at 7:56 AM, rhutson2 <rhutson2@...> wrote:

>

> They are not perspective. A*B*C* is also the circumcevian triangle of

> X(56). If mixtilinear excircles are used instead, touching the circumcircle

> at A**,B**,C**, then A**B**C** is the circumcevian triangle of X(55). The

> triangle bounded by the Simson lines of A**,B**,C** is also not perspective

> to A'B'C' or A"B"C".

>

> Randy

>

>

> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...>

> wrote:

> >

> > Let ABC be a triangle and A'B'C' the cevian triangle of I.

> >

> > The three mixtilinear incircles touch the circumcircle at A*,B*,C*

> > respective to A,B,C.

> >

> > The Simson lines of A*,B*,C* bound a triangle A"B"C".

> >

> > Are the triangles A'B'C', A"B"C" perspective?

> >

> > APH

Let ABC be a triangle and IaIbIc the antipedal triangle of I

(excentral triangle)

.

Denote:

IaaIabIac = the pedal triangle of Ia (wrt triangle ABC)

IbbIbcIba = the pedal triangle of Ib

IccIcaIcb = the pedal triangle of Ic.

Fa, Fb, Fc= the Feuerbach points relative to (Ia), (Ib), (Ic)

(ie the points of contact of the NPC with the excircles (Ia), (Ib), (Ic),

resp.)

sFa, sFb, sFc = the Simson lines of Fa, Fb, Fc wrt triangles

IaaIabIac,IbbIbcIba, IccIcaIcb.

A*B*C* = the triangle bounded by sFa, sFb, sFcI think the triangles IaIbIc, A*B*C* are parallelogic. and the parallelogic center

(IaIbIc, A*B*C*) is the O of ABC = N of IaIbIc(ie the parallels to sFa, sFb, sFc through Ia, Ib,Ic, resp. concur at O)The other parallelogic center if exists ?(quite complicated, I guess :-)Note:

sFa, sFb, sFc are not concurrent.See Hyacinthos, messages ##21663, 21664Antreas- [APH]Let ABC be a triangle and IaIbIc the antipedal triangle of I

(excentral triangle)

.

Denote:

IaaIabIac = the pedal triangle of Ia (wrt triangle ABC)

IbbIbcIba = the pedal triangle of Ib

IccIcaIcb = the pedal triangle of Ic.

Fa, Fb, Fc= the Feuerbach points relative to (Ia), (Ib), (Ic)

(ie the points of contact of the NPC with the excircles (Ia), (Ib), (Ic),

resp.)

sFa, sFb, sFc = the Simson lines of Fa, Fb, Fc wrt triangles

IaaIabIac,IbbIbcIba, IccIcaIcb.

A*B*C* = the triangle bounded by sFa, sFb, sFcI think the triangles IaIbIc, A*B*C* are parallelogic. and the parallelogic center

(IaIbIc, A*B*C*) is the O of ABC = N of IaIbIc(ie the parallels to sFa, sFb, sFc through Ia, Ib,Ic, resp. concur at O)The other parallelogic center if exists ?(quite complicated, I guess :-)Note:

sFa, sFb, sFc are not concurrent.See Hyacinthos, messages ##21663, 21664[Telv Cohl]:

In general:

Let P, Q be isogonal conjugate of ABC

Let P' be the orthopole of OP with respect to ABC

Then the Simson line of P' WRT the pedal triangle of Q is parallel to OP

(O is the circumcenter of triangle ABC ) - [APH]:

Let ABC be a triangle and IaIbIc the antipedal triangle of I

(excentral triangle)

.

Denote:

IaaIabIac = the pedal triangle of Ia (wrt triangle ABC)

IbbIbcIba = the pedal triangle of Ib

IccIcaIcb = the pedal triangle of Ic.

Fa, Fb, Fc= the Feuerbach points relative to (Ia), (Ib), (Ic)

(ie the points of contact of the NPC with the excircles (Ia), (Ib), (Ic),

resp.)

sFa, sFb, sFc = the Simson lines of Fa, Fb, Fc wrt triangles

IaaIabIac,IbbIbcIba, IccIcaIcb.

A*B*C* = the triangle bounded by sFa, sFb, sFcI think the triangles IaIbIc, A*B*C* are parallelogic. and the parallelogic center

(IaIbIc, A*B*C*) is the O of ABC = N of IaIbIc(ie the parallels to sFa, sFb, sFc through Ia, Ib,Ic, resp. concur at O)The other parallelogic center if exists ?(quite complicated, I guess :-)Note:

sFa, sFb, sFc are not concurrent.See Hyacinthos, messages ##21663, 21664[PETER MOSES]:>The other parallelogic center if exists ?2 a^13+2 a^12 b-12 a^11 b^2-10 a^10 b^3+29 a^9 b^4+21 a^8 b^5-36 a^7 b^6-24 a^6 b^7+24 a^5 b^8+16 a^4 b^9-8 a^3 b^10-6 a^2 b^11+a b^12+b^13+2 a^12 c-2 a^11 b c-18 a^10 b^2 c+6 a^9 b^3 c+47 a^8 b^4 c-5 a^7 b^5 c-54 a^6 b^6 c-a^5 b^7 c+30 a^4 b^8 c+3 a^3 b^9 c-8 a^2 b^10 c-a b^11 c+b^12 c-12 a^11 c^2-18 a^10 b c^2+24 a^9 b^2 c^2+54 a^8 b^3 c^2+6 a^7 b^4 c^2-48 a^6 b^5 c^2-42 a^5 b^6 c^2+30 a^3 b^8 c^2+18 a^2 b^9 c^2-6 a b^10 c^2-6 b^11 c^2-10 a^10 c^3+6 a^9 b c^3+54 a^8 b^2 c^3+32 a^7 b^3 c^3-34 a^6 b^4 c^3-47 a^5 b^5 c^3-28 a^4 b^6 c^3+6 a^3 b^7 c^3+24 a^2 b^8 c^3+3 a b^9 c^3-6 b^10 c^3+29 a^9 c^4+47 a^8 b c^4+6 a^7 b^2 c^4-34 a^6 b^3 c^4-28 a^5 b^4 c^4-18 a^4 b^5 c^4-22 a^3 b^6 c^4-10 a^2 b^7 c^4+15 a b^8 c^4+15 b^9 c^4+21 a^8 c^5-5 a^7 b c^5-48 a^6 b^2 c^5-47 a^5 b^3 c^5-18 a^4 b^4 c^5-18 a^3 b^5 c^5-18 a^2 b^6 c^5-2 a b^7 c^5+15 b^8 c^5-36 a^7 c^6-54 a^6 b c^6-42 a^5 b^2 c^6-28 a^4 b^3 c^6-22 a^3 b^4 c^6-18 a^2 b^5 c^6-20 a b^6 c^6-20 b^7 c^6-24 a^6 c^7-a^5 b c^7+6 a^3 b^3 c^7-10 a^2 b^4 c^7-2 a b^5 c^7-20 b^6 c^7+24 a^5 c^8+30 a^4 b c^8+30 a^3 b^2 c^8+24 a^2 b^3 c^8+15 a b^4 c^8+15 b^5 c^8+16 a^4 c^9+3 a^3 b c^9+18 a^2 b^2 c^9+3 a b^3 c^9+15 b^4 c^9-8 a^3 c^10-8 a^2 b c^10-6 a b^2 c^10-6 b^3 c^10-6 a^2 c^11-a b c^11-6 b^2 c^11+a c^12+b c^12+c^13::Best regards,Peter.