- Dear Quim,

in the general case, if P={u,v,w} (barycentric coordinates), then

X={a^2*b^2*u*w^2, b^2*c^2*u^2*v, a^2*c^2*v^2*w}

Y={a^2*c^2*u*v^2, a^2*b^2*v*w^2, b^2*c^2*u^2*w}

From this, your statement follows.

--- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer" <qcastell@...>

wrote:>

´

> Dear friends,

>

> I have noticed this fact:

> P, Q being two isogonal points, the cevians of P and the cevians of Q

> intersect at P, Q and six more points. The lines (other than the

> cevians) joining each vertex with these points are concurrent at two

> isogonal points X, Y. When P, Q are G, K the new points X, Y are the

> Brocard points.

>

> Greetings from Bsrcelona,

>

> Quim Castellsaguer

>

- Dear All,

I have been away for a bit, and am just catching up with this

thread. As Kostas indicates, the "isogonal" case is just a

special case of a more general result. This apears in my paper

http://forumgeom.fau.edu/FG2006volume6/FG200613.pdf

See especially Lemma 11 and Theorems 12, 13, 14. I have

summarized some of the relevant bits below Kostas's result

--- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>

wrote:>

of

> On Aug 4, 2006, at 11:17 AM, Quim Castellsaguer wrote:

> >

> > I have noticed this fact:

> > P, Q being two isogonal points, the cevians of P and the cevians

> > Q intersect at P, Q and six more points. The lines (other than

the

> > cevians) joining each vertex with these points are concurrent at

the

> > two isogonal points X, Y. When P, Q are G, K the new points X, Y

> > are the Brocard points.

> >

>

> Dear Quim, Steve, Jeff, Francois and all.

>

> Let PaPbPc, QaQbQc be the cevian triangles of two arbitrary

> points P, Q, in the plane of a given triangle ABC. We denote the

> intersection points of the lines connecting the vertices of ABC (

> other than cevians of P, Q ), with the six , furthermore P, Q,

Note that much of result c) appears in the glossary for ETC -

> intersection points of their cevians, as follows:

>

> - As D, the intersection point of BQb, CPc.

> - As D', the intersection point of BPb, CQc.

> - As E, the intersection point of CQc, APa.

> - As E', the intersection point of CPc, AQa.

> - As F, the intersection point of AQa, BPb.

> - As F', the intersection point of APa, BQb.

>

> RESULTS:

> a)- The lines AD,BE,CF, are concurrent at a point so be it, X.

> b)- The lines AD',BE',CF', are concurrent at a point so be it, Y.

> c)- The lines DD',EE',FF',XY, are concurrent at a point so be it S.

>

the point S is the crosspoint of P and Q.

We also observe that the triangles ABC, DEF, D'E'F' are all

triply perspective - the other perspectors are P and Q in each

case. We have a desmic structure containing A,B,C,D,E,F,D',E',F'

with perspectors X,Y,S. Indeed, any such desmic structure with

triply perspective triangles arises this way from a pair P,Q.

An alternative proof uses barycentrics :

Suppose that P = p1:p2:p3, and Q = q1:q2:q3.

Then X, Y are given by

x1:x2:x3 = 1/p2q3:1/p3q1:1/p1q2 and

y1:y2:y3 = 1/q2p3:1/q3p1:1/q1p2

In fact, we also have that the barycentric products P*Q and X*Y

are equal. If we write W for the common value, then we can say

that P,Q are W-isoconjugates, as are X,Y. W = p1q1:p2q2:p3q3.

This is obvious from the coordinates.

Further S = x1+y1:x2+y2:x3+y3 has the properties in c), and

patently lies on XY.

The point R = x1-x2:x2-y2:x3-y3 is also on XY - it is the

"harmon" of the desmic structure. It can be described as

the intersection of the tripolars of P and Q or as the

perspector of C(R), the circumconic through P and Q.

Alternatively, given R and W, we can recover P and Q as the

intersections of C(R) and the tripolar of R*, the W-isoconjugate

of R. Of course, P and Q may be complex. We can also recover S

as the R*-ceva conjugate of R.

X and Y are the intersections (other than A,B,C and the points

fixed by W-isoconjugation) of the pivotal isocubics pK(W,R)

and pK(W,S).

The points D,E,F,D',E',F' are intersections (other than A,B,C)

of pK(W,S) and nK0(W,R).

That's probably enough for now ....

Regards,

Wilson