- Dear Quang Tuan Bui,

> [QTB] Given triangle ABC, one point P with pedal triangle PaPbPc. X

Don't you think it's better to say that the three signed distances

> is any point with Cevian triangle XaXbXc. We consider to find the

> locus of X such that three segments XaPa, XbPb, XcPc hold a

> property: one is sum of other two.

sum up to zero, the triangle ABC being arbitrarily orientated.

> [QTB] The result:

You obtain the same cubic K200 for any point P on the line IO = X1X3

> If barycentrics of P = (p : q : r), X = (x : y : z) then the locus

> is one cubic with equation:

>

> CyclicSum[a*b*r*((SA*x - SB*y)*(x*y + z^2) + (a - b)*c*x*y*(x + y)

> - 2*sc*(sb*x^2 - sa*y^2)*z - 2*sc*(a - b)*x*y*z)] = 0

>

> If P = incenter then the locus is one well known interesting cubic

> Z(X(8), X(31)) (by Clark Kimberling, ETC) or

> K200 (by Bernard Gibert, Cubics In The Triangle Plane)

and this is the only case when the cubic is a pK.

Now, when PaPbPc is the cevian triangle of P (instead of pedal), the

locus is a cubic again which turns out to be a pK for P on K200 and

the pK is always K200.

In other words, for any points P, Q on K200 the cevian triangles

PaPbPc, QaQbQc of P, Q verify the property above.

One particular interesting case : the Soddy points X175, X176.

Best regards

Bernard

[Non-text portions of this message have been removed] - Dear Bernard,

Thank you very much for your advices.

Yes, I think that it's better to say that the three signed distances sum up to zero, the triangle ABC being arbitrarily orientated. In fact, I use somehow "signed distances" in all my calculations related "one is some of other two" my messages. I keep "one is some of other two" phrase because historical reason only.

Your remarks about K200 are very interesting for me which I can not see. Thank you for it!

Best regards,

Bui Quang Tuan

Bernard Gibert <bg42@...> wrote: Dear Quang Tuan Bui,

Don't you think it's better to say that the three signed distances

sum up to zero, the triangle ABC being arbitrarily orientated.

You obtain the same cubic K200 for any point P on the line IO = X1X3

and this is the only case when the cubic is a pK.

Now, when PaPbPc is the cevian triangle of P (instead of pedal), the

locus is a cubic again which turns out to be a pK for P on K200 and

the pK is always K200.

In other words, for any points P, Q on K200 the cevian triangles

PaPbPc, QaQbQc of P, Q verify the property above.

One particular interesting case : the Soddy points X175, X176.

Best regards

Bernard

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

Fussy? Opinionated? Impossible to please? Perfect. Join Yahoo!'s user panel and lay it on us.

[Non-text portions of this message have been removed]