- Dear friends,

Let A1,B1,C1 be the feets of perpendiculars

from point P(x:y:z) to the sides BC,CA,AB

and F1=area(PBA1),F2=area(PA1C),

F3=area(PCB1),F4=area(PB1A),

F5=area(PAC1),F6=area(PC1B).

Then we have

1.F1+F3+F5=F2+F4+F6

iff P is on Stammler hyperbola.

2.F1*F3*F5=F2*F4*F6

iff P is on Darboux cubic.

3.F1*F3+F3*F5+F5*F1=F2*F4+F4*F6+F6*F2

iff P is on conic

Cyclic sum(bc(b^2-c^2)(bcxx-2aayzcosA))=0.

As I remember,I certainly saw one of first two

propositions in Hyacinthos message.

Maybe,all of them are known.

It could be of interest to find some other

relations for these six areas which could

give some interesting properties of curves.

Best regards

Sincerely

Milorad R.Stevanovic

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

[MS]> Let A1,B1,C1 be the feets of perpendiculars

Are you sure about this one ?

> from point P(x:y:z) to the sides BC,CA,AB

> and F1=area(PBA1),F2=area(PA1C),

> F3=area(PCB1),F4=area(PB1A),

> F5=area(PAC1),F6=area(PC1B).

> Then we have

> 1.F1+F3+F5=F2+F4+F6

> iff P is on Stammler hyperbola.

> 2.F1*F3*F5=F2*F4*F6

> iff P is on Darboux cubic.

> 3.F1*F3+F3*F5+F5*F1=F2*F4+F4*F6+F6*F2

> iff P is on conic

> Cyclic sum(bc(b^2-c^2)(bcxx-2aayzcosA))=0.

I found the McCay cubic...

> It could be of interest to find some other

4. F1^2+F3^2+F5^2= F2^2+F4^2+F6^2

> relations for these six areas which could

> give some interesting properties of curves.

gives a beautiful cubic with asymptotes parallel to those of McCay and

concuring at O which is the center of symmetry of the curve.

I usually call this type of cubic a $\mathcal{K}_{60}^{++}$.

O is an inflexion point with tangent the Euler line.

it passes through the in/excenters with tangents concuring at G (the

polar conic of G is the Stammler hyperbola) and their reflections in O.

it meets the circumcircle at the second intersections of the bisectors.

barycentric equation :

b^2c^2[(b^2-c^2) x^3 + a^2 y z(y-z)] + cyclic = 0

Best regards

Bernard

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

[MS]> Let A1,B1,C1 be the feets of perpendiculars

[BG]

> from point P(x:y:z) to the sides BC,CA,AB

> and F1=area(PBA1),F2=area(PA1C),

> F3=area(PCB1),F4=area(PB1A),

> F5=area(PAC1),F6=area(PC1B).

> Then we have

> 1.F1+F3+F5=F2+F4+F6

> iff P is on Stammler hyperbola.

> 2.F1*F3*F5=F2*F4*F6

> iff P is on Darboux cubic.

> 3.F1*F3+F3*F5+F5*F1=F2*F4+F4*F6+F6*F2

> iff P is on conic

> Cyclic sum(bc(b^2-c^2)(bcxx-2aayzcosA))=0.

>Are you sure about this one ?

Yes,in third case the solution is

>I found the McCay cubic...

the McCay cubic.The solution is not so easy to obtain.

When I passed from case 2 to case three I lost some

factors.Sorry about mistake.

> It could be of interest to find some other

4. F1^2+F3^2+F5^2= F2^2+F4^2+F6^2

> relations for these six areas which could

> give some interesting properties of curves.

>gives a beautiful cubic with asymptotes parallel to those of McCay and

.it passes through the in/excenters with tangents concuring at G (the

>.concuring at O which is the center of symmetry of the curve.

>.I usually call this type of cubic a $\mathcal{K}_{60}^{++}$.

>O is an inflexion point with tangent the Euler line.

>polar conic of G is the Stammler hyperbola) and their reflections in O.

.barycentric equation :

>it meets the circumcircle at the second intersections of the bisectors.

>b^2c^2[(b^2-c^2) x^3 + a^2 y z(y-z)] + cyclic = 0

It seems that this cubic could be very important.

I am happy about this discovery.

Congratulations.

Best regards

Sincerely

Milorad R.Stevanovic

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