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• 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),
Message 1 of 3 , May 12, 2003
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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

[Non-text portions of this message have been removed]
• Dear Milorad, [MS] ... Are you sure about this one ? I found the McCay cubic... ... 4. F1^2+F3^2+F5^2= F2^2+F4^2+F6^2 gives a beautiful cubic with asymptotes
Message 2 of 3 , May 16, 2003
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[MS]
> 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.

I found the McCay cubic...

> It could be of interest to find some other
> relations for these six areas which could
> give some interesting properties of curves.

4. F1^2+F3^2+F5^2= F2^2+F4^2+F6^2
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] ... [BG] ... Yes,in third case the solution is the McCay cubic.The solution is not so easy to obtain. When I passed from case 2 to case
Message 3 of 3 , May 16, 2003
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Dear Bernard,

[MS]
> 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.
[BG]
>I found the McCay cubic...

Yes,in third case the solution is
the McCay cubic.The solution is not so easy to obtain.
When I passed from case 2 to case three I lost some

> It could be of interest to find some other
> relations for these six areas which could
> give some interesting properties of curves.

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

It seems that this cubic could be very important.
Congratulations.

Best regards
Sincerely