HYACINTHOS 10935

Dear Jean-Pierre,

[MS]

> I have found new points on the incircle
>
> 1.P1((b-c)^2/(s-a):(c-a)^2/(s-b):(a-b)^2/(s-c)).
>
> 2.P2((a(b-c))^2/(s-a):(b(c-a))^2/(s-b):((c(a-b))^2/(s-c)).
>
> 3.P3((b+c-2a)^2/(s-a):(c+a-2b)^2/(s-b):(a+b-2c)^2/(s-c)).
>
> 4.P4((b^2-c^2)^2/(s-a):(c^2-a^2)^2/(s-b):(a^2-b^2)^2/(s-c)).
>
> 5.P5((s-a)[2a^2-a(b+c)+(b-c)^2]:(s-b)[2b^2-b(c+a)+(c-a)^2]:
> (s-c)[2c^2-c(a+b)+(a-b)^2])
>
> This point is the antipode of Feuerbach point Fe=X(11)
> X(11)=((b-c)^2(s-a):(c-a)^2(s-b):(a-b)^2(s-c))
> on the Euler or nine-point circle.
>
> 6.P6(a^2(b-c)^2(s-a)^3:b^2(c-a)^2(s-b)^3:c^2(a-b)^2(s-c)^3).

[JPE]

> Note that if u+v+w=0 then the point u^2/(s-a):v^2/(s-b):w^2/(s-c)
> lies on the incircle.
> There is a typo in P5 : your point lies on the line at infinity and,
> in any case, the antipode of the Feuerbach point on the 9P-circle
> doesn't lie on the incircle.

Thanks for your comment and corrections.
First about comment.
I know about it and for some other
similar relations.
Correct coordinates for P5 are

P5((s-a)[2a^2-a(b+c)+(b-c)^2]^2:(s-b)[2b^2-b(c+a)+(c-a)^2]^2:
(s-c)[2c^2-c(a+b)+(a-b)^2]^2).

I omitted the squares of the expressions in square brackets
This point is the antipode of Feuerbach point wrt to incircle.
Interesting are the coordinates of P1 compared with
coordinates of Feuerbach point.
It could be of interest to find some geometric property
of point P1.
I suppose that the point P1 is very important
in the triangle.
I see two reasons for that
1.Its position - it is on the incircle.
In the triangle with a>b>c both points X(11) and P1
are on the incircle arc A1C1 nearer to B(points
A1 and C1 are the points of tangency of the incircle
and of sides BC and AB of triangle ABC).
2.Its coordinates are almost as simple as coordinates
of X(11).

Best regards

Sincerely

Milorad R.Stevanovic