HYACINTHOS 10936

Dear Milorad,


I think you'll find that P1 is X(1358), P2 is X(1357)
and P4 is X(1365)

But P3, P5(corrected) and P6 seem new....

Best wishes

Wilson Stothers

>
> 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).
>
> Best regards
>
> Milorad R.Stevanovic