Dear Darij and other Hyacinthists
it seems that your point is barycentric
x = a^2(x1 + 2d x2/root(3)) where
d =area(ABC)
x1 = 4a^2SA^3+(SB+2SC)(SC+2SB)SA^2-7SBC^2-5a^2SABC
x2 = 3SA^3+7a^2SA^2-6a^2SBC-5SABC
May be, there are nice nice coordinates in trigonometric form but I
didn't find.
I am very happy for Professor Jean-Pierre Serre who winned the first
Abel prize of Mathematics.
Friendly. Jean-Pierre Ehrmann
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