Dear Hyacinthists, Darij Grinberg wrote:
>X(????) = CIRCUMCENTER OF KOSNITA TRIANGLE
>Trilinears f(A,B,C) : f(B,C,A) : f(C,A,B),
> where f(A,B,C) = cos(B-C)
> (tan A + tan B + tan C - 4 sin A sin B sin
C)
> + 4 sin A sin B sin C
> (sec A - 2 cos A) (4 cos A cos B cos C - 1)
>
>X(????) lies on the following lines: 2,3 ...
>X(????) = midpoint of X(3) and X(26)
> does anybody have simpler trilinears??
Since tanA+tanB+tanC = tanA*tanB*tanC
the trilinears become
f(A,B,C) = cos(B-C)+4cos(2A)*cosB*cosC
Best regards
Nikolaos Dergiades
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