<Tri-2024-2089>, hint

Inradius r  of triangle ABC: a+c-b=2r  r=(bsinC+bcosC-b)/2=b/2(sinC+cosC-1)

Since  triangles ABC , BHC, AHB are similar

r'=b sinC/2 (sinC+cosC-1), r"= bcosC/2 (sinC+cosC-1))

rxr=r'x r'+r"x r"

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