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Re: [EMHL] Monotonicity

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  • Nikolaos Dergiades
    ... predefined, ... reversed... ... If m and t are the b-bisector and C-bisector respectively then 4m^2 = 2(a^2+c^2)-b^2 hence c^2 = (4m^2-2a^2+b^2)/2 t^2
    Message 1 of 1 , Sep 2, 2002
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      >Dear Hyacinthians,
      >
      >May be some of you will find the following interesting:
      >
      >In a triangle ABC with a-side and b-bisector m_b
      predefined,
      >consider the relation between B-angle an C-bisector t_c:
      >t_c = F(B). Prove that F is monotonic.
      >
      >I believe I have a simple argument in case m_b >= a.
      >Unfortunately, it breaks up when the condition is
      reversed...
      >
      >Regards,
      >Barukh.


      If m and t are the b-bisector and C-bisector respectively
      then
      4m^2 = 2(a^2+c^2)-b^2 hence c^2 = (4m^2-2a^2+b^2)/2

      t^2 = a*b*(a+b+c)*(a+b-c)/(a+b)^2=
      = a*b*(4a^2-4m^2+4ab+b^2)/(a+b)^2 = f(b)
      where a, m are constants. This function has positive
      derivative and hence is increasing.

      If O is the reflection of C about B then BA = 2m and the
      locus of A is the circle with center O and radius 2m.
      If OC meets this circle at A1, A2 (A1 closer to C) then
      it is known that the distance b = AC is increasing as A
      moves on the circle from A1 to A2 and also the angle B
      is increasing and conversely. Hence
      B1 > B2 => b1 > b2 => f(b1) > f(b2) => t1 > t2 =>
      => F(B1) > F(B2).
      Hence the function F(B) is increasing.

      Best regards
      Nikos Dergiades
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