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18472Re: [EMHL] a problem from the Monthly

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  • Nikolaos Dergiades
    Dec 3, 2009
      Dear Pierre,
      now I found a proof without
      restriction for p, q, r.
      P + Q = cot(p) + cot(q) = sin(p+q)/sin(p).sin(q)
      = -sin(r)/sin(p).sin(q).
      Since 0 < sin(p).sin(q) = (1/2)[cos(p-q)-cos(p+q)] <
      < (1/2)[1-cos(p+q)] = (1/2)[1-cos(r)] = sin^2(r/2)
      we have P + Q <= -sin(r)/sin^2(r/2) = -2.cot(r/2)
      Similarly working we get
      P + Q + R <= -(cot(p/2) + cot(q/2) + cot(r/2))
      <= -3cot[(p+q+r)/6] = -sqr(3)
      since the function cot(x) is convex in the
      interval (0, pi/2).

      Best regards
      Nikos Dergiades



      > Dear Pierre,
      > Sorry I said nonsense that the
      > function cot(x) is convex
      > and that P + Q + R >= -sqrt(3).
      > So my proof is not correct.
      > My proof is correct in the direction
      > P + Q + R <= -sqrt(3)
      > and hence f >= 4S.sqrt(3)
      > if p, q, r > pi/2 since the function
      > cot(x) is concave in the interval
      > (pi/2, pi).
      >
      > Best regards
      > Nikos Dergiades

      >





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