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Re: fermat point (more)

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  • Darij Grinberg
    In Hyacinthos message #7961, I proved the following result by Antoine Verroken: If a point P lies in the interior of a circle M, and three lines intersecting
    Message 1 of 4 , Jan 22, 2004
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      In Hyacinthos message #7961, I proved the following
      result by Antoine Verroken:

      If a point P lies in the interior of a circle M, and
      three lines intersecting at P at angles of 60 degrees
      meet the circle M at the points A, B, C, D, E, F such
      that A and D lie on the first line, B and E on the
      second line, and C and F on the third line, and these
      points A, B, C, D, E, F form a convex hexagon, then
      AP + CP + EP = BP + DP + FP.

      In Hyacinthos message #7969, I conjectured that the sum
      AP + CP + EP depends on the radius of M and the distance
      from P to the center of M only, and not of the concrete
      position of the three lines through P. I asked if
      anybody could prove this. In fact, nobody could or can:
      the conjecture turned out to be wrong!

      Since the falseness of the conjecture can hardly be
      observed on any geometric sketch, here is an algebraic
      way to disprove it:

      If XY is any chord of the circle M such that XY passes
      through P, and O and R are the center and the radius of
      M, then

      XY = sqrt(R^2 - OP^2 * sin^2 phi)/2,

      where phi is the angle between OP and XY.

      Now, if t is the angle between OP and AD, then t + 60 is
      the angle between OP and BE, and t - 60 is the angle
      between OP and CF. [Here, 60 means 60 degrees.]

      Hence,

      AD + BE + CF = sqrt(R^2 - OP^2 * sin^2 t)/2
      + sqrt(R^2 - OP^2 * sin^2 (t + 60))/2
      + sqrt(R^2 - OP^2 * sin^2 (t - 60))/2.

      Since AP + CP + EP = BP + DP + FP and obviously
      AP + CP + EP + BP + DP + FP = AD + BE + CF, we have
      AP + CP + EP = (AD + BE + CF)/2, and

      AP + CP + EP = 1/4 * (sqrt(R^2 - OP^2 * sin^2 t)
      + sqrt(R^2 - OP^2 * sin^2 (t + 60))
      + sqrt(R^2 - OP^2 * sin^2 (t - 60))).

      If we abbreviate x = R^2 / OP^2, then

      AP + CP + EP = OP/4 * (sqrt(x - sin^2 t)
      + sqrt(x - sin^2 (t + 60))
      + sqrt(x - sin^2 (t - 60))).

      Now, the conjecture we wanted to disprove stated that
      this term is independent of t. It is enough to show that

      sqrt(x - sin^2 t) + sqrt(x - sin^2 (t + 60))
      + sqrt(x - sin^2 (t - 60))

      depends on t.

      The number x can take any value >1; for instance, we can
      take x = 50. Then, for t = 10 degrees, the above sum is
      21.1067364..., while for t = 30 degrees, it is
      21.1067359... (calculated by Maple). Qed..

      Darij Grinberg
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