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Tangents to the circumcircle

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  • Darij Grinberg
    A little warm-up problem for 2004: Let a line l touch the circumcircle of a triangle ABC at a point on the arc BC. Denote by x, y, z the distances from A, B, C
    Message 1 of 3 , Dec 31, 2003
      A little warm-up problem for 2004:

      Let a line l touch the circumcircle of a triangle
      ABC at a point on the arc BC. Denote by x, y, z
      the distances from A, B, C to l. Prove, without the
      use of coordinate systems, that

      a * sqrt x = b * sqrt y + c * sqrt z.

      Happy New Year to all!

      Darij Grinberg
    • Vijaya Prasad Nalluri
      Dear Darij Grinberg, Let line l touch circle ABC at P.Let Q be the antipode of P.From similar triangles PAX,QPA we have x/AP = AP/2R giving AP^2 = 2xR,
      Message 2 of 3 , Dec 31, 2003
        Dear Darij Grinberg,
        Let line l touch circle ABC at P.Let Q be the antipode of P.From similar triangles PAX,QPA we have
        x/AP = AP/2R giving AP^2 = 2xR,
        similarly, we get BP^2 = 2yR and CP^2 = 2zR
        So BP/AP = sqrt (y/x) and CP/AP = sqrt (z/x)
        From Ptolemy's theorem applied to the cyclic quadrilateral ABPC,
        BC * AP = AC* BP + AB * CP
        Dividing either side by AP
        BC = AC * (BP/AP) + AB * (CP/AP)
        a = b * sqrt (y/x) + c * sqrt (z/x)
        a * sqrt x = b *sqrt y + c * sqrt z
        With warm NY Gtngs to all,
        Vijay Prasad Nalluri.

        Darij Grinberg <darij_grinberg@...> wrote:
        A little warm-up problem for 2004:

        Let a line l touch the circumcircle of a triangle
        ABC at a point on the arc BC. Denote by x, y, z
        the distances from A, B, C to l. Prove, without the
        use of coordinate systems, that

        a * sqrt x = b * sqrt y + c * sqrt z.

        Happy New Year to all!

        Darij Grinberg



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      • Darij Grinberg
        Dear Vijaya Prasad Nalluri, ... Yes, thanks. This was nearly the same way I proved my formula. It is related to trilinear coordinates, namely: If a line has
        Message 3 of 3 , Dec 31, 2003
          Dear Vijaya Prasad Nalluri,

          In Hyacinthos message #8891, you wrote:

          >> Dear Darij Grinberg,
          >> Let line l touch circle ABC at P.Let Q be the
          >> antipode of P.From similar triangles PAX,QPA we
          >> have x/AP = AP/2R giving AP^2 = 2xR,
          >> similarly, we get BP^2 = 2yR and CP^2 = 2zR
          >> So BP/AP = sqrt (y/x) and CP/AP = sqrt (z/x)
          >> From Ptolemy's theorem applied to the cyclic
          >> quadrilateral ABPC,
          >> BC * AP = AC* BP + AB * CP
          >> Dividing either side by AP
          >> BC = AC * (BP/AP) + AB * (CP/AP)
          >> a = b * sqrt (y/x) + c * sqrt (z/x)
          >> a * sqrt x = b *sqrt y + c * sqrt z

          Yes, thanks. This was nearly the same way I proved
          my formula. It is related to trilinear coordinates,
          namely: If a line has trilinear equation
          kx + ly + mz = 0, then it is tangent to the
          circumcircle of triangle ABC if and only if

          (ak)^2 + (bl)^2 + (cm)^2
          = 2 ( blcm + cmak + akbl ),

          or, equivalently,

          ( sqrt(bl) + sqrt(cm) - sqrt(ak) )
          * ( sqrt(cm) + sqrt(ak) - sqrt(bl) )
          * ( sqrt(ak) + sqrt(bl) - sqrt(cm) )
          * ( sqrt(ak) + sqrt(bl) + sqrt(cm) ) = 0.

          Now, since the last factor never vanishes, we get
          the condition that one of the values sqrt(ak),
          sqrt(bl), sqrt(cm) equals the sum of the two
          others.

          Now, if x, y, z are the distances from A, B, C to
          the line, the line coordinates k, l, m are
          proportional to ax, by, cz, hence we get our
          formula above.

          Sincerely,
          Darij Grinberg
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