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Mathesis Problem

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  • Deoclecio
    Dear Geometers, In Mathesis problem number 2376, Victor Thébault wrote: Une transversale D(delta) rencontre les côtés BC, CA, AB d un triangle en M, N, P.
    Message 1 of 8 , Aug 12, 2004
      Dear Geometers,

      In Mathesis problem number 2376, Victor Thébault wrote:

      Une transversale D(delta) rencontre les côtés BC, CA, AB d'un
      triangle en M, N, P. Si l'on désigne par a(alfa), b(beta), g(gama)
      les angles des diagonales AM, BN, CP du quadrilatère complet (D
      (delta), ABC) et par t(theta) l'angle de D(delta) avec la droite de
      Newton de ce quadrilatère, on a la relation

      Cotgt(theta) = cotga(alfa) + cotgb(beta) + cotgg(gama)

      Somebody can translate into the English? Any proof?

      Sincerely,

      Deoclecio Gouveia Mota Jr.
    • Steve Sigur
      ... A line D meets sides BC, CA, AB of a triangle in M, N, P. If we call alpha, beta, gamma the angles of the diagonals AM, BN, CP of the complete
      Message 2 of 8 , Aug 16, 2004
        On Aug 12, 2004, at 8:36 AM, Deoclecio wrote:

        > Une transversale D(delta) rencontre les côtés BC, CA, AB d'un
        > triangle en M, N, P. Si l'on désigne par a(alfa), b(beta), g(gama)
        > les angles des diagonales AM, BN, CP du quadrilatère complet (D
        > (delta), ABC) et par t(theta) l'angle de D(delta) avec la droite de
        > Newton de ce quadrilatère, on a la relation
        >
        > Cotgt(theta) = cotga(alfa) + cotgb(beta) + cotgg(gama)

        A line D meets sides BC, CA, AB of a triangle in M, N, P. If we call
        alpha, beta, gamma the angles of the diagonals AM, BN, CP of the
        complete quadrilateral made from D and the three sides of ABC and by
        theta the angle D makes with the Newton line of this quadrilateral, we
        have the relation

        cot theta = cot alpha + etc. I assume cotgx means the cotangent but I
        might mean something different. I am not sure what the angles alpha,
        etc are from this definition, but it looks like there is a triangle out
        there for which theta is the Brocard angle.

        Steve, friendly from usa
      • Darij Grinberg
        Dear Steve, dear Deoclecio, I m sorry for my sporadic presence on Hyacinthos, but let me make a remark here... ... The definition of alpha, beta, gamma is not
        Message 3 of 8 , Aug 17, 2004
          Dear Steve, dear Deoclecio,

          I'm sorry for my sporadic presence on Hyacinthos, but
          let me make a remark here...

          In Hyacinthos message #10270, Steve Sigur translated:

          >> A line D meets sides BC, CA, AB of a triangle in
          >> M, N, P. If we call alpha, beta, gamma the
          >> angles of the diagonals AM, BN, CP of the
          >> complete quadrilateral made from D and the three
          >> sides of ABC and by theta the angle D makes with
          >> the Newton line of this quadrilateral, we have
          >> the relation
          >>
          >> cot theta = cot alpha + etc.

          The definition of alpha, beta, gamma is not really
          clear here - what does "angles of the diagonals"
          mean? But if alpha, beta, gamma are meant to be
          the angles *between* the diagonals, i. e. the angles
          of the triangle formed by the lines AM, BN, CP, then
          the result cannot be true in this form. In fact, the
          angles alpha, beta, gamma are all defined
          *symmetrically* with respect to the four lines BC,
          CA, AB and D of the complete quadrilateral
          (BC, CA, AB, D), while the angle theta is defined
          *asymmetrically*, since its definition involves the
          line D. Hence, cot theta cannot be generally equal to
          cot alpha + cot beta + cot gamma. So I believe there
          must be an error in the problem.

          Friendly,
          Darij Grinberg
        • Eric Danneels
          Dear Darij, Steve and Deoclecio, ... I think it works (in some way) when: - alpha = oriented angle (0 = 180 degrees) between AM and BC; - beta = oriented
          Message 4 of 8 , Aug 17, 2004
            Dear Darij, Steve and Deoclecio,

            Darij wrote:

            > Dear Steve, dear Deoclecio,
            >
            > I'm sorry for my sporadic presence on Hyacinthos, but
            > let me make a remark here...
            >
            > In Hyacinthos message #10270, Steve Sigur translated:
            >
            > >> A line D meets sides BC, CA, AB of a triangle in
            > >> M, N, P. If we call alpha, beta, gamma the
            > >> angles of the diagonals AM, BN, CP of the
            > >> complete quadrilateral made from D and the three
            > >> sides of ABC and by theta the angle D makes with
            > >> the Newton line of this quadrilateral, we have
            > >> the relation
            > >>
            > >> cot theta = cot alpha + etc.
            >
            > The definition of alpha, beta, gamma is not really
            > clear here - what does "angles of the diagonals"
            > mean? But if alpha, beta, gamma are meant to be
            > the angles *between* the diagonals, i. e. the angles
            > of the triangle formed by the lines AM, BN, CP, then
            > the result cannot be true in this form. In fact, the
            > angles alpha, beta, gamma are all defined
            > *symmetrically* with respect to the four lines BC,
            > CA, AB and D of the complete quadrilateral
            > (BC, CA, AB, D), while the angle theta is defined
            > *asymmetrically*, since its definition involves the
            > line D. Hence, cot theta cannot be generally equal to
            > cot alpha + cot beta + cot gamma. So I believe there
            > must be an error in the problem.


            I think it works (in some way) when:
            - alpha = oriented angle (0 => 180 degrees) between AM and BC;
            - beta = oriented angle between BN and CA;
            - gamma = oriented angle between CP and AB.
            - theta = angle between line D and Newton line. I don't know how to
            define an oriented angle between these lines because the order of M,
            N and P on D can be different from the order of the corresponding
            midpoints of AM, BN and CP

            So my best restatement of the problem is

            ABS(cot(theta)) = ABS(cot(alpha) + cot(beta) + cot(gamma))

            where ABS means "absolute value" and cot means "cotangent"

            Greetings from Bruges

            Eric
          • Darij Grinberg
            Dear Eric Danneels, ... Indeed, you needn t consider the order. When you define a directed angle modulo 180 degrees, you consider it as an angle between two
            Message 5 of 8 , Aug 17, 2004
              Dear Eric Danneels,

              In Hyacinthos message #10273, you wrote:

              >> I think it works (in some way) when:
              >> - alpha = oriented angle (0 => 180 degrees)
              >> between AM and BC;
              >> - beta = oriented angle between BN and CA;
              >> - gamma = oriented angle between CP and AB.
              >> - theta = angle between line D and Newton line.
              >> I don't know how to define an oriented angle
              >> between these lines because the order of M, N
              >> and P on D can be different from the order
              >> of the corresponding midpoints of AM, BN and
              >> CP

              Indeed, you needn't consider the order. When you
              define a directed angle modulo 180 degrees, you
              consider it as an angle between two *lines*,
              not depending from the order of the points which
              define the lines.

              So you say that

              cot theta = cot alpha + cot beta + cot gamma.

              This would be nice (I have no geometry software
              at hand to test it, but it can be true). Just
              one little remark: This result is not directly
              related to the Brocard angle, since the angles
              alpha, beta, gamma generally aren't the angles
              of a triangle.

              Darij Grinberg
            • Alexey.A.Zaslavsky
              Dear Deoclecio! In Mathesis problem number 2376, Victor Thébault wrote: Une transversale D(delta) rencontre les côtés BC, CA, AB d un triangle en M, N, P.
              Message 6 of 8 , Aug 29, 2004
                Dear Deoclecio!

                In Mathesis problem number 2376, Victor Thébault wrote:

                Une transversale D(delta) rencontre les côtés BC, CA, AB d'un
                triangle en M, N, P. Si l'on désigne par a(alfa), b(beta), g(gama)
                les angles des diagonales AM, BN, CP du quadrilatère complet (D
                (delta), ABC) et par t(theta) l'angle de D(delta) avec la droite de
                Newton de ce quadrilatère, on a la relation

                Cotgt(theta) = cotga(alfa) + cotgb(beta) + cotgg(gama)

                Somebody can translate into the English? Any proof?

                I don't understand what are the angles alfa, beta, gamma.

                Sincerely Alexey
                .
              • Darij Grinberg
                Dear Alexey, ... In fact, Eric Danneels cleared this up in ... Sincerely, Darij Grinberg
                Message 7 of 8 , Aug 30, 2004
                  Dear Alexey,

                  In Hyacinthos message #10324, you wrote:

                  >> I don't understand what are the angles alfa,
                  >> beta, gamma.

                  In fact, Eric Danneels cleared this up in
                  Hyacinthos message #10273:

                  >> - alpha = oriented angle (0 => 180 degrees)
                  >> between AM and BC;
                  >> - beta = oriented angle between BN and CA;
                  >> - gamma = oriented angle between CP and AB.
                  >> - theta = angle between line D and Newton
                  >> line.

                  Sincerely,
                  Darij Grinberg
                • Deoclecio
                  ... Dear Alexey, See message number 10273. Sincerely, Deoclecio
                  Message 8 of 8 , Aug 31, 2004
                    --- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@c...>
                    wrote:
                    > Dear Deoclecio!
                    >
                    > In Mathesis problem number 2376, Victor Thébault wrote:
                    >
                    > Une transversale D(delta) rencontre les côtés BC, CA, AB d'un
                    > triangle en M, N, P. Si l'on désigne par a(alfa), b(beta), g(gama)
                    > les angles des diagonales AM, BN, CP du quadrilatère complet (D
                    > (delta), ABC) et par t(theta) l'angle de D(delta) avec la droite de
                    > Newton de ce quadrilatère, on a la relation
                    >
                    > Cotgt(theta) = cotga(alfa) + cotgb(beta) + cotgg(gama)
                    >
                    > Somebody can translate into the English? Any proof?
                    >
                    > I don't understand what are the angles alfa, beta, gamma.
                    >
                    > Sincerely Alexey

                    Dear Alexey,

                    See message number 10273.

                    Sincerely,

                    Deoclecio
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