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Generalized Gossard triangles

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  • Darij Grinberg
    According to Clark Kimberling s treatise in http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html in 1999, Paul Yiu generalized Gossard s theorem,
    Message 1 of 6 , Apr 12 4:19 AM
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      According to Clark Kimberling's treatise in

      http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html

      in 1999, Paul Yiu generalized Gossard's theorem,
      triangle and perspector to arbitrary lines through
      the centroid; some additional properties of the
      configuration were later given by John Conway and
      Steve Sigur, and finally Wilson Stothers made some
      important observations in Hyacinthos message #8383.

      Let G be the centroid of a triangle ABC and P an
      arbitrary point, and let the line PG meet the
      sidelines BC, CA, AB of triangle ABC at X, Y, Z.
      Also, define a point T by the vectorial equation

      --> --> -->
      --> GX + GY + GZ
      GT = - ---------------.
      3

      Obviously, this point T lies on the line PG; it is
      also easily seen by vector addition that

      (1) This point T lies on the line joining the
      midpoints of the segments AX, BY, CZ, i. e. on
      the Gauss line of the quadrilateral formed by
      the lines BC, CA, AB and PG.

      Okay, this was not a big deal yet. But now we can
      denote by Ga, Gb, Gc the centroids of triangles AYZ,
      BZX, CXY, respectively; then,

      (2) The points Ga, Gb, Gc lie on the reflections of
      the lines BC, CA, AB in the point T.

      This is simple using vectors, too. But life is not
      that easy (in general), and we can get rather
      difficult theorems. If the parallel to CP through Y
      meets the parallel to BP through Z at Pa, and
      similarly we define the points Pb and Pc, then

      (3) The lines GaPa, GbPb, GcPc are parallel to the
      lines BC, CA, AB, respectively.

      I have found only an awkward proof of this, so I
      would appreciate a rather simple synthetic proof if
      there is one.

      Now, (2) and (3) together yield:

      (4) The triangle formed by the lines GaPa, GbPb,
      GcPc is the reflection of triangle ABC in the
      point T.

      These four results already contain most of what is
      referred to as "Gossard stuff". I would be mainly
      interested in a good proof of (3).

      Darij Grinberg
    • ben_goss_ro
      ... Hi! I m a bit confused here. T is supposed to be the symmetric of the centroid of the system of points X, Y, Z with respect to G, right? I didn t make the
      Message 2 of 6 , Apr 12 5:33 AM
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        --- In Hyacinthos@yahoogroups.com, "Darij Grinberg"
        <darij_grinberg@w...> wrote:
        > According to Clark Kimberling's treatise in
        >
        > http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html
        >
        > in 1999, Paul Yiu generalized Gossard's theorem,
        > triangle and perspector to arbitrary lines through
        > the centroid; some additional properties of the
        > configuration were later given by John Conway and
        > Steve Sigur, and finally Wilson Stothers made some
        > important observations in Hyacinthos message #8383.
        >
        > Let G be the centroid of a triangle ABC and P an
        > arbitrary point, and let the line PG meet the
        > sidelines BC, CA, AB of triangle ABC at X, Y, Z.
        > Also, define a point T by the vectorial equation
        >
        > --> --> -->
        > --> GX + GY + GZ
        > GT = - ---------------.
        > 3
        >
        > Obviously, this point T lies on the line PG; it is
        > also easily seen by vector addition that
        >
        > (1) This point T lies on the line joining the
        > midpoints of the segments AX, BY, CZ, i. e. on
        > the Gauss line of the quadrilateral formed by
        > the lines BC, CA, AB and PG.
        >
        > Okay, this was not a big deal yet. But now we can
        > denote by Ga, Gb, Gc the centroids of triangles AYZ,
        > BZX, CXY, respectively; then,
        >
        > (2) The points Ga, Gb, Gc lie on the reflections of
        > the lines BC, CA, AB in the point T.
        >
        > This is simple using vectors, too. But life is not
        > that easy (in general), and we can get rather
        > difficult theorems. If the parallel to CP through Y
        > meets the parallel to BP through Z at Pa, and
        > similarly we define the points Pb and Pc, then
        >
        > (3) The lines GaPa, GbPb, GcPc are parallel to the
        > lines BC, CA, AB, respectively.
        >
        > I have found only an awkward proof of this, so I
        > would appreciate a rather simple synthetic proof if
        > there is one.
        >
        > Now, (2) and (3) together yield:
        >
        > (4) The triangle formed by the lines GaPa, GbPb,
        > GcPc is the reflection of triangle ABC in the
        > point T.
        >
        > These four results already contain most of what is
        > referred to as "Gossard stuff". I would be mainly
        > interested in a good proof of (3).
        >
        > Darij Grinberg


        Hi! I'm a bit confused here. T is supposed to be the symmetric of the
        centroid of the system of points X, Y, Z with respect to G, right? I
        didn't make the vector calculations, but I drew a dynamic sketch and
        I don't get T to be on the Gauss line of the quadrilateral formed by
        BC, CA, AB and PG. However, T' is on that line, where GT'=
        (GA+GB+GC)/6 (they are all vectors). What am I doing wrong?
      • ben_goss_ro
        ... Here s what I did: Let Y and Z be 2 points on CA and AB respectively. Let P a variable point on YZ. It s natural to ask th question: if L is the
        Message 3 of 6 , Apr 12 7:37 AM
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          --- In Hyacinthos@yahoogroups.com, "Darij Grinberg"
          <darij_grinberg@w...> wrote:
          > According to Clark Kimberling's treatise in
          >
          > http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html
          >
          > in 1999, Paul Yiu generalized Gossard's theorem,
          > triangle and perspector to arbitrary lines through
          > the centroid; some additional properties of the
          > configuration were later given by John Conway and
          > Steve Sigur, and finally Wilson Stothers made some
          > important observations in Hyacinthos message #8383.
          >
          > Let G be the centroid of a triangle ABC and P an
          > arbitrary point, and let the line PG meet the
          > sidelines BC, CA, AB of triangle ABC at X, Y, Z.
          > Also, define a point T by the vectorial equation
          >
          > --> --> -->
          > --> GX + GY + GZ
          > GT = - ---------------.
          > 3
          >
          > Obviously, this point T lies on the line PG; it is
          > also easily seen by vector addition that
          >
          > (1) This point T lies on the line joining the
          > midpoints of the segments AX, BY, CZ, i. e. on
          > the Gauss line of the quadrilateral formed by
          > the lines BC, CA, AB and PG.
          >
          > Okay, this was not a big deal yet. But now we can
          > denote by Ga, Gb, Gc the centroids of triangles AYZ,
          > BZX, CXY, respectively; then,
          >
          > (2) The points Ga, Gb, Gc lie on the reflections of
          > the lines BC, CA, AB in the point T.
          >
          > This is simple using vectors, too. But life is not
          > that easy (in general), and we can get rather
          > difficult theorems. If the parallel to CP through Y
          > meets the parallel to BP through Z at Pa, and
          > similarly we define the points Pb and Pc, then
          >
          > (3) The lines GaPa, GbPb, GcPc are parallel to the
          > lines BC, CA, AB, respectively.
          >
          > I have found only an awkward proof of this, so I
          > would appreciate a rather simple synthetic proof if
          > there is one.
          >
          > Now, (2) and (3) together yield:
          >
          > (4) The triangle formed by the lines GaPa, GbPb,
          > GcPc is the reflection of triangle ABC in the
          > point T.
          >
          > These four results already contain most of what is
          > referred to as "Gossard stuff". I would be mainly
          > interested in a good proof of (3).
          >
          > Darij Grinberg



          Here's what I did: Let Y and Z be 2 points on CA and AB respectively.
          Let P a variable point on YZ. It's natural to ask th question: if L
          is the intersection between the parallel through Z to BP and the
          parallel through Y to CP then what's the locus of L?

          We can how that the locus is a line parallel to BC by using some
          projective results. The transformation CP->BP (from the pencil of
          lines with center C to the pencil with center B) is a homographic one
          (in fact, it's a perspectivity). Since YL || CP and ZL || BP it means
          that the transformation YL->ZL is also a homographic one, and it's
          well-known that in this case the locus of L is a conic which
          degenerates if the image of YZ is YZ. Since this is obviously the
          case here, it means that the locus of L is a line. By putting P in
          the positions Y and Z we can see that this line is also parallel to
          BC.

          We apply all of this to our situation, where L is Pa. The locus of Pa
          is thus a line || BC, and all we need to do is show that Ga is on
          this line. We do this by putting P in the positions Y and Z (we take
          Y and Z to be fixed s.t. G is on YZ), and by using the following: If
          ABC is a triangle, G its centroid and S and T are on AB and AC
          respectively, then G is on ST iff SB/SA+TC/TA=-1 (the segments are
          oriented). We can thus show that GaPa is parallel to BC for a fixed
          line GP passing, and this obviously suffices in order to prove the
          whole thing.
        • Darij Grinberg
          Dear Grobber, ... I m sorry; my formula -- -- -- -- GX + GY + GZ GT = - --------------- 3 should have actually been -- -- -- -- GX +
          Message 4 of 6 , Apr 12 8:12 AM
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            Dear Grobber,

            In Hyacinthos message #9668, you wrote:

            >> Hi! I'm a bit confused here. T is supposed to
            >> be the symmetric of the centroid of the system
            >> of points X, Y, Z with respect to G, right?

            I'm sorry; my formula

            --> --> -->
            --> GX + GY + GZ
            GT = - ---------------
            3

            should have actually been

            --> --> -->
            --> GX + GY + GZ
            GT = ---------------.
            6

            We also have

            --> --> --> -->
            3 TS = XT + YT + ZT .

            >> However, T' is on that line, where
            >> GT'= (GA+GB+GC)/6 (they are all vectors).

            (I guess you mean GX + GY + GZ rather than
            GA + GB + GC here.)

            In Hyacinthos message, #9669, you wrote:

            >> Here's what I did: [...]

            Thanks, I'll have to learn projective
            geometry at last... Here is the main idea
            of my proof for (3):

            Let's prove that the line GaPa is parallel
            to BC. Obviously, this is equivalent to
            saying that the parallels to CP through Y,
            to BP through Z, and to BC through Ga
            concur. Now, after the well-known theorem
            on reciprocal triangles ("If the lines
            through the vertices of one triangle T1
            parallel to the respective sidelines of
            another triangle T2 concur, then the lines
            through the vertices of T2 parallel to the
            sidelines of T1 also concur"), the
            assertion we have to prove is equivalent
            to the following one: The parallels to ZGa
            through B, to YGa through C, and to YZ
            through P concur. In other words, the
            parallels to ZGa through B, and to YGa
            through C, and the line YZ concur. This
            should be easier than the initial
            statement, since the point P is not
            involved anymore; however, I have no proof
            of this other than an ugly proof using
            Menelaos. Do you have something
            elementary?

            Sincerely,
            Darij Grinberg
          • ben_goss_ro
            ... Yeah, I mean GX+GY+GZ, of course. I guess I couldn t help myself and I wrote A, B, and C being more familiar with these letters :). ... Not really. I tried
            Message 5 of 6 , Apr 12 10:15 AM
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              --- In Hyacinthos@yahoogroups.com, "Darij Grinberg"
              <darij_grinberg@w...> wrote:
              > (I guess you mean GX + GY + GZ rather than
              > GA + GB + GC here.)

              Yeah, I mean GX+GY+GZ, of course. I guess I couldn't help myself and
              I wrote A, B, and C being more familiar with these letters :).

              > Do you have something
              > elementary?

              Not really. I tried something with Menelaus as well, but I abandoned
              it quickly..
            • Darij Grinberg
              Dear Grobber, ... Sorry, another typo. It should be -- -- -- -- 3 TG = XT + YT + ZT . But the main property is still that T lies on the Gauss line
              Message 6 of 6 , Apr 15 8:19 AM
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                Dear Grobber,

                In Hyacinthos message #9671, I wrote:

                >> --> --> --> -->
                >> 3 TS = XT + YT + ZT .

                Sorry, another typo. It should be

                --> --> --> -->
                3 TG = XT + YT + ZT .

                But the main property is still that T lies on
                the Gauss line of the quadrilateral formed by
                the lines BC, CA, AB and PG; even more can be
                said: T is the centroid of the points X', Y',
                Z', where X', Y', Z' are the midpoints of the
                segments AX, BY, CZ. (The best way to prove this
                is to define T as the centroid of the points X',
                Y', Z'; then, all the rest follows by simple
                vector calculation.)

                >> In other words, the parallels to ZGa
                >> through B, and to YGa through C, and
                >> the line YZ concur. [...] I have no
                >> proof of this [...]

                Now I have quite a very artificial proof:

                I rename Ga as D, so that D is the centroid of
                triangle AYZ, and I have to show that the
                parallels to ZD through B, and to YD through C,
                and the line YZ concur.

                Well, let the parallel to YZ through A meet the
                parallel to ZD through B at B', the parallel to
                YD through C at C', the parallel to ZD through Y
                at M, the line ZD at K, and the line YD at L.
                Then, YZKM is a parallelogram (trivial), and
                AZYK and AYZL are parallelograms (since the
                lines ZD and YD pass through the midpoints of
                the segments AY and AZ, respectively). Hence, we
                have MK = YZ, KA = YZ and AL = YZ with directed
                segments. Now,

                LC' LC' YC
                --- = --- = -- (Thales)
                YZ AL AY

                and

                B'K B'K BZ
                --- = --- = -- (Thales).
                YZ KA ZA

                However, with d(R; l) denoting the directed
                distance from a point R to a line l, we have

                YC BZ d(C; PG) d(B; PG)
                -- + -- = - -------- - -------- = 1,
                AY ZA d(A; PG) d(A; PG)

                since it is well-known that
                d(A; PG) + d(B; PG) + d(C; PG) = 0 (this is a
                property of any line passing through the
                centroid G of triangle ABC).

                Hence,

                LC' B'K
                --- + --- = 1,
                YZ YZ

                and LC' + B'K = YZ, or LC' + B'K = MK. Therefore,

                MB' = MK - B'K = LC',

                -->
                so that the translation by the vector MB' maps M
                to B' and L to C'. This translation maps the
                point Y to a point W such that B'W || MY,
                -->
                C'W || LY, and, of course, YW || MB' (since MB'
                is the translation vector). From B'W || MY and
                C'W || LY, it follows that W lies on the
                parallel to ZD through B and on the parallel to
                YD through C. From YW || MB', it follows that W
                lies on the line YZ. Hence, the parallel to ZD
                through B, the parallel to YD through C, and the
                line YZ concur. Qed.!

                How do you like this proof? :-)

                Sincerely,
                Darij Grinberg
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