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Re: Areal center

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  • Jeffrey Brooks
    Oops. I accidentally reversed a couple of midpoints. Sorry. Sincerely, Jeff
    Message 1 of 5 , Mar 10, 2009
      Oops. I accidentally reversed a couple of midpoints.
      Sorry.
      Sincerely, Jeff


      --- In Hyacinthos@yahoogroups.com, "Jeffrey Brooks" <cu1101@...> wrote:
      >
      > Can someone please remind me the definition of areal center? Something is not right here; that is I construct the areal center following Quim Castellsaguer's The Triangles Web
      >
      > 1) Let A'B'C', A"B"C" be the two inscribed triangles, La the line through A parallel to the line joining the midpoints of B'C" and B"C'.
      > 2) Construct analogously Lb, Lc.
      > 3) The lines La, Lb, Lc intersect at the areal center S.
      >
      > Then I try Francois' choo-choo way from a previous message. Both methods give a center S where Area(S,A',A")=Area(S,B',B")=Area(S,C',C").
      >
      > What am I doing wrong?
      > Sincerely, Jeff
      >
    • Francois Rideau
      You are not doing wrong. Area(S, A , A ) = Area( S, B , B ) = Area(S, C , C ) ( where S is a signed area) is just the definition of the areal center S of
      Message 2 of 5 , Mar 11, 2009
        You are not doing wrong.
        Area(S, A', A") = Area( S, B', B") = Area(S, C', C") ( where S is a signed
        area) is just the definition of the areal center S of triangles A'B'C' and
        A"B"C".
        Friendly
        Francois

        On Wed, Mar 11, 2009 at 1:33 AM, Jeffrey Brooks <cu1101@...> wrote:

        > Oops. I accidentally reversed a couple of midpoints.
        > Sorry.
        > Sincerely, Jeff
        >
        >
        > --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>, "Jeffrey
        > Brooks" <cu1101@...> wrote:
        > >
        > > Can someone please remind me the definition of areal center? Something is
        > not right here; that is I construct the areal center following Quim
        > Castellsaguer's The Triangles Web
        > >
        > > 1) Let A'B'C', A"B"C" be the two inscribed triangles, La the line through
        > A parallel to the line joining the midpoints of B'C" and B"C'.
        > > 2) Construct analogously Lb, Lc.
        > > 3) The lines La, Lb, Lc intersect at the areal center S.
        > >
        > > Then I try Francois' choo-choo way from a previous message. Both methods
        > give a center S where Area(S,A',A")=Area(S,B',B")=Area(S,C',C").
        > >
        > > What am I doing wrong?
        > > Sincerely, Jeff
        > >
        >
        >
        >


        [Non-text portions of this message have been removed]
      • Jeffrey Brooks
        Dear Francois, Next time I ll screw up better, I promise. :-) I reversed two sets of midpoints and still got an S with
        Message 3 of 5 , Mar 11, 2009
          Dear Francois,
          Next time I'll screw up better, I promise. :-)
          I reversed two sets of midpoints and still got an S with Area(S,A',A")=Area(S,B',B")=Area(S,C',C")
          Sincerely, Jeff

          --- In Hyacinthos@yahoogroups.com, Francois Rideau <francois.rideau@...> wrote:
          >
          > You are not doing wrong.
          > Area(S, A', A") = Area( S, B', B") = Area(S, C', C") ( where S is a
          > signed area) is just the definition of the areal center S of
          > triangles A'B'C' and
          > A"B"C".
          > Friendly
          > Francois
          >
          > On Wed, Mar 11, 2009 at 1:33 AM, Jeffrey Brooks <cu1101@...> wrote:
          >
          > > Oops. I accidentally reversed a couple of midpoints.
          > > Sorry.
          > > Sincerely, Jeff
        • Jeffrey Brooks
          Dear Francois, S is simply a point but I suppose it could be viewed as an area in a dual sense? Sincerely, Jeff
          Message 4 of 5 , Apr 1, 2009
            Dear Francois,
            S is simply a point but I suppose it could be viewed as an area in a dual sense?
            Sincerely, Jeff

            > You are not doing wrong.
            > Area(S, A', A") = Area( S, B', B") = Area(S, C', C") ( where S is a
            > signed area) is just the definition of the areal center S of
            > triangles A'B'C' and A"B"C".
            > Friendly
            > Francois
            >
            > On Wed, Mar 11, 2009 at 1:33 AM, Jeffrey Brooks <cu1101@...> wrote:
            >
            > > Oops. I accidentally reversed a couple of midpoints.
            > > Sorry.
            > > Sincerely, Jeff
            > >
            > >
            > > --- In Hyacinthos@yahoogroups.com <Hyacinthos%
            > > 40yahoogroups.com>, "Jeffrey Brooks" <cu1101@> wrote:
            > >
            > > > Can someone please remind me the definition of areal center?
            > > > Something is not right here; that is I construct the areal
            > > > center following Quim Castellsaguer's The Triangles Web
            > > >
            > > > 1) Let A'B'C', A"B"C" be the two inscribed triangles, La the
            > > > line through A parallel to the line joining the midpoints of
            > > > B'C" and B"C'.
            > > > 2) Construct analogously Lb, Lc.
            > > > 3) The lines La, Lb, Lc intersect at the areal center S.
            > > >
            > > > Then I try Francois' choo-choo way from a previous message.
            > > > Both methods give a center S where Area(S,A',A")=Area(S,B',B")>
            > > > =Area(S,C',C").
            > > >
            > > > What am I doing wrong?
            > > > Sincerely, Jeff
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