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17322Re: NPC and Kiepert hyperbola

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  • chris.vantienhoven
    Mar 1, 2009
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      Dear Francisco,

      I can follow your reasoning and it looks like you are quite right.
      So that's why I constructed your way and my way together.
      After doing this I saw immediately the origin of the appararent
      When P is on NPC, the line Q.R is parallel to DP', etc. indeed.
      However it still doesn't pass through P.
      Actually there is no ground to go through P.
      However this line can be constructed.
      0. As a matter of fact when P is on NPC the points Q and R are
      infinitypoints and QR intersects The BrocardAxis in a special point T.
      This point T is derived by a mapping from NPC on the BrocardAxis.
      1. P is a point on NPC.
      2. S1 and S2 are different points on any radius of NPC.
      3. Construct QR-lines as introduced by you before for S1 and S2.
      These lines intersect in the same point T (on the BrocardAxis) for
      all different S1 and S2 (unequal to X5, P, antipode of P).
      So you could say that for every line through X5 there is a mapping on
      a point on the BrocardAxis. Because of continuity-considerations this
      is also valid for the intersectionpoint P of this line with NPC.
      4. So draw the line through T parallel to DP'.
      This is the line QR when Q and R are infinitypoints.
      This was confirmed by my drawings in Cabri.
      I hope it still is understandable.
      Best regards,

      Chris van Tienhoven

      --- In Hyacinthos@yahoogroups.com, "garciacapitan"
      <garciacapitan@...> wrote:
      > Dear Chris, I guess that you make some error in your construction,
      > please do the following:
      > ABC is a triangle, DEF its orthic triangle, LMN its medial triangle.
      > P any point on the nine point circle of ABC.
      > We have that altitude AD is the internal bisector of angle EDF. So
      > call P' the reflection of P with respect to line AD.
      > Call P'' the reflection of P with respect the internal angle
      > of angle MLN.
      > You can see that (isogonal) lines DP' and LP'' are parallel. If you
      > construct the isogonals with respect the other vertices of the two
      > triangles you we'll see that these lines are also parallel to DP'
      > LP'', meaning that both isogonal conjugates of P with respect the
      > triangles are the same infinity point.
      > Hence if P belongs to NPC, the three points P, Q, R are in fact only
      > two points, always collinear.
      > --- In Hyacinthos@yahoogroups.com, "chris.vantienhoven"
      > <van10hoven@> wrote:
      > >
      > >
      > > Dear Francisco,
      > >
      > > --- In Hyacinthos@yahoogroups.com, "garciacapitan"
      > > <garciacapitan@> wrote:
      > > > If P is on the NPC then Q and R are the same infinity point, so
      > > can
      > > > say that P, Q and R are collinear (if P is an infinity point we
      > > > also that Q=R, hence P, Q and R can be considered collinear in
      > > > case).
      > >
      > > Take any point P on any line through Center X5 from NPC and
      > > Q and R as described earlier.
      > > Draw the line QR.
      > > Let P run over this line. Then you can see that P is only on line
      > > when it crosses the KiepertHyperbola.
      > > At least this can be done in Cabri.
      > > When it crosses NPC you can see also where the line from one
      > > InfinityPoint to another InfinityPoint tends to.
      > > This is a line not through P.
      > > So as far as I can see it, when P is on NPC it does not
      > > with Q and R.
      > > Best regards,
      > >
      > > Chris van Tienhoven
      > >
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