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PRIZE (Re: CONCURRENT CIRCLES - ORTHOCENTERS - ISOGONAL CONJUGATE POINTS)

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  • Antreas Hatzipolakis
    Let me report on what I did for this. It looked like a routine barycentric calculation, but the intermediate results were such long formulas that further
    Message 1 of 2 , Sep 12, 2013
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      Let me report on what I did for this.

      It looked like a routine barycentric calculation, but the intermediate results
      were such long formulas that further calculation was impossible. I assume that
      others encountered the same problem.

      So I tried complex numbers instead, taking A,B,C on the unit circle. The
      intermediate results (the orthocenters and circumcenters) were fairly long, but
      never used polynomials with more than 100 terms, so I could finish the
      calculation. This verified the concurrency, but the final formula for the first
      point of concurrency uses a polynomial with 242 terms.

      Some details: If the circumcenters of H1HbHc, H2HcHa and H3HaHb are O1, O2 and
      O3, then (O1) meets (O2) at Hc and at the reflection of Hc across the line O1O2,
      which is
      O2 + (Hc#-O2#)*(O1-O2)/(O1#-O2#)
      where X# means the complex conjugate of X. To prove the first concurrency, it
      suffices to show that this is equal to
      O2 + (Ha#-O2#)*(O3-O2)/(O3#-O2#)
      which is the reflection of Ha across the line O2O3.

      Incidentally, there have beem many posts in this forum with a result like:
      (1) The circumcircles of A1BC, B1CA and C1AB concur.
      and (2) The circumcircles of AB1C1, BC1A1 and CA1B1 concur.

      It should be mentioned that (2) follows from (1) by the six circles theorem.

      Antreas, keep your book prizes for the younger members of this forum. I retired
      from the University of Manitoba a few years ago.

      --- In Anopolis@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
      >
      > This "theorem" (conjecture) is still unproved (quoted below).
      >
      > Seiichi Kirikami has computed the coordinates for (P, P*) = (G, K)
      >
      > Available here:
      > http://tech.groups.yahoo.com/group/Hyacinthos/message/21992
      >
      > I offer the following books for proofs:
      >
      > 1. For a analytic proof by computing the homogeneous
      > coordinates of the concurrence points:
      > RICHARD HEGER: ELEMENTE DER ANALYTISCHEN GEOMETRIE IN
      > HOMOGENEN COORDINATEN (1872)
      >
      > 2. For a synthetic proof:
      > FRANK MORLEY & F. V. MORLEY: INVERSIVE GEOMETRY
      >
      > 3. For awny other proof (by complex numbers, etc):
      > C. ZWIKKER: THE ADVANCED GEOMETRY OF PLANE CURVES
      > AND THEIR APPLICATIONS.
      >
      > Antreas
      >
      >
      > --- In Anopolis@yahoogroups.com, "Antreas" <anopolis72@> wrote:
      > (Anopolis #521)
      > >
      > > Let ABC be a triangle and P,P* two isogonal conjugate points.
      > > Denote: H1,H2,H3 = the orthocenters of PBC, PCA, PAB, resp. and
      > > Ha,Hb,Hc = the orthocenters of P*BC, P*CA, P*AB, resp.
      > >
      > > The circumcircles of:
      > > (1) H1HbHc, H2HcHa, H3HaHb
      > > (2) HaH2H3, HbH3H1, HcH1H2
      > >
      > > are concurrent.
      > >
      > > Figure:
      > >
      > > http://anthrakitis.blogspot.gr/2013/07/concurrent-circles-orthocenters.html
      > >
      > > If P = (x:y:z), which are the points of concurrences?
      > >
      > > APH

      --
      Barry Wolk





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