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Re: [EMHL] a small problem

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  • Nikolaos Dergiades
    Sorry a correction. It is obvious that the circumconic that passes through P is not pyz - qzx + rxy = 0 but pyz - 2qzx + rxy = 0 Best regards Nikos Dergiades
    Message 1 of 26 , Feb 10, 2011
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      Sorry a correction.
      It is obvious that the circumconic
      that passes through P is not
      pyz - qzx + rxy = 0
      but
      pyz - 2qzx + rxy = 0

      Best regards
      Nikos Dergiades



      > Dear Sung Hyun,
      >
      > What you wrote is true:
      >
      > > Here is a simple projective generalization:
      > >
      > > Given ABCD, Let BC.AD = P. Take E on BC. Take
      > harmonic
      > > conjugate of P wrt BE
      > > and let it be I. Take AE.CD=F. Take IF.DE=M
      > >
      > > Locus of M as E moves around BC is a circumconic of
      > ABCD.
      >
      > If we use other names
      > P for D
      > A1 for P
      > X for E
      > Xb for I
      > Bx for F
      > then if P = (p:q:r) is a point in barycentrics relative to
      > ABC
      > A1B1C1 is the cevian triangle of P
      > A2B2C2 is the tripolar of P (A2 = harm_conj(A1,B,C))
      > X is an arbitrary point on BC
      > Xb = harm_conj(X,A1,B)
      > Bx = CP.AX  and
      > M = BxXb.PX
      >
      > then the locus of M is the circumconic
      > pyz - qzx + rxy = 0 that passes through P
      > and the envelope of BxXb is the conic
      > xx/pp + yy/qq + zz/rr + 2yz/qr - 6zx/rp + 2xy/pq = 0
      > that is tangent
      > to BC at A2, to BA at C2, to AP at A3 and to CP at C3.
      > Construction of A3: The parallel from A to BC meets CP at
      > C'.
      > If C" is the midpoint of AC' then CC" meets AP at A3.
      >
      > Best regards
      > Nikos Dergiades
      >
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