## Re: [EMHL] a small problem

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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
Message 1 of 26 , Feb 10, 2011
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

> 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
>
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